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.

Output resistance of a current mirror

Status
Not open for further replies.

CAMALEAO

Full Member level 4
Joined
Jul 29, 2016
Messages
201
Helped
2
Reputation
4
Reaction score
2
Trophy points
18
Activity points
1,868
Hi guys,

I am trying to measure the output and the input resistance of a current mirror.

Currently I am applying a vdc voltage at the output to set the operating point and then an AC voltage source with 1VAC. After simulating I compute V/I.

I am getting around 80Gohm. Is this normal? 5V devices, 130nm, vdd 3.8V, 500nA.

Is this correct? Is there any other method to measure the output resistance? Maybe I can use a current source at the output with 1A AC, but how do I set the operating point?

Finally I read somewhere that we can measure Iout at Vout_min and Vout_max and compute R(ac) using V/I. For this I need to do the simulation in DC?

Finally, is there any logic in simulating the output resistance DC? Because in the books we read output resistance, and in simulation if we use an AC voltage source we are measuring the impedance.

Regards.
 
Last edited:

Really? I thought is was in the order of Mohms.

What about measuring the input or output resistance using a DC voltage source? Then do V/I? Does it make sense? Since the current mirror is basically to be used as DC current source.. So which one?
 

I presume you are asking about measurement in simulation?

There should be no difference between low frequency output impedance in AC measurement and differential DC impedance. Of course the operation point matters.

It would be much easier to rate the 80 GOhm number if you tell the current mirror topology. It might be possible in a cascaded current mirror.
 

Really? I thought is was in the order of Mohms.
?
Its a bit like asking what is the impedance of a dead short.
The answer would be theoretically zero.
And if someone says "I thought is was in the order of milliohms" that would not be unreasonable.
 

Hi FVM, sorry. I forgot that. I am talking about the simple cascade current mirror (afterwards, I will try to do the same with the high swing current mirror and compare both, knowing that I have to set the same biasing conditions).

Yes, through simulation.

Regarding your comment about the DC o AC impedance. You are saying that both should give the same. How?

1) If you are talking about plotting the current vs voltage and then compute the deltaV / deltaI, that's the same as using the AC analysis right?

2) If you are talking about sweeping the dc voltage source at the output and divide that voltage by the current, we should be expecting the same resistance as the AC method? Don't think so. Right?

I think I did that last week and the results are not the same as the ones I am getting for the AC method.

Regards.
 

sweeping the dc voltage source at the output and divide that voltage by the current, we should be expecting the same resistance as the AC method?
Yes if you divide dV/dI, not V/I.

The results should be roughly the same. A certain variation with Vdd can be expected.

If the results are completely different, you have probably a fault in AC measurement setup.
 

Alright, it makes sense then. I thought, when you referred differential dc impedance, that you meant divide the DC voltage by the DC current.

Thanks for the clarification.

I guess that I am getting this sort of very high output resistance because I am using a very low current.

Regards.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top