Q1
Find vE, vC1, vC2 and the current source internal resistance R in the circuit of Fig. Q1.
Assume that VBE = 0.7 V, vB1 = 1 V, vB2 = 0 V VCC = -VEE = 5 V, I = 1, mA, RC = 4 kΩ
and α ≈ 1.
Figure Q1 the basic BJT differential-pair configuration.
SOLUTION
Q1 is on while Q2 is off.
VBE1 = vB1 – vE1
vE = vE1 = vE2 = vB1 - VBE1 = 1 – 0.7 = +0.3 V
iE1 = I = 1 mA
iC1 = αI = 1 mA
vC1 = VCC – (iC1* RC) =5 - (1* 4) = 1 V
iC2 = αiE2 = 0 mA
vC2= 5 V
R = (VE – VEE)/ I = (0.3 +5)/1 = 5.3 kΩ
Ans. +0.3 V; 1V; 5 V, 5.3 kΩ
1
Q2
Find vE, vC1, vC2 and the current source internal resistance R in the circuit of Fig. Q2.
Assume that VBE = 0.7 V, vB1 = -1 V, vB2 = 0 V VCC = -VEE = 5 V, I = 1, mA, RC = 4 kΩ
and α ≈ 1.
Figure Q2 the basic BJT differential-pair configuration.
SOLUTION
Q1 is off while is Q2 is on.
VBE2 = vB2 – vE2
vE = vE1 = vE2 = vB2 - VBE2 = = 0 – 0.7 = - 0.7 V
iC1 = αiE1 = 0 mA
vC1= 5 V
iE2 = I = 1 mA
iC2 = αI = 1 mA
vC2 = VCC – (iC2* RC) =5 - (1* 4) = 1 V
R = (VE – VEE)/ I = (- 0.7 +5)/1 = 4.3 kΩ
Ans. -0.7 V; 5 V; 1 V, 4.3 kΩ
2
Q3
Find vE, vC1, vC2 and the current source internal resistance R in the circuit of Fig. Q3.
Assume that VBE = 0.7 V, vB1 = vB2 = vCM = 0 V VCC = -VEE = 5 V, I = 1, mA, RC = 4
kΩ and α ≈ 1.
Figure Q3 the basic BJT differential-pair configuration.
SOLUTION
Q1 is on and Q2 is on.
VBE1 = VBE2 = VCM = vB – vE
vE = vE1 = vE2 = vCM - VBE = = 0 – 0.7 = - 0.7 V
iE1 = I/2 = 0.5 mA
iC1 = αiE1 = αI/2 = 0.5 mA
vC1 = VCC – (iC1* RC) = 5 - (0.5 1* 4) = 3 V
iE2 = I/2 = 0.5 mA
iC2 = αiE2 = αI/2 = 0.5 mA
vC1 = VCC – (iC2* RC) = 5 - (0.5 1* 4) = 3 V
R = (VE – VEE)/ I = (- 0.7 +5)/1 = 4.3 kΩ
Ans. -0.7 V; 3 V; 3 V, 4.3 kΩ
3
Q4
Find vE, vC1, and vC2 in the circuit of Fig. Q4. Assume that |vBE| of a conducting
transistor is approximately 0.7 V and that α ≈ 1.
Figure Q.4 The BJT differential-pair configuration.
SOLUTION
Q1 is off while Q2 is on
vEB2 = vE2 – vB2
vE = vE1 = vE2 = vEB2 + vB2 = +0.7 V
vC1 = – 5 V
iC2 = iE2 = (VEE – vE2)/RE == (5 – 0.7)/1 = 4.3 mA
vC2 = iC2* RC2 + VCC = (4.3* 1 – 5 = -0.7 V
Ans. +0.7 V; −5 V; −0.7 V
4
Q5
Determine the input common-mode range for a bipolar differential amplifier operating
from ± 2.5 V power supplies and biased with a simple current source that delivers a
constant current of 0.4 mA and requires a minimum of 0.3 V for its proper operation.
The collector resistances RC = 5 kΩ. (Hint: VCE,SAT = 0.3 V).
Figure Q5 The basic BJT differential-pair configuration.
SOLUTION
The allowable range of VCM is determined at the upper end by Q1 and Q2 leaving the
active mode and entering saturation. To operate Q1 (or Q2) in active mode:
VCMmax = VBE + VEEMAX = VBE + VC -VCE,SAT = VC+ 0.7 – 0.3 = VCC - α(II2)RC + 0.4
VCMmax = 2.5 - (0.4I2)5 + 0.4 = 1.9 V
The lower end of the VCM range is determined by the need to provide a certain
minimum voltage VCE,SAT across the current source I to ensure its proper operation.
Thus,
VCMmin = VEEmin + VBE = – VEE + VCE,SAT + VBE
VCMmin = – 2.5 + 0.3 + 0.7 = - 1.5 V
Ans. - 1.5 V to +1.9 V
5
Q6
For large signal operation of the basic BJT differential-pair configuration in figure Q5,
derive an expression for IE1 of Q1 and IE2 of Q2.
Figure Q6 the basic BJT differential-pair configuration
SOLUTION
For large signal operation, IE1 of Q1 and IE2 of Q2 can be expressed as:
6
Figure 6 Transfer characteristics of the BJT differential pair assuming α = 1.
7
Q7
Briefly, explain how the transfer characteristics of the BJT differential pair can be
linearized.
SOLUTION
Figure Q7 The transfer characteristics of the BJT differential pair (a) can be linearized (b) (i.e., the
linear range of operation can be extended) by including resistances in the emitters.
The linear range of operation of the large-signal operation of the differential BJT
pair can be extended employing an effective technique. It consists of including two
equal resistances Re in series with the emitters of Q1 and Q2, as shown in Fig. Q7 (a).
The resulting transfer characteristics for three different values of Re are sketched in Fig.
Q7(b). Observe that expansion of the linear range is obtained at the expense of reduced
Gm (which is the slope of the transfer curve at vid = 0) and hence reduced gain. This
result should come as no surprise; Re here is performing in exactly the same way as the
emitter resistance Re does in the CE amplifier with emitter degeneration.
8
Q8
For the basic BJT differential pair of Fig. Q8, find the value of input differential signal
that is sufficient to cause iE1 = 0.99I.
Figure Q8 The basic BJT differential-pair configuration.
SOLUTION
vB1 – vB2 = VT ln(IE1/IE2) = VT ln(0.99I/0.01I) = 25 ln(99) = 115 mV
Ans. 115 mV
9
Q9
Find an expression of IREF and IO (in terms of IREF) for the basic BJT current mirror
circuit depicted in Figure Q9, in the following cases:
a)
b)
c)
d)
e)
Infinite β and AQ1 = AQ2.
Infinite β and AQ2= mAQ1.
Finite β and AQ2 = AQ1.
Finite β and AQ2= mAQ1.
Finite β, AQ2 = mAQ1, and include Early effect.
Figure Q9 The basic BJT current mirror.
10
Q10
Find the value of IO for the basic BJT current mirror circuit depicted in Figure Q10,
given that VCC = 5 V, R = 4.3 kΩ and VBE = 0.7 V, for the following cases:
a)
b)
c)
d)
e)
Infinite β and AQ2 = AQ1.
Infinite β and AQ2= 5AQ1.
β = 100 and AQ2 = AQ1.
β = 100 and AQ2 = 5AQ1.
β = 100, VO = 2.7 V, VA =100 V and AQ2 = 5AQ1.
Figure Q10 The basic BJT current mirror.
SOLUTION
IREF = (5 – 0.7)/4.3 = 1mA
a)
b)
c)
d)
e)
IO = IREF = 1 mA
IO = mIREF = 5 mA
IO = IREF/(1 +2/β) = 1/1.02 = 0.98 mA
IO =m IREF/[1 +[(m+1)/β] = 5/1.06 = 4.7 mA
IO = {m IREF/[1 +[(m+1)/β]}{1 + (VO – VBE)/VA2}
= (5/1.06)(1.02)= (4.7)(1.02)= 4.794 mA
11
Q11
Consider a BJT current mirror with a nominal current transfer ratio of unity. Let the
transistors have IS = 10-15 A, β = 100, and VA = 100 V. For IREF = 1 mA, find IO when VO
= 5 V Also, find the output resistance.
.
SOLUTION
IREF = IS exp(VBElVT)
VBE = VT ln(IREF/IS) = 25ln(1012)= 25*12*2.3 =0.7 V
IO = {IREF/[1 +(2/β)]}{1 + (VO – VBE)/VA2}
= (1/1.02)(1.043)= (0.98)(1.043)= 1.02 mA
RO = VA/IREF =100/1 = 100 kΩ
Ans. 1.02 mA; 100 kΩ
12
Q12
Assuming the availability of BJTs with scale currents IS = 10-15 A, β = 100, and VA = 50
V, design the current-source circuit of Fig. Q12 to provide an output current ΙΟ = 0.5
mΑ at VO = 2 V. The power supply VCC = 5 V. Give the values of IREF, R, and VOmin.
Also, find IO at VO = 5 V.
Figure Q12 A simple BJT current source.
SOLUTION
Io= IS exp(VBElVT)
VBE = VT ln(Io/IS) = 25ln(1012)= (25)(12)(2.3) = 0.67 V
IO = {IREF/[1 +(2/β)]}{1 + (VO – VBE)/VA2}
IREF = {Io [1 + (2/β)]}/{1 + (VO – VBE)/VA2}
IREF = {0.5[1.02]}/{1.026} = (0.51)/(1.026)= 0.497 mA
R = (VCC – VBE)/IREF = 4.33/0.497 = 8.71 kΩ
VOmin = VCE2,SAT = 0.3 V to keep Q2 in active mode
IO = {IREF/[1 +(2/β)]}{1 + (VO – VBE)/VA2}
= (0.497 /1.02)(1.086)= (0.487)(1.086)= 0.53 mA
Ans. 0.497 mA; 8.71 kΩ; 0.3 V; 0.53 mA
13
Q13
Figure Q2 shows the basic BJT differential pair biased with a current mirror.
a) Perform an approximate dc analysis (assuming β ˃> 1, VBE ≈ 0.7 V and neglect
the early effect) to calculate dc currents and voltages everywhere in the circuit.
b) Calculate the quiescent power dissipation in this circuit.
c) If the transistors Q1 and Q2 have β = 100, calculate the input bias current of the
op amp.
d) What is the input common-mode range of this op amp?
14
Q13
15
16
VCC = 15 V
SOLUTION of EXAMPLE Q13
a) The values of all dc currents and voltages are
indicated on the circuit diagram Fig. Q13a. These
values were calculated by ignoring the base current
of every transistor—that is, by assuming β to be very
high.
VB3 = VC9 = VB9 = VBE + VEE = 0.7 -15 = - 14.3 V
IC9 = (0 – VC9)/RC9 = (0 + 14.3)/28.6 = 0.5 mA
IO = IC3 = IC9 = 0.5 mA
IO = IE1 = IC1 = IE2 = IC2 = IO/2 = 0.25 mA
VC1 = [VCC – (IC1)(RC1)]
= [+15 – 0.25 × 20] = +10 V.
VC2= [VCC – (IC2)(RC2)] = [+15 – 0.25 × 20] = +10 V.
VE1 = VE2 = 0 – VBE1 = -0.7 V
b) To calculate the power dissipated in the circuit in the
quiescent state (i.e., with zero input signal) we
simply evaluate the dc current that the circuit draws
from each of the two power supplies.
PD = VCC(IC1 + IC2) + VEE(IE9 + IE3)
= 15(2.5 +2.5) + 15(0.5 + 0.5) = 22.5 mW
c) The input bias current of the differential pair is the
average of the dc currents that flow in the two input
terminals (i.e., in the bases of Q1 and Q2). These
two currents are equal (because we have assumed
matched devices); thus the bias current is given by
IB = IB1 = IB2 = IE1/(β + 1) ≈ 2.5 μA
d) At edge of saturation of Q1 (or Q2)
VE1 = VC1 - VCE1,sat = 10 -0.3 = 9.7 V
VCM,max = VE1 + VBE1 = 9.7 + 0.7 = VC1 + 0.4 = 10 +
0.4 = 10.4 V
At edge of saturation of Q3
VE1 = VC3 = VCE3,sat + VEE = 0.3 - 15 = - 14.7 V
VCM,min = VE1 + VBE1 = - 14.7 + 0.7 = - 14 V
17
VEE = - 15 V
Figure Q13a. The basic IC BJT
Differential Pair Biased with current
mirror
18