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(ELEC202)a890e6 - exam1_MT_soln_2009.pdf
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Name: ID No.


Chinese Name:

1
7


2
8


3
11


4
4


Total
30





The following equations may be useful:
Id = Is(

kn = nCox
kn' = nCox
For Vds < Vgs C Vtn Id = nCox
[(Vgs C Vtn)Vds C
Vds2]
For Vds > Vgs C Vtn Id =
nCox
(Vgs C Vtn) 2(1 + nVds)
For ax2 + bx + c = 0 x1,2 =



(1) Fig. P1 shows a diode circuit with an ideal op amp EA. 7 marks















Fig. P1

(1a) If M1 and M2 have the same threshold voltage and the same (W/L) ratio, 2 marks
argue that I1 = I2. What do M1 and M2 together call?

The EA forces V+ equal to VC, and makes M1 and M2 having the same source, gate and drain voltages. Hence, I1 is equal to I2 as M1 and M2 have the same Vtn and W/L. M1 and M2 together are called a current mirror.

(1b) The diode equation of D2 could be approximated as Id = Is. 1 mark

Express VD2 in terms of I2.


Id = Is

. Vd=VTln(Id/Is).
. VD2=VTln(I2/Is).


(1c) The diode D1 is N times larger than D2, and the diode equation could 1 mark
be approximated as Id = NIs. Express VD1 in terms of I1.



Id = NIs

. Vd=VTln(Id/NIs)
. VD1=VTln(I1/NIs)


(1d) The ideal op amp forces V+ = VC. Using KVL or otherwise, express 2 marks
I1 in terms of the other parameters.


I1 = (V+ C VD1)/R1
= (VC C VD1)/R1
= (VD2 C VD1)/R1
Now, VD2 C VD1 = VTln(I2/Is) C VTln(I1/NIs)
= VTln(N)
. I1 = VTln(N)/R1.


(1e) If Is = 210C15A, N = 4, VT = 26mV, compute R1 such that I1 = 5A. 1 marks


VTln(N) = 26mVln(4) = 36.04mV
. R1=36.04mV/5A = 7.21 k.



(2a) Compute the drain current of M1 (I1) in Fig. P2 using the following 4 marks
parameters (show your steps):
Vdd = 5V, RB = 160k, Vtn = 0.75V, nCox = 50A/V2,
(W/L)1 = (W/L)2 = 16/1, n1 = 0/V, n2 = 0.05/V.












Fig. P2


M1 is diode connected and works in saturation region:
RB : I1 = (VddCV1)/RB
M1: I1 = .nCox (W/L)1(V1CVtn)2

RB : I1 = (5CV1)/160k
M1: I1 = .50(16/1)(V1C0.75)2
On solving,
V1 = 1V or V1=0.484 (<Vtn, rejected)
I1 = 25 A


(2b) The channel length modulation parameter of M2 is n2 = 0.05/V. 4 marks
Plot the drain current I2 for V2 = 0V to 5V using the graph paper
provided. To have an accurate curve, you should compute I2 for
V2 = 0.0V, 0.1V, 0.2V, 0.3V, 0.5V, 1V, 3V and 5V. Show your
calculations.

V2=0.1 . I1=800[ 0.250.1 C .(0.1)2] = 16A
V2=0.2 . I1=800[ 0.250.2 C .(0.2)2] = 24A
V2=0.3 . I1=4000.252(1+0.050.3) = 25.4A
V2=0.5 . I1=25(1+0.050.5) = 25.6A
V2=1 . I1=25(1+0.051) = 26.2A
V2=3 . I1=25(1+0.053) = 28.8A
V2=5 . I1=25(1+0.055) = 31.3A


























(3) Fig. P3 shows an am