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I've had this question for a while.

Assume you have an otherwise perfect op-amp with a gain-bandwidth product of 5 MHz. You input a signal of 50mVp-p and amplify it by 10x. This limits your bandwidth to 500 kHz. Now, say you stack another op-amp on the output and configure that as a 10x amplifier. Your overall bandwidth is 500 kHz but you have amplified by 100x, so your GBWP is 50 MHz. Where is the flaw in this logic?

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  • \$\begingroup\$ The bandwidth of two 500 kHz filters in series would be a little less than 500 kHz, though. \$\endgroup\$ – endolith Nov 5 '10 at 2:08
  • \$\begingroup\$ Dave Jones demonstrates that your assumptions are (mostly) true in this video: youtube.com/watch?v=ZvT9hHG17tQ . \$\endgroup\$ – user36113 Jan 25 '14 at 11:14
  • \$\begingroup\$ GBWP (as I stated in my answer) doesn't have a meaningful usage for multiple opamps. In your cascaded op-amp case, let's say you change them to a 5x amplifier, with each opamp having 1MHz bandwidth; total gain is 25x, so your "overall GBWP" calculation is 25MHz. Not constant -- GBWP only has meaning for a single op-amp. \$\endgroup\$ – Jason S Oct 13 '16 at 15:40
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Your logic is sound. More opamps means more gain for a given bandwidth.

Compensated op-amps are made with a single dominant pole so the gain/bandwidth product is constant. If this isn't working for your application, use a decompensated amp and do the compensation network yourself.

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Gain-bandwidth product only has meaning w/r/t one op-amp: when you multiply the gain and bandwidth, you get a constant because of the way the op-amp is internally compensated.

When you have more than one stage, the overall gain times the overall bandwidth is not constant, so an overall gain-bandwidth product has no meaning.

But your analysis of overall gain and overall bandwidth is correct, or at least mostly correct: it wouldn't be 500kHz, it would be slightly less. Bandwidth is usually measured in terms of a -3db point, so when you cascade two stages you get -6dB at 500kHz, and therefore the -3dB point is somewhere below that, probably in the 400-450kHz range.

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  • \$\begingroup\$ The bandwidth of an opamp is not specified in terms of a -3db point. The GBP of an opamp is specified as the point where the gain of the opamp has fallen back to one. Cascading two opamps doesn't change that. The bandwith of the total stage where the gain has fallen back to one is still the GBP of the opamps. (if you use opamps with the same GBP.) \$\endgroup\$ – Hendrik Jun 4 '11 at 5:54
  • \$\begingroup\$ I'm not talking about the gain-bandwidth; I'm talking about the bandwidth with gain = 10. \$\endgroup\$ – Jason S Jun 4 '11 at 14:38
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I would be in op amp heaven if that would be true. Sorry, but the other answers are both wrong. You can't multiply the GBP's of op amps (and amplifier stages in general) because the gain of the op amp is one at the specified GBP.

The actual bandwidth of cascaded amplifiers is limited by the amplifier with the smallest bandwidth. (And it doesn't has to be the one with the smallest GBP.)

Regards

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    \$\begingroup\$ You don't explain where the flaw is in his logic. \$\endgroup\$ – stevenvh Apr 8 '12 at 5:49
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When you have a single op amp, you'll have a single pole at the break frequency and the output voltage will roll off at 6dB/octave, but when you have two op amps, you have two poles at the break frequency (break frequency of a single op amp, which we assumed is the same for both op amps) and the output voltage will thus roll of at 12dB/octave (since the transfer functions are multiplied), meaning that the system will reach its overall break frequency sooner (as seen from the point at which it starts to roll off) than wit a single op amp.

More precisely at f3dB_overall = f3db*sqrt(2^(1/2) - 1) ~= 0.64fd3B, where f3dB is the common break frequency for each individual op amp.

More generally, for n cascaded op amps, f3dB_overall = f3db*sqrt(2^(1/n) - 1).

Also, as markrages points out, to get around this problem you can use uncompensated op amps and add the compensation capacitor over the whole cascade by yourself.

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enter image description here

I will try to answer to answer in a simple way. When we cascade to systems its gain will be multiplied in frequency domain. Assume that the maximum gain of the system is 1. So its 3db frequency is '.707'. Let's call this frequency F' which is the cutoff frequency of the system.

Let's check the value of gain at F' for the cascaded system. This is were it gets interesting. For the cascaded system the gain at F' will be .707×.707=0.499. So F' is not the cutoff frequency of the cascaded system. Thus the new cutoff frequencies shifted from the old value and the new bandwidth will be lower than previous one. I have tried to illustrate this in the above drawing. Hope you will get my point.

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Your logic is perfect correct! so is the output. Only thing you got to remember is that when cascaded the gain multiplies but not the frequency response. So in your case the gain-bandwidth of both the op-amps cascaded would be gain1 * gain2 *frequency(the smallest of the two)

so the answer would be 10 * 10 * 500khz = 50MHz which is perfectly correct.

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You are not getting more bandwidth by cascading these amps. Remember, there is a looming limitation of 5Mhz. You just get more of the existing 5Mhz bandwidth at a given gain.

There is still a roll-off of gain to unity at the same frequency, but that roll-off is faster, similarly to how you get faster roll-off as you add more poles to a filter. So it's like you are getting a better approximation of a "brick wall".

At the frequency where the open loop gain is unity, you cannot get any more gain by cascading the amplifiers. Beyond that frequency where there is less than unity gain, you get less gain by cascading the amps.

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