電路與電子一 Final Exam Date:6/14/2022
1. (12%) For the circuit shown in Fig. 1 with VPS=4V, R=1KΩ, and Vγ =0.4V,
(a) For DC analysis find ID, Vo (b) For small signal analysis, assume vi=0.1sin(wt), find
small signal current, small signal output voltage and small signal output diffusion
resistance.
Fig. 1
2. (12%) Assume each diode in the circuit shown in Fig.2 has a cut-in voltage of Vγ =
0.65 𝑉. (a) The input voltage VI = 5 V. Determine the value of R1 required such that
ID1 is one half the value of ID2. What are the values of ID1 and ID2? (b) If V1 = 8V and R1
= 2kΩ, determine ID1 and ID2.
Fig. 2
3. (12%) Consider the circuit in Fig. 3. Let Vγ = 0. Plot vo versus vI over the range
-10≤ vI ≤10V.
15kΩ
4V
15kΩ
Fig. 3
4. (12%) Sketch vo versus time for the circuit in Fig. 4 with the input shown. Assume
Vγ = 0.
+20V
R1=4kΩ
RL = 6kΩ
-20V
R2=4kΩ
Fig. 4
5. (12%) In the circuit in Fig. 5 the diodes have the piecewise linear parameters of
Vγ = 0.6 𝑉 𝑎𝑛𝑑 𝑟𝑓 = 0. Calculate the output voltage VO and the currents ID1, ID2
and I for the following input conditions: (a) V1 = V2 = 15 V; (c) V1 = 15 V, V2 = 5 V.
10kΩ
1kΩ
1kΩ
Fig. 5
6. (12%) In the circuit in Fig. 6, the transistor parameters are VTN = 0.6 V and Kn =
0.5mA/V2, Calculate VGS, ID and VDS.
Fig. 6
7. (12%) The parameters of the transistors in the circuit in Fig. 7 are VTND = VTNL = 0.6
V, KnD = 2mA/V2, KnL = 0.5 mA/V2, and λD = λL = 0. At IDQ = 0.15mA, find the smallsignal voltage gain Av = Vo/Vi = dVO/dVI.
Fig. 7
8. (16%) (a) Consider the circuit shown in Figure below with transistor parameters
𝐾𝑛1 = 250μA/𝑉 2 , 𝐾𝑛2 = 100μA/𝑉 2 , 𝑉𝑇𝑁1 = 𝑉𝑇𝑁2 = 1V, and 𝜆1 = 𝜆2 = 0.
Design the circuit of finding the values for all unknown resistors such that 𝐼𝐷𝑄1 =
0.1mA, 𝐼𝐷𝑄2 = 0.4mA, 𝑉𝐷𝑆𝑄1 = 𝑉𝐷𝑆𝑄2 = 5V, and 𝑅𝑖 = 80kΩ. 𝑅𝑆𝑖 = 2kΩ. (b)
determine the small signal gain.
Bonus (2%) What is the purpose of the capacitors used in the circuit of problem
8? (a) AC short (b) AC open (c) DC short (d) DC open
I = −v  W  h  n  q
Drift current
1


=
=
1

=
q is equivalent to e in the above two equations
Cj =
s
Wdep
=
C j0
1−
VR
V0
 si q N A N D 1
C j0 =
2 N A + N D V0

=11.7x8.85x10^(-14) F/cm
I tot = I s (exp
I s = Aqni (
2
VF
− 1)
VT
Dp
Dn
+
)
N A Ln N D L p
ni = 52 1015 T 3 2 exp
− Eg
2kT
electronscm3
Full-wave rectifier
V
1
Vr = M where f =
2 fRC
2TP
t 1
=
T 
2Vr
VM
1
qnn + qp p
i C = −CVM t
iC , peak = +CVM t
t =
2Vr
VM
P.
NMOS Transistor
g m = nCox
W
(VGS − VTH )
L
g m = 2nCox
gm =
W
ID 
L
2I D

VGS − VTH
VTH = VTH 0 +  (  2F + VSB  −  2F  )
Common source amplifier small signal gain
Av = Vo Vi = − g m (ro RD )
Common source amplifier with source resistor small signal gain
Av =
− g m RD
1 + g m RS
Source follower gain
Av =
RS ro
Ri
)
1
R
+
R
i
Si
+ RS ro
gm
(
Common gate amplifier small signal gain
Av =
g m ( RD RL )
1 + g m RSi