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1. Chapter 17
Problem Solutions
17.1
a. vI = −1.5 Vi Q1 off , Q2 on
−0.7 − (−3.5)
iE = ⇒ iE = 0.56 mA
5
iC1 = 0 ⇒ v01 = 3.5 V
iC 2 = iE ⇒ v02 = 3.5 − iE RC 2 = 3.5 − ( 0.56 )( 2 )
or
v02 = 2.38 V
b. vI = 1.0 Vi Q1 on, Q2 off
(1 − 0.7 ) − ( −3.5)
iE = ⇒ iE = 0.76 mA
5
iC 2 = 0 ⇒ v02 = 3.5 V
c. logic 0 at v02 (low level) = 2.38 V
Then
v01 = 2.38 = 3.5 − (0.76) RC1
or
RC1 = 1.47 kΩ
17.2
3−0
(a) iC 2 = I Q = 0.5 = ⇒ RC 2 = 6 K
RC 2
3 −1
(b) iC1 = I Q = 0.5 = ⇒ RC1 = 4 K
RC1
⎛V ⎞
I S exp ⎜ BE1 ⎟
(c)
iC1
= ⎝ VT ⎠
IQ ⎡ ⎛V ⎞ ⎛ V ⎞⎤
I S ⎢ exp ⎜ BE1 ⎟ + exp ⎜ BE 2 ⎟ ⎥
⎣ ⎝ VT ⎠ ⎝ VT ⎠ ⎦
1
=
⎛ VBE 2 − VBE1 ⎞
1 + exp ⎜ ⎟
⎝ VT ⎠
vI = VBE1 − VBE 2
So
iC1 1
=
IQ ⎛ −v ⎞
1 + exp ⎜ I ⎟
⎝ VT ⎠
0.1 1
= = 0.2
0.5 ⎛ −v ⎞
1 + exp ⎜ I ⎟
⎝ VT ⎠

2. ⎛ −v ⎞ 1
exp ⎜ I ⎟ = −1 = 4
⎝ VT ⎠ 0.2
( −vI ) = ( 0.026 ) ln (4)
vI = −0.0360 V
17.3
(a) vI = 0.5 V, Q1 on, Q2 off = v02 = 3 V
v01 = 3 − (1)(0.5) = 2.5 V
(b) vI = −0.5 V Q1 off, Q2 on ⇒ v01 = 3 V
v02 = 3 − (1)(0.5) = 2.5 V
17.4
(a) Q2 on, vE = −1.2 − 0.7 = −1.9V
−1.9 − ( −5.2 )
iE = iC 2 = = 1.32 mA
2.5
v2 = −1V = −iC 2 RC 2 = − (1.32 )( RC 2 )
RC 2 = 0.758 k Ω
(b) Q1 on, vE = −0.7 − 0.7 = −1.40 V
−1.4 − ( −5.2 )
iE = iC1 = = 1.52 mA
2.5
v1 = −1V = −iC1 RC1 = − (1.52 )( RC1 )
RC1 = 0.658 k Ω
(c) For vin = −0.7 V , Q1 on, Q2 off
⇒ vO1 = −0.70V
vO 2 = −1 − 0.7 ⇒ vO 2 = −1.7 V
For vin = −1.7 V , Q1 off , Q2 on
⇒ vO 2 = −0.7 V
vO1 = −1 − 0.7 ⇒ vO1 = −1.7 V
(d) (i) For vin = −0.7V , iE = 1.52 mA
−1.7 − ( −5.2 )
iC 4 = = 1.17 mA
3
−0.7 − ( −5.2 )
iC 3 = = 1.5 mA
3
P = ( iE + iC 4 + iC 3 )( 5.2 ) = (1.52 + 1.17 + 1.5 )( 5.2 )
or P = 21.8 mW
(ii) For vin = −1.7V , iE = 1.32 mA
−0.7 − ( −5.2 )
iC 4 = = 1.5 mA
3
−1.7 − ( −5.2 )
iC 3 = = 1.17 mA
3
P = (1.32 + 1.5 + 1.17 )( 5.2 )
or P = 20.7 mW

3. 17.5
3.7 − 0.7
a. I3 = = 1.5 mA
0.67 + 1.33
VR = I 3 R4 + Vγ = (1.5 )(1.33) + 0.7
or
VR = 2.70 V
b. logic 1 level = 3.7 − 0.7 ⇒ 3.0 V
For v X = vY = logic 1.
3 − 0.7
iE = = 2.875 mA = iRC1
0.8
vB 3 = 3.7 − ( 2.875 )( 0.21) = 3.10 V
⇒ v01 ( logic 0 ) = 2.4 V
For v X = vY = logic 0, QR on
2.7 − 0.7
iE = = 2.5 mA = iRC 2
0.8
vB 4 = 3.7 − ( 2.5 )( 0.24 ) = 3.1 V
⇒ v02 ( logic 0 ) = 2.4 V
17.6
1.1 − 0.7 − ( −1.9 )
(a) I REF = = 0.115 mA
20
I Q = I REF = 0.115 mA
(b) vo1 (max) = 1.1 − 0.7 = 0.4 V
vo1 (max) − ( −1.9 )
i5 (max) =
R5
0.4 + 1.9
R5 = R6 = = 11.5 K
0.2
(c) vo1 = −0.4 V ⇒ vB 5 = +0.3 V
1.1 − 0.3
RC1 = = 6.96 K
0.115
(d) vo 2 = −0.4 V ⇒ vB 6 = +0.3V
1.1 − 0.3
RC 2 = = 6.96 K
0.115
17.7
logic 1 + logic 0 1 + 0
VR = = = 0.5 V
2 2
For i2 = 1 mA
0.5 − ( −2.3)
R5 = ⇒ R5 = 2.8 kΩ
1
For QR on,
VR − VBE − ( −2.3)
iE =
RE
or

4. 0.5 − 0.7 + 2.3
RE = ⇒ RE = 2.1 kΩ
1
VB 2 = VR + VBE = 0.5 + 0.7 = 1.2 V
1.2 − 1.4 − ( −2.3)
i1 =
R2
or
1.2 − 1.4 + 2.3
R2 = ⇒ R2 = 2.1 kΩ
1
1.7 − 1.2
R1 = ⇒ R1 = 0.5 kΩ
1
1 − (−2.3)
i3 = = 3 ⇒ R3 = 1.1 kΩ
R3
0 − (−2.3)
i4 = = 3 ⇒ R4 = 0.767 kΩ
R4
For QR on,
v0 R = logic 0 = 0 V ⇒ vB 3 = 0.7 V
iE = iCR = 1 mA
So
1.7 − 0.7
RC 2 = ⇒ RC 2 = 1 kΩ
1
For vI = logic 1 = 1 V,
1 − 0.7 − ( −2.3)
iE = = 1.238 mA
2.1
For vNOR = logic 0 = 0 V,
vB 4 = 0.7 V
Then
1.7 − 0.7
RC1 = ⇒ RC1 = 0.808 kΩ
1.238
17.8
Maximum iE for vI = logic 1 = 3.3 V
3.3 − 0.7
Then iE = 5 mA = ⇒ RE = 0.52 kΩ
RE
For v02 = logic 1 = 3.3 V
3.3 − 0
iE 3 = = 5 mA
R3
or
R3 = 0.66 kΩ
By symmetry,
R2 = 0.66 kΩ
For Q1 on,
4 − ( 2.7 + 0.7 )
iE = iRC1 = 5 mA =
RC1
So
RC1 = 0.12 kΩ

5. For QR on,
3 − 0.7
iE = = 4.423 mA = iRC 2
0.52
4 − ( 2.7 + 0.7 )
and iRC 2 = 4.423 =
RC 2
⇒ RC 2 = 0.136 kΩ
17.9
Neglecting base currents:
(a) I E1 = 0, I E 3 = 0
5 − 0.7
I E5 = ⇒ I E 5 = 1.72 mA
2.5
Y = 0.7 V
5 − 0.7
(b) I E1 = ⇒ I E1 = 0.239 mA
18
I E3 =0
5 − 0.7
I E5 = ⇒ I E 5 = 1.72 mA
2.5
Y = 0.7 V
5 − 0.7
(c) I E1 = I E 3 = ⇒ I E1 = I E 3 = 0.239 mA
18
I E 5 = 0, Y = 5 V
(d) Same as (c).
17.10
(a) VR = −(1)(1) − 0.7 ⇒ VR = −1.7 V
(b) QR off , then vO1 = Logic 1 = −0.7 V
QR on, then vO1 = −(1)(2) − 0.7 ⇒
vO1 = Logic 0 = −2.7 V
QA / QB − off , then vO 2 = Logic 1 = −0.7 V
QA / QB − on, then vO 2 = −(1)(2) − 0.7 ⇒
vO 2 = Logic 0 = −2.7 V
(c) A = B = Logic 0 = −2.7 V , QR on,
VE = −1.7 − 0.7 ⇒ VE = −2.4 V
A = B = Logic 1 = −0.7 V , QA / QB on,
VE = −0.7 − 0.7 ⇒ VE = −1.4 V
(d) A = B = Logic 1 = −0.7 V , QA / QB on,

6. −2.7 − (−5.2)
iC 3 = = 1.67 mA
1.5
−0.7 − (−5.2)
iC 2 = = 3 mA
1.5
P = (1.67 + 1 + 1 + 1 + 3)(5.2) ⇒ P = 39.9 mW
A = B = Logic 0 = −2.7 V
iC 3 = 3 mA, iC 2 = 1.67 mA
P = 39.9 mW
17.11
a. AND logic function
b. logic 0 = 0 V
5 − (1.6 + 0.7)
Q3 on, i = = 2.25 mA
1.2
V2 = (2.25)(0.8) ⇒ logic 1 = 1.8 V
5 − 0.7
c. iE1 = ⇒ iE1 = 1.65 mA
2.6
5 − (0.7 + 0.7)
iE 2 = ⇒ iE 2 = 3 mA
1.2
iC 3 = 0, iC 2 = iE 2 = 3 mA
V2 = 0
5 − (1.8 + 0.7)
d. iE1 = ⇒ iE1 = 0.962 mA
2.6
5 − (1.6 + 0.7)
iE 2 = ⇒ iE 2 = 2.25 mA
1.2
iC 2 = 0, iC 3 = iE 2 = 2.25 mA
V2 = 1.8 V
17.12
3.5 + 3.1
a. VR = − 0.7 ⇒ VR = 2.6 V
2
b. For Q1 on, v X = vY = logic 1 = 3.5 V
3.5 − (0.7 + 0.7)
iE = = 0.175 mA
12
0.175 0.4
Want iRC1 = = ⇒ RC1 = 4.57 kΩ
2 RC1
2.6 − 0.7
c. For Q2 on, iE = = 0.158 mA
12
0.158 0.4
Want iRC 2 = = ⇒ RC 2 = 5.06 kΩ
2 RC 2
d. For vY = logic 1 = 3.5 V
3.5 − 0.7
iR1 = = 0.35 mA, i E = 0.175 mA
8
P = ( iR1 + iE )(VCC ) = ( 0.35 + 0.175 )( 3.5 )
⇒ P = 1.84 mW

7. 17.13
a. logic 1 = 0.2 V
logic 0 = −0.2 V
( 0 − 0.7 ) − ( −3.1)
b. iE = = 0.8
RE
So
RE = 3 kΩ
0.8 0.4
c. Want iR1 = = ⇒ R1 = 1 kΩ
2 R1
d. For v X = vY = logic 1 = 0.2 V
( 0.2 − 0.7 ) − ( −3.1)
iE = = 0.867 mA
3
0.4
iR 2 = ⇒ iR 2 = 0.4 mA
1
iD 2 = 0.467 mA
e. iE = 0.867 mA
0.2 − (−3.1)
i3 = = 1 mA
3.3
−0.2 − (−3.1)
i4 = = 0.879 mA
3.3
P = (0.867 + 1 + 0.879)(0.9 − (−3.1))
or P = 11.0 mW
17.14
a.
( −0.9 − 0.7 ) − ( −3)
i1 = ⇒ i1 = 1.4 mA
1
( −0.2 − 0.7 ) − ( −3)
i3 = ⇒ i3 = 0.14 mA
15
( −0.2 − 0.7 ) − ( −3)
i4 = ⇒ i4 = 0.14 mA
15
i2 + iD = i1 + i3 = 1.4 + 0.14 = 1.54 mA
0.4
i2 = ⇒ i2 = 0.8 mA
0.5
iD = 0.74 mA
v0 = −0.4 V
b. i1 = 1.4 mA
(0 − 0.7) − (−3)
i3 = ⇒ i3 = 0.153 mA
15
i4 = i3 ⇒ i4 = 0.153 mA
i2 + iD = i4 ⇒ i2 = 0.153 mA
iD = 0
v0 = −(0.153)(0.5) ⇒ v0 = −0.0765 V

8. ( 0 − 0.7 − 0.7 ) − ( −3)
c. i1 = ⇒ i1 = 1.6 mA
1
(−0.2 − 0.7) − (−3)
i3 = ⇒ i3 = 0.14 mA
15
i4 = i3 ⇒ i4 = 0.14 mA
i2 + iD = i3 ⇒ i2 = 0.14 mA
iD = 0.0
v0 = −(0.14)(0.5) ⇒ v0 = −0.07 V
(0 − 0.7 − 0.7) − (−3)
d. i1 = ⇒ i1 = 1.6 mA
1
(0 − 0.7) − (−3)
i3 = ⇒ i3 = 0.153 mA
15
i4 = i3 ⇒ i4 = 0.153 mA
i2 + iD = i1 + i4 = 1.6 + 0.153 = 1.753 mA
0.4
i2 = ⇒ i2 = 0.8 mA
0.5
iD = 0.953 mA
v0 = −0.40 V
17.15
a. i. A = B = C = D = 0 ⇒ Q1 − Q4 cutoff
So
VDD = 2 I E R 1 + VEB + I B R2
and
I I
IB = E = E
1 + β P 51
Then
I
2.5 = 2 I E (2) + 0.7 + E ⋅ (15)
51
⎛ 15 ⎞
2.5 = 0.7 = I E ⋅ ⎜ 4 + ⎟
⎝ 51 ⎠
So
I E = 0.419 mA
and
Y = 2.5 − 2(0.4192)(2) ⇒ Y = 0.823 V
ii. A = C = 0, B = D = 2.5 V
Now
vB 5 = vB 6 = 2.5 − 0.7 = 1.8 V
and
Y = vB 5 + 0.7 ⇒ Y = 2.5 V
b. Y = ( A OR B ) AND (C OR D)
17.16
a. logic 1 = 0 V
logic 0 = −0.4 V

9. b. v01 = A OR B
v02 = C OR D
v03 = v01 OR v02
or
v03 = ( A OR B ) AND (C OR D)
17.17
a. For CLOCK = high, I DC flows through the left side of the circuit.. If D is high, I DC flows through
the left R resistor pulling Q low. If D is low. I DC flows through the right R resistor pulling Q low.
For CLOCK = low, I DC flows through the right side of the circuit maintaining Q and Q in their previous
state.
b. P = ( I DC + 0.5 I DC + 0.1I DC + 0.1 I DC )( 3)
P = 1.7 I DC ( 3) = (1.7 )( 50 )( 3) ⇒ P = 255 μ W
17.18
(a)
vI = 0 ⇒ V1 = 0.7 V
3.3 − 0.7
i1 = = 0.433 mA
6
iB = iC = 0
vo = 3.3 V
(b)
vI = 3.3 V v1 = 0.7 + 0.8 = 1.5 V
3.3 − 1.5
i1 = = 0.3 mA
6
0.8
iR = = 0.04 mA
20
iB = 0.3 − 0.04 = 0.26 mA
3.3 − 0.1
iC = = 0.8 mA
4
vo = 0.1 V
17.19
i. For v X = vY = 0.1 V ⇒ v ′ = 0.8 V
5 − 0.8
i1 = ⇒ i1 = 0.525 mA
8
i3 = i4 = 0
ii. For v X = vY = 5 V,
v′ = 0.8 + 0.7 + 0.7 ⇒⇒ v′ = 2.2 V
5 − 2.2
i1 = ⇒ i1 = 0.35 mA
8
0.8
i4 = i1 − ⇒ i4 = 0.297 mA
15
5 − 0.1
i3 = ⇒ i3 = 2.04 mA
2.4

10. 17.20
a. For v X = vY = 5 V , both Q1 and Q2 driven into saturation.
v1 = 0.8 + 0.7 + 0.8 ⇒ v1 = 2.3 V
5 − 2.3
i1 = ⇒ i1 = iB1 = 0.675 mA
4
5 − (0.8 + 0.7 + 0.1)
i2 = ⇒ i2 = 1.7 mA
2
i4 = iB1 + i2 ⇒ i4 = 2.375 mA
0.8
i5 = ⇒ i5 = 0.08 mA
10
iB 2 = i4 − i5 ⇒ iB 2 = 2.295 mA
5 − 0.1
i3 = ⇒ i3 = 1.225 mA
4
v0 = 0.1V
5 − (0.1 + 0.7)
b. ′
iL = = 1.05 mA
4
′
iC (max) = β iB 2 = NiL + i3
(20)(2.295) = N (1.05) + 1.225
So
N = 42
17.21
DX and DY off, Q1 forward active mode
v1 = 0.8 + 0.7 + 0.7 = 2.2 V
5 = i1 R1 + i2 R2 + v1 and i1 = (1 + β )i2
So 5 − 2.2 = i2 [ (1 + β ) R1 + R2 ]
Assume β = 25
5 − 2.2
i2 = ⇒ i2 = 0.0589 mA
(26)(1.75) + 2
i1 = (1 + β )i2 = (26)(0.05895) ⇒ i1 = 1.53 mA
i3 = β i2 ⇒ i3 = 1.47 mA
0.8
iBo = i2 + i3 − = 0.0589 + 1.47 − 0.16 ⇒
5
iBo = 1.37 mA
Qo in saturation
5 − 0.1
iCo = ⇒ iCo = 0.817 mA
6
17.22
(a) vI = 0 V, Q1 forward actions
5 − 0.7
iB = = 0.717 mA
6
iC = (25)(0.71667) = 17.9 mA
iE = (26)(0.71667) = 18.6 mA

11. (b) VI = 0.8 V
5 − (0.8 + 0.7)
iB = = 0.583 mA
6
Because of the relative doping levels of the Emitter and collector, and because of the difference in B-C and
B-E areas, we have −iC ≈ iB = 0.583 mA and iE = small value.
(c) vI = 3.6 Q1 inverse active.
5 − (0.8 + 0.7)
iB = = 0.583 mA
6
iE = − β R iE = −(0.5)(0.583) = −0.292 mA
iC = −iB − iE = −0.583 − 0.292 ⇒ iC = −0.875 mA
17.23
a. i. v X = vY = 0.1 V, so Q1 in saturation.
5 − (0.1 + 0.8)
i1 = ⇒ i1 = 0.683 mA
6
⇒ iB 2 = i2 = i4 = iB 3 = i3 = 0
ii. v X = vY = 5 V, so Q1 in inverse active mode.
Assume Q2 and Q3 in saturation.
5 − (0.8 + 0.8 + 0.7)
i1 = ⇒ i1 = iB 2 = 0.45 mA
6
5 − (0.8 + 0.1)
i2 = ⇒ i2 = 2.05 mA
2
0.8
i4 = ⇒ i4 = 0.533 mA
1.5
iB 3 = ( iB 2 + i2 ) − i4 = 0.45 + 2.05 − 0.533
or
iB 3 = 1.97 mA
5 − 0.1
i3 = ⇒ i3 = 2.23 mA
2.2
b. For Q3 :
i3 2.23
= = 1.13 < β
iB 3 1.97
For Q2 :
i2 2.05
= = 4.56 < β
iB 2 0.45
Since ( I C / I B ) < β , then each transistor is in saturation.
17.24
(a) v X = vY = Logic 1

12. v ′ = 0.8 + 2 ( 0.7 ) = 2.2 V
5 − 2.2
i1 = = 0.35 mA
8
0.8
i4 = i1 − = 0.35 − 0.0533 = 0.2967 mA
15
5 − 0.1
i3 = = 2.04 mA
2.4
5 − ( 0.1 + 0.7 )
′
iL = = 0.525 mA
8
Assume β = 25
Then ( 25 )( 0.2967 ) = 2.04 + N ( 0.525 )
So N = 10.2 ⇒ N = 10
(b) Now
5 = 2.04 + N ( 0.525 )
So N = 5.64 ⇒ N = 5
17.25
a. For v X = vY = 5 V, Q, in inverse active mode.
5 − ( 0.8 + 0.8 + 0.7 )
iB1 = = 0.45 mA
6
iB 2 = iB1 + 2 β R iB1 = 0.45(1 + 2 [ 0.1]) = 0.54 mA
5 − ( 0.8 + 0.1)
iC 2 = = 2.05 mA
2
0.8
iB 3 = ( iB 2 + iC 2 ) − = 0.54 + 2.05 − 0.533
1.5
or
iB 3 = 2.06 mA
Now
5 − (0.1 + 0.8)
′
iL = = 0.683 mA
6
Then
′
iC 3 (max) = β F iB 3 = NiL
or (20)(2.06) = N (0.683)
⇒ N = 60
b. ′
From above, for v0 high, I L = (0.1)(0.45) = 0.045 mA. Now
⎛ 5 − 4.9 ⎞ (21)(0.1)
′
I L (max) = (1 + β F ) ⎜ ⎟ =
⎝ R2 ⎠ 2
= 1.05 mA
So
′
I L (max) = NI L
or 1.05 = N (0.045)
⇒ N = 23
17.26

13. (a) Vin = 0.1V : QI , Sat : Qs , Qo , Cutoff
5 − (0.1 + 0.8)
iI = = 1.025 mA
4
P = iI (5 − 0.1) = (1.025)(4.9) ⇒
P = 5.02 mW
(b) Vin = 5V , QI , Inverse Active; QS , Qo , Saturation
vBI = 0.7 + 0.8 + 0.7 = 2.2V
5 − 2.2
iI = = 0.7 mA
4
iEI = β R ⋅ iI = (0.1)(0.7) = 0.07 mA
Vout = 0.7 + 0.1 = 0.8 V
5 − 0.8
i2 = = 4.2 mA
1
P = (iI + iEI + i2 )(5) = (0.7 + 0.07 + 4.2)(5) ⇒
P = 24.9 mW
17.27
a. v X = vY = vZ = 0.1 V
5 − (0.1 + 0.8)
iB1 = ⇒ iB1 = 1.05 mA
3.9
Then
iC1 = iB 2 = iC 2 = iB 3 = iC 3 = 0
b. v X = vY = vZ = 5 V
5 − (0.8 + 0.8 + 0.7)
iB1 = ⇒ iB1 = 0.692 mA
3.9
Then
iC1 = iB 2 = iB1 (1 + 3β R ) = (0.692)(1 + 3[0.5])
⇒ iC1 = iB 2 = 1.73 mA
5 − (0.1 + 0.8)
iC 2 = ⇒ iC 2 = 2.05 mA
2
0.8
iB 3 = iB 2 + iC 2 − = 1.73 + 2.05 − 1.0
0.8
⇒ iB 3 = 2.78 mA
5 − 0.1
iR 3 = = 2.04 mA
2.4
5 − (0.1 + 0.8)
′
iL = = 1.05 mA
3.9
′
iC 3 = iR 3 + 5iL = 2.04 + (5)(1.05)
⇒ iC 3 = 7.29 mA
17.28
a. v X = vY = vZ = 2.8 V, Q1 biased in the inverse active mode.

14. 2.8 − (0.8 + 0.8 + 0.7)
iB1 = ⇒ iB1 = 0.25 mA
2
iB 2 = iB1 (1 + 3β R ) = 0.25(1 + 3 [0.3])
⇒ iB 2 = 0.475 mA
vC 2 = 0.8 + 0.1 = 0.9 V
0.9 − (0.7 + 0.1) 0.1
iB 4 = =
(1 + β F )(0.5) (101)(0.5)
= 0.00198 mA (Negligible)
5 − 0.9
iR 2 = = 4.56 mA
0.9
⇒ iC 2 = 4.56 mA
0.8
iB 3 = iB 2 + iC 2 − = 0.475 + 4.56 − 0.8
1
⇒ iB 3 = 4.235 mA
b. v X = vY = vZ = 0.1 V
5 − (0.1 + 0.8)
iB1 = ⇒ iB1 = 2.05 mA
2
From part (a),
′
iL = β R ⋅ iB1 = (0.3)(0.25) = 0.075 mA
Then
5iL′ 5(0.075)
iB 4 = = ⇒ iB 4 = 0.00371 mA
1+ βF 101
17.29
a. v X = vY = vZ = 0.1 V
2 − (0.1 + 0.8)
iB1 = + iB 3
RB1
where
(2 − 0.7) − (0.9) 0.4
iB 3 = =
RB 2 1
⇒ iB 3 = 0.4 mA
Then
1.1
iB1 = + 0.4 ⇒ iB1 = 1.5 mA
1
iB 2 = 0 = iC 2
′
Q3 in saturation iC 3 = 5iL For v0 high,
′ ′
vB1 = 0.8 + 0.7 = 1.5 V ⇒ Q3 off
2 − 1.5
′
iB1 = = 0.5 mA
1
′ ′
iL = β R iB1 = (0.2)(0.5) = 0.1 mA
Then

15. iC 3 = 0.5 mA
b. v X = vY = vZ = 2 V
From part (a),
⇒ iB1 = 0.5 mA
iB 3 = 0 = iC 3
iB 2 = iB1 (1 + 3β R ) = (0.5)(1 + 3 [0.2])
iB 2 = 0.8 mA
′ ′
iC 2 = 5iL , and from part (a), iL = 1.5 mA
So
iC 2 = 7.5 mA
17.30
5.8 − 0.7
(a) IB + ID = = 0.51 mA
10
5 − (0.7 − 0.3)
IC − I D = = 4.6 mA
1
Now
I I
I D = 0.51 − I B = 0.51 − C = 0.51 − C
β 50
Then
⎛ I ⎞ ⎛ 1 ⎞
I C − I D = I C − ⎜ 0.51 − C ⎟ = I C ⎜ 1 + ⎟ − 0.51 = 4.6
⎝ 50 ⎠ ⎝ 50 ⎠
So I C = 5.01 mA
IC 5.01
IB = = ⇒ I B = 0.1002 mA
β 50
I D = 0.51 − 0.1002 ⇒ I D = 0.4098 mA
VCE = 0.4 V
(b) I D = 0, VCE = VCE ( sat ) = 0.1 V
5.8 − 0.8
IB = ⇒ I B = 0.5 mA
10
5 − 0.1
IC = ⇒ I C = 4.9 mA
1
17.31
a. v X = vY = 0.4 V
vB1 = 0.4 + 0.7 ⇒ vB1 = 1.1 V
5 − 1.1
iB1 = ⇒ iB1 = 1.39 mA
2.8
vB 2 = 0.4 + 0.4 ⇒ vB 2 = 0.8 V
iB 2 = iC 2 = iB 0 = iC 0 = iB 5 = iC 5
= iB 3 = iC 3 = 0 ( No load )

16. 5 = iB 4 R2 + VBE + (1 + β )iB 4 R4
5 − 0.7
iB 4 = ⇒ iB 4 = 0.0394 mA
0.76 + (31)(3.5)
iC 4 = β F iB 4 ⇒ iC 4 = 1.18 mA
vB 4 = 5 − (0.0394)(0.76) ⇒ vB 4 = 4.97 V
b. v X = vY = 3.6 V
vB1 = 0.7 + 0.7 + 0.3 ⇒ vB1 = 1.7 V
vB 2 = 1.4 V
vB 0 = 0.7 V
vC 2 = 1.1 V
5 − 1.7
iB1 = ⇒ iB1 = 1.1786 mA
2.8
iB 2 = iB1 (1 + 2 β R ) = 1.18(1 + 2 [0.1])
iB 2 = 1.41 mA
1.1 − 0.7
iB 4 = ⇒ iB 4 = 0.00369 mA
(31)(3.5)
5 − 1.1
iR 2 = = 5.13 mA ⇒ iC 2 ≈ 5.13 mA
0.76
iB 0 ≈ iB 2 + iC 2
iB 0 = 6.54 mA
17.32
(a) vI = 0, v1 = 0.3 V
1.5 − 0.3
i1 = = 1.2 mA
1
iB = iC = 0, vO = 1.5 V
(b) vI = 1.5 V v1 = 0.7 + 0.3 = 1 V
1.5 − 1
i1 = = 0.5 mA
1
0.7
iR = = 0.035 mA
20
iB = 0.5 − 0.035 = 0.465 mA
1.5 − 0.4
iC = = 0.917 mA
1.2
vO = 0.4 V
17.33
a. Assuming the output transistor Q2 is a Schottky transistor, then
2.5 − (0.4 + 0.3)
′
v0 = 0.4 V, iL = = 0.5
RB1
Then

17. RB1 = 3.6 kΩ
Then
2.5 − (0.7 + 0.8) 1.0
iB1 = = = 0.278 mA
RB1 3.6
0.7
iB 2 = 0.5 mA, iE1 = 0.5 + = 1.50 mA
0.7
iE1 = iB1 + iC1 ⇒ iC1 = 1.50 − 0.278 = 1.222 mA
2.5 − (0.7 + 0.1)
and iC1 = = 1.222 mA
RC1
⇒ RC1 = 1.39 kΩ
b. v X = vY = 0.4 V, vB1 = 0.7 V
vC 2 = 2.5 − 0.7 ⇒ vC 2 = 1.8 V
All transistor currents are zero.
c. vB1 = 1.5 V, vC1 = 0.8 V
Currents calculated in part (a).
d. ′
iB 2 = 0.5 mA, iL = 0.5 mA
′
iC 2 (max) = β iB 2 = NiL or (50)(0.5) = N (0.5)
So
N = 50
17.34
a. For v X = vY = 3.6 V
5 − 2.1
vB1 = 3(0.7) = 2.1 ⇒ iB1 = = 0.29 mA
10
5 − 1.8
vC1 = 0.7 + 0.7 + 0.4 = 1.8 V ⇒ iC1 = = 0.32 mA
10
1.4
iB 2 = iB1 + iC1 − = 0.29 + 0.32 − 0.0933
15
So
iB 2 = 0.517 mA
vC 2 = 0.7 + 0.4 = 1.1 V
5 − 1.1
iC 2 = = 0.951 mA
4.1
0.7
iB 5 = iB 2 + iC 2 − = 0.517 + 0.951 − 0.175
4
or iB 5 = 1.293 mA
′
For v0 = 0.4 V, vB1 = 0.4 + 0.7 = 1.1 V
Then
1.1 − 0.7
′
iB1 = = 0.00086 mA
(31)(15)
5 − 1.1
′
iL = ′
− 0.00086 or iL ≈ 0.39 mA
10
So iC 5 (max) = β iΒ 5 = NiL′
(30)(1.293) = N (0.39) ⇒ N = 99

18. b. P = (0.29 + 0.32 + 0.951)(5) + (99)(0.39)(0.4)
P = 7.805 + 15.444 or P = 23.2 mW
(Assumming 99 load circuits which is unreasonably large.)
17.35
a. Assume no load. For v X = logic 0 = 0.4 V
5 − (0.4 + 0.7)
iE1 = = 0.0975 mA
40
Essentially all of this current goes to ground from VCC .
P = iE1 ⋅ VCC = (0.0975)(5) ⇒ P = 0.4875 mW
5 − (3)(0.7)
b. iR1 = = 0.0725 mA
40
5 − (0.7 + 0.7 + 0.4)
iR 2 = = 0.064 mA
50
5 − (0.7 + 0.4)
iR 3 = = 0.26 mA
15
P = (0.0725 + 0.064 + 0.26)(5)
P = 1.98 mW
c. For v0 = 0, vC 7 = 0.7 + 0.4 = 1.1 V
5 − 1.1
iR 7 = ⇒ iR 7 = 78 mA ≈ iSC
0.050
17.36
(a) vl = vO = 25 V ; A transient situation
vDS ( M N ) = 2.5 − 0.7 = 1.8 V
vGS ( M N ) = 2.5 − 0.7 = 1.8 V ⇒ M N in saturation
vSD ( M P ) = 5 − (2.5 + 0.7) = 1.8 V
vSG ( M P )5 − 2.5 = 2.5 V ⇒ M P in saturation
iDN = K n (vGSN − VTN ) 2 = (0.1)(1.8 − 0.8) 2 ⇒ iDN = 0.1 mA
iDP = K P (vSGP + VTP ) 2 = (0.1)(2.5 − 0.8) 2 ⇒ iDP = 0.289 mA
iC1 = β iDP = (50)(0.289) ⇒ iC1 = 14.45 mA
iC 2 = β iDN = (50)(0.1) ⇒ iC 2 = 5 mA
Difference between iE1 and iDN + iC 2 is a load current.
(b) Assume iC1 = 14.45 mA is a constant
1 i ⋅t (V )(C )
VC = ∫ iC1dt = C1 ⇒ t = C
C C iC1
(5)(15 × 10−12 )
t= ⇒ t = 5.19 ns
14.45 × 10−3
(5)(15 × 10−12 )
(c) t= ⇒ t = 260 ns
0.289 × 10−3

19. 17.37
(a) Assume R1 = R2 = 10 kΩ; β = 50
0.7
Then iR1 = iR 2 = = 0.07 mA
10
NMOS in saturation region; vGSN = 2.5 − 0.7 = 1.8 V
iDN = K n ( vGSN − VTN ) = ( 0.1)(1.8 − 0.8 )
2 2
iDN = 0.10 mA
Then iB 2 = 0.03 ⇒ iC 2 = (50)(0.03) = 1.5 mA
iE1 = 1.53 mA ⇒ iB1 = 0.03 mA ⇒ iC1 = 1.5 mA
So iDP = 0.10 mA
Now, M P biased in non-saturation region
vSGP = 2.5 V
iDP = 0.10 = 0.10 ⎡ 2(2.5 − 0.8)vSD − vSD ⎤
⎣
2
⎦
0.10 vSD − 0.34 vSD + 0.10 = 0
2
0.34 ± (0.34) 2 − 4(0.10)(0.10)
vSD =
2(0.10)

20. Or
vSD = 0.325 V
Then vo = 5 − 0.325 − 0.7
vo = 3.975 V
1 i ⋅t
(b) v=
C ∫ idt = C
Cv (15 × 10−12 )(5)
t= =
i 1.53 × 10−3
t = 49 ns
(c)
Cv (15 × 10−12 )(5)
t= =
i 0.1× 10−3
t = 0.75 μ s