Please explain the gain function for given circuit

[PratapKollu]

**broken link removed**​

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.


Thanks ans Regards
Pratap

 

[ericgibbs]

**broken link removed**​

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.


Thanks ans Regards
Pratap

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.

Does this help.

Whats the supply to the opa?

 

[PratapKollu]

Respected Eric
Supply voltage +/-15 V DC
May i know, why I cannot get 10V(p-p) when feedback resistor is 20K and Rin is 1K and capacitor value is 104?

How can i get exactly 20 times gain when feedback resistor is 10k and Rin is 420 ohm and capacitor 103?

How can I calculate/know input reactance of capacitor?

Thanks for ur reply
pratap

 

[ericgibbs]

Respected Eric
Supply voltage +/-15 V DC
May i know, why I cannot get 10V(p-p) when feedback resistor is 20K and Rin is 1K and capacitor value is 104?

How can i get exactly 20 times gain when feedback resistor is 10k and Rin is 420 ohm and capacitor 103?

How can I calculate/know input reactance of capacitor?

Capacitive Reactance Xc= 1/( 2 * pi * freq * Cap)

Xc= 1/( 6.28 * 100,000 * 104) Ohms



Thanks for ur reply
pratap

Also look at the graph for large signal gain/output swing from the opa datasheet.

Eric

 

[Optikon]

Respected Eric
Supply voltage +/-15 V DC
May i know, why I cannot get 10V(p-p) when feedback resistor is 20K and Rin is 1K and capacitor value is 104?

How can i get exactly 20 times gain when feedback resistor is 10k and Rin is 420 ohm and capacitor 103?

How can I calculate/know input reactance of capacitor?

Thanks for ur reply
pratap

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.

 

[PratapKollu]

Also look at the graph for large signal gain/output swing from the opa datasheet.

Eric

thnx eric
Capacitive Reactance Xc= 1/( 2 * pi * freq * Cap)

Xc= 1/( 6.28 * 100,000 * 104) Ohms

means

Xc= 1/( 6.28 * 100,000 * 10 x10e-9) Ohms, (Is it correct?)

 

[Optikon]

thnx eric
Capacitive Reactance Xc= 1/( 2 * pi * freq * Cap)

Xc= 1/( 6.28 * 100,000 * 104) Ohms

means

Xc= 1/( 6.28 * 100,000 * 10 x10e-9) Ohms, (Is it correct?)

Yes, your Xc in this case ~ 159 Ohms

 

[PratapKollu]

thanks every one for the replies
i am grateful to you for clearing my basic doubt

May i know, why I cannot get 10V(p-p) when feedback resistor is 20K and Rin is 1K and capacitor value is 104???
Please clarify

 

[audioguru]

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.