KCC's Quizzes: AQQ239 about a High gain & high input resistance inverting amplifier

So, and here are the official (and hopefully) correct answers:

1. Rin = 100k. Justification: by assuming the opamp is well in its linear mode, the potentials at V- and V+ are equal. Since there is no current in the 51k resistor, V+ is 0 voltTherefore, Vin/In is 100k.

2. At node V’, the current flows must be balanced:

Vin/100k = V’/1k +(V’-Vout)/100k (eq 1)

V+ being equal to V- we have then V’ = -Vin (eq 2)

By injecting eq2 in eq 1 we get : Vin/100k = -Vin/1k – Vin/100k – Vout/100k

By multiplying by 100 the whole expression and by simplifying: 102 Vin = -Vout

We have the gain = Vout/Vin = -102

So congratulation to all the ones having found the solutions and big applause for the 4 first ones:

 from Cavite, Philippine

Linux11 from Germany

 from Bangalore, India

gpiarino from Milan, Italy

Now, be prepared for the next coming challenge!

Kuo-Chang

So, and here are the official (and hopefully) correct answers:

1. Rin = 100k. Justification: by assuming the opamp is well in its linear mode, the potentials at V- and V+ are equal. Since there is no current in the 51k resistor, V+ is 0 voltTherefore, Vin/In is 100k.

2. At node V’, the current flows must be balanced:

Vin/100k = V’/1k +(V’-Vout)/100k (eq 1)

V+ being equal to V- we have then V’ = -Vin (eq 2)

By injecting eq2 in eq 1 we get : Vin/100k = -Vin/1k – Vin/100k – Vout/100k

By multiplying by 100 the whole expression and by simplifying: 102 Vin = -Vout

We have the gain = Vout/Vin = -102

So congratulation to all the ones having found the solutions and big applause for the 4 first ones:

 from Cavite, Philippine

Linux11 from Germany

 from Bangalore, India

gpiarino from Milan, Italy

Now, be prepared for the next coming challenge!

Kuo-Chang