cascode and wide swing current mirror output impedance

Status
Not open for further replies.
J

Junus2012

Advanced Member level 5
Joined
Jan 9, 2012
Messages
1,552
Helped
47
Reputation
98
Reaction score
53
Trophy points
1,328
Location
Italy
Activity points
15,235
Hello

My question is very basic

I am comparing the performance of two current mirror,the first one is Cascode mirror as shown in the lift side of my image and the other is the Wide swing mirror

from any resource I read the consider both having the same output impedance and the difference between the is only the minimum output voltage needed for proper mirroring

but when I am simulating it I am finding that the cascode mirror has much more output impedance than the wide swing and this what i want to discuss with you



thank you
 

Both circuits will have equal output impedance if and only if the upper transistors have equal Vds respectively the same gate potential. Seeing a lower output impedance suggests a too low bias potential. This is up to the dimensioning of your bias circuit.
 
Thanks FvM

I working with the same current condition for cascode and wide swing, the I am getting less ouput impedance for the swing circuit. The only way to increase the output impedance of the swing is by reducing the ratio of the biasing transistor ( like 1 / 6 ) but still it is not comparable for the Cascode
 

senan said:
I am getting less ouput impedance for the swing circuit.
Click to expand...

I think this is because M6 (in the wide swing version) operates in deep triode region and thus has a rather low output impedance. Actually the cascode advantage is nearly lost. You trade output impedance against wide voltage swing.
 
could you please tell me how M6 is working in the triode region

erikl said:
I think this is because M6 (in the wide swing version) operates in deep triode region and thus has a rather low output impedance. Actually the cascode advantage is nearly lost. You trade output impedance against wide voltage swing.
Click to expand...
 

senan said:
could you please tell me how M6 is working in the triode region
Click to expand...

See here its Vds - which surely is unrealisticaly low - but I guess it will still be low enough for triode region in your simulation:

 

by the way I am simulating the output impedance of the cascode with the frequency as in the same method I use to simulate the output impedance of the op-amp
but I am not getting a logical result, I attached you the image of the setup



please

- - - Updated - - -

I set the A.C current signal (not source)to magnitude of 1 and i run the A.C simualtion
 

senan said:
I am not getting a logical result
Click to expand...

Of course not: R1 (10kΩ) is parallel to the output impedance which you want to measure. So you see essentially the value of R1. You need a huge inductance in between, see my testbench above!
 
Ok erikl, but I dont have an inductor in my simulator, what should I do ???

- - - Updated - - -

Oh I am really sorry for this mistake, I have an inductor with my simulator I found it,

I did the simulation as exactly as you told and I got the following graph



I think at the beginning of the graph L was still has low value impedance although I have chosen 10KH, then I will reach the real value of the output impedance which is appropriately 153 db,

but this has shifted the frequency behavioural of the circuit, what you say ? am I getting the right graph ?

- - - Updated - - -

Yes , I think I have to use very large inductor like 100KH
 

Hello FvM
for me it also hard to believe that, any way I attached you the result with using 1MH and still there is some drop at the beginning

I am using Eldo simualtor from Mentorgraphic


 

I am using Eldo simualtor from Mentorgraphic
Click to expand...
Then I don't understand your confusing statements. ELDO is a SPICE like simulator and of course has inductors.

It's also uses SPICE syntax, thus 1 MH is a pretty small inductance (1e-3). If you mean 1e6, you're expected to write 1MegH. I think, this explains your simulation result.
 

Yes, FvM

I have told before that I did a mistake when I said I dont have a inductor, I have corrected it... yes I mean to say Meg (1e6) , but dont you see it is very large ??
 

senan said:
... dont you see it is very large ??
Click to expand...

Yes! Your last plot looks more like you inserted 100MegH (1e8) -- if your dB scale is referenced to 1V resp. 1Ω .

Your output impedance would be nearly 700 MegΩ in this case.
 

Oh no, if we take the linear value from the db value the output impedance will be 40Meg, by ( ant log (152/20) ). I am comparing this result with the result I got with the D.C analyses ant it is the same


erikl said:
Yes! Your last plot looks more like you inserted 100MegH (1e8) -- if your dB scale is referenced to 1V resp. 1Ω .

Your output impedance would be nearly 700 MegΩ in this case.
Click to expand...
 

senan said:
if we take the linear value from the db value the output impedance will be 40Meg, by ( ant log (152/20) ).
Click to expand...

Yes, you're right, sorry! My simulation (above) showed 2MegΩ only (for your wide swing version).
 
A large henry value is associated with small current. (With capacitors a large Farad value is associated with large current.)

Your schematic has '5 uA' next to a couple of wires. Therefore to obtain 1 kHz pole frequency, it is plausible that you'll need a henry value in the thousands or millions.
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…