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Differential pair questions

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STOIKOV

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I executed some simulations in a MOS differential pair in order to verify some theory I read, and have some doubts.

If (Vin+ = Vin- = 0 volts) the current in each drain is Is/2. Isn't it supposed that with 0 volts in gates they are off ?

Also, when apply any DC value to the gates (Vin+ = Vin- = any volts) the current in each transistor is Is/2, no matther what the gates' DC voltages are. Why happens this instead off the quadratic law Id=K(Vgs-Vt)²

In summary, the drain currents are independient of the DC value in gates. The sum in both networks is always Is. These are the doubts I have and believe is due to the current source but do not explain well why.
 

STOIKOV said:
I executed some simulations in a MOS differential pair in order to verify some theory I read, and have some doubts.

If (Vin+ = Vin- = 0 volts) the current in each drain is Is/2. Isn't it supposed that with 0 volts in gates they are off ?

Also, when apply any DC value to the gates (Vin+ = Vin- = any volts) the current in each transistor is Is/2, no matther what the gates' DC voltages are. Why happens this instead off the quadratic law Id=K(Vgs-Vt)²

In summary, the drain currents are independient of the DC value in gates. The sum in both networks is always Is. These are the doubts I have and believe is due to the current source but do not explain well why.
Click to expand...


the DC value u put at the gate, will determine either the input in off or linear saturation region.. only in saturation, the circuit work as amplifire. For example, if the input is 0 v, there is no current flow and the output will pull to VDD if u use resistor as the load.
 

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    STOIKOV

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STOIKOV said:
IAlso, when apply any DC value to the gates (Vin+ = Vin- = any volts) the current in each transistor is Is/2, no matther what the gates' DC voltages are. Why happens this instead off the quadratic law Id=K(Vgs-Vt)²
Click to expand...


The square law Id=K(Vgs-Vt)² here is talking about Vgs(not only Vg). You didn't consider that source voltage will move with common input voltage.
 

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    STOIKOV

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bageduke is right! Vgs voltage is important, not Vg. if you apply same dc voltage into gates, vgs voltages should be same. this is common input voltage and diff.amp. rejects it. (CMRR)
 

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    STOIKOV

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You've connected an ideal current source, that's why you have the same current at any Vcommon. In real circuit the current will change.
 

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    STOIKOV

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STOIKOV said:
If (Vin+ = Vin- = 0 volts) the current in each drain is Is/2.
Click to expand...

where u have read this from does not mean that their respective gate-voltages are zero

actualy speaking Vin+=Vcommon+vac(+), Vin-=Vcommon+vac(-), the equation actually is

vac(+)=vac(-)=0:

ie both gate voltages are at 'Vcommon' and both transistors are on and both currents are equal to IS/2: it doesnt violate the I=K(vgs-vt)^2 as it explained by
"bageduke"
[/i]\]
 

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    STOIKOV

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Hi,
I am satisfy with above fact. but I dont get that how to write syntax in spice for Vin+, Vin-, sweep, print or plot. Can any body help me?
Thanks in advance!
 

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    STOIKOV

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actually in simulation I'm not using ac, only dc values. And when I set 0V in both gates there is ID/2 flowing in each drain and expected the transistors to be off. It's my main doubt.
 

actually in simulation I'm not using ac, only dc values. And when I set 0V in both gates there is ID/2 flowing in each drain and expected the transistors to be off. It's my main doubt.
Click to expand...

As mentioned, remove the ideal current source. Replace by a current mirror. You will see that if your inputs=0, no current will flow through.
 

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    STOIKOV

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Here is a manual of amp simulation method!
Read it, you will know how to simulate an amp!
 

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    STOIKOV

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