Problem 1.
An AC generator with an internal resistance of 50 Ω is connected to a step-down
transformer loaded with a 1 Ω resistor (see the figure below). Find the voltage
across the primary if N1 = 50, N2 = 10 and the open circuit voltage of the
generator is 1 V (peak-to-peak).
Rin=50 Ω
Rtr
N1
N2
RL = 1 Ω
Vopen = 1 V
AC Generator
Solutions:
Input impedance of the transformer: Rtr = (N1/ N2)2 RL = 25 Ω.
Voltage across the primary: V = VopenRtr / (Rtr +Rin ) = 1/3 (V).
Problem 2.
A diode is biased in the forward direction. Find the voltage drop across the diode
if the current is 1 mA. Assume the following parameters:
•
•
•
saturation current density Js = 10-10A/cm2;
cross–section area A = 1 mm2 (10mm=1cm)
non-ideality factor n = 1.2.
Solution:
A = 1mm 2 = 10 −2 cm 2
I s = J S * A = 10 −12 A
I = I s (exp(V / nVth ) − 1) ≈ I s * exp(V / nVth )
V = nVth * ln( I / I s ) ≈ 1.2 * 0.026V * ln(10 9 ) ≈ 1.2 * 0.026V * 20.7 ≈ 0.65V
Problem 3.
The reverse-saturation currents of a Schottky diode and a pn junction diode at 300K are
5*10-8A and 10-12A, respectively. The diodes are connected in parallel and are driven by
a constant current of 0.5mA. Determine the current and current density in each diode and
voltage across each diode. Assume that pn junction diode has nonideality factor equal to
1. Cross-sectional area of pn junction diode is 0.2cm-2 and of Schottky diode is 1cm-2.
Solution:
Since diodes are connected in parallel they should have the same voltage drop and share
total current of 0.5mA. Then: I pn + I Schottky = 0.5mA
⎛ I pn ⎞
⎛ I Schottky ⎞
Using diode IV equation we can write that: ln⎜⎜ pn ⎟⎟ = ln⎜⎜ Schottky ⎟⎟ .
⎝ IS ⎠
⎝ IS
⎠
⎛ I pn
Hence ISchottky=0.49999mA and Ipn=0.00001mA. Voltage is V = Vth ⋅ ln⎜⎜ pn
⎝ IS
⎞
⎟ ≈ 0.239V .
⎟
⎠
Current densities: pn junction diode ≈5*10-8A/cm2; Schottky ≈5*10-4A/cm2.
Problem 4.
Input signal is sinusoid with 30V peak-to-peak
voltage. Sketch the waveforms of current in diode and voltage across it. Find the peak
values of the current through and voltage across diode.
.
Vin (t)
R=100Ω
10V
Solution:
Diode conducts only when forward bias voltage around 0.7 drops across it. So during the
positive half cycle of the input signal the diode conducts only when input voltage is
above 10.7V. The peak value of the diode current is (15V-10.7V)/100Ohm=43mA.
During negative half cycle of the input signal diode is under reverse bias and current in
circuit is just diode reverse saturation current (nearly zero). Peak values of the voltage
across diode are: +0.7 V for positive half cycle and - 25V for negative one.
Problem 5.
Determine the bias point (ICQ and VCEQ) for the circuit below. Assume β=100. For
simplicity, neglect base current for bias point calculations. Determine the voltage gain for
output signal taken from collector and input applied to base terminal (assume infinite
transistor output resistance). Determine input resistance for this amplifier. Determine the voltage
gain and input resistance when RE is shunted by bypassing capacitor for AC signal.
Sketch DC and AC load lines.
10V
R1=8kΩ
R2=2kΩ
RC=3.3kΩ
RE=1kΩ
Solution:
Voltage at BJT base with respect to ground can be found from voltage divider (this is true
only when base current can be neglected otherwise Thevenin’s theorem should be used to
convert R1 and R2 into equivalent resistor connected between equivalent voltage supply
R2
and base terminal): VB =
⋅10V = 2V . Emitter current can be found assuming that
R1 + R 2
0.7V drops between base and emitter for transistor to operate in forward active:
2V - 0.7V
IE =
= 1.3mA and ICQ≈IE=1.3mA.
RE
IC
≈ 13uA and indeed it is very small and can be
β
neglected when compared with ~1mA flowing through R1 and R2.
Voltage drop between collector and emitter can be estimated as:
VCEQ≈10V-4.3kOhm*1.3mA≈5.59V.
To determine voltage gain we observe that input voltage when applied to base terminal is
distributed between RE and rπ, hence v i = i B ⋅ rπ + i E ⋅ R E . The output signal
*Base current can be estimated as I B =
vo
iC ⋅ R C
RC
. We
=−
≈−
rπ
vi
i B ⋅ rπ + i E ⋅ R E
+ RE
β
r
Vth 0.026V
=
≈
= 2kOhm . Since π ≈ 20 Ohm one can
β
IB
13uA
v o = −i C ⋅ R C . Hence the voltage gain is: A V =
can estimate rπ from rπ =
v BE
iB
RC
= − 3 .3 .
RE
Circuit input resistance will be composed of R1, R2 and rπ+β*RE connected in parallel,
hence Rin≈1.6kOhm.
For the case when RE is shunted for AC signal we have to replace RE by short circuit in
R
gain and input resistance calculation, hence A V ≈ − C = −165 and Rin≈0.9kOhm.
rπ
β
write that A V ≈ −
Problem 6.
A MOS differential pair operated at bias current of 0.8mA (current through RSS) employs transistors with
W/L=100 and µn*COX=0.2mA/V2, using RD=5kOhm and RSS=25kOhm.
(a) Find the differential gain, the common mode gain, and CMRR if the output is taken
single-endedly and the circuit is perfectly matched. (b) Repeat (a) when the output is
taken differentially.
VDD
RD
RD
vo1
vo2
vi1
vi2
RSS
VSS
Solution:
VO1
V
R
= O2 = g m ⋅ D . MOSFET transconductance in
2
VID
VID
saturation can be found from:
[V − VT ]2 = W ⋅ C ⋅ µ ⋅ VOV 2 ;
W
⋅ C OX ⋅ µ ⋅ GS
I DS =
OX
2
2
L
L
dI
2 ⋅ I DS
W
⋅ C OX ⋅ µ ⋅ VOV =
g m = DS =
dVGS
L
VOV
(a) Differential gain A D =
2
0.8mA
mA V
= 100 ⋅ 0.2 2 ⋅ OV ,
2
2
V
2 ⋅ 0.4mA
mA
≈4
. Single ended
hence VOV≈0.2V. Transconductance is g m =
0.2V
V
differential gain is AD=10V/V.
V
V
RD
5
Common mode gain A C = O1 = O2 =
=
= 0.1V/V .
VIC
VIC
2 ⋅ R SS 2 ⋅ 25
CMRR=AD/AC=100. In dB CMRR=40dB.
(b) AD will be twice larger, i.e. AD=20V/V. For fully differential output for perfectly
symmetric circuit AC=0, hence CMRR is infinite.
Overdrive voltage can be estimated from
Problem 7:
Find the voltage gain for the circuit below.
10kΩ
20kΩ
4kΩ
5kΩ
Vout
Vin
Solution:
Net voltage gain is equal to product of the voltage gains of the individual cascades
Vout
V
⎛ 20 ⎞ ⎛ 10 ⎞
= A V1 ⋅ A V2 = ⎜ − ⎟ ⋅ ⎜1 + ⎟ = −14 .
AV =
Vin
4⎠
V
⎝ 5 ⎠ ⎝
Problem 8:
Find the voltage gain for the circuit below.
R
Vin
Vout
Solution:
Using virtual ground concept:
⎛
⎛V
Vin
= − I S ⋅ ⎜⎜ exp⎜⎜ out
R
⎝ Vth
⎝
⎡ ⎛ V ⎞ ⎤
Hence Vout = Vth ⋅ ⎢ln⎜⎜ in ⎟⎟ + 1⎥ .
⎢⎣ ⎝ IS ⋅ R ⎠ ⎥⎦
⎞ ⎞
⎟⎟ − 1⎟ .
⎟
⎠ ⎠