Chapter 1
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Chapter 1
1.1
We will now explain where the 1.728MHz number for DECT comes from. This will lead
into the discussion of possible pulse shapes limited by the channel bandwidth to represent
the data sequence..
In DECT, many users share a common transmission medium. Thus, it is important to
provide access between any user without major interference degradation. This process is
commonly known as multiple access, where in DECT the messages are separated in time
(TDMA = time division multiple access), ensuring that the different users transmit at
different times; and in frequency (FDMA = frequency division multiple access), ensuring
that the different users use different frequency band. There are twelve channels time
division multiplexed per carrier in DECT. Furthermore, time duplex (TDD) is used,
where the two possible transmission directions share a single channel. From basic signal
processing we know, that to avoid aliasing during sampling of a voice signal, which is
assumed in practice to be band-limited to roughly 3.4kHz, the sampling rate has to be at
least two times the signal bandwidth. Hence, one usually uses a sampling frequency of
8kHz. In a standard PCM signal, eight bits per sample are used resulting in a PCM stream
of 64kbps. However, in DECT we are using ADPCM (adaptive differential PCM)
resulting in a reduction of the bit rate (=32kbps). This stream from each user are now to
be multiplexed by interleaving the stream of each user at the output to obtain a higher
channel bit rate. The idea is shown in fig.a.
2
Figure: a) Principal of TDMA and b) Spectrum of a raised cosine pulse
Each stream of each user is divided continuously into groups of bits, known as time-slots,
and then these time-slots are interleaved to form the output bit stream. In the output bit
stream, the collection of bits corresponding to precisely one time-slot from each user is
known as a time frame. Accordingly, in DECT, where twelve users are time division
multiplexed
(time
duplex),
the
number
of
bits
in
the
output
frame
is
2 12 32kbps 768kbps . Comparing with the channel bit rate of the DECT standard it seems
that there is something missing in our approach. Indeed, in order to realize the
demultiplexing at the receiver, the boundaries of the time-slots and the different frames
must be known. For this purpose the multiplex typically inserts additional bits into the
frame known as framing bits. Furthermore, we need some further information to perform
tasks such as carrier and timing recovery, stored in a so called preamble before the actual
message stream. Accordingly, 384kbps are needed to accomplish this task. Hence, the
signaling interval for the baseband sequence is simply T = 1/R = 0.868us.
Now, we have to choose a pulse shape, which maximizes the channel rate over a bandlimited channel. For the time being we assume a noiseless channel. We observe that the
Gaussian pulse is strictly band-limited, but it exceeds the ideal minimal bandwidth
(Nyquist bandwidth = 1/T) by a certain amount, called the excess bandwidth. Under the
3
assumption that one uses a Gaussian pulse in GMSK DECT and that we need a rolloff
factor of 1 / 2 to represent the baseband data sequence, then B (1 0.5) R where R is
the channel rate (R=1.152Mbps). Now we get the baseband channel bandwidth of B =
1.728MHz.
1.2
Smsk(t)
Smsk(t)
Adapted from the text “RF microelectronics”, B. Razavi, Prentice Hall, 1998.
4
1.3
For a PSK signal, we have Eb/N0=SNR(fN/R), where fN is the effective noise bandwidth,
R is the effective bit rate. For an equivalent BPSK system which has the same data (bit)
rate, RQPSK=1/2RBPSK. Now the nearest constellation points for QPSK are closer to one
another (dQPSK = (1/2 ) dBPSK), the signal power is halved. Therefore the Pe versus Eb/N0
curve is identical for BPSK and QPSK.
1.4
For circuit designers their jobs are that given different standards, they have to design
different receiver front ends. This means given any standard, which automatically
specifies the modulation scheme and required BER, they want to figure out the SNR
required at the input of the demodulator. There is no need to do comparisons among
different standards and so there is no need to compare among different modulation
schemes. They simply use the SNR required to calculate the NF of the front end. Hence a
BER versus SNR plot is more appropriate.
For communication system engineers, they are interested in comparing the BER of
different modulation scheme under the constraints that they each take up the same
channel capacity, or same bit rate. Using SNR as an independent variable is not an
absolutely fair comparison since the bit rates of the different modulation schemes are
different and have not been taken into consideration. This means, e.g. for two different
modulation schemes, for the same BER, let say scheme 1 has a lower SNR than scheme
2, this does not necessarily mean scheme 1 is better. Why? Because scheme 1 may need
lower bit rate than scheme 2 (a case in point is scheme 1 is BPSK and scheme 2 is
5
QPSK; then scheme 1 needs half the bit rate as scheme 2). In this case, the energy per
bit/noise of scheme 1 may turn out to be higher than that of scheme 2, which makes it a
worse scheme. Now a plot of BER versus Eb/N0 takes that into account and therefore is
more appropriate.
1.5
Effect of slight local movements and separation.
In the first instance, the carriers between direct path and reflected path are 180 degrees
out of phase. Due to extra 180 degree phase shift because of reflection, the two signals
through the 2 paths interfere constructively. This results in a larger than normal eye
(envelope).
In the second instance, we can see that only a slight movement of the mobile causes the
carrier of the reflected wave to shift 180 degrees. Hence now the direct path and reflected
6
path are 540 degree out of phase and interfere destructively. This results in a smaller
than normal eye.
Consequently, the eye opening changes rapidly with slight movements of the reflector,
causing great fluctuation in the envelope’s amplitude and also ISI problems.
1.6
1/d12 – 1/d22 = (d22 - d12)/( d12 d22). For d1 almost same as d2, this becomes almost equal
to constant/d24. Here the numerator is a constant, because if d1 almost same as d2, then as
d1 and d2 change, the difference remains almost constant. This means the difference of
their squares and hence the numerator stays constant. Hence the exponent n in (1.24) is 4.
1.7
Now how do we explain the dependency of attenuation on delay? In fig.1.25 path 2 has
longer path -> larger delay, path2 has longer path -> stronger attenuation. Therefore
delay and attenuation is correlated. Well, let us assume, very crudely, that all the delayed
7
signal come from one simple reflection. Hence the delay is directly proportional to length
of the delay path. Hence a larger delay would mean the reflected signal must come
through a longer path. So for example, in fig.1.25, one can see that path 2’s distance of
d’’ is larger than path 3’s distance of d’’’. This means path 2 arrives later and has larger
delay. Comparing fig.1.26 and fig.1.23a that is indeed the case. One can justify this path
loss because from path loss (yes, we are reverting to the pass loss formula again), it
seems that the larger the separation between Tx and Rx antenna, the received signal is
attenuated more.. Again comparing fig.1.26,fig.1.23a this is true. Also this is because
there are more reflections as delay increases. Hence attenuation increases.
1.8
In an FIR filter, the delay are same for each tap, in (1.59), the delays i for different taps
are different.
1.9
In general as a circuit designer, we can assume the coherence bandwidth is given to us
from people doing channel characterization. Indeed sweep measurements have been
performed for different environments to get the power spectrum profile during multi-path
reception in real life.
This will help us to have a rough idea if the channel is flat/frequency selective and hence
allow us to decide if diversity/equalization are needed. This will have a big impact on the
circuit designer in, among other things, his choice of receiver architecture, complexity of
8
the receiver (specifically how much DSP needed to be incorporated in the demodulation
block, its impact on the power budget etc.).
1.10
For fast fading, this is due primarily to the fact that perturbation caused by fast fading is
multiplicative in nature:
xR,baseband(t) = (t) s1(t) exp(-j(t)), as opposed to being addtive in nature (like the AWGN
case). Hence the error is insensitive to signal amplitude and there is an irreducible BER.
For frequency selective fading, irreducible BER is caused by ISI. This occurs when the
main (undelayed) signal is removed due to multi-path cancellation.
9
Chapter 2
2.1
Here, with front end SNRdemo_in = Sdemod_in (worse case) - Ndemod_in = -77dbm10log10(4kTRsB) = -77dbm-10log10(4kTRs) –10log10(B/1Hz) = -77dbm – (-174dbm) –
92db=5db
2.2
a) Fix IF :
IF
f IF 100
Image
f IF 10
LO
Image LO
IF
894
IF
Image
LO
894
IF
Image LO
894
894
b) Fix LO :
IF
f lo 760
Image
f lo 850
IF
LO
Image LO
IF
894
LO
894
894
IF
Image
LO Image
894
10
c) Final answer is fimage for both schemes fall inside the band and scheme (b) does not
help.
2.3
(a) Harmonic distortion is usually generated by the active components in a receiver chain.
Hence BPF1 typically does not play any role. For BPF2,3 if desired signal is modulated
using amplitude modulation scheme, fully or partially, (AM, QAM, combination of
QAM/PSK etc.) to obtain received signal, then if the received signal is too large,
distortion will occur on the received signal itself. After the received signal get
demodulated into the desired signal (even with acceptable final SNR in AM and
acceptable BER in QAM), the desired signal may remain distorted and unacceptable. For
phase modulated signal, (FM, PSK etc.), distortion can also mean phase got distorted as
well. This depends on the phase linearity of the receiver front end as opposed to the
amplitude linearity. Again after the received signal get demodulated into the desired
signal(even with acceptable final BER in QPSK), the desired signal may remain distorted
and unacceptable.
(b) Basically the interferers is so large that it moves the operating point to a place on the
input/output curve where there is not much gain. The desired signal is very small in
amplitude and its gain now goes to zero and so the output goes practically to zero. Of
course intermodulation product from interferers itself also has gain drop to zero (we can
now treat interferers itself as signal and there is gain compression for the interferes as
well). However noise from subsequent stages start to become important. As desired
signal goes down in amplitude but noise from subsequent stages stay constant, the SNR
11
goes down and eventually goes below minimum SNR and we say signal got blocked. We
can then see blocking affects SNR differently from intermodulation of interferes in that
blocking reduces S whereas intermodulation of interferes increases N.
(c) Yes. From (2.34) it is seen that if 3A2 >> 1 then IM3 goes up. Applying the same
condition to (2.59), but assuming 3 and 1 are both positive, it is seen that there is no
blocking
2.4
(a) This assumption means we assume intermodulation of interferers completely
characterises the effect of non-linearity on the receiver front end’s performance. Strictly
speaking one has to see whether the gain compression/blocking mechanism is more or
less dominant in degrading the SNRrec_front_out..
(b) This is strictly speaking, not true. Noise at the antenna which is outside of this
bandwidth can be mixed down into this bandwidth. Hence this is an underestimation.
Noise bandwidth is usually slightly larger than filter bandwidth, but the difference is
quite small and can usually be neglected.
(c) Since we assume BPF3 filter out all interference sources, then BPF3 and IF
AMPLIFIER non-linearity does not matter since there are no interferers being applied to
the non-linearity part of BPF3 and IF amplifier to generate intermodulation product. (We
are assuming that BPF3 filter first, then the resulting signal is applied to the nonlinear
12
part of BPF3). In other words, at the receiver front end output at 100Mhz, there is still the
same ID3’’. In addition at the output there will be harmonic distortion components that lie
at 200Mhz, 300Mhz etc. They are generated when the minimum desired signal and ID3’
are applied at the input to the BPF3_IF block. However these harmonic distortion’s
amplitude are small relative to the minimum desired signal at the output of the front end.
Hence the resulting SNRrec_out remains practically the same.
2.5
We are going to determine how the available power gain of each stage is determined with
respect to its voltage gain and its input, output impedance respectively. It is important to
note that several power gain equations appear in the literature and are used in the design
of microwave sub-components. They are called the transducer gain, the operating gain
and the available power gain. For our application we are interested in the available power
gain defined as the ratio of the power available from the network to the power available
from the source. The power available from the network is the power delivered by the
network to a conjugate matched load. Similarly, the power available from the source is
the power delivered by the source to a conjugate matched input impedance. Accordingly,
the available power gain of the two-port network can be defined as follows (refer to the
figure below):
Psource
Rin
V2
Rin Vin2
Rin
Rin Rs
2
Rin
1
Vin2
Rin Rs 2
Rin
(S2.1)
13
Pnetwork
Rin
V2
RL Vin2
RL
Rin Rs
2
RL
Av2
Rout RL
2
1
RL
(S2.2)
where Av denotes the voltage gain.
By definition, the input of the two-port network is matched to the source resistance and
the output of the two-port network is matched to the load. Hence,
Psource
Vin2
4 Rs
(S2.3)
Pnetwork
1 2 1 1 Vin2 Av2
V Av
4
4 RL
RL
2
in
(S2.4)
Finally, the available power gain G can be expressed by the source resistance, the load
and the voltage gain of the subcomponent.
G
Pnetwork Vin2 Av2 4 Rs
R
2 Av2 s
Psource
16 RL Vin
4 RL
(S2.5)
Notice that equation (S2.3) is correct for one stage. However, if we have a cascade of
stages, equation (S2.2) could be written as follows (for the ith stage)
2 1
Rin,i
Av ,i
Pnetworks Vin2,i
R
R
RL
out ,i 1
in,i
14
(S2.6)
If we compare (S2.6) with (S2.2) we observe that there is one term missing in equation
(S2.6). This is due to the fact that the power dissipated in the output resistance of the twoport network has already been taken into consideration (in our case Rout,i-1 for the
previous stage). Accordingly, equation (S2.6) becomes:
Gi Av2,i
R s ,i
Rout,i
Av2,i
Rin,i
Rout,i
(S2.7)
Note that equation (S2.7) assumed that the input of the ith stage is matched to the output
of the previous stage (the (i-1) th stage).
2.6
Substituting the value from row 3,4 of Table 2.1 into the Friis’s formula, and carrying out
calculation in ratio (not in decibels) we have the noise contribution calculated for
individual components as follows:
1st component=NF1=1.58
2nd component=(NF2-1)/G1 =1.587
3rd component=(NF3-1)/G1G2=0.058
4th component=(NF4-1)/G1..G3=2.368
15
5th component=(NF5-1)/G1..G4=3.66
6th component=(NF6-1)/G1..G5=3.073
2.7
Note that the overall noise figure of the above set of values mainly depends on the
isolated noise figure of BPF1 and the LNA, and on the gain of the LNA. One can see that
varying the gain of the LNA over a range of 10dB to 20dB has a large impact on the
overall noise figure of the system. These effects are shown in the following two figures.
Accordingly, we get better noise figure performance of the receiver with a high gain, low
noise LNA subcomponent.
16
Overall noise figure
14
13.5
13
12.5
12
Overall NF [dB] 11.5
11
10.5
10
9.5
9
NF1
1
2
3
4
5
6
7
8
9
10
Isolated NF of the LNA [dB]
Effect of varying the isolated NF of the LNA on the overall noise performance for a fix
LNA voltage gain (=12dB).
Overall noise figure
13
12
11
10
Overall NF [dB]
9
NF2
8
7
6
10
11
12
13
14
15
16
17
18
19
20
Voltage gain of the LNA [dB]
Effect of varying the voltage gain of the LNA on the overall noise performance for a fix
LNA noise figure (=3dB).
2.8
In wireless communication design one usually talks about “spurious-free dynamic range
(SFDR)”. The spurious-free dynamic range represents the range over which a
17
receiver/component can span while producing an acceptable SNR. A receiver or
subcomponent with a larger SFDR is better.
Physically one thinks of the front end or a sub-component with larger SFDR as being
able to take a larger input signal + interference before the maximum allowable signal +
interference level is reached. This maximum input signal is dictated by the reduced SNR,
caused by reducing signal, due to compression and blocking. The minimum input signal
is dictated by the reduced SNR, caused by increasing noise. This noise increase is caused
by circuit noise as well as inter-modulated interference and is shown to be related to IIP3,
not OIP3 (refer to (2.95)). Accordingly for a subcomponent with a better IIP 3, the SFDR
is larger and hence the input signal can go lower (in the presence of large interference)
before the noise (from circuit noise as well as from inter-modulated interference) makes
the output SNR unacceptable.
Now in comparing the mixer and IF amplifier, the IF AMPLIFIER has a smaller IIP3.
Therefore, in the presence of large interferers, the noise generated by intermodulation
will be larger. Hence for a given input signal, the SNR is smaller. Hence its SNR drops
below SNRmin at a larger input signal, thus reducing its SFDR and is thus poorer than the
mixer.
Alternatively one can start by realising that a receiver/component which can accept a
larger input range for the desired signal (rather than a larger output range), in the
presence of interference, is a better receiver/component. Since IIP3 determines the input
range and OIP3 determines the output range, it is the IIP3 that is important. A common
misconception is to compare the effect of IIP3, OIP3 on the desired signal. This is the
18
wrong approach as it is the IIP3, OIP3’s effect on the interferers (generating noise) which
then indirectly affects the desired signal (reducing SNR), that matters.
To quantify the above discussion can refer to [Ref #2 of Chapter 2] where we have the
following formula for (SFDR):
SRDR
2IIP3 F
SNRmin dB
3
(S2.8)
where F is the total noise power (referred to the input) and IIP3 is the input referred third
order intercept point. SNRmin is the minimum SNR required of the receiver front end. For
example, in section 2.6 for the whole front end IIP3= -20dbm (2.106), NF=10.6db
~=10db (2.105), since it is about the same as NF derived in equation (2.84), then F will
be same
as
Nrec_front_in, which is
given
by equation (2.81)
as
-112dbm.
SNRmin=SNRrec_front_out, which is given by (2.78) as 25db. Substituting SFDR=36db.
Equation (S2.8) can also be applied to determine the SFDR of a sub-component of the
receiver, like a mixer and an IF amplifier.
In general from (S2.8) one can see that a front end or a sub-component with larger IIP3
gives better SFDR, and is better. This means mixer is better.
2.9
(a) Analysing the requirements we observe that the –104 dBm specifies the range
restricted by the overall noise performance (the sensitivity) and that the –10 dBm
19
specifies the range restricted by the overall blocking and compression performance of the
receiver front end shown in Figure P.2(a).
Refer to equation (2.81) – (2.84):
S rec _ front _ in 174dBm NF|dB 10 log 10 ( B) SNRde mod_ in|dB 104dBm
(S2.9)
The band of interest is the channel spacing associated with each user (B = 200kHz). The
required SNRdemod_in for a BPSK modulation scheme for a BER of 10-3, assuming only
AWGN and no fading, turns out to be approximately 7dB (see chapter 1). Thus, we can
evaluate the required overall noise figure performance from relation (S2.9).
NF|dB 104dBm 174dBm 10 log 10 (200k ) 7dB 10dB
(S2.10)
(b) The bandwidth depends on the amount of subsampling. One choice is we use the
largest subsampling factor as possible. In this case fs=400kHz and fb=200kHz.
The gain of the A/D converter is assumed to be 1, or A4=G4=1.
The noise figure of the A/D converter is specified by the required peak-to-peak input
signal. Since the gain of the A/D converter can be assumed to be unity, we can use NF
equation to calculate the noise figure:
NF
Vn2,out
A 2 4kTRs
Vn2,out
4kTRs
20
(S2.11)
Now V2n,out can be written as follows:
3.13
12
0.45 2 11 V
12 2 12
2
Vn ,out 0.2 2 22 V 2
Vn ,out
(S2.12)
Since equation (S2.11) defines the spot noise figure (noise figure for 1Hz bandwidth), we
have to determine V2n,out per Hertz. Recall that the band of interest is the channel
bandwidth B = 200kHz. Accordingly, the result from (S2.12) becomes:
Vn2,out
Hz
23.8 10
14
V2
Hz
(S2.13)
Remember that the noise figure has to be defined with respect to a source resistance,
which is taken as reference for each isolated noise figure specification of the different
sub-components. In this example, we assume a source resistance of Rs = 1k has been
used. Thus, equation (S2.12) yields the following noise figure:
NF
23.8 10 14 V 2 / Hz
14970
16 10 18 V 2 / Hz
NF 42 dB
Hence NF4=42db
(c) Now we can make use of the Friis formula, which relates the overall noise figure to
the noise figure of each subcomponent.
21
NF 10 NF1
NF2 1 NF3 1
NF4 1
G1
G1 G2 G1 G2 G3
(S2.14)
Gi the power gain, can be derived from Ai , the voltage gain, which are given.
Now we can use the additional information to write equation (S2.14) as follows:
NF 10 1.585
NF2 1 NF3 1
1500 1
0.63
0.63 G2 0.63 G2 G3
(S2.15)
This allows us to solve for the isolated noise figure requirement of the LNA. We have
observed that the gain and noise requirements of the LNA stage have a major impact on
the overall noise performance. Thus, inserting the given NF3 into (S2.15) yields,
NF 10 1.585
NF2 1 10 1
1500 1
0.63
0.63 5 0.63 5 35 2
NF2 3 dB
(S2.16)
The possible specifications of the subcomponent for the receiver front end architecture
described in Figure P.2(a) are summarised in the following table.
Summary
A1 [dB]
-2
NF1 [dB]
2
A2 [dB]
20
NF2 [dB]
3
22
A3 [dB]
32
NF3 [dB]
10
A4 [dB]
0
NF4 [dB]
42
Table: Specification of the different sub-components.
23
Chapter 3
3.1
a) Shunt-Shunt feedback
b)
(i) From now on we deal only with single ended portion of the differential amplifier
(1) Without feedback
ieq 0
2
veq Noise from first stage Noise from second stage Noise from Ro
2
2
2
4kT f
g m 4kT g m
}
1
2
3
3 Ro g m 2
1
* important because cascode does not provide current gain
(2) With feedback
{4kT
veq
ieq
(ii)
*
*
2
2
veq
2
ieq
2
veq
R
2
2
F
4kTf
RF
(1) Without feedback (i.e. without Rf, Cf): 1/T at drain node of M1:
g m2
C gd 1 C gs 2
,
1
1
1
. Use zero value time constant: f 3dB
Ro C gd 2
T T1 T2
(2) With Rf in feedback: First find f-3dB open loop: Without feedback (but with
loading from Rf)
1/T at drain node of M2:
24
Use zero value time constant:
1/T at Node 1:
1/T at Node 2:
Therefore f 3dB
g m2
C gd 1 C gs 2
1
( R f // Ro ) C gd 2
1
T
1
g m2
1
( R f // Ro ) C gd 2 C gd1 C gs 2
Then find f-3dB close loop:
Need to find loop gain (with loading from Rf)
Loop gain = af, to calculate a, find forward amplifier with loading:
Note with shunt-shunt feedback, input is current and output is voltage.
vo (ii R f ) g m ( Ro // R f ) , where Ro//Rf means Ro in parallel with Rf.
1
v
Therefore, a o ( R f g m ) ( Ro // R f )
1
ii
if
1
By inspection: f
vo
Rf
25
loop gain = af=gm1(Ro//Rf)
f -3dB closed loop f -3dB open loop loop gain
1
( R f // Ro )c gd2
g m1 ( Ro // R f )
g m2
c gd1 c gs2
c) Looking at single-ended portion only of this circuit.
i)
without feedback capacitor Cf and assuming source is in series with L and Rf is
connected to ground.
Q
ii)
o L
can be quite large and causes peaking
Rf
with Cf,Rf feedback
Miller effect at input
R m ille r
C miller
2R
f
1 g m1 R o
(1 g m1 Ro )C f
2
26
Therefore effective Q is now reduced.
Circuit schematic and part (c) is adapted from text, “Non-linear integrated circuits”,
course notes, R. Meyer, 1986, U.C. Berkeley
3.2
3
NF 1
2
3 g m Rs
1
2 g m ro
2
2
1
2
3 g m Rs
1
2g m
2
2
2
3 g m g m Rs
1
3
2
2 g m ro
2g m
2
1
2
2 C gs 2 Rs
2
2
3g m1
3g m
3g m g m
1
1
3
3
2
2
27
3.3
a) For circuit in (i):
ii
*2
vi 2 4kT
ii
f
RF RF
2
vi *2 vi 2
For circuit in (ii):
ii *2 ii 2
vi *2 vi 2 ii 2 RE 2 4kTRE f
b)
For circuit in (i):
3A2
1
IM 3
2
32(VGS Vt )
(1 g m RF ) 2
1
1
IM 3 with open loop
reduction by open loop gain
For circuit in (ii):
3A2
1
IM 3
2
32(VGS Vt ) (1 g m RE ) 2
1
Here A is the amplitude of Vi or ii.
3.4
28
a) Zin cannot be made resistive. This is because Io in Figure 3.6b will no longer be 90o
leading Vin.
b) C1 can be used in conjunction with L3 to form a bandpass filter and serves to act as the
anti-imaging filter for the subsequent mixer and also for output matching.
3.5
We can use the classical formula for transconductance with source degeneration where
gm
Gm
1 gmZ E
(S3.1)
where ZE is the source degeneration impedance. (If the source is degenerated by a
gm
resistor RE this becomes the familiar formula Gm
.) In the case when the
1 g m RE
magnitude of the loop gain g m Z E 1 , (S3.1) becomes
Gm
1
1
1
under matching, which agrees with (3.106).
j c L 2
ZE
j L 2
3.6
8kTg m
8kTR 2 4C 4
8kTg m1 Rs 2 2C12
8kT 2C12
s
1
1
f
f
veq 4kTRs f
2
2
3 | Gm1 |2
3 | Gm1 |2
3
|
G
|
g
3
|
G
|
g
m1
m2
m1
m2
Q
where Gm g m
1
1 j
*2
3.7
Wideband
1-stage CS
Wideband
CS-cascode
Narrowband
CS; inductor
Degenerated
Gain
poor
good
Good
NF
good
poor
Good
Matching
poor
poor
Good
3.8
29
(a) Shunt-Shunt
(b) “a” and “f” circuit
vout iin ro g m (ro // R1 ) g m (ro // R2 )
3
1
2
1 2
R
R1
2
v
a out ro ( g m R1 )( g m R2 )
3
1
2
iin
i
f
gm
3
vout
(c) Rin
Rin
OL
1 af
ro
3
ro ( g m R1 )( g m R2 ) g m
3
1
2
3
1
( g m R1 )( g m R2 ) g m
1
(d) Without feedback, (but with loading from M3)
2
3
30
1/T at Node 1:
R1 (C gd C db
1
1/T at Node 2:
1
C gs g m R2 C gd )
1
2
2
2
1
R2 (C gd C db C gd C gs )
2
2
3
3
W W
Since , Cgs, Cgd of M1 Cgs, Cgd of M2
L 1 L 2
Since further R1R2, then pole at Node 1 dominates if we assume g m R2 C gd dominates.
2
3dB
1
1
T R1 (C gd C db C gs g m R2 C gd )
1
Then:
2
1
R1 g m R2 C gd
2
1
2
2
2
2
Otherwise:
3dB
1
1
T R1 (C gd C db C gs g m R2 C gd ) R2 (C gd C db C gd C gs )
1
1
Now with feedback:
3dB (1 af ) 3dB
withoutfeedback
If 1/T at node 1 dominates:
2
2
2
2
2
3
3
31
3dB
1
ro ( g m R1 )( g m R2 ) g m
3
1
2
3
R1 g m R2 C gd
2
2
g m g m ro
1
3
C gd
2
3
Circuit schematic is adapted from text “RF microelectronics”, B. Razavi, Prentice Hall,
1998.
32
Chapter 4
4.1
In practice, since Vlo is large (on the order of volts) and Vrf is small (on the order of
millivolts) there is a difference between the two. In particular in Figure 4.4, if there is
mismatch, Vlo+, Vlo- feed through (through Cgd1 , Cgd 2 ) will not be cancelled. In the
second case, since Vlo is applied at the gate, they will be insulated somewhat. Since Vrf is
small the feedthrough via M1, M2 will be relatively small. For re-radiation, if there is no
mismatch the coupling cancels out. Assume there is mismatch between M1 and M2, in
first case, Vlo+ and Vlo- both got coupled through M1, M3 to a low impedance mode (tail
node of SCP) with a gain of 1 (source follower). Due to mismatch they do not completely
cancel and the residue then got coupled through Cgd to the RF input and cause reradiation problem. Similarly in case 2, Vlo is coupled into a low impedance mode (tail
node of SCP) with a gain of 1 (cascode configuration) first before coupled to the input
signal Vrf+ and Vrf-. Due to mismatch they do not completely cancel and the residue
causes re-radiation problem. The re-radiation is roughly the same in both cases.
4.2
I rf I rf
I rf
2
I rf I rf
I rf I rf
I rf
2
I rf
2
, then
, and
33
I rf I rf
I rf
2
I rf I rf I rf I rf
I rf I rf
I if I if
Hence the swapping shown in Figure 4.8 is completely symmetrical with respect to yaxis. Therefore the output current does not contain any RF feedthrough and hence is
balanced, as we have claimed.
4.3
Now (4.9) does not take into consideration the dynamics of the circuit (R, L, C of internal
nodes). It is assumed the poles associated with these nodes are at much higher
frequencies than the LO frequency. The equation will not be true any more when the LO
frequency becomes comparable to these poles frequencies. Taking them into
consideration will be very complicated, since there are many nodes in the Gilbert mixer.
In the next chapter when we talk about passive mixer, where the circuit is a lot simpler,
an example of the conversion gain that takes this into consideration will be given.
4.4
IIP3 | dBm Pi | dBm
IM 3 | dB
means 20log AIP3=20log Ainterference+1/2(20logID1-20logID3)
2
->20logID1-20logID3=20log AIP32-20log Ainterference2
->ID1/ID3= AIP32/ Ainterference2 -> 1/IM3= AIP32/ Ainterference2 -> AIP32=Ainterference2/IM3.
34
4.5
10dbm 10 log
Setting IIP3 to -10dbm means
A IP 2 50 10
A IP 3 |dbmV
10
3
2 50 10
10dBmV
10
A IP
2
3
2 50
2 50 10
or solving we have
40dBV
10
0.1V . Substituting
into the equation (4.33) we have (Vgs1 Vt ) 0.03V , Now with P=5mW, Vdd=3.3V,
I=1.5mA. Since there are 2 branches substituting half of this into the square law current
equation, with a given k’ = 100uA/V2 (same value as used in LNA chapter) we solve for a
W
W
16700 that is
16700
L
L
This is a rather large number and is easily satisfied.
4.6
Case 1: 1 input sinusoid
So far we have treated input exponents, let us put in a real sinusoid
x(t ) A cos 0 t
1
Ae j 0t Ae j 0t
2
(S4.1)
which can be considered as a special case of two input exponents, where the two
exponents have exactly the opposite frequency.
Taking the transform, substitute in (4.49) and taking the inverse transform we have
35
2
2
A
A
y t H 2 j 0 , j 0 e 2 j 0t H 2 j 0 , j 0 e 2 j 0t
2
2
2
2
A H j , j A H j , j
2 0
2
0
0
0
2
2
Case 2: 2 input sinusoids
Finally let us turn to the 2-input sinsuoids case
x(t ) A1 cos 1t A2 cos 2t
1
1
A1e j1t A1e j1t A2 e j 2t A2 e j 2t
2
2
(S4.2)
Again taking the transform, substitute in (4.49) and taking the inverse transform we can
get the outputs. Rather than expressing the total output, let us do it by superposition. First
identify each term by the frequency component associated with it e.g. t terms
2t terms etc. Then we go and apply pairs of input frequencies to this system and see
what terms are generated.
First from input equation, (S4.2), collect all positive t terms in the input: .
e
e
j1t
j 21t
, e j 2 t . Substitute them into (4.49). It is obvious we generate 3 terms:
,e
j 2 2 t
,e
j (1 2 ) t
36
Next let us repeat by going to the input equation, (S4.2) again, and start by collecting all
negative t terms in the input: . e
j1t
, e j 2 t . Substitute them into (4.49). It is
obvious we generate 3 terms:
e
j 21t
, e j 2 2 t , e j (1 2 ) t
Then let us repeat by going to the input equation, (S4.2) again, and collect pairs of
positive and negative t terms in the input: .For e
into (4.49). It is obvious we generate 3 terms: e
for e
j 1t
for
e
j 21t
for
e
j 2 2 t
,e
e
j1t
,e
e
j 2 t
j 21t
j 2 t
,e
. we generate 3 terms: e
,e
j1t
.
j 1t
j 21t
j 21t
we
, e j 2 t . Substitute them
, e j 2 2 t , e
,e
generate
j 2 2 t
j (1 2 ) t
,
, e j (1 2 ) t
3
terms:
3
terms:
, e j (1 1 ) t 0( dc )
,e
j 2 2 t
j 2 t
.
we
generate
, e j ( 2 2 ) t 0( dc )
Let us label the terms just by its frequency components. We can see if we eliminate the
overlaps
between
these
terms
(e.g.
e j(1 1 ) t 0(dc) and
e j( 2 2 ) t 0(dc) are the same term) there are 7 distinct terms: 2 1 , 2 2 ,
1 2 , dc
We can represent all 7 terms by using the shorthand notation as a b
a = 1, 2,
b = 1, 2
37
4.7
(a) Pick
a=1,
b=2
and
apply
to
a 2 Y j a Y j b Siy
2
and
we
have
a 2 Y j1 Y j 2 S1S2 cos((1 2 )t Y( j1 ) Y( j 2 )) . Compared this with
(4.60) one can see that under the second
this
term.
Next
if
we
a2
expression, the first term agrees with
2
put
a=2,
b=2,
then
we
have
a 2 Y j 2 Y j 2 S2 cos(( 2 2 )t 2Y( j 2 )) . Compared this with (4.60) one
2
can see that under the third
a2
expression, the first term agrees with this term. Hence
2
the shorthand notation is consistent.
(b) Suppose
we
have
3
frequency
components
as
inputs,
i.e.
Si S1 cos 1 t S2 cos 2 t S3 cos 3 t , but we are still interested in the second order
term, a 2 Y j a Y j b Siy , then because there are three possibilities a=1, b=2 is
2
not the only choice. We can have a=1,b=2 a=2,b=3 a=3,b=1 . Hence for a=1,b=2 we
have
the
same
term
as
described
above:
a 2 Y j1 Y j 2 S1S2 cos((1 2 )t Y( j1 ) Y( j 2 ))
But
we
also
should
include
a
term
a 2 Y j 3 Y j 2 S3S2 cos(( 3 2 )t Y( j 3 ) Y( j 2 ))
and a term for a=3,b=1:
a 2 Y j 3 Y j1 S3S1 cos(( 3 1 )t Y( j 3 ) Y( j 2 ))
for
a=2,b=3:
38
(c) Suppose we are interested in the third order term
Third order term:
a 3 Y j a Y j b Y j c Siy
3
So a 1Y j a Siy a 2 Y j a Y j b Siy a 3 Y j a Y j b Y j t Siy ...
3
Suppose there are 3 input frequencies, then a=b=c=1 gives:
a 3 Y j1 S1 cos(31 t 3Y( j1 ))
3
3
and a=1,b=2,c=3 gives:
a 3 Y j1 Y j 2 Y j 3 S1S2S3 cos((1 2 3 )t Y( j1 ) Y( j 2 ) Y( j 3 ))
and a=1,b=2,c=1 we have
a 3 Y j1 Y j 2 Y j1 S1S2S1 cos((1 2 1 )t Y( j1 ) Y( j 2 ) Y( j1 ))
However
suppose
we
only
have
2
frequency
component
as
inputs
i.e.
Si S1 cos1 t S 2 cos 2 t , let set a=1,b=2,c=1 then we have
a 3 Y j1 Y j 2 Y j1 S1S2S1 cos((1 2 1 )t Y( j1 ) Y( j 2 ) Y( j1 ))
Note this is a subset of 3 frequency component as inputs with a 3rd order system.
4.8
2
From (4.99): the first term is v
Now vin is a first order term.
in
v
s
2
39
For vs let us repeat (4.61) here:
v t v v v ... H v H , v 2 . ..H , , v 3 .....
s
s1
s2
s3
1
in
2 1 2
in
3 1 2 3 in
Paying attention to the expansion after the second equality sign it is obvious then v s
v
consists of sum of vin terms of first order and above. Since in v
s
2
2
consists of the
square of the difference of vin and vs, then upon squaring, each expanded term will
consists of cross-product between vin and vs and it is obvious that each expanded term
2
will consists of terms that are vin terms of second order and above. Hence v
in v
s
2
does not contribute to any vin terms of first order.
Next, how about the vin
term? Again expanding vs term, but paying
2
v V
V
s GS
t
2
attention to expansion after the first equality sign and we have:
v t v v v .......
s
s1
s2
s3
40
The only term in this expansion that will contribute to a first order term is vs1, and so
upon substituting the term that has contribution to first order the term becomes
v
.
2 in v V
V
s 1 GS
t
2
4.9
V -> with this approximation
VGS-VT is quite small -> j C d 2k V
1
2
GS
t
(4.80)
becomes
k
4
->
H 2 1 , 2
j1 2 C d
this
makes
k H H 2 1, 2 H 2 2 , 1
G ( , , ) 2 1 1
3 1 1 2
2
3
k2 1
1
1
24 jC d
1
1
2
2
1
1
is defined to be at , substituting and G3 becomes
2
1
k2
3
G ( , , )
3 1 1 2
24 jC d 2
1
1
kI
From (4.116) G
1 2
SS
(4.129)
becomes
41
From (4.121) IM
3
G ( , , ) A 3 cos t cos t 3
3 1 1 2
1
2
G A cos t
1
1
A 3 cos t cos t 3
3
1
2
In addition
A2
A cos t
4
1
Substituting we have | IM 3 |
9
k 1.5 A 2
96 C I SS
1 d
2
Now I SS 2k V
V 18.72 10 8 A
GS
t
1
Substituting we have IM3=8.98 or 19db.
IM3 is substantially worse than IM3 calculated in numerical example 4.4.
IM3 will be much larger than HD3.
4.10
I
1
kI
From (4.116) G
. From (4.23) a 1 SS n . From equation following (4.16)
1 2
SS
2
I SS
2I
. From (4.20) a 1 i rf / rf
I SSn SS . Substituting a 1
k
k
n
42
i
i
Now G d rf . Therefore upon comparison we should pick the normalising factor
1 v in v rf
to
be
i rf / i rf
k/2.
n
1
kI
G1
SS
2
k/2
k/2
I SS
k
If
we
use
this
to
normalise
G1
we
have:
a1
4.11
a)
u
observe
launch
v
observe
u
v
T
v+T
launch
h is different because h is periodic
in launch time, NOT OBSERVE
time. As a side note, had the system
been a LTI system, since the observe
time is different (even launch time is
the same), h will still be DIFFERENT
Notice, one common misconception about an LPTV system is that if the observe time is
delayed by one period, then h is the same. This is INCORRECT, as shown above.
Therefore when one talks about LPTV, one refers to getting the same response if one
delayed by one period in the launch time BUT NOT the observe time.
b) From (4.146) -> Y( if )
H(
if
, s )X( s )d s
)
And from (4.142) -> H , H n ( if ) ( if n
if s n
s
LO
Substituting (4.142) into (4.146) ->
43
Y( if )
H
n
n
( if ) ( if n
)X( s )d s
s
LO
Swapping the order of integration and summation ->
Y( if )
H
n
H
if
Y
H
n
n
( if ) ( if n
)X( s )d s
s
LO
( if ) ( if n
)X( s )d s
s
LO
n
n
n
2
( if )X( if n
LO
)
Y( if )Y( if ) H n ( if )X( if n
) H n ( if )X( if n
)
LO n
LO
n
*
*
H n ( if )X( if n
) X * ( if n
)H n ( if )
LO
LO
n
n
2
H n ( if ) X if n LO
n
2
if nLO X if mLO 0 for
The last step occurs because any cross term X
*
nm as the narrowband noise from different frequency bands are assumed to be
uncorrelated.
4.12
In general h
is more than a simple gain function and indeed can have memory.
Nevertheless the impulse response is usually short.
a) mixer loaded with Rtermination
*
44
Here Vif
Vrf RTer min ation
RTer min ation RM 1
Vrf RTerm
RM 1
(as Rtermination typically = 50 and is << RM1 (typically ~1k))
Vrf k 'VLO RTerm Vrf h
where h is the transfer function k ' V R
. h behaves like a gain function which is
LO term
periodically varying.
Then
H 1 amp. _ of _ 1st _ harmonics _ of _ Vlo 2
H0
dc _ of _ LO _ waveform
Where system is described by h k 'VLO Rterm and is memoryless.
b) mixer loaded with C
45
Now 1st approach: treat the system as a LTI system and incorporate the time varying
feature by replacing the component R by R(t).:
V h(t )V ( )d
if
rf
t
e
t
he
R
R
C
M 1 V ( )d
rf
C
M1 , but RM1=1/(k’VLO). Substituting:
tk ' V
LO
C
he
Now we can see h is not periodic in T. However we know from Figure 4.20 that h for a
LPTV system should be periodic in launch time. Hence the above expression is wrong
and the approach is wrong. Actually one cannot find h by assuming a linear system and
replacing the component R by R(t). One has to find impulse response h(v,u) which
depends on both launch and observation time.
Even for such a simple circuit, closed form solution of such an impulse response is not
trivial to find. First it will be obvious that h is not VLO and
46
H 1 amp. _ of _ 1st _ comp. _ of _ Vlo
H0
dc _ of _ VLO
2
To derive Hn basically we start by looking at the two phases of the LO signal and sum the
response to a tracking system and a holding system. During tracking, one is doing natural
sampling of a signal and the response is found by multiplying the Fourier transform of
switching function (1/) to the Fourier transform of the impulse response of the LTI
system (an RC network). During the hold mode, the switch is turned off and the output
held. The response during this phase is found by applying the impulse samples, weighted
by the sampled and held values, to the RC filter (an LTI system). The final expression is:
out
sin
2
j
lo
H (
, ) H (
)
1 out lo
0 out
out
lo
out
2
lo
exp
1
j out Cload
H 0 ( out )
1
Ron
j out Cload
Here out is the output frequency of the mixer and is equal to if. For a downconversion
out=if =rf-lo
Now let us look at a more detail derivation of the above expression for H1. We start with
(4.149), rewritten below:
j v jn u
1
if
lo du
H T hv u, u e
dv e
n if T 0
47
The conversion gain, or H1 can then be written by setting n=1:
j v j u
1 T
if
H hv u, u e
dv e lo du
1 if T 0
With t=v+u and for IF << LO frequency
j u
1
H 0T ht , u dt e lo du
1 if T
For the circuit under consideration, there are only two launch phases that matter. The first
launch phase corresponds to launching an impulse so that the response dies down before
the switch is turned off. The second launch phase corresponds to launching an impulse so
that the response is still on when the switch is turned off. Since we assume the RC time
constant is small compared to the period, we assume the first launch phase has its impulse
launched at T=0. The network is a simple RC network. The second launch phase has its
impulse launched just right before T/2. The network is a held RC network. It is
equivalent to a cascade of two networks: a RC network followed by a hold network.
(The impulse responses for each launch phase are similar to that obtained for the circuit
drawn in fig.5.1 of Chapter 5, except that now instead of a resistive terminated load we
have a capacitive load.) Because there are only two launch phases the expression for H1
can be best evaluated by breaking it into two integrals:
j u
j u
1
1
H 0T / 2 ht , u dt e lo du T ht , u dt e lo du
1 if T
T T /2
From 0 to T/2, no matter when you launch the circuit configuration remains constant.
Hence h(t,u) is independent of u and we denote it h1(t). From just before T/2 to T again
no matter when you launch the circuit configuration remains constant. h(t,u) is again
48
independent of u. However it is different from h(t,u) from 0 to T/2. We denote it h2(t).
Therefore
j u
j u
1
1
H 0T / 2 h1 t dt e lo du T h2 t dt e lo du
1 if T
T T /2
Now the integral of h1 is zero beyond T/2 and the integral of h2 is zero beyond T.
Furthermore we can time shift h2 by T/2 Hence H1 can be written as:
j u
j u
1
1
H 0T h1 t dt e lo du 0T h2 t T / 2dt e lo du
1 if T
T
Due to natural sampling we have h1 is the impulse response of the network when switch
is on. The network is a simple RC network. Therefore h1 is given as:
t
R C
M1
h e
1
h t dt is simply the first coefficient of the Fourier
In the expression of H1 the term
1
transform of h1. It is denoted as H . Using the above expression of h1 it is evaluated to be
0 if
1
j if C load
H 0 ( if )
1
Ron
j if C load
Due to sample and hold we have h2 is the impulse response of the network when the
impulse is launched just before switch is off. The network is a hold RC network. It is
equivalent to finding the impulse response of a cascade of two networks: a RC network
followed by a hold network. The holding, however, always occur at u=T/2. Hence the
total response is just the multiplication of the impulse response to the RC network and
this holding function. The impulse response to the RC network is the same as h1, except
since the impulse is launched at T/2, the response is delayed by T/2. The holding function
of the
hold network is given by a rectangular function and is denoted as
(u /(T / 2)) where T/2 denote how long it is being held for. Since the hold function starts
at T/2 we actually has ((u T / 2) /(T / 2)) Hence h2 is given as:
(t T / 2)
R C
M 1 (u T / 2) /(T / 2))
h2 e
49
In the expression of H1 the term h2 t dt is simply the first coefficient of the Fourier
transform of h2. Using the above expression for h2 it is evaluated to be
if
2
lo (u T / 2) /(T / 2))
With these substitutions H1 becomes
H ( ) exp
0
if
if
j u
2
1
1
lo
H T H ( )e lo du T H ( ) exp
0
1 if T 0 0 if
0
if
T
j lo u
(u T / 2) /(T / 2)) du
e
Factoring out the constant in the integral and H1 becomes
if
1
j
u
2
j
u
lo 1 T e lo (u T / 2) /(T / 2)) du
H H ( ) 0T e lo du exp
1 if
0 if T
T 0
The two integrals are just the first Fourier coefficients of a dc and a rectangular function.
sin
The first integral evaluates to j/ and the second integral evaluates to
2
if
lo
if
lo
if
if
sin
2
j
lo exp 2 lo
Hence H1 becomes H ( , ) H ( )
1 if lo
0 if
if
lo
This is the same as the expression derived qualitatively above, with out=if
50
4.13
Using (4.171)
S
2
2
2
H (100MHz ) S (100MHz ) H (100MHz ) S (1.9GHz ) H (100MHz ) S (1.7GHz )
n3
0
n2
1
n2
1
n2
H LTI (100 MHz )
2
5 pA2 0.63 H
Hz
LTI (100 MHz )
2
50 pA2 2
Hz
2
=20.25(5pA)
Comparing this to (4.175), this is 20.25/0.45 or 45 times larger, hence Ndev is also 45
times larger. Hence NF equation, (4.177), becomes NF=1+4510.12=456 or 26.5dB. This
is a lot worse than the 8.2dB calculated without taking the switching effect into
consideration.
51
Chapter 5
5.1
Let us check the assumptions. R=50 , W/L=762; VGS-VT=4V means RON=3.3 ;
R>>RON, and assumption (5.13) is satisfied; for assumption (5.14), W/L=762 means
1
1
C=9.1pF. Now
5.3GHz . At 1.9GHz, the assumption is
2R
C 2x3.3x9.1pF
ON
still not violated but is less accurate. A more complete analysis may be needed.
5.2
In triode region Cgd=Cgs=1/2WLCox=1/2762u1u2fF/u2=0.76pF, C=Cgd in series with
Cgs=0.35pF, Ron=3.3ohm, R=50ohm
1=0.078ns, 2=0.001ns, since 1>>2 Gc depends on 1 only and can be approximated
from (5.29) to become
1
2 f lo 1
1 f lo
1 e
1
1
R
G
2
c R
R
1 2f lo
ON
1
2
2
Substituting Gc=0.29 or –10db
5.3
The original switching mixer with resistive load as shown in Figure 5.1 is redrawn in
Figure S5.1a, where it is shown that in hold mode, Vif is reset to ground. The distortion
formula is shown in (5.10) for low frequency (Taylor series expansion) and (5.15) for
high frequency (Volterra series analysis), where the non-linear junction capacitance is
included.
In Figure 5.8, we have modified this switching mixer by deliberately
introducing a capacitor Cext at the output of M1 to filter out the RF feedthrough. This is
52
redrawn in Figure S5.1b. It is seen that in the hold mode, Vif is held. The high frequency
distortion formula is still the one given in (5.15) with R=0 and C=Cext. Next we redraw
Figure 5.10 and more importantly, the relevant waveforms in Figure 5.1c, where the held
waveforms are shown in bold. The high frequency distortion formula is essentially the
same as (5.15) with R=0 and C=Cload, also the value is reduced by 1/. One subtle
difference between (b) and (c) is that Cext is introduced in (b) for filtering out the RF
feedthrough whereas Cload is introduced in (c) to hold the IF signal. Because there are
different requirements the value of C can be different. Typically Cload is larger than Cext
and so (c) has poorer IM3.
Now at this point in time the reader may discover the circuits in Figure S5.1b and Figure
S5.1c are very similar. The circuits in (b) and (c) are structurally same, but that due to
different design requirements, arising out of different applications, the design
methodologies are quite different in some aspects.
53
Figure S5.1a
Figure S5.1b
54
Figure S5.1c
5.4
Note transistor M2, has a non-linear Ron that has to be considered. However as noted in
section 5.7 if Vrf can be assumed to be almost constant in the S/H circuit case, then the
voltage at the bottom plate of Cload is almost constant and distortion introduced by M2 is
negligible. For equal second harmonic distortion, the device constant k2 of the bottom
transistor is related to the top switch k1 as:
k 1C
k
2
2(VGS Vt )
Increasing the switch size may not always help. This is because this may increase the
distortion since an increase switch size results in an increase in the load capacitance of
the mixer buffer. This causes an increase in the fall-time of the LO resulting in higher
time-varying and sampling distortion.
55
Providing a virtual ground via the use of an opamp is not feasible for high frequency
mixer.
5.5
a) Threshold Modulation
Body effect VT modulation 1 1 1
2
0.56
2 2 VSB
f 0.393
VSB 2.5V
FBodyEff 1 1 1.3332 1.69 4.5dB
2
b) Velocity Saturation
g ds
Fvel = Velocity saturation factor = 1 L
1 2 g ds R
2 f
2
C ox W 2 RVsat
1
, where C=Cload=0.352pf,
C
1
W 110u
then is calculated. Next gds=1/Ron, Ron
. If we assume
L
0 .8 u
k (VGS VT V DS )
1
47.5 . We
VDS is small and therefore neglect VDS2 term, then R on
k (VGS VT )
arbitrarily set this to 50 for ease of calculation. Substituting we get:
Fvel 1.5dB
Now assume f 0.5 . With a capacitive load R
c) Mobility Modulation from Normal Field
IM 1 V
V IM , where is the parameter that relates the effective mobility
3
GS
T
3
due to the normal field eff to the normal mobility as follow:
eff
1 V V
GS
T
.
56
Substituting we have IM 1 0.051.45IM 0.942IM 3 . Hence FNormal=-0.654db.
3
3
5.6
To calculate distortion one go through the following procedure:
Step 1: assume LO is on all the time, then IM 3 IM 3C where C stands for continuous
time.
A2
C
IM 3
4 k VG Vt 3
Next put into 2nd order effect, and modify IM 3C to IM 3C Mo d where C-Mod stands for
continuous time modified due to threshold modulation velocity saturation and mobility
modulation.
g ds
2
1 VGS VT IM 3C
IM 3CMod 1 1 1 L
1 2g ds R
(S5.1)
Step 2: LO is falling with zero tf, with a falling LO, one has to include distortion due to
charge injection - IM 3CI . Now IM 3CI can come from a few sources. Let us concentrate
on IM 3CI due to body effect (due to body effect, impedance is different and signal
dependent , therefore Qerror is a function of signal which causes distortion.)
IM 3CI
3
128
A 2 COX L
C VBS 2 f
5
2
Then one calculate Total IM 3TOT C ,where TOT_C means Total continuous time
Distortion, as IM 32CMod IM 32CI (neglecting phase relationship between the 2 terms).
Step 3: Incorporate finite tf with finite fall time, one has 3 different sources of distortion.
This is further divided into 3 regions, depending on value of tf, let us look at different
region of tf,
57
in region (1) IM 3final IM 3TOT C
in region (2) IM 3final IM 32TV Mod IM 32CI where TV-Mod stands for Time varying
Modified.
IM 3TV Mod is modified from IM 3TV like in (1).
To see how this is done let us start from original equation for time varying distortion.
3
IM 3TV
3A 2 3 2 C
j
8 4 k
1
2
Tf
VG
5
2
0.0913
Basically modification can be carried over like in equation (S5.1) except all the indices
will be modified by a factor ½. This is because IM3TV is modified with the short channel
1
C 2
effects via the dependence of K. Now with k appearing as . Therefore the factor is
k
1
also modified by
.
1
2
IM 3TV Mod IM 3TV
K NewK IM 3TV
1
1 VGS VT 2
1
g ds
1 L
1 2 gds R
1
2
1 1
where k ' C ox is modified now since (the mobility) is modified.
*This is true even though the original differential equation has to be modified but the
resulting additional non-linear term is small and can be neglected.
In region (3) IM3 final is dominated by IM3sample and is set equal to
2
1 AT f
IM 3 final IM 3sample
8 2VG
Notice in the case, IM3sample is not modified by the second order effect. This can be
understood in the case of velocity saturation and mobility degradation mechanism by
referring to Figure S5.2
58
VT
tcut off
Figure S5.2
In Figure S5.2 the cutoff point tcutoff does not depend on mobility or velocity saturation.
It does depend on VT which can be modified, but again the effect can be shown to be
small for the level of distortion experienced in this region. Finally no charge injection
effect is included because with a slow fall time, the channel is almost always in
equilibrium and most (all) of the charge flows back to the source. Consequently there is
no error charge on C. In Step 3 for a give tf we compare IM 3TOT C IM 3TV MOD IM 3sample .
Whichever is the largest dominates and is the final IM 3 IM 3 final .
100MHz
1.8GHz
Signal 1.9GHz
+
LNA
interference
gain=20dB
mixer 1
IF amplifier
gain=13dB
mixer 2
A/D converter
: Example Architecture
Figure S5.3
The following is an example that applies the above procedure first to a 100 MHz IF
sampling mixer and then to a 1.9 GHz RF sampling mixer in a DECT application. The
architecture is assumed to be shown in Figure S5.3.
To calculate distortion, from numerical example 5.2 we know that the input consist of a 83 dBm signal together with a -33 dBm interference. Therefore Pinterference= -33 dBm.
First let us assume the following parameters:
W 110
Cload = 0.352pf
L
0.8
A
k 0.0137 2
V
59
Therefore from step 1
A 2 C
IM 3
3
4k VGS Vt
Furthermore from question 5.5 we have calculated:
g ds
Fvel = Velocity saturation factor = 1 L
1
2
g
R
ds
2 f
2
C ox w 2 RVsat
with f 0.5 , g ds 1 / 50 . For
W 110
1
,R
C
L
0.8
Fvel factor 1.5dB
Body effect VT modulation 1 1 1
2
0.56
2 2 VSB
f 0.393
VSB 2.5V
FBodyEff 1 1 1.69 4.5dB
2
Mobility modulation from normal field 1 VGS Vt
FNormal 0.654dB
100 MHz Mixer 2 design
Step 1:
A 2C
IM 3C
3
4k VGS Vt
Here amplitude at the input to mixer 2 = -33+20+13=0dBm
0.05V 1
60
Pint erference 0dbm A 0.316V
2 100 10 6 Hz
C 0.352pf
A
k 0.0138 2
V
VGS Vt 2.5 1.05 1.45V
IM 3C 77.64db
Put in second order effect
IM 3 IM 3C Mod 77.6 FVel FBodyEff FNormal
= - 77.64 + 1.5 + 4.50 - 0.65
= - 72.3 dB
Step 2
IM 3CI
3 A2COX w L
128C V X
5
3.05 x10 5
2
VX VSB 2 2.5 2 0.393V
VX 3.286V
0.561V
1
2
C OX 2fF / u 2
W 100u
L 0.8u
C 0.352pf
IM 3TOT C IM 3CMOD IM 3CI 72.2dB
2
2
Step 3 with finite fall time depends on fall time
In region 1, small fall time IM 3 IM 3TOT C 72.2dB
In region 2, let us first calculate IM 3TV
61
3A 2 3 C
8 4 k
32
IM 3TV
0.0913
120dB
120
2
Tf
VG
5
2
120.3dB
Tf 0.4 ns
VG 2.5V
IM 3T .V . MOD 120dB
1
1
FVel FBodyeff FNorm al
2
1
1.5 2.5 0.654
2
1
x 3.346dB 121.7dB
2
IM 3 IM 3TV MOD 121.7dB
In region 3
IM3-sample
A2 T f
:
32 VG
2
Tf = 0.4 ns
: -90 dB
IM 3 IM 3sample
Comparing essentially for Tf = 0.4 ns IM3 is dominated by IM3-C so IM 3final 72.2dB
Next mixer 1 at 1.9GHz is designed. Amplitude A at input to mixer 1 = -33+20=-13dBm
Step 1
A 2 C o
Pint erference 13dBm
IM 3C
90.9dB
3
4k VG Vt
2 1.9GHz
FVel
g ds
1 L
1 g ds R
g ds
1 g ds R
1
50
g ds
1
1
1
9
50 2 1.9 10
1
50
1
R
jC
g ds
62
FVel
dB
same as previous case = 1.5 dB
Fbody_ Eff
dB =
same as previous case = 4.5 dB
FNormal = same as previous case = -0.654dB
Therefore IM 3CI Mod 90.9 1.5 4.5 0.654 85.6dB
Step 2
Since IM 3TOT C IM 3CMod IM 3CI
2
2
IM 3CI same 3.05x10 5
IM 3TOT C 84.4dB
Step 3
Region 1
IM 3 IM 3C 84.4dB
3
IM 3TV
3A 2 3 2 C
8 4 k
IM 3TV IM 3TV
1
FVel
2
dB
Fbodyeff
1
2
dB
Tf
VG
5
Fnormal
2
69.6dB for T f 0.4ns
dB
69.6 12 3.346 71.3dB
Region 3
Sampling Distortion:
IM 3 sampling
A 2 Tf
32 VG
2
IM 3 sampling
dB
64.8dB
At this frequency IM 3TV MOD and IM 3 sampling dominate and they add together so the
2
2
worse case IM 3 final IM 3TV
Notice even though A has
MOD IM 3 sampling 63.1dB
gone down, but has gone up and IM 3T V and IM 3sampling increases as 3 and 2 so
they finally become dominant.
5.7
To calculate the NF including both n_lo_output,S2 and intrinsic noise with switching
(general LPTV) we have to interpret the NF formula differently. This is because NF as
given by
63
NF
N dev N source _ resis tan ce _ output
N source _ resis tan ce _ output
is calculated from PSD of the individual noise (such as Ndev is the PSD of the intrinsic
device noise). The PSD of the noise process, Nlo_output, is not even defined, as it is a
cyclostationary process.. We only know the integrated noise or the variance n_lo_output2.
Hence we have to reinterpret the NF formula:
NF
N dev N source _ resis tan ce _ output
N source _ resis tan ce _ output
so that Ndev, Nsourcce_resistance_output are interpreted as integrated noise power, and not as
PSD. Also this is being calculated at the output. Now to calculate Nsource_resistance_output in
the integrated sense, you need to find the conversion process from input to output of the
mixer, under general LPTV condition. How about Ndev? There are two independent
sources: Ron and Nlo_input. To find integrated noise at output due to Ron, again you need to
find the conversion process from input to output of the mixer, under general LPTV
condition. To understand the conversion process from input to output, under general
LPTV condition, we need to know the gain and the bandwidth involved with the
corresponding conversion gain.
Now the conversion process from input to output, under special LPTV condition, and the
accompanying gain and bandwidth involved with the corresponding conversion has
already been described via (5.117). We will refresh the readers memory by going through
the description once more. Remember (5.117) deals with the case when we have
switching, but under special LPTV. It differs from (5.112) in the sense that switching
(special LPTV) introduces both a gain (1/) factor and a noise folding factor (fb/flo),
64
(provided fb > flo). This folding factor is calculated by assuming that all the noise at
multiples of flo will alias, each with a bandwidth of flo. We can assume all the noise
within a bandwidth of flo will pass because when we refer to (5.46), we see that the –3db
bandwidth of the sampling mixer is given by 1/RonC. By design this bandwidth has to be
larger than frf, so that Vrf can pass through the mixer. However frf is larger than flo and so
the mixer has a –3db bandwidth larger than flo, which means all the noise within flo will
pass.
How does the output noise derived under the present conversion process, (general LPTV
condition), differs from (5.117)? When we deal with switching, general LPTV, the mixer
conversion gain has additional filtering effect, which is not there in the special LPTV
case. This filtering effect first modifies the gain, as the gain now has a different
frequency dependency. It also filters the noise, so that not all noise within a bandwidth of
flo is passed and aliased. Both of these effects are evident when we compare the new
conversion gain, Gc given by
out
sin
2
j
lo
G (
, ) G (
)
c out lo
c0 out
out
lo
G c 0 ( out )
out
2
lo
exp
1
out C load
1
R s R on
out C load
to the old conversion gain, given in (5.46) and repeated below:
65
1
1
G
c
C
rf
1
C
rf
R
on
This new conversion gain can be interpreted as having a new gain and a new –3db
bandwidth when compared to the old conversion gain.
Now we can substitute the proper values given in numerical example 5.2 into this Gc
formula and get a numerical value for this conversion gain. To make things more
interesting we will show the reader yet another alternative and simpler approach. Let us
again take a look at this Gc formula. First, the overall-3dB bandwidth is a product of the
bandwidth in the first term and the bandwidth in the second term of that formula. One can
calculate that the second term falls to –3db at roughly around 1GHz. The first term in the
formula, Gc0, has a -3db bandwidth of 1/(2RonCload). From the numerical example 5.2 we
calculate Ron=50ohm, hence the Gc0 -3db bandwidth=1(250ohm0.352p)=9Ghz.
Hence the overall -3db bandwidth is dictated by the second factor, and is around 1Ghz.
Secondly the low frequency gain of Gc is obtained by setting if in the formula to be zero.
|Gc| becomes |j/-1/2|=0.6. Notice this is slightly different from the gain of 1/ in (5.46).
One more difference occurs when we deal with switching, general LPTV, which is not
there in the special LPTV case. This difference is that the conversion gain for noise at flo,
2flo … are all different. (In special LPTV case, they are all the same and equal G c). In
general LPTV case, the conversion gain for noise at flo is denoted as H1, and is the same
as Gc given above. How do we find H2, H3 …? It turns out they are difficult to find and
that is why in the problem you are asked to simply set them equal to H1, which in turn
equals Gc. In summary we have calculated Gc using two approaches.
66
With this in mind we proceed to calculate integrated Sn2 due to Ron:
Following the conversion process from input to output, under special LPTV condition,
which leads to (5.117), we repeat that for the present case, under general LPTV
condition: First due to new low frequency gain of 0.6 we have:
Sn2=4kTRon0.6(fb/flo)
Notice the (fb/flo) is due to all the extra noise contribution at LO, 2LO … frequencies.
They are the same because we assume all the Hn are the same.
Next because of filtering only 1G (out of flo=1.8G) of this noise is admitted and folded.
Hence
we
have
Sn2=Sn21G=4kTRon0.6(fb/flo)1G=41.38e-
Integrated
23300500.6(9G/1.8G) 1G=2.5e-9V2
Now we can apply the above argument to determine the integrated noise due to source
resistance
of
50ohm
at
the
output
of
the
mixer:
Nsource_reisstance_output=4kTRs0.6(fb/flo)1G =2.5e-9V2
Then Ndev, which is the integrated device noise contribution at the output of the mixer,
comes from roughly n_lo_output,S2 + integrated Sn2 (this is rough because you can only sum
when the circuit under consideration is a time varying LINEAR circuit. The n_lo_output,S2
formula, however, has been derived for a time varying NON-LINEAR circuit).
a) case 1: For n_lo_input2 of 1kohm
n2 _ lo _ output,S A 2 Vin2 0
1
2
3
k Tf
C load VG
3
2 2 2
in n _ lo _ input 0.443
0.0138 0.4n 2
2
2 1.9G 4kTR 0.443
0.352p 2.45
4 10 7 V 2
67
n_lo_output,S2 dominates and Ndev becomes 410-7V2
NF
N dev N source _ resis tan ce _ output
N source _ resis tan ce _ output
4 10 7 2.5 10 9
22db
2.5 10 9
case 2: Now for n_lo_input2 of 50ohm,
n_lo_output,S2 = 210-8. This dominates.
NF
N dev N source _ resis tan ce _ output
N source _ resis tan ce _ output
2 10 8 2.5 10 9
9db
2.5 10 9
As a final note, when n_lo_input2 comes from a 50 ohm resistor, n_lo_output,S2 becomes
comparable to integrated noise due to Ron. Hence one can see that the sampled LO noise
for this fall time can become significant because ideally the system are all terminated by
the same 50ohm resistance.
b) case 1: n_lo_input2 of 1kohm
n_lo_output_dc2=4.5e-9 and with Integrated Sn2=2.5e-9V2 Ndev becomes 7e-9, NF=5.8db
Case 2: n_lo_input2 of 50ohm
n_lo_output_dc2=2.2e-10 and with Integrated Sn2=2.5e-9V2 Ndev becomes 2.7e-9, NF=3.2db
68
Chapter 6
6.1
a)
Adopted from “Oversampled Delta-Sigma Data Converters: Theory, Design and
Simulation”, James C. Candy and Gabor C. Temes. Figures 1.5.
b)
Adopted from “Oversampled Delta-Sigma Data Converters: Theory, Design and
Simulation”, James C. Candy and Gabor C. Temes. Figures 1.15b.
c)
Adopted from “Oversampled Delta-Sigma Data Converters: Theory, Design and
Simulation”, James C. Candy and Gabor C. Temes. Figures 1.15a.
69
6.2
Let us denote H1 as the transfer function of integrator in the forward loop and H2 as that
of the integrator in the feedback loop. Eadc is the quantization noise introduced by
quantizer. Thus the signal X and output Y can be expressed in the following:
( X Yz 1 Yz 1 H 2 ) H 1 E adc Y
(S6.1)
Rearrange the above expression,
H1
1
Y
X
Eadc
1
1
1
1 H1 z H1 H 2 z
1 H 1 z H 1 H 2 z 1
(S6.2)
Since
H1
1
1 z 1
(S6.3)
and
H2
1
1 z 1
(S6.4)
Then substituting H1 and H2 into (S6.2), we get
Y (1 z 1 ) X (1 z 1 ) 2 E adc
(S6.5)
Hence STF 1 z 1
NTF (1 z 1 ) 2
Adapted from textbook McGraw Hill, Analog VLSI, (editor. M. Ismail, T. Fiez)
“Oversampled A/D Converter”, Chapter 10 author B. Leung.
6.3
In general, the noise shaping function for a L-th order sigma-delta modulator is
the following:
NTF (1 z 1 ) L
The spectral distribution of the quantization noise power after shaped is
2
2
2
S ee ( f ) NTF e rms (2 sin( f / f s )) 2 L e rms
where erms2 is the quantization noise power of the uniform quantizer, which is
2 / 12 if white noise is assumed.
70
The total quantization noise in the band of interest [-fN/2, fN/2] is given by
fN / 2
nbw
2
S
ee
( f )df
fN / 2
erms
fN / 2
e
2
rms
fN / 2
2
df
e
rms
fs
2
fN / 2
(2 sin( f / f
s
)) 2 L df
fN / 2
2 L 1
2L f N
erms 2
2L 1 f s
Adapted from textbook McGraw Hill, Analog VLSI, (editor. M. Ismail, T. Fiez)
“Oversampled A/D Converter”, Chapter 10 author B. Leung.
6.4
First let us denote OSR as the oversampling ratio.
(a) In P6.2,
U 1 k1a
z 1
( X Y1 )
1 z 1
z 1
k1bU 1 2k1a k1bY1
1 z 1
z 1 z 1
2 z 1
k1a k1b
X
Y1
1 z 1 1 z 1
1 z 1
In a single bit quantizer the choice of gain or slope of the transfer characteristics is
optional; hence
1
Y1
U 2 E1 or U 2 k1a k1b Y1 E2
k1a k1b
U2
Y1 E1
1 z
1 2
z 1 z 1
2 z 1
X
Y1
1 z 1 1 z 1
1 z 1
2
z 1 2 z 1 Y1 z 2 X 1 z 1 E1
2
Y1 z 2 X 1 z 1 E1
The signal fed to the second stage is
j1U 2 j1 k1a Y1 E1 j1 k1a k1b z 2 X 1 z 1 E1 E1
also
Y2 j1 z 1U 2 1 z 1 E2
After scaling by g1
2
71
V2 g1 j1 z 1U 2 g1 1 z 1 E 2
2
g1 j1 k1a z 1 z 2 X 1 z 1 E1 E1 g1 1 z 1 E 2
1
V3 V2 V1 V2 z Y1
zz 3 X z 1 1 z 1 1 z 1 E1 g1 1 z 1 E 2
where 1 g i ji k1a k1b . Double differentiation of V3 results in
2
V4 z 3 1 z 1 X z 1 1 z 1
Y V1 V4
2
1 z 1 z E
4
1 2
1
1
3
g1 1 z 1 E2
z 3 1 1 z 1 X z 1 1 z 1 z 1 1 z 1 E1 g1 1 z 1 E 2
Neglecting the fourth order noise and slight frequency distortion in X, we obtain
2
2
4
3
Y(z) z 3 X(z) z 1 1 z 1 E1 (z) g1 1 z 1 E 2 (z)
For the case of perfect matching (i.e. 0 or g1 1 / j1k1a k1b )
2
3
Y ( z ) z 3 X ( z ) g1 (1 z 1 ) 3 E2 ( z )
(b) PSD of noise in a quantizer is . After shaping by the L-th order differentiator
(1 z 1 ) L has a psd
( )
Q2 T
(2 sin(
T
2
)) 2 L
and its integration over baseband yields
/ OSR T
T 2 L
2
2 T
N Q
(2 sin(
)) d
0
2
2L 1
2 L 1
2
Q for OSR
2L 1 OSR
where OSR is the oversampling ratio. From the results of part (a) for the nonideal case
( 0)
4
6
2
2
g
Q2 2
Q1
1
5
7
5OSR
7OSR
2
where is the standard deviation of and Q1 , Q2 2 , are variances of E1(z) and E2(z)
respectively.
2N 2
The increase in quantization noise due to the component mismatch is the ratio of N2
with and without mismatch.
7 2 OSR 2 2
Q1
quantization _ noise (dB) 10 log 10 1
5 g1 Q 2
Adapted from textbook McGraw Hill, Analog VLSI, (editor. M. Ismail, T. Fiez)
“Oversampled A/D Converter”, Chapter 10 author B. Leung.
72
6.5
Following similar steps as in solution 6.4(a),
z 1
I1 ( z ) X ( z ) I1 ( z ) 2Y1 ( z ) E1 ( z)
Y1 ( z )
1 z 1
substituting for I1(z) and simplifying the result
z 2
(1 )(1 z 1 ) (1 z 1 )
Y1 ( z )
X
(
z
)
E1 ( z )
1 1 z 1 z 2
1 1 z 1 z 2
Proceeding to obtain Y(z)
U 2 ( z) Y1 ( z) E1 ( z)
Y2 ( z ) z 1 (Y1 ( z ) E1 ( z )) (1 z 1 ) E2 ( z )
Y3 ( z ) z 1 E1 ( z ) 1 z 1 E 2 ( z )
Y
z 3
z 3 (1 z 1 ) (1 z 1 ) 2
X
(
z
)
E1 ( z ) g1 (1 z 1 ) 3 E 2 ( z )
1
2
1
2
1 1 z z
1 1 z z
Having known that (1 z 1 ) 1 over the baseband, and ignoring 2 terms, frequency
distortion and gain distortion of signal
Y(z) z 3 X(z) z 3 (1 z 1 )E1 (z) g1 (1 z 1 ) 3 E 2 (z)
Following steps similar to solution 6.4(b)
() 2 3
6
2
2
N
Q1 g1
Q2 2
3
7
3OSR
7OSR
7 2 OSR 4 2
Q1
quantization(dB) 10 log 10 1
3 g1 Q 2
Adapted from textbook McGraw Hill, Analog VLSI, (editor. M. Ismail, T. Fiez)
“Oversampled A/D Converter”, Chapter 10 author B. Leung.
6.6
We first find Vo :
Vo ( z ) X ( z ) Y ( z ) B1V1 ( z ) B2V2 ( z ) B3V3 ( z )
X ( z ) Y ( z ) ( B1 I B2 I 2 B3 I 3 )Vo ( z )
X ( z) Y ( z)
1 B1 I B2 I 2 B3 I 3
Then the output Y(z) is found as follows
73
Y ( z ) z 1V4 ( z ) E ( z )
z 1 ( AoVo ( z ) A1V1 ( z ) A2V2 ( z ) A3V3 ( z )) E ( z )
z 1 ( Ao A1 I A2 I 2 A3 I 3 )Vo ( z ) E ( z )
z 1 Ao A1 I A2 I 2 A3 I 3
X ( z) Y ( z) E ( z)
1 B1 I B2 I 2 B3 I 3
H x ( z) X ( z) H e ( z)E ( z)
where
3
H x ( z)
z 1 Ai I i
i 0
3
3
1 Bi I I z 1 Ai I i
i 1
i 0
3
H e ( z)
1 Bi I i
i 0
3
3
i 1
i 0
1 Bi I I z 1 Ai I i
1 1
or putting I (1 z )
3
H x ( z)
Ai ( z 1)
3i
i 0
z ( z 1) 3 Bi ( z 1) 3i Ai ( z 1) 3i
i 0
i 0
3
3
and
3
H x ( z)
( z 1) 3 Bi ( z 1) 3i
i 0
z ( z 1) 3 Bi ( z 1) 3i Ai ( z 1) 3i
i 0
i 0
3
3
Adapted from textbook McGraw Hill, Analog VLSI, (editor. M. Ismail, T. Fiez)
“Oversampled A/D Converter”, Chapter 10 author B. Leung.
6.7
74
There is more than one solution, depending on the order of modulator chosen. We will
show the result if we choose a first order modulator. Here OSR=512, NTF=1-z-1, op-amp
gain=2560, fs=4Mhz, ft=20Mhz, Cs=0.5pf, CI=2.34pf
6.8
a) Adapted from Journal of Solid State Circuits, December 1997, p.1920, Figure 4(a)
b) Adapted from Journal of Solid State Circuits, December 1997, p.1920, Figure 5
c) Adapted from Journal of Solid State Circuits, December 1997, p.1920, Figure 4(b)
75
6.9
fs=4fIF=80MHz, OSR=80MHz/8kHz=10000, therefore L=2 will meet DR requirement.
This means NTF=1+z-2. Compared to the example in chapter, in the present case since
order goes from 4 to 2, NTF’s order goes down from 4 to 2. However since PSD is in
terms of power, therefore its order goes from 8 to 4. In any case slope of PSD rises half
as fast when compared to Design Example 6.2. Let us assume original PSD’s (PSD1)
value at frequency fo+ fbw is denoted as y and the new PSD’s (PSD2) value at fo+ fbw is
denoted by y’. Now we want quantization noise, in the original case, from fo+ fbw to fo
+fbw+ fo to be the same as the quantization noise, in the new case, from fo+ fbw to fo’
+fbw+ fo’. This will allow both case to have the same DR reduction. Assuming fbw is
large compared to fo, fo’. Then y and y’ stays constant from fo+ fbw to fo +fbw+ fo and
fo+ fbw to fo’ +fbw+ fo’, respectively . The excess quantization noise is then simply
calculated by yfo and y’fo’, respectively. Again for same DR reduction we equate
them and we have:
yfo =y’fo’ -> y 0.2% fo = y’ fo’.
(S6.6)
Now how is y related to y’? Since y near fo is approximated by a straight line, therefore
y=slope of PSD1fbw and y’=slope of PSD2fbw. But from above we note that slope of
PSD1=2 slope of PSD2. This means y’=slopefbw. Therefore
y’=1/2y.
(S6.7)
Substitute in (S6.6)
and we get f o’ =2 0.2% fo =0.4%fo. Therefore the fractional change is two time that
as given in Design Example 6.2. With fo/fo=0.4%, from (6.40) we have op-amp
gain=500. From (6.45) g goes up by 2 to 0.0032 -> =0.087T. But fs is still 80Mhz. This
means T still =12.5ns so =1.08ns(does not change much), so ft =318Mhz, Cs=0.0256pf,
CI=0.11pf
76
Chapter 7
7.1
a)
The output U and D will respond only to the positive-going edges of the inputs R and V.
Therefore, the input duty cycles do not have any effect on the outputs. When the two
frequencies are equal, one of the outputs has a duty cycle that is a function of the
difference between the input transition times while the other output remains inactivated
or low. The active output depends on the initial conditions. Hence, the time average of
the differential output, (U-D)ave, is a function of the input phase difference. Figure S7.1(a)
shows the input and output waveforms and Figure S7.1(b) shows the phase detector
characteristics for low frequencies at which the gate delays are neglected.
R
V
T
U
D
RESET
Figure S7.1(a)
R
DPFD waveforms for equal input frequencies
77
(U-D)AVE
1.0
-4
2
4
PHASE
(rad)
-1.0
Figure S7.1(b)
DPFD output as a phase detector at low frequencies
b)
i)
One can define
fV
f fV
1
and R
1
fR
fV
(S7.1)
If fR is greater than fV, then is between 0 and 1, is always positive. In this case, the
output, U, is set high by R and set low by V.
The output D ideally stays low.
Furthermore, there is either a single or no V transition between the two successive
transitions of R. Figure S7.1(c) shows the input and output waveforms for this case.
78
R
V
U
D
RESET
R
Figure S7.1(c)
ii)
DPFD waveform for not equal input frequencies fR>fV
Let us assume that R has two successive transitions at time t and t+TR. Then, one
can define the following probabilities:
P(0)=probability of no V transition in [t, t+TR]=1-
(S7.2)
and
P(1)=probability of a single V transition in [t, t+TR]=
(S7.3)
(Note: A V transition can occur randomly over TV. Probability that this transition occurs
in TR is TR/TV=fV/fR=.) If there is no transition in [t, t+TR], the output U is set to high at
time t but never reset in that time interval (e.g. 3rd period of R in Figure S7.1(c)).
Therefore (U-D)AVE normalized with respect to the logic swing is equal to 1. If there is a
single V transition in [t, t+TR], the output U is set to high at time t and set to low when V
transition appears (e.g. 2nd period of R in Figure S7.1(c)). If one assumes that the
probability density function of a single V transition within that interval is uniformly
79
distributed, (U-D)AVE normalized with respect to the logic swing is equal to 0.5.
Therefore one has
U DAVE P(0) 1 P(1) 0.5 1 0.5
from (S7.2) and (S7.3)
Using (S7.1) from part (i) to replace by :
U D AVE 0.5
1.0
Due to the symmetry of the circuits with respect to R and V this time U will stay at 0 and
D will vary from being high all the time to having a 0% duty cycle. A similar analysis for
fV>fR yields
U D AVE 0.5
1.0
where
fV f R
fR
Figure S7.1(d) shows the frequency-discriminator characteristics of the DPFD at low
frequencies.
Hence checking U D AVE will allow us to determine whether fV is
larger/smaller than fR. As can be seen from this figure, the DPFD circuits considered
above have unlimited frequency acquisition capacity when the frequency operation is low
enough.
(U-D)AVE
1.0
0.5
-6
-4
2
-2
-0.5
-1.0
4
6
80
Figure S7.1(d)
DPFD output as a frequency detector at low frequencies
Adapted from paper, “Frequency limitations of a conventional phase frequency detector”,
M. Soyuer, R. Meyer, Journal of Solid-state Circuits, August 1990, pg. 1019.
81
7.2
a)
Q (U3)
Q (U4)
82
b)
Q (U3)
Q (U4)
83
_
i pd
½
2
e
2
-½
Adapted from paper, “A 155 MHz Clock Recovery Delay and Phase Locked Loop”, Lee,
T. H., Bulzacchelli, J. F., Journal of Solid-state Circuits, Vol 12, December 1992.
7.3
a)
Phase detector output
+ +
2
+
U1
Delayed
data
U4
CK
Clock
U3
U2
Q
D
-
Clock.inv
Retimed
data
Q
D
U5
CK
Clock
Q
D
U6
CK
84
b)
Delayed Data
Clock
Q (U4)
Q (U5)
Q (U6)
Output U1
Output U2
Output U3
Phase detector
output
2
1
0
-1
-2
Loop integrator
Output
T/2
Adapted from paper, “A 155 MHz Clock Recovery Delay and Phase Locked Loop”, Lee,
T. H., Bulzacchelli, J. F., Journal of Solid-state Circuits, Vol 12, December 1992.
85
7.4
Figure P7.5 shows the functional block diagram of a dual-modulus frequency divider,
which includes a divide-by-3-or-4 synchronous counter as the first (high-frequency) stage
followed by a divide-by-4 asynchronous counter as the second (low-frequency) stage.
The input signal, amplified by a logic inverter configuration, clocks the first stage. The
first stage output clocks the second stage. Depending on the signal value at MC1, then
the first stage division ratio is 3 (MC1=0) or 4 (MC1=1).
a)
For MC=0, the OR gate in the figure forces the first stage to divide by 3 during
one of the four states of the second stage, changing the total division ratio of the
divider to 15.
b)
If there is a logic 1 on MC, the first stage always divides by 4, resulting in a total
division ratio of 16 for the divider
Circuit schematic adapted from text, “Bipolar and MOS analog integrated circuit design”,
Grebene, A., J. Wiley, 1984.
7.5
Q
100
// RL // ro
// 2 10 3 // 100 10 3
R
9
12
(2 1.9 10 ) (5.26 10 )
o C1
1.59 K // 2 K // 100 K 878
C2
g m R 40 0.878 35.4
C1
C2 35.4 5.26 10 12 186 pF
86
1
1
0.137 10 8 1.37nH
9 2
12
C
C
(2 1.9 10 ) 5.26 10
o 2 1 2
C1 C 2
Circuit schematic is adapted from text, “Microelectronic Circuits”, Sedra A. S. and Smith
K. C., Oxford University Press, 1998.
L
7.6
a)
The oscillator section is formed by the emitter-coupled differential pair Q1 and
Q2. The positive feedback is obtained by directly short-circuiting the collector of
Q1 to the base of Q2. The negative feedback from the collector of Q1 to the base
of Q1 is through the external LC-tank circuit. This configuration forces Q1 to
operate at zero collector-base bias.
b)
i) Voltage control of the frequency of oscillation is achieved either by a varactor
diode connected in series with the tank circuit, as shown in figure below, or by
using two back to back varactor diode to replace CT in the tank circuit. The
frequency stability of the circuit is primarily determined by the quality of the LCtank circuit.
87
ii) o
K vco
1
(CT CV ) L
, CV
CV 0
Vtune
1
d o dCV
dCV dVtune
-1
2 L (CT CV )
3
2
CV 0
3
V 2
2 1 tune
1
2
, Fermi potential
88
Circuit schematic is adapted from text, “Bipolar and MOS analog integrated circuit
design”, Grebene, A., J. Wiley, 1984.
7.7
a) Redraw CE Colpitts oscillator (Figure P7.6) as CS Colpitts oscillator (Figure 7.38),
but with Q1 replaced by M1. The position of RL is also slightly different but it has no
major impact as f does not depend on RL and a(s) does not have major dependency on
RL. Next redraw CS Colpitts to its equivalent CG Colpitts oscillator and it becomes
Figure 7.36. Now we know Figure 7.36 is equivalent to Figure 7.35 (except for the
omission of CL, which is not important in the present discussion). Figure 7.35 is in
turn equivalent to the transformer coupled LC tank oscillator shown in Figure 7.27. In
summary the transformer coupled LC tank oscillator given in Figure 7.27 is
equivalent to Figure P7.6, except the M1 is replaced by Q1 transistor and the
transformer is replaced by the capacitor divider C1, C2. Hence a(s) is the same and the
feedback factor f changes from
C1
1
to
, o of course is
n
C1 C 2
1
L(C1 // C 2 )
. Now
from (7.47) we know “a” for the transformer coupled LC oscillator is:
g R
a( s) m
n
L
s
R
1
L
s LCs 2
R
(S7.4)
This then is the a(s) in (P7.1). What are R, C? From solution to problem 7.5 R=878 ohm.
Also C=C1C2/(C1 + C2)
f in (P7.1) is:
89
f
C1
C1 C 2
(S7.5)
First we multiply (S7.4) by (S7.5), differentiating with respect to and evaluate the
expression at =2fo, where fo is the oscillation frequency and obtain the result. We then
substitute the result, together with the square of the absolute value of (S7.4) (also
evaluated at s=j=j×2fo) in (P7.1) and we have:
2
j 2f o g m L
2
vo ( f i )
L
2
2
1 j 2f o 2f o LC
1
f
R
2
vn ( f i )
fi
2
f
jg m L j 2f o g m L2 C
C1
2
C C
L
2
2
1
1
j
2
f
2
f
LC
o
o
R
2
(S7.6)
2
v
4kT 2kT
b) From question (b) n is represented by 4kTR n
. Therefore
2g m
gm
f
2
vn
2kT
f
gm
(S7.7)
Substituting (S7.7) in (S7.6) we have:
90
2
j 2f o g m L
L
2
2
1 j 2f o 2f o LC
vo2 ( f i ) 1
R
f
fi
2
jg m L j 2f o g m L2 C
C1
2
C C
L
2
2
1
1 j 2f o 2f o LC
R
2
2kT
gm
(S7.8)
2
c) What is the value of vcarrier
? Let us denote vcarrier as the carrier signal with amplitude
given by
1
2
(amplitude _ of _ vcarrier ) 2 vcarrier . Now from Figure P7.6 vcarrier is the
2
signal across the load RL and is therefore given by:
vcarrier vbe g m R vbe
IC
RL .
Vt
(S7.9)
Here vbe is the small signal voltage arcoss BE junction. But what is vbe? vbe, if we
describe the oscillator as an LTI system, is in turn related to vcarrier through the capacitor
divider formula. However because the oscillation condition: loop gain=1, must be
satisfied, these two equations are identical. Hence we still have only 1 equation and 2
unknowns, which can satisfied by an arbitrary set of vcarrier and vbe. Hence this does not
give us the amplitude of the carrier.
Why is this the case? Remember back in the chapter, we never find this amplitude. As a
matter of fact, in the chapter we treat this oscillator as an LTI circuit and at resonance, the
poles are on imaginary axis, which means theoretically this amplitude goes to infinity. In
91
reality, as soon as the amplitude goes beyond a certain value, transistor Q1 behaves
nonlinearly. This translates into a decrease in its transconductance gm, and hence the loop
gain, so that eventually the amplitude stabilizes.
We try to make a guess of the carrier amplitude by assuming Q1 goes into nonlinear
region when:
amplitude of vbeVt (the thermal voltage)
(S7.10)
Given the Ic vs Vbe equation this is actually the criteria most people use to separate the
linear and nonlinear region of operation for a BJT. Then substituting (S7.10) in (S7.9)
we have
amplitude of vcarrier =ICRL.
(S7.11)
It should be noted, in steady state, vbe, vcarrier are not sinusoidal but consists of pulse; so
the above equation only relates the amplitude of the fundamental component of this pulse
function. (S7.11) can be made more accurate by using the nonlinear analysis. Specifically
it can be refined by using a technique called the describing function. This technique will
allow us to calculate a new transconductance for Q1, denoted as Gm, which represents the
transconductance corresponding to this fundamental components. As stated above, this
Gm would be smaller than gm of Q1 so as to stabilize the amplitude of oscillation. If the
analysis is carried out, where amplitude of vcarrier is denoted as vtank, we finally obtain:
C1
amplitude _ of _ vcarrier 2 I C RL 1
C
C
1
2
(S7.12)
92
From solution to Problem 7.5,
C1
5.2 pF
0.027 , Ic=1mA
C1 C 2 186 pF 5.2 pF
(S7.13)
Substituting (S7.13) in (S7.12) we have:
amplitude _ of _ vcarrier 2 1mA 2k1 0.027
3.89V 4V
(S7.14)
Finally we have:
vcarrier 12 (amplitude _ of _ vcarrier ) 2 8V 2
2
(S7.15)
Now from section 7.8, phase noise's PSD can be interpreted in both phase and amplitude
domain. We choose to use the amplitude representation. Hence VCO's phase noise's
PSD, denoted as S_vco(fi), is simply the ratio of the (noise voltage)2/Hz to the (carrier
voltage)2, both obtained at the output of the oscillator, or across the resistor R .
S _ vco
vo2 ( f i )
f
vcarrier
2
(S7.16)
Substituting (S7.8), (S7.15) into (S7.16) we have:
93
S _ vco
2
j 2f o g m L
L
2
2
1 j 2f o 2f o LC
1
R
fi
2
jg m L j 2f o g m L2 C
C1
2
C C
L
2
2
1
1 j 2f o 2f o LC
R
2
8V
2
2kT
gm
(S7.17)
7.8
K VCO
fo
o
o
Vif Vtune
I
2CVN
f o
I tune
f o
2I tune
2( g m9 g m1 )
, KVCO 2
2CloadVswing N
Vtune 2CloadVswing NVtune
2CloadVswing N
Cload Cdb7 Cdb13 C gd 7 C gd13 C gd 8 C gd14 Cdb8 Cdb14
Capacitance from next stage (ignored in first iteration)
Note:
I tune
g m1 g m9 because there are 2 paths.
Vtune
94
Now Cgs8, Cgd8, Cgs14 and Cgd14 are usually small because M8, M14 are tuning transistors
and are usually small compared to main transistors (M7, M13).
To include capacitance from next stage, let us redraw the circuit:
Then Cload will have added value = Cgs7 + Cgs13 + CM8 + CM7 + CM14 + CM13 -- all from
the next stage.
Note: Cgs8, Cgs14 of next stage are not included because M8, M14 are source degenerated
by ro19, ro4. With these large resistances, M8, M14 behave almost like a source follower.
Since this means there is no voltage change across Cgs8, Cgs14 and so they are not
included.
CM8, CM7, CM14, CM13 are the Miller capacitances where in general CM=AVCgd=gmRoutCgd.
Specifically:
C M 8 ( g m8 )( Rup // Rdown )C gd 8 , Rup g m8 ro ro , Rdown g m14 ro ro
8
C M 14 ( g m14 )( Rup // Rdown )C gd14
4
14
19
95
C M 7 ( g m7 )( ro // ro )C gd 7
7
14
C M 14 ( g m14 )(ro // ro )C gd14
7
14
96
Chapter 8
8.1
It can be shown that if voltage is being selected as the input, a PLL behaves like a
bandpass filter, in the sense, that it will lock only on to frequencies centered around the
free running oscillation frequency 0 . Any input frequency outside of the capture range
(which is on the order of magnitude as well as being directly related to the loop filter
bandwidth) will not be locked and hence can essentially be viewed as being rejected. In
this sense a PLL has a bandpass frequency response.
8.2
In a frequency synthesizer, when one wants to select a new channel via, for example,
changing the prescaler or feedback divider, momentarily the error e changes. The PLL
must respond to this rather fast (it must settle within a time slot in a TDMA system like
DECT). Let us look at the case when we change the divider. As an example, from (8.5),
for a second order loop (first order active filter) we have:
He
e (s)
s2
2
r (s) s 2 n n 2
For a second order loop (first order passive filter) we have
97
1
s s
1
e (s) s s n / K VCO K pd
He
2
2
2
2
r (s)
s 2n s n
s 2n s n
2
Both have a response similar to one shown in Figure S8.1 where a highpass characteristic
is obtained. What that means is if there is a step change in r (due to a change in the
divider division ratio), the loop initially will follow the high frequency response. Because
of the high pass nature this means e is large. This means the synthesizer is not tracking.
As time progresses the loop will follow the low frequency response. Again because of the
high pass nature, its low frequency response is small or that means e is small. This
means eventually the synthesizer will track.
|He|
0
1
i
n
Figure S8.1
8.3
The reason why we have a shift in frequencies of -o in the argument of H VCO can be
explained by the mixing operation going on in the PLL.
98
Referring to Figure 8.11, notice that portion of the phase noise from i _ VCO at
2
frequencies around o is first divided down by M (by the divider) and then mixed down
by the phase detector. The mixing frequency is r. Therefore that portion of the phase
noise from i _ VCO at frequencies around o is frequency translated by Mr.
2
remember in lock,
o
r and so that portion of the phase noise from
M
But
i _ VCO 2 at
frequencies around o is essentially frequency translated to baseband, before being
applied to F(s). Therefore the transfer functions such as F(s), G(s), H VCO , Href all operate
on baseband signals. Accordingly it is their baseband representations that can be used.
Notice in the above discussion we have already invoked the time varying (to be specific,
periodic time varying) nature of the PLL by noting that the phase detector acts as a mixer
and hence the PLL is no longer a LTI system. Therefore strictly speaking the concept of
transfer function (originally derived for LTI system) is not applicable here. In other
words, strictly speaking one cannot talk about G(s), Hvco(s) or Href(s).
To get around this limitation and still use the concept of transfer function, we include
these frequency translation processes by translating the baseband transfer functions up in
frequency to become bandpass transfer fucntions. That is one replaces H VCO ( i ) by
H VCO ( o ) . This is essentially why the correct equation for (7.119), chapter 7 should
be:
So _ VCO ( ) S _ VCO ( ) H n _ VCO o
2
Finally one may want to go back to the original discussion on loop filter where we derive
H ref and H VCO as shown in (8.31), (8.32). Note that even though the phase detector
99
generates a baseband term and a term around 2 times the carrier frequency (in this case
2r), we have neglected these terms at 2r in the derviation of (8.31), (8.32). In that
sense we have derived only the baseband component of H ref and H VCO . Because of
mixing (or sampling if the phase detector is a sampling type) these transfer functions
should have replica responses centered around multiples of o (just like what sampling
would have done) and we are just using one of these replicas around o for our present
calculation.
8.4
a) If we reduce Ibias then S_vco shifts up and c increases.
b) In section 8.4 we design loop filter using the approach adopted in section 8.3.1. This
means we pick u = c. Since from solution in (a) we know c increases this means
u also increases. According to (8.68) this will lead to an increase in spur and the
spur specs may not be met.
c) If we change section 8.4 so that this time we design loop filter using section 8.3.2,
then u no longer equals c. As a matter of fact u will be designed by using the spur
specs to set the allowable spur. This allowable spur will then be substituted in (8.68)
to find the allowable u. This u will remain fixed as S_vco shifts up and c increases,
which we know is happening according to the solution given in (a). Basically with the
increase in VCO’s phase noise, we have not try to suppress the resulting synthesizer
phase noise increase due to it by redesigning the loop filter (notice u remains fixed).
Hence the phase noise specs may not be met.
100
8.5
One potential problem is as follows:
Section (8.4.2) determines the KVCO necessary to get all channels. If instead we do
section (8.4.3) first then at the end of this section we would have finished designing G(s).
Using |G(ju)|=1 we would then have solved for Kvco. The Kvco so designed may be quite
different from the value as designed according to (8.91). If the KVCO so designed is a lot
larger than the value given in (8.94), this will cause phase noise coupling problem as
discussed in section 7.7.2, chapter 7. If the KVCO so designed is a lot smaller than the
value given in (8.94), (8.91) cannot be satisfied with any reasonable Vtune_min and
Vtune_max. In other words we may end up having the following problem: for a given
frequency range (as needed by a standard, here it is DECT),
Vtune
f
is too large and get outside the power supply rail.
K VCO
8.6
Noise contribution from R2: With b large, C3 can be neglected and so the filter consists of
a RC section driven by a current source. It turns out we have the following equation that
quantifies such noise, So_R2, which is obtained by simply reapplying (8.127), except this
time the excitation is noise from R2:
4K vco kTR2
So _ R 2 ( i ) 20 log
i 2 R2 C
8.7
.
101
a) The forward path from noise source to output has been approximated as shown in
Figure P8.1:
The forward path noise from loop filter to output can be described by:
o
4K vco KTBR 2 1 /( R2 C3 )
2K vcoVn 1 /( R2 C3 )
s
s 1 /( R2 C3 )
s
s 1 /( R2 C3 )
b) This noise is of greatest interest in the adjacent channel region (>>1/R2C3). Thus,
the expression can be simplified:
o
4K vco KTBR 2
s 2 R2C3
c) So _ R 2 ( i )
dBc / Hz
4K vco KTR2
20 log | o | 20 log
2R C
i
2 3
At i=25 kHz this expression can be evaluated:
So_R2(i) =-139 dBc/Hz @ 25 kHz offset
8.8
Let us label the original synthesizer with the following characteristics: (M1, fo1, fc1 (=fu1)
). The new frequency synthesizer we are designing has the following characteristics: (M2,
fo2, fc2 (=fu2) ). To calculate fc and hence fu we rewrite (8.107) using these two set of
characteristics:
2
f o1
kTav 2
KM 12
f c1 I bias V gs Vt
102
f o2
f c2
2
kTav 2
KM 22
I bias V gs Vt
Dividing the two above equations we have:
2
f o1 f c 2
M
1
M2
f o 2 f c1
2
Now we further know M1= fo1/fchannel
and M2= fo2/fchannel and so M1/M2= fo1/fo2.
Substituting this in the above equation we have
f c2
1
f c1
or
f u2
1
f u1
Substituting (8.113) we get fu2 remains at 51.6kHz.
In other words, changing carrier frequency does not change the optimal unity gain
frequency. Hence the loop filter design remains the same (same poles/zeroes location and
scaling factor).
8.9
From solution to P7.7, chapter 7 we have:
103
S _ vco
2
j 2f o g m L
L
2
2
1 j 2f o 2f o LC
1
R
fi
2
jg m L j 2f o g m L2 C
C1
2
C C
L
2
2
1
1 j 2f o 2f o LC
R
2
8V
2
2kT
gm
(S8.1)
At the crossover frequency fc
M2S_ref(fc)=S_vco(fc)
(S8.2)
From (8.28) M2S_ref(fc)=KM2
(S8.3)
From (8.105)
-20dBc= 2KM2fc
(S8.4)
Substitute (S8.4) into (S8.3) we have:
M2S_ref(fc)= -20dBc/2fc=0.005/fc
(S8.5)
Substituting (S8.5), (S8.1) (evaluated at fi=fc) into (S8.2) we have:
104
2
j 2f o g m L
L
2
2
1 j 2f o 2f o LC
1
R
fc
2
jg m L j 2f o g m L2 C
C1
2
C C
L
2
2
1
1 j 2f o 2f o LC
R
0.005
2
fc
8V
2
2kT
gm
(S8.6)
For f o 1.9G Hz , C2=190pF, C1=5.2pF, L=1.37nH, gm=40mA/V, R=878 ohm
0.005 1
fc
fc
2
0.44
(S8.7)
Substituting:
fc=88Hz
(S8.8)
This is a very small number and is due to the fact that the gm of the BJT is very large.
This makes I) the noise source very small and II) carrier quite large and so the
oscillator phase noise quite small, resulting in a very small fc. Hence fu following the
optimal condition will be very small too.
8.10
105
a) F ( s)
R1 (1 sR2 C )
1 s( R1 R2 )C
R1
R1//R2
b)
Z1
1
, R2=240, C=0.33uF,
R2 C
Z1=-12.6 Krad/s
(S8.9)
u
1
Z 1 4.2 Krad / s
3
(S8.10)
G( s)
K pd K vco R1 (1 sR2 C )
Ms (1 s( R1 R2 )C )
(S8.11)
From definition G( ju ) 1
(S8.12)
Substitute (S8.11), with proper values, in (S8.12) we have:
j
1
K pd K vco R1
1
3
1 , where p 2
( R1 R2 )C
M j u
1 j u
p2
R1 can be solved
106
Use approx. 1
|
K pd K vco R 1
M u
R 1 |
j
1 , assume p 2 u 1 j u 1
3
p2
| 1
M u
141 4.2Krad / s
|
264
K pd K vco 1mA / rad 2.24 10 6 rad / s / V
(S8.13)
p2
1
1
6 Krad / s NOT u , this can be iterated
( R1 R2 )C (264 240)0.33u
further
If we do not use any approx., we can solve for R1:
R1=318.
(S8.14)
c)
phase margin 360 o 180 o G2 ( j u )
phase margin 180 o G2 ( j u )
Note: -180o due to negative feedback
(This can also be arrived at by writing down phase
margin = how far the phase G2 is from the -180o line =G2 – (-180o)= G2 +180o )
phase margin = 180o + (-90o + tan-1(u /z )-tan-1(u /p )) -90o due to the pole, 1/s
of VCO, tan-1 term due to zero and pole of F2(s).
phase margin = 90o+tan-1(1/3)-tan-1(4.2/6)=90o+18o-35o =73o, stable
d) Repeating (8.68)
107
p ...p
Spur amplitude = 20 log 2 d M u 3 n 1 n |dBc
spur
We can simplify by noting for the present filter we only have p2 but no p3 … pn. Then we
have:
Spur amplitude = 20 log u M 2 d
spur
Further more, fspur=10 KHz, substituting
4.2 K
141 2 0.001
Spur amplitude = 20 log
10 K 2
4.2 141 0.001
= 20 log
10
= -24 dBc
e)
TL
f step
1
ln
u
f error
1
10 KHz
ln
4.2 Krad/s
1Hz
2.19ms
108
Chapter 9
9.1
(a) The oscillator frequency is offset from the carrier frequency and the LO-pulling is
reduced.
(b) At first glance this architecture seems to be similar to the indirect conversion
architecture, because there are still two LO signals. Also whereas in indirect conversion
architecture, after each of the two LO, there is a filter, here there are also two filters for
similar purposes. However, the filtering requirements of the upconverted LO (fLO2 fLO1)
are different from the IF filter requirement.
References: Adapted from “Integrated Radio Transmitter Circuits and Architectures”
Graduate Student project for Intelect, S. Lindfors, H. Olsson, Radio Electronics Lab
KTH, April, 2000.
9.2 (a)
109
I/Q Modulator
BB1
Non-Linear
PA
LO
90
PA
BB0
3LO-BB
LO-BB
LO
LO+BB
LO-3BB
3LO-BB
LO-BB
LO
LO+BB
3rd Order
Intermodulation
As shown in the diagram, the desired signal, which has a component at
LO+BB(LO+baseband) intermod with its third harmonic, which has a component at
3LO-BB(3LO-baseband) and generate the component LO-3BB. This intermod is due to
nonlinearity in the PA. This intermod is generated from 2 (component at LO+BB) and
1 (component at 3LO-BB) i.e. (3LO-BB)-[2( LO+BB)], which gives LO-3BB. Thus at
the output of the PA, you have an unwanted third harmonic of the baseband i.e. 3BB. The
pre-PA filter reduces the LO+BB component, and hence the resulting intermodulation
product.
References: Figure adapted from “A 1.75-GHz Highly Integrated Narrow-Band CMOS
Transmitter with Harmonic-Rejection Mixers” IEEE Journal of Solid State Circuits, Vol.
36, No. 12, pg. 2003 2015, December 2001.
110
(b)The advantages are:
a) LO feedthrough is reduced because of the lower frequency used in the I/Q
modulator; single balanced mixer can be used
b) LO pulling is reduced because neither oscillator is operating at the transmitted
frequency.
c) Quadrature LO can be generated at a lower frequency and/or in the digital
domain, which would result in a better amplitude and phase tracking between
them
References: adapted from “A 1.75-GHz Highly Integrated Narrow-Band CMOS
Transmitter with Harmonic-Rejection Mixers” IEEE Journal of Solid State Circuits, Vol.
36, No. 12, pg. 2003 2015, December 2001.
9.3
In an indirect upconversion transmitter, energy is generated at three times the carrier
frequency from the 3LO23LO1 mixing product. However, in a HRT, due to the
rejection of the third IF harmonic, third order IM is a less significant problem.
References: adapted from “A 1.75-GHz Highly Integrated Narrow-Band CMOS
Transmitter with Harmonic-Rejection Mixers” IEEE Journal of Solid State Circuits, Vol.
36, No. 12, pg. 2003 2015, December 2001
9.4
Drawbacks include both phase and amplitude frequency dependency, phase error from
either component mismatch or process variations, and loss in LO carrier power.
9.5
111
For equal amplitudes this means (2R2/ R1) ALO=LO (2R1C) ALO. This means
R2=LOCR12
9.6
Loss mechanisms include:
a) polyphase filters generate quadrature from a single phase by passing the signal
through multiple stages of an RC network when the signal frequency is near the 3dB frequency of the RC network. Thus there is a loss close to 3dB.
b) Parasitics associated with the bottom plate capacitor used to create the capacitors
contribute to loss.
c) If the output is directly coupled to the mixer input, then an ac-coupling capacitor
might be needed, which introduces additional parasitics and loss in LO carrier
power.
The circuit shown in the problem reduces one mechanism of loss by using a buffer
which drives the input of the filter with signals roughly in quadrature and thus
eliminates the inherent 3dB loss associated with converting a single phase to
quadrature phases.
In addition, the input impedance looking into the buffer is roughly capacitive, thereby
minimizing the reduction in the VCO tank Q and loading.
The mixer LO input was desgned to have a common mode at Vdd and thus the need
for an ac-coupling capacitor has been eliminated as well as its associated loss.
112
References: adapted from “A 1.75-GHz Highly Integrated Narrow-Band CMOS
Transmitter with Harmonic-Rejection Mixers” IEEE Journal of Solid State Circuits, Vol.
36, No. 12, pg.
9.7
a)
The output is given simply by multiplying the inputs to the I and Q mixers and summing
the result:
Output
cos1 t cos2 t 1 sin 1 t sin 2 t
cos1 2 t 1 2 1 2 cos sin 1 2 t 1 2 sin
cos1 2 t 1 2 1 2 cos sin 1 2 t 1 2 sin
Collecting terms of 1 2 and 1 2 the unwanted sideband is at frequency
1 2 and the wanted sideband is at frequency 1 2
2
2
1 1 cos 1 sin
S ,
1 1 cos 2 1 sin 2
b)
.
113
Plot the error of wanted to unwanted sideband at the output of the SSB generator:
Gain
error
80
0db
60
0.1db
Gain error
0.25db
40
0.5db
1db
2db
20
3db
0
0.01
0.1
1
10
10
100
Phase error
References: adapted from Frank Carr-ISCAS 1996, special session on “VLSI for wireless
applications”
114
9.8
a) 500MHz b) 250MHz c) Q=2 d) bandpass, asymmetrical about center frequency
a) 500MHz b) 500MHz c) Q=1 d) bandpass, asymmetrical about center frequency
References: G. Gonzalez, Microwave Transistor Amplifiers, 2nd ed., Prentice Hall, 1997
115
9.9
From chapter 4
2
2
1 vˆi
1 vˆi
0.01
vˆi 0.566VGS Vt
HD3
32 VGS Vt
32 VGS Vt
?? W
I D 50A VGS Vt 2 15 10 6 1.5 VGS Vt 2
2 L
VGS Vt 1.5V vˆi 844mV
If W/L is increased 10000 fold, then v̂i will be reduced to 8.4mV, which is impractical.
That is the reason why the complete class A amplifier in fig 9.27 has the differencial pair
driving
subsequent
stages.
References: “nonlinear integrated circuits”, course notes, R. Meyer, 1986, U.C. Berkeley
116
9.10
M5 W/L=20/2
a) I D5
I D5
3V
25K
300 A , R51
1.25K
10 K
20
Cox W
2
L
VGS 5 Vt 5 2 .
Internal VGS 5 1.626V plus 300A 1.25K 3.75mV
External VGS 5 2V
VGS 2 1V VGS1 4V I D1 6.3A VGS 2 0.9V VGS1 4.1V
Vt1 0.7 0.5 0.9 0.6 0.6 0.9V
70 10 6 3
4.1 0.9 7 A I D 2 I D3 I D 4
2
150
Vt 3 0.7 0.5 2.6 0.6 1.1V
I D1
VGS 3 1.3V
117
b)
VDD = 5V
10K
M1
3/150
M3
10/2
+
Vi
-
Vo
M5
20/2
V1
M2
1/2
M4
10/2
RS3
does
Vi ViB vi
not
affect
distortion-current
;
fed
2I D4
k
,
VGS 3 Vt 3 2 VGS 3 Vt 3
2
k
Vt 3 Vt 0 8 V1 2 f 2 f
I D3 I D 4
2I D4
Vt 3 .
k
V1 ViB vi
2I D4
Vt 0 V1B v1 2 f 2 f
k
1 1
1 1 3
1
2
2 2 2 2 2 3
1 x 1 x
x
x ...
2
1 2
1 2 3
V1B v1 ViB vi
2I D4
Vt 0 V1 2 f 2 f
k
2I D4
Vt 0 V1B 2 f 2 f
k
constant.
so:
v1
1
v1
1
V1B 2 f
2 V1B 2 f 8 V1B 2 f
But V1B ViB
I D4
V1 V1B v1 ,
Put
V1B v1 ViB vi
and
in
source.
V1 Vi VGS 3
2
v1
1
16 V1B 2 f
3
...
118
v 1 v 2 1 v 3
Subtract: v1 vi
V1B 2 f 1 1 1
2
V 4 V 8 V
Where V V1B 2 f 2V 0.6V 2.6V
Let K
2 f V1B
2
0.5
2.6 0.4
2
vi b1v1 b2v12 b3v13
K
1 K
1 K
Where b1 1 , b2 2 , b1 3
V
4 V
8 V
Reverse series:
v1 a1vi a2vi 2 a3vi 3
1
1
1
a1
0.867
b1 1 K 1 0.4
V
2.6
b2
K
1
0.4
1
a2 3
9.63 103
2
3
2
3
4V K
4 2.6 0.4
b1
1
1
V
2.6
a3 2
b2 2
b15
2
b3
b14
2
0.42
16 2.6
4
K2
16V
4
1
K
1
V
1
0.4
1
2.6
5
5
K
8V
0.4
8 2.6
3
3
1
K
1
V
4
1
0.4
1
2.6
4
2.14 10 4 1.60 103
1.39 103
V1 Distortion
io 28.2A ,
v1
g m5
3.58 10 4
1 g m5 R5
28.2 106
78.8mV peak.
3.58 10 4
1 a3 2 1 1.39 103
HD3
So
78.8 103
3
3
4 a1
4 0.867
HD2
(see
2
3.3 106 109.6dB
1 a2
1 9.63 103
S
78.8 103 5.05 10 4 65.9dB
o
2 a12
2 0.867 2
Vo Distortion
next
page),
119
id 5 c1v1 c2v12 c3v13
FET with R5.
gm
c1
1 g m R5
Cox W
1
c2
2
L 1 g m R5 3
2
C W
2 ox R5
L
2
c3
1 g m R5 5
I D5 300A , R5 1.25K
W
g m5 Cox VGS Vt 70 10 6 10 1.626 0.7 6.48 10 4
L
g m5 R5 6.48 104 1250 0.81
gm
c1
3.58 10 4
1 g m R5
c2
70 106
1
10
59 106
3
2
1.81
2
70 10 6
2
10 1250
2
c3
16.8 106
5
1.81
Sub (1) in (2)
id 5 d1vi d 2vi 2 d3vi 3
Where
d1 c1a1
d2 c1a2 c2a12
d3 c1a3 2c2a1a2 c3a13
d1 3.58 104 0.867 3.1104
d2 3.58 104 9.63 103 59 106 0.8672 47.8 106
d3 3.58 10 4 1.39 103 2 59 106 0.867 9.63 103 16.8 106 0.8673
9.42 106
In Vo.
HD3
1 d3
1 9.42 106
2
S
28.2 106
om
4 d13
4 3.110 4 3
2
6.29 105 84dB
120
HD2
1 d2
1 47.8 106
S
28.2 106 7 103 43.1dB
om
2 d12
2 3.110 4 2
121
c)
9-3) part c
K K old 1000
RD
R D _ old
1000
RS 0
I I old 1000
2 I
V Eff . _ old
K
b1 VEff . K R D 4.59
V Eff .
b2
K RD
9
2
b3 0
a1 , a 2 , a3 same
c1 4.1
c 2 7.3
c3 0.1
7.3
HD 2 20 log
0.068 -24.4dB
2 4.1
For HD3 Calculation
If we used the feedback effect (Rs) only as a source of nonlinearity it will give us -90 dB
because Rs in this is negligible (due to high W)
Therefore we’ve calculated nonlinearity due to the channel length modulation effect
which can be given by the formula
Starting from the equation…
1
iDS k (VGS 0 vin Vt ) 2 1 VDS 0 vout
2
We have proven that…
IR
A2
1 VDS 0
HD3 20 log
4 VGS Vt 2
And By substitution
3
HD3 20 log
0.0682 39dB
4 (0.558)
122
References: “nonlinear integrated circuits”, course notes, R. Meyer, 1986, U.C. Berkeley
