A 800nA
B 99.2μA
C 49.6μA
D 0
輸入抵補電流 = IB1 - IB2
= 50μ - 49.2μ = 0.8μA = 800nA
A 200mV
B 225mV
C 250mV
D 300mV
CMRR = `20 log ((Ad)/(Ac))`
80 = `20 log ((4500)/(Ac))`
4 = ` log ((4500)/(Ac))`
10000 = `(4500)/(Ac)`
Ac = `4500/10000`
Ac = 0.45
Vo = Ad Vd +Ac Vc
= 4500 * (0.5m-0.45m) +0.45 * `(0.5m+0.45m)/2`
= 4500 * 0.05m +0.45 * 0.475m
= 4500 * 0.05m +0.45 * 0.475m
= 225m +0.21m
≒ 225mV
A 20kHz
B 50kHz
C 200Hz
D 2MHz
Av(dB) = 40dB =20 log Av
Av = 100
操作頻寬 = 2M / 100 =20kHz
A 2cos(2π400t)+0.1cos(2π60t)
B 1cos(2π400t)+0.2cos(2π60t)
C 2cos(2π400t)+0.2cos(2π60t)
D 4cos(2π400t)+0.1cos(2π60t)
Vd = `V_+ - V_-` = 0.02cos(2π400t)
Vc = `(V_++ V_-)/2` = 0.2cos(2π60t)
Vo =Ad.Vd + Ac.Vc
= 100.0.02cos(2π400t)+0.5.0.2cos(2π60t)
= 2cos(2π400t)+0.1cos(2π60t)
A Vi / RL
B Vi / 2R1
C Vi / R1
D Vi / 2RL
因為有負回授,所以V+ = V- = Vi
因為 I+ = I- = 0 所以 R 跟 R2 沒有消耗電壓
所以 流過RL電阻的電流等於 流過 R1電流
I = `V_i / R_1`
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