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:
Good luck and try to be among the first!
And communicate this around you (colleagues, partners, customers and friends)!
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:
from Cavite, Philippine
Linux11 from Germany
rkg 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:
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
rkg from Bangalore, India
gpiarino from Milan, Italy
Now, be prepared for the next coming challenge!
Kuo-Chang
