I made a fully differential opamp. In open loop ac analysis, the DC gain is 100db, and the GBW is 2G, but the phase margin is only 30 degrees. But when I put the opamp into a closed loop with gain 8, the 3db bandwidth is only 1.4M. So, what's the problem?
And there's another question. In the real application, if only 8 times amplification is needed, and the input signal is only 100MHz, so there's no problem of the poor phase margin?
Re: A question about closed loop ac of fully differential op
liqiyue said:
I made a fully differential opamp. In open loop ac analysis, the DC gain is 100db, and the GBW is 2G, but the phase margin is only 30 degrees. But when I put the opamp into a closed loop with gain 8, the 3db bandwidth is only 1.4M. So, what's the problem?
And there's another question. In the real application, if only 8 times amplification is needed, and the input signal is only 100MHz, so there's no problem of the poor phase margin?
1.) The term "phase margin" is connected with the loop gain, which means that the opamp alone has no phase margin. Therefore, the mentioned value of 30 deg belongs to 100% feedback (unity gain configuration of your opamp).
2.) Of course, the bandwidth is smaller than GBW for higher gains (because GBW is a fixed value for universal compensation).
3.) When your margin in unity gain configuration is 30 deg., it will be larger for a gain of 8. Did you simulate it for this configuration already?
Re: A question about closed loop ac of fully differential op
LvW said:
1.) The term "phase margin" is connected with the loop gain, which means that the opamp alone has no phase margin. Therefore, the mentioned value of 30 deg belongs to 100% feedback (unity gain configuration of your opamp).
2.) Of course, the bandwidth is smaller than GBW for higher gains (because GBW is a fixed value for universal compensation).
3.) When your margin in unity gain configuration is 30 deg., it will be larger for a gain of 8. Did you simulate it for this configuration already?
The gain of 100db I've mentioned above is open loop gain, and GBW of 2G is also measured in open loop. So, I think when the closed loop gain is 8, then the 3db bandwidth should be 2G/8=250MHz.
Any problems here?
Re: A question about closed loop ac of fully differential op
Yes, that´s approximately correct - if your open loop gain response equals a single pole response (20 db/dec within the active range) and when you have a single feedback path.
But in reality, when the closed loop gain is 8, the bandwidth is only 1.4M.
That's very strange. I don't know what's wrong with it. Do you have any idea?
I'll put the simulation results on later.
Thanks!
Yes, I've verified that the GBW is 2G in simulation.
There's another strange problem.
Resistors with ratio of 8:1 are used to form the closed loop feedback in order to achieve 8 times amplification. But the 3db bandwidth is related to the value of the resistors. The greater the resistor is, the smaller the bandwidth is. I've tested several values from 1K ohm to 100M ohm.
Can anyone give any reasons?
Thanks
How big is your input parasitic capacitance (due to the input pair, for instance)? That might explain why higher resistor values decrease your bandwidth.
Re: A question about closed loop ac of fully differential op
Another question: How large/small is approximately the output resistance of your opamp ? Please note thats it´s really hard to help you without some more detailed information about your design.
Re: A question about closed loop ac of fully differential op
LvW said:
Another question: How large/small is approximately the output resistance of your opamp ? Please note thats it´s really hard to help you without some more detailed information about your design.
How big is your input parasitic capacitance (due to the input pair, for instance)? That might explain why higher resistor values decrease your bandwidth.
Yes, that makes sense. The input capacitance is about 1.1pf.
The output resistance is about 900 ohm. So, the feedback resistors should be set to at least 100K & 800K. So the dominant pole will be determined by feedback resistor and input cap which is about 9M.
Is that correct?
Thank you all guys!