Electronics - Operational Amplifiers - Discussion
Discussion Forum : Operational Amplifiers - General Questions (Q.No. 17)
17.
If a noninverting amplifier has an RIN of 1000 ohms and an RFB of 2.5 kilohms, what is the RIN voltage when 1.42 mV is applied to the correct input?
Discussion:
16 comments Page 1 of 2.
Christie said:
1 decade ago
how to solve this problem?...
Ashok said:
1 decade ago
r in=1000:
i=v/r:
v=1.42mA
so i=1.42A
i=v/r:
v=1.42mA
so i=1.42A
Bhoopendra said:
1 decade ago
Negative terminal of op amp is at virtual ground therefore at this point the voltage will be equal to applied voltage at positive terminal.
Vishal said:
1 decade ago
Ashok how will be vtg 1.42mA ?
Mahesh said:
1 decade ago
Other end is grounded so due to virtual ground concept the voltage at non inverting pin will be zero so drop will be across 1kohm.
Vicky said:
1 decade ago
I think mahesh and bhoopendra is right. Thanks friends.
ASHOK KUMAR.R said:
1 decade ago
According to the Virtual Ground concept if the inverting terminal is Grounded then the Rin will looks like parallel to that of the source.
So the voltage drop across the Rin is equal to the applied voltage.
So voltage across Rin is 1.42mA.
So the voltage drop across the Rin is equal to the applied voltage.
So voltage across Rin is 1.42mA.
Dharm said:
1 decade ago
How to solve this ?
Daks said:
1 decade ago
Vin = Vo((Rf/R1)+1)).
Vin = 1.42mv*3.5
Vin = 4.97mv.
The current passing throughout RF and R1.
RF+R1 = 3.5K.
I = Vout/RT.
I = 4.97mV/3.5k.
I = 1.42X10-6 amps.
Vr = I*R1.
Vr = 1.42x10-6 x 1k.
Vr = 1.42mv.
Vin = 1.42mv*3.5
Vin = 4.97mv.
The current passing throughout RF and R1.
RF+R1 = 3.5K.
I = Vout/RT.
I = 4.97mV/3.5k.
I = 1.42X10-6 amps.
Vr = I*R1.
Vr = 1.42x10-6 x 1k.
Vr = 1.42mv.
(1)
Caro said:
1 decade ago
2.5/1000+1x1.42 = 1.42.
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