Dear Members,
For the circuit shown in figure above, input - 0.5V (p-p),100 kHz sine wave required output - 10 V (p-p), 100 kHz sine wave
op amp used is TL082 Result: I cannot get 10 V, I can only obtain 5 V (p-p) signal
I changed C7 = 103,R15 = 420 Ohms,Near J3 = 10K(feed back resistor)
Now I can get 12 V(p-p)signal, please explain what is the design equation thats is governing the gain here.
Dear Members,
For the circuit shown in figure above, input - 0.5V (p-p),100 kHz sine wave required output - 10 V (p-p), 100 kHz sine wave
op amp used is TL082 Result: I cannot get 10 V, I can only obtain 5 V (p-p) signal
I changed C7 = 103,R15 = 420 Ohms,Near J3 = 10K(feed back resistor)
Now I can get 12 V(p-p)signal, please explain what is the design equation thats is governing the gain here.
hi,
The basic gain is 20K/R15,, Gain= Rfeedback/Rinp. [inverted]
To determine the gain over the full frequency range, as its a ac amplifier, the reactance of the input capacitor should be included with Rin resistance value.
The gain EQ is correct, make sure Reactance of C7 is much less than 1k.
1/(2*pi*f*C)
Since it worked for 10k & 420 Ohm, the Opamp is capable of doing it. Go double check the values for your resistors you are using. You may have read one wrong.
The gain of the opamp at 100kHz without any negative feedback is only about 30, so trying to get a gain of 20 with resistor calculations is difficult.
The source impedance must be very low or its value must be included in the gain calculation to reduce the amount of gain.