ELG 2136 / 2536
Electronics I / Électronique I
Winter / Hiver xxxx
FINAL EXAMINATION (3 Hours)
EXAMEN FINAL (3 Heures)
Question 1
For each question, put in the box the answer which is closest to the correct answer. YOU MUST
JUSTIFY YOUR ANSWER.
Mettre la lettre qui correspondant à la réponse la plus appropriée dans l’espace encadré. VOUS
DEVEZ JUSTIFIER VOTRE RÉPONSE.
Part 1.
(1) The gain of the voltage follower is
Le gain d’un suiveur de tension est
a)
b)
c)
d)
e)
1

0
0 < gain  
-
(a)
(2) Given the following information for a particular diode’s model: VD0 = 0.7 V, rD = 10 .
If vD = 0.5V, as shown in Figure 1, the current in the diode is
Soit un modèle particulier de diode dont les paramètres sont: VD0 = 0.7 V, rD = 10 . Si vD = 0.5 V
tel que montré à la Figure 1, le courant parcourant la diode est :
a)
b)
c)
d)
e)
-20 mA
0 mA
20 mA
0.12 mA
0.7 mA
Figure 1
(b)
VD < VD0 : inverse
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ELG2136/2536
Final Exam/Examen Final
Winter/Hiver xxxx
Part 2.
Consider the circuit shown in Figure 2, and assume that
Soit le circuit de la Figure 2. Nous avons

|VtP| = |VtN| = 1V

kn’ (W/L)N = N Cox(W/L)N = kp’ (W/L)P = P Cox (W/L)P
(1) If vI = 0.5 V, then ISp is equal to
Si vI = 0.5 V, alors ISp est égal à
a)
b)
c)
d)
e)
2 mA
6 mA
0
-10 mA
3.5 mA
(c)
vI < Vt . FET in cutoff mode / FET en mode de blocage.
Figure 2
(2) If RS = 0, vI = 3.5V and VDD = 5V then QN is biased in the
Si RS = 0, vI = 3.5V et VDD = 5V alors QN est polarisé dans la région de
a)
b)
c)
d)
e)
Saturation
Triode
Cutoff
Saturation/Triode
Triode/Cutoff
(b)
1
5VDD  2Vt  = 2.875 V
8
 2.875 V  v I  VDD  Vt  4 V
VIH 
VIH
(3) If RS = 0, vI = 1.5V and VDD = 5V then vO is equal to
Si RS = 0, vI = 1.5V et VDD = 5V alors vO est égale à
a)
b)
c)
d)
e)
0V
0 < vO < 1.5 V
5V
2.5 V
3.5 < vO < 5 V
(e)
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Page No 2/7
ELG2136/2536
Final Exam/Examen Final
Page No 3/7
Winter/Hiver xxxx
Question 2
The BJT in the circuit shown in Figure 3 has  = 100 and VA = .
The FET parameters are kn’ (W/L)N = N Cox(W/L)N = 0.5 mA/V2, Vt = 1V,  = 0 V-1.
Le BJT du circuit de la Figure 3 a un  = 100 et VA = .
Les paramètres du FET sont kn’ (W/L)N = N Cox(W/L)N = 0.5 mA/V2, Vt = 1V,  = 0 V-1.
VCC = + 10 V
R1 = 25 k
RE = 1 k
RD = 2 k
V3
V1
V2
R2 = 75 k
RC = 4 k
RS = 1 k
Figure 3
a) Using DC analysis, calculate the following DC voltages: V1 and V2.
En utilisant l’analyse DC, calculer les tensions DC suivantes : V1 et V2.
VBB  7.5 V ; RBB  R1 // R2  18.75 kΩ
IB 
IE
 0.015 mA
 1

VCC  VBB  I B RBB  VEB  RE I E
V1  VCC  VEB  RE I E  7.782 V
et
V2   I B RC  6 V
b) Show that the BJT is on the active mode
Montrer que le BJT travaille en mode actif.
VEC  VE  VC  2.482 V  VEB  0.7 V
c) Compute the value of V3.
Calculer la valeur de V3.
VG  V2  6 V
ID 
*) I D  11.9 mA

1 'W
VGS  Vt 2 (1  VDS )
k
2 L
VS  I D RS  11.9 V

*) I D  2.1 mA



11.9 mA
ID  
2.1 mA
VGS  Vt
Solution to be rejected / Solution doit être rejetée
V3  V D  5.8 V
d) Show that the FET is on the Saturation mode.
Montrer que le FET travaille en mode de saturation.
VDS  VD  VS  V3  RS I S  5.8  1 * 2.1  3.7 V

VDS  VGS  Vt
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: Saturation
ELG2136/2536
Final Exam/Examen Final
Page No 4/7
Winter/Hiver xxxx
Question 3
1. Calculate the value of the current I if the diodes are ideal.
Calculer la valeur du courant I si les diodes sont idéales.
Figure 4
Assumption: D1 OFF, D2 ON /
Supposition : D1 non conductrice, D2 conductrice
I = 0 mA
I2 
2
5
mA  0  Vo   V  0
3
3
non-conducting diode (D1) with a positive voltage  Wrong assumption
diode non conductrice (D1) avec tension positive  Supposition fausse
Assumption: both diodes are conducting / les deux diodes sont conductrices
I1  1 mA , I 2  0.5 mA
 I = I1 – I2 = + 0.5 mA
QUESTION 4
Consider the circuit shown in Figure 5, and assume that / Soit le circuit de la Figure 5, avec
R1 = R2 = 5M RS =RD =RL =4k, VDD = 12V, Rsig = 1kΩ
k’n W/L = nCox W/L = (15/40) mA/V2, Vt = 1V
Figure 5
(a) Using DC analysis, and assuming that  = 0, calculate the following DC voltages and currents:
VD, VS, and ID (Write their values in the boxes on the next page)
En utilisant l’analyse DC avec  = 0, calculer les tensions et courants DC suivants : VD, VS, et ID
(écrire leurs valeurs dans les espaces encadrés situés à la page suivante)
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ELG2136/2536
Final Exam/Examen Final
VG  VDD
R2
 6V
R1  R2
*) I D  2.08 mA

ID 
1 'W
VGS  Vt 2
k
2 L
VS  8.32 V

*) I D  0.75 mA

VDS  VGS  Vt

3
 4  0.75 mA
ID  
 25  2.08 mA
12

VGS  Vt
Solution to be rejected / Solution doit être rejetée
VGS  6  3  3 V
:
Page No 5/7
Winter/Hiver xxxx

V DS  6 V
Saturation
(b) Let  = 0.01 V-1. Calculate gm and ro
Soit  = 0.01 V-1. Calculer gm et ro
1

 VA  100 V
ro 
VA
 133.33 k
ID
gm  2 k '
W
I D  0.75 mS
L
(c) Draw CLEARLY and PRECISELY the AC small-signal equivalent circuit for the whole circuit
indicating the small-signal equivalent circuit of the transistor.
Dessiner CLAIREMENT et PRECISEMENT le circuit équivalent AC petits signaux pour tout le
circuit, en montrant le circuit équivalent petit signaux du transistor.
RS
vi
+
-
vi
R1
gmvgs
R2
vgs
ro
RD
(d) Calculate the AC small-signal voltage gain ratio : AV = vo / vi
Calculer le gain en tension AC en petits signaux : AV = vo / vi
AV 
vo vo

  g m RD // ro // RL   1.477 V/V
vi v gs
(e) Calculate the AC small-signal input resistance : Rin = vi / ii
Calculer la résistance d’entrée AC en petits signaux : Rin = vi / ii
Rin  R1 // R2  2.5 M
(f) Calculate the overall small-signal voltage gain : Gv = vo / vsig
Calculer le gain total en tension AC en petits signaux : Gv = vo / vsig
Gv  Av
vin
Rin
 Av
 Av  1.477 V/V
v sig
Rin  Rs
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RL
vo
ELG2136/2536
Final Exam/Examen Final
Winter/Hiver xxxx
Page No 6/7
(g) Calculate the AC small-signal current gain : Ai = io / ii
Calculer le gain en courant AC en petits signaux : Ai = io / ii
Ai 
io vo Rin
R

 AV in  923.12 A/A
ii RL vi
RL
(h) Calculate the output resistance Rout seen by the load resistance
Calculer la résistance de sortie Rout vue par la résistance de charge
Rout  ro // RD  3.88 k
Question 5
The transistor in the circuit shown in Figure 6, has  = 100 and VA =  (i.e., no Early Effect).
Le transistor du circuit de la Figure 6 a un  = 100 et VA =  (i.e., pas d’effet Early).
Figure 6
a) Using DC analysis, calculate the following DC voltages and currents: VC, VB, VE, IC, and IE.
En utilisant l’analyse DC, calculer les tensions et courants DC suivants : VC, VB, VE, IC, et IE.
VBB  3 V ;

VBB  VBE
 1.673 mA
RE  RBB /   1
VE  2.175 V
VC  4.879 V
I C   I E  1.656 mA
R BB  R B1 // R B 2  7.5 kΩ
V B  V BE  R E I E  2.875 V
Verify active mode / Vérifier mode actif

VCE
IE 
 VC  VE  2.704 V  VBE  0.7 V
b) Draw CLEARLY and PRECISELY the AC small-signal equivalent circuit for the whole circuit
indicating the small-signal equivalent circuit of the transistor.
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ELG2136/2536
Final Exam/Examen Final
Page No 7/7
Winter/Hiver xxxx
Dessiner CLAIREMENT et PRECISEMENT le circuit équivalent AC petits signaux pour tout le
circuit, en montrant le circuit équivalent petit signaux du transistor.
Pi Equivalent small-signal circuit / Schéma équivalent en petits signaux en Pi
is
vb
ib
RS
vs
+
-
RBB
r
i1
ib
ro
RC
io
vo
RL
c) Calculate the AC small-signal voltage gain ratio : AV = vo / vi
Calculer le gain en tension AC en petits signaux : AV = vo / vi
gm 
IC
 66.24 m -1
VT
Av 
vo vb   RC // RL 

 280.5 V/V
vb vi
r
ro 
VA

IC
r 
VT
 1.47 k
IB
d) Calculate the AC small-signal input resistance : Rin = vi / ii
Calculer la résistance d’entrée AC en petits signaux : Rin = vi / ii
Rin  r // RBB  1.23 k
e) Calculate the overall voltage gain : Gv = vo / vsig
Calculer le gain total en tension AC en petits signaux : Gv = vo / vsig
Gv  Av
vin
Rin
1.23
 Av
 280.5 *
 154.4 V/V
vsig
Rin  Rs
1.23  1
f) Calculate the AC small-signal Current gain : Ai = io / ii
Calculer le gain en courant AC en petits signaux : Ai = io / ii
Ai 
io
R
Av in  3.44 A/A
ii
RL
g) Calculate the output resistance Rout seen by the load resistance
Calculer la résistance de sortie Rout vue par la résistance de charge
Rout  RC  4.3 k
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re 
r
 14.55 
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