fully balanced opamp closed loop problem

[jimito13]

Hi,i have designed a fully differential opamp.Let's say that my unity gain freq is x, after the open-loop simulations.When i close the loop with resistive feedback elements (R=100K at the feedback loop,R=10K at the input -- so closed loop gain to be 20dB) and i run ac analysis i notice at the bode diagram of the gain that the unity gain freq is much more under the x value...Below i show you my testbench.The vsin sources are set up as :

vsin for the plus input --> vdc=vdd/2 and ac magnitude=500mV

vsin for the minus input --> vdc=vdd/2 and ac magnitude=500mV and ac phase=180deg

Thanks in advance for any helpful answer.

 

[LvW]

Re: fully balanced opamp cloded loop problem

What is the value of CL?
Open loop simulation with or without CL?

 

[erikl]

fully balanced opamp cloded loop problem

You cannot expect comparable unity gain band widths with different loading:
Adding the feedback devices changes the impedances at the respective nodes, and so the poles associated with them.
In reality even more than in schematics, as real devices - apart from resistance - always add reactances.

 

[jimito13]

Re: fully balanced opamp cloded loop problem

Cload=1pF and the open loop simulation done with load present.

 

[LvW]

Re: fully balanced opamp cloded loop problem

Cload=1pF and the open loop simulation done with load present.

OK, no influence to be expected.

Of course, I agree with the comments made by erikl - however, only in principle!
A "good" voltage amplifier (and only such a "good" amplifier should be called "opamp") has an output impedance and an input impedance which is so small resp. large, that resistive loads which are not extreme shouldn't change the unity gain frequency remarkable. Not extreme means, for example: between 1k and 100kohms.
What are your input/output impedances?

 

[jimito13]

Re: fully balanced opamp cloded loop problem

My opamp is a folded cascode OTA...that is i have a very big output impedance...some MΩ's i suppose...the input impedance is large.What do you suggest me to do?Maybe place buffers at the output?

 

[LvW]

Re: fully balanced opamp cloded loop problem

My opamp is a folded cascode OTA...that is i have a very big output impedance...some MΩ's i suppose...the input impedance is large.What do you suggest me to do?Maybe place buffers at the output?

Aha! I see, you have designed an OTA and NOT an opamp as mentioned in your first posting. Of course, such an amplifier has a gain which is determined mostly by the load only!! And a feedback loop acts as a load. Thus, your result is no surprise at all!
The gain is gm*ZL||rout (gm=mutual transconductance; ZL=effective load impedance)

 

[jimito13]

Re: fully balanced opamp cloded loop problem

So,the solution is to place buffers to fix my problem??Thanks a lot for your help

 

[erikl]

Re: fully balanced opamp cloded loop problem

So,the solution is to place buffers to fix my problem?

Yes, I think so; so you'd make an opAmp out of your OTA. But it will also change the fT (UGBW) a little bit.

BTW: to test if our/my assumption is correct, you could run an open loop analysis with the 2 feedBack resistors connected to GND (keep the 10kΩ's in series to your input stimuli) and check the resulting UGBW.

Thank you for having clicked the Helped me button!

 

[LvW]

Re: fully balanced opamp cloded loop problem

So,the solution is to place buffers to fix my problem??Thanks a lot for your help

I don't know what your problem is. What was your task resp. goal?
To design an OTA (high output impedance) or an opamp (low Zout) ?
Both amplifier types are in use and have their specific properties (advantages/disadvantages).

 

[jimito13]

Re: fully balanced opamp cloded loop problem

i am intending to place the OTA in a Tow Thomas biquad to implement a filter.what would you suggest?i think that i do not need the high output impedance that's why i thought to use buffers at the output of the i circuit i designed.

 

[LvW]

Re: fully balanced opamp cloded loop problem

The original TT-biquad consists of a non-inverting integrator (realized with a classical MILLER integrator and an inverter stage) and an inverting damped integrator (active low pass) - connected in a closed loop via a voltage divider.
This topology can be realized with opamps (that's the classical solution) or with OTAs.
I would recommend the opamp version because of easier dimensioning.
Thus you need for each OTA an additional buffer stage.

Added after 5 hours 1 minutes:

If you are interested in an active filter - based on the TT structure - but realized with two OTAs, here is the reference:
Fig. 7(a) in:
**broken link removed**

 

[jimito13]

Re: fully balanced opamp cloded loop problem

BTW: to test if our/my assumption is correct, you could run an open loop analysis with the 2 feedBack resistors connected to GND (keep the 10kΩ's in series to your input stimuli) and check the resulting UGBW.

Erikl i tried the above and the unity gain frequency is far from the desired...I must also add that only with the input resistors,as you propose,things are ok (like my openloop initial simulations without any feedback or something at the inputs) only if R=10Ω..I cannot understand how these input resistors affect the fT...This also happens with the resistive feedback + resistive input + buffers at the output of the OTA...any idea?

 

[erikl]

Re: fully balanced opamp cloded loop problem

BTW: to test if our/my assumption is correct, you could run an open loop analysis with the 2 feedBack resistors connected to GND (keep the 10kΩ's in series to your input stimuli) and check the resulting UGBW.

Erikl i tried the above and the unity gain frequency is far from the desired...I must also add that only with the input resistors,as you propose,things are ok (like my openloop initial simulations without any feedback or something at the inputs) only if R=10Ω..I cannot understand how these input resistors affect the fT...This also happens with the resistive feedback + resistive input + buffers at the output of the OTA...any idea?

Yes, I think so. I simply forgot, that the feedBack reduces the output resistance by the loop gain: Let's say your OTA open loop gain is 60dB and its original output resistance is 1MΩ. A 100kΩ load resistance will change the overall output impedance to about this value.

In a configuration for 20db closed loop gain, the original output resistance of 1MΩ will be reduced - in 1st approximation - by 40dB, i.e. to about 10kΩ, which will dominate and therefore determine the output node's impedance. Hence the output impedances of both configurations - and so the poles associated with them - are not the same, in contrary to what i assumed in my former posting, sorry! I hope LvW can confirm this.