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3.38

to find small signal reponse

v0, open the dc current source I

and short the capa c1 and c2

also replace the diode with its small signla resistance:

rd = n vt / I n=2

vo = vs * rd / rd+Rs

=vs * n vt / I // n vt / I + Rs = vs n vt / n vt + I Rs

vo = 10mv * 0.05 / 0.05 + 10^3 I

= 0.476mv ~ I = 1 mA

3.333 mv ~ I = 0.1 mA

9.804 mv ~ I = 1 mu A

for vo - 1/2 vs = vs * 0.05 / 0.05 + 10^3 I

I= 50 mu A

3.57

(a)

vo / vi = R / R + (2rd // 2rd) = R / R+rd

where rd = Vt / I / 2 = 2vt / I = 0.05 V / I

I vo / vi

0 0

10^-3 0.167

10 ^ -2 0.667

10 ^ -1 0.952

1.0 0.995

10 0.9995

(b) if the signal current is to be lemited to +- 10I , the chanse in diode voltage @vd can be found from

id / I = e @vd / n Vt = 0.9 to 1.1

thus, for n=1

@ vd = -2.63mv to + 2.38 mv

or approximately +-29.5 mv

fod the diode current to remain within +-10% of their dc bias currents the signal voltage across each diode must be limited to 2.5mv

now if vi_peak = 10mv

we can obtain the following situation

figure

we see that vo = 5mv and i = 5mv / 10k ohm = 0.5 mu A

thus, each iode is carrying a current signal of 0.25 mA for this to e at most 10% of the dc current, the dc current in each diode must be al least 2.5 mu A it follows that the minimum value of i must be 5 mu A

(c)

for i = 1m A, id = 0.5 mA, and for maximum signal of 10%, Id = 0.05ma

thus id = 2id = 0.1mA and the corresponding maximum vd is 0.1 mA * 10 k ohm = 1v

the corresponding peak input can be found by dividing vo by the transmisiion factor of 0.995, thus

vi_max = 1 v / 0.995 v = 1. 005 v

3.43

opening the current cource we get the following samll-siganl circuit : (n=1)

figure

vo / vi = 1 / sc // 1/sc + rd = 1/ 1+ sc rd

phase shift = - arctan ( wcrd / 1)

= - arc tan (2 pi 10 ^ 5 * 10 * 10 ^ -9 * 0.025 / I )

for a phase shift of -45도 we have

2 pi 10 ^5 * 10 * 10 ^-9 * 0.025 / I = 1

I = 157 mu A

range of phase shift for I -= 15.7 mu A to 1570 mu A

io
: -84.3 도 to - 5.71도

 

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