problem with differential pair

[ahmad1954]

I want to calculate linear range of a simple mos differential pair with hspice. this is my netlist:

Vdc1 1 6 0
vcm 6 0 3
VCC 11 0 DC 5
VDD 12 0 DC -5
M1 3 1 5 5 NMOS1 w=10u l=.1u
M2 4 6 5 5 NMOS1 w=10u l=.1u
RC1 11 3 1k
RC2 11 4 1k
RE 5 12 7.2K
.MODEL NMOS1 Nmos level=2
.print dc id(m1) id(m2)
.dc vdc1 -3 3 1m

and this is output:


my question is why 2 graphs are not symmetric?

 

[Usama Siddiqui]

did you made a proper mirror i.e all the same values of resistors and transistors ??

 

[ahmad1954]

if you look at netlist you see that all values for resistors and transistors are same.

 

[RCinFLA]

By process of elimination I would say issue is in gate connections. Does Vdc1 have a default source resistance?

 

[keith1200rs]

Because you have fixed M2 gate and varied M1 gate. You have not applied a symmetric differential input signal.

Keith

 

[ahmad1954]

so what should I do for symmetric output? should I apply another source to M2 gate? in this case which source should vary for calculating linearity range?

 

[keith1200rs]

You should increase one input as you decrease the other so as one goes from 3V to -3V to other goes from -3V to +3V.

Keith

 

[ahmad1954]

please tell me if I'm wrong. I should apply 2 sources to transistors gate and perform 2 dc sweep with conditions that you said. is this true?

 

[keith1200rs]

I would probably use a voltage controlled voltage source to produce the other voltage - you want the results in a single sweep. See attached.

Keith

 

[ahmad1954]

What software did you use?
I tried your idea in hspice and I received this error: **error** inductor/voltage source loop found containing 0:vdc1 defined in subckt 0

this is new code:
Vdc1 1 6 0
vcm 6 0 3
VCC 11 0 DC 5
VDD 12 0 DC -5
M1 3 1 5 5 NMOS1 w=10u l=.1u
M2 4 7 5 5 NMOS1 w=10u l=.1u
RC1 11 3 1k
RC2 11 4 1k
RE 5 12 7.2K
E1 1 6 6 7 1
.MODEL NMOS1 Nmos level=2
.print dc id(m1) id(m2)
.dc vdc1 -3000m 3000m 1m

 

[keith1200rs]

I think your E1 is incorrect. HSPICE has output nodes first then input nodes. Try E1 7 6 6 1 1

I use SIMetrix, but the simulator shouldn't matter - it should work any any SPICE simulator (with slight syntax differences).

Keith

 

[ahmad1954]

Thank you very much for your help. It's work.

 

[keith1200rs]

And hopefully symmetrical now.

Keith

 

[ahmad1954]

now I want to calculate GM. by definition I should calculate d(Iout). the output is shown below:



my question is why the red rectangle in the right of the figure is this form?

 

[keith1200rs]

It depends how you calculated it. The current drops to zero and you may have a divide by zero in there.

Keith

 

[ahmad1954]

are there any other ways to calculate gm?

 

[LvW]

Hi ahmad1954,
just for a better understanding of the differential pair:

Your output characteristics would exhibit better symmetry - even if you apply one single input voltage only - if you reduce the common mode gain by increasing the common emitter resistance.
An ideal diff. pair has a current source in the emitter path (resistance approaching infinite) with a nearly ideal common mode rejection ratio and symmetrical output curves.

 

[keith1200rs]

What equation have you plotted to get gm? I am not sure it looks correct.

Keith

 

[ahmad1954]

I plot d(idm1). is this true?

 

[keith1200rs]

For a differential pair I would have thought you would want diff(Id1-Id2)/diff(Vg1-Vg2) but it depends on what your objective is. Your syntax might be slightly different to mine (DERIV for HSPICE I think). I don't see any strange discontinuities doing that and the gm tails off rather than making an abrupt drop to zero which is what yours seems to do.

Keith.