Institut für Integrierte Systeme
Integrated Systems Laboratory
Communications Electronics
Solutions Exercise 6
Xu Han(Benjamin Sporrer)
25.04.2016
in
dra
IS
M1
so
Vin
M1
ga
VS
−gmVin
te
M1
RS
urc
e
Problem 1
Figure 1: Small signal equivalent circuit of the mixer input stage.
(a) The input impedance of the common gate NMOS is simply the inverse of the transconductance
M1 .
Vin
Vin
1
Zin =
=
=
Iin
gm ·Vin gm
(b) For a power match:
Zin = R∗S
Therefore, to have an input power match the transconductance has to be as follows:
gm =
1
1
1
=
=
= 20 mS
Zin RS 50Ω
(c) The signal current IS as a function of the voltage VS is,
1
IS = −gm ·Vin = −gm ·
Zin
−gmVS
gm
·VS = −gm ·
·VS =
1
RS + Zin
gm RS + 1
RS + g
m
(d) If the switching transistors of the mixer M2 and M3 perfectly conduct the whole current IS to the
one side or the other, then the switching function is a rectangular function g(t) which goes from
+1 to −1 with the oscillator frequency fLO (see Fig. 2). In that case, the output differential current
Iout = IO1 − IO2 can be considered as the multiplication of the signal current IS with the rectangular
waveform.
The Fourier series of the rectangular function is:
4
1
1
g(t) = sgn[cos(ωLOt)] = · cos(ωLOt) − cos(3ωLOt) + cos(5ωLOt) − · · ·
π
3
5
g(t)
+1
1
2fLO
0
1
fLO
3
2fLO
t
−1
Figure 2: Rectangular function.
Since we are interested only in the frequency component at ωLO we find for the output signal;
Iout = IO1 − IO2 = −IS ·
4
VS
4
ARF · cos(ωRFt) 4
· cos(ωLOt) =
· · cos(ωLOt) =
· · cos(ωLOt)
1
π
π
RS + gm π
RS + g1m
= ARF ·
gm
4
· · cos(ωLOt) · cos(ωRFt)
1 + gm RS π
For the desired output frequency (ωIF = ωRF − ωLO ) the output voltage is:
Vout = RL · Iout (ω = ωIF ) =
gm RL
4 1
· ARF · · · cos[(ωRF − ωLO )t]
1 + gm RS
π 2
(e) The voltage gain is:
Vout (ω = ωIF )
GV =
=
Vin (ω = ωRF )
For the input power match Zin =
1
gm
gm RL
2
1+gm RS · ARF · π
1
1+gm RS · ARF
= RS the voltage gain is:
GV =
2
2 RL
·
π RS
=
2
· gm RL
π
(f) The output power is:
Pout =
2 (ω
Vout
= ωIF )rms
=
Rout
2 (ω
Vout
= ωIF )rms
=
2 · RL
gm RL
1+gm RS
· π2 · A√RF2
2 · RL
2
RL
A2RF
=
2 · 2
π
RS + g1m
The input power is:
Pin =
Vin2 (ω = ωRF )rms
=
Rin
1
gm
RS + g1m
·
A√RF
2
2
=
1
gm
A2RF
2
2 · gm · RS + g1m
Therefore the power gain of the mixer is:
GP =
Pout (ω = ωIF )
2
= 2 · gm · RL
Pin (ω = ωRF )
π
and for an input power match:
GP =
2 RL
·
π 2 RS
(g) For Rout = 2RL and Rin = RS the voltage gain is:
GV =
2 RL
1 Rout
·
= ·
π RS
π Rin
and the power gain is:
GP =
2 RL
1 Rout
Rin
·
= 2·
= G2V ·
2
π RS
π Rin
Rout
Therefore the voltage gain is equal to the power gain only if the input and output impedance are
equal, that is:
Rout = 2 · RL = Rin
Problem 2
(a) The output voltage is:
Vout = RL · (I3 + I5 ) − RL · (I4 + I6 ) = RL · [(I3 − I4 ) + (I5 − I6 )]
and:
VLO
I3 − I4 = I1 · tanh
2 ·Vt
VLO
I5 − I6 = −I2 · tanh
2 ·Vt
3
we finally get:
VLO
Vout = RL · tanh
2Vt
VLO
· (I1 − I2 ) = RL · IB · tanh
2Vt
VRF
· tanh
2Vt
(b) Since all three differential pairs act as small signal amplifiers, we can replace the tanh function of
each differential pair with the linear part of the Taylor expansion. Then Vout is:
Vout = RL IB ·
VLO VRF RL IB
·
=
·VLO ·VRF
2Vt 2Vt
4Vt2
Therefore, under these conditions the mixer functions as an ideal multiplicator between the
oscillator signal and the input signal!
For
VRF = ARF · cos(ωRFt)
VLO = ALO · cos(ωLOt)
we get:
Vout =
RL IB
RL IB
·VLO ·VRF =
· ALO · ARF · cos(ωLOt) · cos(ωRFt)
4Vt2
4Vt2
Vout (ω = ωIF ) =
RL IB
1
· ALO ARF · cos[(ωRF − ωLO )t]
2
4Vt2
The voltage gain in this case is:
GV =
Vout (ω = ωIF )
RL IB
=
· ALO
VRF (ω = ωRF )
8Vt2
(c) For very large oscillation voltages you can neglect the time during which the switch transistors
work as a differential amplifier. Similar to Problem 1, you can assume a square function for the
switch transfer characteristic. Since VRF Vt the transfer characteristic can be replaced again by
the linear terms of the Taylor series. Then we get for Vout
Vout = RL IB ·
VRF
· sign [cos(ωLOt)]
2Vt
Since we are again interested only in the basic component at ωLO we get:
Vout = RL IB ·
ARF
4
· cos(ωRFt) · · cos(ωLOt)
2Vt
π
At the desired output frequency (ωIF = ωRF − ωLO ) the output voltage is:
Vout (ω = ωIF ) = RL IB ·
ARF 2
· · cos [(ωRF − ωLO )t]
2Vt π
and finally the voltage gain of the mixer is:
GV =
Vout (ω = ωIF )
RL · IB 1
=
·
VRF (ω = ωRF )
Vt
π
4