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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