Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

how to measure open loop gain in cadence

Status
Not open for further replies.

pekachoo007

Member level 2
Joined
Jul 29, 2013
Messages
45
Helped
1
Reputation
2
Reaction score
1
Trophy points
8
Activity points
284
I want to measure the open loop gain of the op amp. Is the test bench i am using is correct
 

Attachments

  • Opamp.PNG
    Opamp.PNG
    20.7 KB · Views: 753
  • Tb opamp.PNG
    Tb opamp.PNG
    7.6 KB · Views: 827

* I don`t see an opamp - it is an OTA, correct?
* More than that, it seems that you are applying POSITIVE feedback; WHY?
* The test signal must be feeded into the loop directly!
 

* I don`t see an opamp - it is an OTA, correct?
* More than that, it seems that you are applying POSITIVE feedback; WHY?
* The test signal must be feeded into the loop directly!

is OTA is not also a type of op-amp. I want to calculate the open loop gain. how to directly feed test signal into the loop?
 

I am not sure what you want to simulate:
a) Open-loop gain - that is the gain of the active device without any feedback
b) Loop gain - that is the gain od the complete feedback loop (amplifier and feedback path in series)
 

Having a large inductor in the negative feedback (as you show in your screenshot) is a common way to simulate open-loop gain. The large value inductor in negative feedback ensures correct DC biasing. For AC analysis it is out of the picture. You should make sure that the feedback is connected in a negative feedback scheme.
 
For AC analysis it is out of the picture. You should make sure that the feedback is connected in a negative feedback scheme.

What do you mean by out of picture for AC analysis.

i think the feedback is connected in negative feedback scheme.

Vin+ might just be a naming convention.

In some analysis I have seen large resistor intesd of inductor.
 

I mean that when you run an AC simulation (i.e. to look at the open loop gain) with a very high value inductor in the feedback, the inductor impedance will be very high at those AC frequencies, so at AC frequencies it will look like you have an open-loop. At DC, the inductor has 0 impedance, so it appears like a short, so for DC solution you will have a feedback ensuring the correct DC voltages.

Having a very large resistor in feedback is also an option, but it will probably affect your AC simulation results.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top