Positive SSL
A Ac / Ad
B Ad / Ac
C Vc / Vd
D Vd / Vc

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 射極隨耦器
B 差動放大器
C 達靈頓放大器
D 電壓隨耦器

A 反相輸入端
B 非反相輸入端
C 兩輸入端
D 接地端點

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 非反相放大
B 比較器
C 加法器
D 反相放大

負回授:放大器

正回授:振盪器

無回授:比較器


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`