IM1002 Tutorial 7
v1 = v2 + IoR2 + (vi=0) --- (1)
vout = v2 + Io(R1 + R2) ----- (2)
Put (1) into (2)
vout = v1 + IoR1 which gives Io = (vout + v1)/R1 (3)
Put (3) into (2) and re-arrange:
vout /v2 = -(R1/R2) + (v1/v2 )(1+R1/R2) (4)
If R1 = R2 : vout /v2 = -1 + (2v1/v2 )
(5)
Rin2 = v2 /I2 = -v2/Io (6)
Put (1) into (6) and short circuit v1 : Rin2 = R2
Rin1 = v1 /I1 = infinite
Ans: vout = 2v1- v2
Rin1 = infinite
Rin2 = R2
I1
Vi
Io
I2
vR = ioR = R*vo/11R ----- (1)
v1 = ii*10R + vR ----- (2)
v2 = i2*R + vR ----- (3)
i1 = -i2 ---- (4)
Put (3) and (4) into (2):
v1 = -i2*10R + vo/11 ----- (5)
Put (3) into (5) and re-arrange:
vo = v1 + 10v2
Rin2 = v2 /i2 and Rin1 = v1 /i1 (6)
Use (6) and short circuit v1 or v2 :
Rin2 = Rin1 =R+10R = 11R
Ans: vout = 2v1- v2
Rin1 = Rin2 = 11R
Io
Use superposition and result in (a):
vo1/v1 = -(100R/10R)
(1)
vo2 /v2 = -(100R/20R)
(2)
vo = v01 + v02 = -(10v1 +5v2)
Rin2 = 20R
Rin1 = 10R
Using superposition:
With v1 = 0 and using result in (c):
vo2 = -10v2
With v2 = 0 and using result in (b):
vo1 = 10v1
vo = v01 + v02 = 10(v1 - v2)
Rin1 = 110R
Rin2 = 10R
(a) Circuit above is the op-amp connected in follower configuration including the DC
power supplies (assuming +/-15V)
(b) When RL is disconnected, IP = IN. With RL connected and Vin = 1V, Vout = 1V and IL =
Vout/RL = 10 mA. This extra 10 mA comes from IP. Likewise, if Vin = -1V, Vout = -1V and
IL = Vout/RL = -10 mA. This extra -10 mA comes from IN.
(c) The additional power drawn is from +15V DC supply when IL is positive and from -15V
DC supply when IL is negative.
Given: G1 = v1/vin = 20
Requirement: vout = 14 when vin = 0.025
Overall Gain = vout/vin = 14/0.025 = 560 = G1*G2
Thus, maximum G2 = 28
If the op-amp is ideal, the voltage across R2 = 0 and hence no current flows through
R1 , R2 and R3 regardless of vin value.
Hence, Vin = VR4 = R4*Vout /(R4 + R5)
Vout = Vin *(R4 + R5)/R4
Extra Questions:
AY2016/17 Semester 2 – Ques 3a.
(i)
(ii)
(v1 – v2)/R = (va – v1)/2R ; Hence va = 3v1 – 2v2
(v1 – v2)/R = (v2 - vb)/R ; Hence vb = 2v2 – v1
(va – vo)/3R = Io ; (vb – 0)/2 = vo + R*Io ; Hence 4vo =3vb – 2va
vo =v2 – 9v1/4
(iii)
Io = 7v1/4R
AY2017/18 Semester 2 – Ques 3a.
(i)
va /20 = I (mA) ; (va – vo)/45 = I (mA) ; Hence vo = va – 45I = -25I
(ii)
(va – vo)/10 = I10K (mA) ; Hence I10K = 4.5I ; I40K = va /40 = I/2
IS = I + I40K + I10K = 6I ; I15K = vo/15 = -5I/3
(iii)
vs = vo +45I + va = vo - 2vo
vo = -vs
AY2018/19 Semester 2 – Ques 3a.
(i)
(ii)
vx = vin/3
vout = 2vy/3
(iii)
(vout – vx)/20 = (vx – vy)/20 ; Hence vout = 2vin/3 - 3vout/2
vout = 4vin/15