problem with differential pair

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ahmad1954

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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?
 

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

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

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?
 

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?
 

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
 

Attachments

  • Diff amp.pdf
    7 KB · Views: 73
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
 

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
 
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?
 

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

Keith
 

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.
 

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.
 

Attachments

  • Diffamp gm.pdf
    5.3 KB · Views: 86

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