High input resistance and high gain Inverting amplifier can be needed in some applications.
In such case an even larger feedback resistor is required. Too large resistors have issues like precision, availability and excessive stray capacitance.
The here below circuit is a way to maintain both a high input resistance, a high gain and still using not too large feedback resistors.
Questions:
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1. 100k
2. -102 / If you want a Gain of -100, change the 1k to 1.02k
seems like KCC likes the AOE book as much as me
The inverting input of the opamp can be considered at virtual ground potential.
The R2 and Rg resistors are in parallel.
R2g= R2Rg/(R2+Rg) = 100/101.
Ro = R3 + R2g = 100 + 100/101 = 101k
Since the input is available at the junction of R2 and R3.
The gain can be evaluated as Av= Ro/R2g
2. Av = - 101/0.99 = -102
1. Rin in this case is R1 = 100k.
So, and here are the official (and hopefully) correct answers:
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:
speedLang360 from Cavite, Philippine
Linux11 from Germany
rkg from Bangalore, India
gpiarino from Milan, Italy
Now, be prepared for the next coming challenge!
Kuo-Chang
