# calculate gain different from gain on oscilloscope

Discussion in 'General Electronics Chat' started by ali8tor, Apr 13, 2013.

1. ### ali8tor Thread Starter Member

May 29, 2012
43
0
Hi

I've made a non inverting preamp circuit and have a calculated gain of 69 but when i go to test it on an oscilloscope i only have a gain of 2?

am i doing something wrong?

2. ### kubeek AAC Fanatic!

Sep 20, 2005
4,670
804
First of all you are asking wrong. Without the schematic, input and output amplitude, and your gain calculation, how is anyone supposed to guess where the problem lies?

3. ### ali8tor Thread Starter Member

May 29, 2012
43
0
I used a given formula for gain calculation 1+r2/r1 my input Amplitude is 10vp i attached a screenshot of my circuit.

Last edited: Apr 13, 2013
4. ### kubeek AAC Fanatic!

Sep 20, 2005
4,670
804
Well, how exactly do you expect an amplifier with +/-15V supply to produce +/-690V on the output? Try putting 10mV on the input instead. Also if you used a scope instead of a lousy multimeter feature, you would see the output clipping hard to the power rails minus a bit.

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5. ### ali8tor Thread Starter Member

May 29, 2012
43
0
Hi thankyou for your input. I changed the value to 10mV and im getting an output of 16V do i keeping play with different figures untill i get my desired output?

6. ### kubeek AAC Fanatic!

Sep 20, 2005
4,670
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With 10mV on the input you should be getting less than a volt on the output. So your multimeter function is lying to you, so use an oscilloscope and look at the actual waveform on the output.

7. ### ali8tor Thread Starter Member

May 29, 2012
43
0
yeah i tried it with the oscilloscope and it reads 16V?

8. ### kubeek AAC Fanatic!

Sep 20, 2005
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can you post a screenshot of the waveform?

9. ### ali8tor Thread Starter Member

May 29, 2012
43
0
blue ones input and reds output

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10. ### kubeek AAC Fanatic!

Sep 20, 2005
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Look at it carefully, the blue one is set to 10V per division, so the blue waveform is 10V peak. This means you didn´t change the input voltage.

11. ### ali8tor Thread Starter Member

May 29, 2012
43
0
oh yes i see i think i changed the dc voltage instead of the inputs. i made the input 10mV and im getting around -18mv

12. ### kubeek AAC Fanatic!

Sep 20, 2005
4,670
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DC offset should not matter (depending on how your simulator treats it) because it is blocked by C2.

13. ### ali8tor Thread Starter Member

May 29, 2012
43
0
oh i see does this look better? i changed the dc voltage instead of the input

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14. ### kubeek AAC Fanatic!

Sep 20, 2005
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Yes that looks correct. So now, why do you think the gain is not 69 with 1kHz signal? The answer lies mostly in C3 and C4.

15. ### ali8tor Thread Starter Member

May 29, 2012
43
0
haha I had a little play around with C3 but cant figure what is the right value i should be using. Is 1KHz signal okay to use or something less then that?

16. ### kubeek AAC Fanatic!

Sep 20, 2005
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Most of all, it depends on what you want from that circuit. Right now, it is a high shelf filter with a gain of 2.4 and 900Hz corner frequency. See the attached frequency plot.

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17. ### ali8tor Thread Starter Member

May 29, 2012
43
0
I just want it to amplify my input and give me a gain of 69. so i need to calculate the frequency im looking for?

18. ### kubeek AAC Fanatic!

Sep 20, 2005
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804
You need to delete C4 and C3, and ground the bottom pin of R2. You might also want to decrease R1 to about 1K, or decrease C2 to about 220nF, because that will give you better response to changes in the DC offset level, but you trade it for gain roll-off at frequencies below 10Hz.

19. ### ali8tor Thread Starter Member

May 29, 2012
43
0
will it still do its job as a mic preamp?

20. ### kubeek AAC Fanatic!

Sep 20, 2005
4,670
804
Yes it should, but really depends on what kind of mic you have in mind. If you are going to use the lm833 or any other high speed opamp, you should add a 22pF capacitor between output and negative input of the opamp.

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