# Calculating gain of an OPAMP

Discussion in 'General Electronics Discussion' started by Govardhan, Dec 12, 2013.

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Oct 1, 2013
2. ### Harald KappModeratorModerator

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Nov 17, 2011
The gain-bandwidth product assumes a linear decrease in gain with frequency.. Basically the term gain*bandwidth is constant. So for a high gain you have a low bandwidth and vice versa. Details are here.

3. ### Govardhan

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Oct 1, 2013
Thanks and now I calculated the gain needed for my OPAMP
MCP6002 has Gain Bandwidth Product of 1 MHz
And I'm using the MCP6002 in the noninverting mode as here http://img853.imageshack.us/img853/3717/2jpp.jpgfor LP filter for a cutoff frequency of 2.3Hz and I'm using a gain of 101 (1+R3/R2=1+680k/6.8K=101).
If want a gain of 200 with the frequency of the signal lie between 0.5 to 3Hz=2.5Hz bandwidth(required).
BW=GBW/gain
2.5Hz=1MHz/gain=>gain=400000
Can I use R2=3.4K so that 1+R3/R2=201.Is it ok for this OPAMP to use a gain of 201.
I guess its nearly simialr for the OPA378 which has bandwidth of 900Khz.

4. ### BobK

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Jan 5, 2010
The word bandwidth here is misleading, it is not the range of frequencies as you have assumed in your calculation, but the upper limit of the frequencies that you need to amplify. In any case, at 3.5Hz you are not going to have any problems with any opamp.

Bob

5. ### Laplace

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Apr 4, 2010
Here is an example of how to use the GBW product. Suppose there is a simple inverting amplifier with a gain of 20 dB (Rf/Ri=10) and the op amp has a GBW as shown. The inverting amplifier will maintain its 20 dB gain out to the corner frequency of 100 KHz. This is an approximation because the 20 dB gain will begin to fall off somewhat before the input frequency reaches the corner.

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