Exam problem

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,685
As I said before, I'm going to write here all the Q-points I found and that matches LTSpice values with slight differences that are perfectly neglectable.

Q1 Q-point = (0.67 V; 4.03 mA)
Q2 Q-point = (12 V; 4.03 mA)
Q3 Q-point = (14.4 V; 3.30 mA)
M1/M2 Q-point = (4.26 V; 2.02 mA)

Now I'll try the Diff and Common gains... I know that @Jony130 already did it but I need to do it by myself!

So, to start, for the current mirror, if I want to find the output impedance seen from Q2 collector branch, when I replace the circuit by it's equivalent (not the hybrid model) after replacing the voltage sources by their internal impedance, what should I consider to be connected to Q2 collector terminal? Is it a voltage supply of -3 V, as the voltage at that node, wrt GND is -3 V?

I mean, in the other side I have +Vcc and R3. After replacing +Vcc by it's internal impedance, typically a short-cut, R3 will be connected to GND... My question is what do I place at Q2 collector terminal? Is it a voltage supply of -3 V???
 

Jony130

Joined Feb 17, 2009
5,435
K = 0.5mA/V² is that Kp parameter you changed to 1m???
Because LTspice is using this formula for Id in saturation region, Id = 0.5*Kp (Vgs - Vt)², and this is why I had to use Kp = 1m instead of 0.5m.
I mean, in the other side I have +Vcc and R3. After replacing +Vcc by it's internal impedance, typically a short-cut, R3 will be connected to GND... My question is what do I place at Q2 collector terminal? Is it a voltage supply of -3 V???
You already forget how to do it ? All supply Voltage are shorted to gnd. And you add one additional voltage source Vx between Q2 collector and ground, and find Zo = Vx/Ix.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,685
Because LTspice is using this formula for Id in saturation region, Id = 0.5*Kp (Vgs - Vt)², and this is why I had to use Kp = 1m instead of 0.5m.

You already forget how to do it ? All supply Voltage are shorted to gnd. And you add one additional voltage source Vx between Q2 collector and ground, and find Zo = Vx/Ix.
I haven't forgot that part... I was just trying to understand what is placed at Q2 collector terminal, between the original circuit and the hybrid model. We use to draw an intermediate step. I'll try to draw it!

Just a sec, I'll be right back with the pictures!
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,685
Ok, here it is...

upload_2016-2-3_23-9-36.png

My question is what is placed at the red arrow??? In the original circuit that terminal had a voltage of -3 V... So, should I place a "virtual temporary" voltage supply of -3 V that will also be replaced by it's internal impedance as was Vcc and -Vcc????
 

Jony130

Joined Feb 17, 2009
5,435
Q2 collector will see the M1 and M2 source pins (two VCCS - voltage controlled current source). And because you are looking for Zo form the thevenin's point of vie you should leave Q2 collector open because we always disconnect the load when we do thevenin.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,685
I don't like my hybrid model, but is it this:
upload_2016-2-4_0-0-3.png

Is that C2 terminal floating correct?? :s

Ok, it is missing there the Vx power supply, but the point is to understand if that floating collector terminal is correct...
 
Last edited:

Jony130

Joined Feb 17, 2009
5,435
Haven't you forgotten about hoe ? C2 is floating because you want to find the impedance seen from M1,M2 sources, when we are looking into Q2 collector.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,685
Ok, I'm back to this problem!

Check if my wording and reasoning is correct:

So, the input impedance of the circuit above is 1/hoe because the input current flows 'towards' E2 and to GND and it does not go anywhere else on the circuit, right? Because there is no return path to the current if it hypothetically it could flow into the rest of the circuit, right?
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,685
I'm not sure if I know how to answer to your questions.

I didn't understood why you said that I have Vin connected to Rs.

About G2 not being shorted, it's because from classes (if my memory is not betraying me) we have an example (made with MOSFETs) where from the Gate to the ground we have a voltage source of value of vid/2 on both sides of the Diff amplifier.

In this case we have more resistors, and that's why I'm not sure how to do this hybrid equivalent circuit.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,685
I think I see the difference.
I think that means that Rg will be out of the hybrid equivalent model, right?
And as the source of the MOSFETs are supposed to be grounded according to the generic hybrid equivalent model, we will have a virtual ground at both MOSFETs sources, making Rs also out of the circuit, no?
 
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