근데 이거 어케 html파일로 따로 저장함?

진범 · 2012-03-21T01:14:48+09:00

메모장에 저장하고 확장자 바꾸면 됨 -_-? 3.38 (∴I = 50 μA) to find small signal reponse v 0 , open the dc current source I and short the capacitive c 1 and c 2 also replace the diode with its small signla resistance: r d = n V t / I (n=2) v o = v s * r d / r d +R s = ( v s * n V t / I ) / ( nV t / I + R s ) = v s n V t / ( n V t + I * R s ) v o = 10 mV * 0.05 / ( 0.05 + 10 3 I ) = 0.476 mV ~ I = 1 mA 3.333 mV ~ I = 0.1 mA 9.804 mV ~ I = 1 μA for v o - 1/2 * v s = v s * 0.05 / (0.05 + 10 3 ) I ∴ I = 50 μA 3.57 (a) (∴ 아래에 있습니다) v o / v i = R / ( R + (2r d // 2r d ) ) = R / ( R + r d ) where r d = V t / ( I / 2 ) = 2 V t / I = 0.05 V / I ∴ I : v o / v i 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 ± 10 I , the chanse in diode voltage △v d can be found from i d / I = e △vd / n Vt => 0.9 to 1.1 thus, for n=1 △v d = -2.63 mV to + 2.38 mv or approximately ±29.5 mV for the diode current to remain within ±10% of their dc bias currents the signal voltage across each diode must be limited to 2.5 mV now if v i peak = 10 mV we can obtain the following situation figure we see that v o = 5 mV and i = 5 mV / 10kΩ = 0.5 μA thus, each iode is carrying a current signal of 0.25 mA for this to at most 10% of the dc current, the dc current in each diode must be al least 2.5 μA it follows that the minimum value of i must be 5 μA (c) for i = 1 mA, i d = 0.5 mA, and for maximum signal of 10%, I d = 0.05 mA thus i d = 2 i d = 0.1 mA and the corresponding maximum v d is 0.1 mA * 10 kΩ = 1 V the corresponding peak input can be found by dividing v o by the transmisiion factor of 0.995, thus v i 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 v o / v i = ( 1 / SC ) / ( 1 / SC + r d ) = 1 / ( 1+ SC * r d ) phase shift = -arctan ( ω c r d / 1) = - arctan (2 π 10 5 * 10 * 10 -9 * 0.025 / I ) for a phase shift of -45° we have 2 π 10 5 * 10 * 10 -9 * 0.025 / I = 1 I = 157 μA range of phase shift for I = 15.7 μA to 1570 μA i o : -84.3° to - 5.71°

메모장에 저장하고 확장자 바꾸면 됨 -_-?

3.38 (∴I = 50 μA)
to find small signal reponse v0 , open the dc current source I and short the capacitive 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 ) / ( nVt / I + Rs ) = vs n Vt / ( n Vt + I * Rs)
 
vo = 10 mV * 0.05 / ( 0.05 + 103 I ) = 0.476 mV ~ I = 1 mA
 3.333 mV ~ I = 0.1 mA
 9.804 mV ~ I = 1 μA
 
for vo - 1/2 * vs = vs * 0.05 / (0.05 + 103) I
∴ I = 50 μA
 
3.57
(a) (∴ 아래에 있습니다) 
 
vo / vi = R / ( R + (2rd // 2rd) ) = R / ( R + rd )
 
where rd = Vt / ( I / 2 ) = 2 Vt / 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 ± 10 I , 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 = -2.63 mV to + 2.38 mv or approximately ±29.5 mV
 
for the diode current to remain within ±10% of their dc bias currents the signal voltage across each diode must be limited to 2.5 mV
now if vi peak = 10 mV
we can obtain the following situation
 
figure
 
we see that vo = 5 mV and i = 5 mV / 10kΩ = 0.5 μA
 
thus, each iode is carrying a current signal of 0.25 mA for this to at most 10% of the dc current, the dc current in each diode must be al least 2.5 μA it follows that the minimum value of i must be 5 μA
 
(c)
for i = 1 mA, id = 0.5 mA, and for maximum signal of 10%, Id = 0.05 mA 
thus id = 2 id = 0.1 mA and the corresponding maximum vd is 0.1 mA * 10 kΩ = 1 V
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 ( ω c rd / 1)
= - arctan (2 π 105 * 10 * 10-9 * 0.025 / I )
 
for a phase shift of -45° we have 
2 π 105 * 10 * 10-9 * 0.025 / I = 1
I = 157 μA
 
range of phase shift for I = 15.7 μA to 1570 μA 
io : -84.3° to - 5.71°

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