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CMOS two stage amplifier 0.8um technology

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thaintrinh

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Good afternoon.
I have been given a circuit like this: note that it has an additional resistor compared to the original one.
1111.PNG
I believe this is a CMOS two stage amplifier in 0.8um technology.
I need to find out the operations, equivalent circuit and Av, Ai, Zin, Zout from this circuit. However, i can do the simulation for this circuit.
None of the documents on the internet seems to help. It would be very helpful if you can help me with this.
Thanks in advance :D
 

thaintrinh said:
I need to find out the operations, equivalent circuit and Av, Ai, Zin, Zout from this circuit. However, i can do the simulation for this circuit.
Click to expand...

An equivalent circuit for CMOS two stage amplifiers can be found in every analog circuit design textbook.
If you can simulate the circuit, you should be able to find the parameters. So what's your problem?
 

It's a "strange" Differential Pair with an Output Stage amplifier.
PMOS and NMOS device cannot have a same dimension for a proper operation.Here, every MOS has 4um/4um..Weird..
 

erikl said:
An equivalent circuit for CMOS two stage amplifiers can be found in every analog circuit design textbook.
If you can simulate the circuit, you should be able to find the parameters. So what's your problem?
Click to expand...

I just don't know where to put that additional resistor in this equivalent circuit to compute all that parameters.
equivalent circuit.PNG
And I have to make comparison between my calculation and simulation.
 

thaintrinh said:
I just don't know where to put that additional resistor in this equivalent circuit to compute all that parameters.
Click to expand...
Between the current source gm1V1 and the output resistance ro2||ro4. But at a branch current of about 2µA this XR2 10kΩ resistor is negligible, I'd say; it is dispensable, anyway.
 
erikl said:
Between the current source gm1V1 and the output resistance ro2||ro4. But at a branch current of about 2µA this XR2 10kΩ resistor is negligible, I'd say; it is dispensable, anyway.
Click to expand...

Thanks a lot :D I am not really good with these kind of circuits and stuff :(.
So, the XR2 will stay in parallel with the current source and the output resistance ro2||ro4 ?
The value of that resistor is not really important since I'm trying to compute everything "in theory" :D.
Thank you for answering these stupid questions of mine.
 

One more small question here,
If i put the resistor in this position
case 2.PNG
I'm trying to determine Rout1 using the impact of the current mirror feedback,
So, is it gonna be: Rout1 = (ro2 + XR2) || ro4 ?
and if the position of XR2 is in the upper position, Rout1 = ro2 || (ro4 + XR2) ?
Thanks a million
 

thaintrinh said:
View attachment 139345 So, is it gonna be: Rout1 = (ro2 + XR2) || ro4 ?
Click to expand...
Yes, I think so.

thaintrinh said:
... and if the position of XR2 is in the upper position, Rout1 = ro2 || (ro4 + XR2) ?
Click to expand...
No, I don't think so: XR2 in this previous position has nothing to do with the (ro2 || ro4) output impedance of the 1st stage. It just diminishes the current of the XM1-XM3 stage (a very little bit: in the order of XR2/(1/gm1)≈XR2/ro1 ). So you could replace the current source (gm1V1) by (V3/((1/gm1)+XR2)) and remove again the XR2 symbol in the equivalent circuit schematic.
 
Hence, for the first position, I just replace (ro2 || ro4) with (ro2 + XR2) || ro4 in this equivalent circuitView attachment 139318
This is the second position that i was talking about in my previous reply
upper right branch.PNG
v3 you mentioned is vg3 = (gm1 / gm3 ) * (Vid/2)
I thought XR2 has nothing to do with this vg3 ?
Thank you for your patience :D
 

thaintrinh said:
Hence, for the first position, I just replace (ro2 || ro4) with (ro2 + XR2) || ro4 in this equivalent circuit
Click to expand...
NO: if this is your first position:
... didn't you read my above explanation (properly)?

erikl said:
replace the current source (gm1V1) by (V3/((1/gm1)+XR2)) and remove again the XR2 symbol in the equivalent circuit schematic.
Click to expand...
thaintrinh said:
v3 you mentioned is vg3 = (gm1 / gm3 ) * (Vid/2)
Click to expand...
What is (Vid/2) ?

thaintrinh said:
I thought XR2 has nothing to do with this vg3 ?
Click to expand...
I think you've mixed that up:
erikl said:
XR2 in this previous position has nothing to do with the (ro2 || ro4) output impedance of the 1st stage.
Click to expand...
Read more thoroughly, please!

thaintrinh said:
This is the second position that i was talking about in my previous reply
View attachment 139364
... I just replace (ro2 || ro4) with (ro2 + XR2) || ro4 in this equivalent circuit
Click to expand...
No, again: you should replace (ro2 || ro4) by (ro2 || (ro4 + XR3))
You changed the out1 connection point!
 

God, i think i just messed everything up.
Thank you for your explanation.
I will run the analysis again thoroughly, can you judge it for me pls?
Thanks a million :(
 

Sorry, I'm quite hopeless now :bang:
This is my circuit, can you derive the equation for Av, Vo1 and Zout1 for me pls,
Thanks a million times case 1.jpg
 

Thank you again for your patience.
I've read all those textbooks.
For all the cases i'm going to consider, I have a differential voltage +Vid/2 applied to XM1 (Gate) and -Vid/2 to XM2 (Gate).
And I this case to consider: the additional resistor is put between XM3 and XM1 like this 7352457800_1497363496.png
Why in this case, as you said, replace (gm1V1) by (V3/((1/gm1)+XR2))?
Correct me if i'm wrong, here's my analysis:
+ The current source that represents XM1 is calculated as gm1*Vgs1 = gm1*Vid/2.
+ Hence, Vg3 = -gm1*((ro1 + XR2) || ro3 || (1/gm3))*Vid/2, since ro1, ro3 >> 1/gm3 => Vg3 = -gm1/gm3 * Vid/2 => XR2 has nothing to do with the voltage gain (Av) of this opamp circuit. I assume that V3 you wrote is Vg3 ?
+ In my equivalent circuit, V1 is Vid through a input impedance Zin (which is infinite) => V1 = Vid ?
+ In my simulation, as XR2 increases, Av increases as well ?
Can you explain for me pls, thank you very much.
 
Last edited:

thaintrinh said:
For all the cases i'm going to consider, I have a differential voltage +Vid/2 applied to XM1 (Gate) and -Vid/2 to XM2 (Gate).
And I this case to consider: the additional resistor is put between XM3 and XM1 like this View attachment 139524
Why in this case, as you said, replace (gm1V1) by (V3/((1/gm1)+XR2))?
Click to expand...
I just replaced your current source by a current source which included XR2 .


thaintrinh said:
I assume that V3 you wrote is Vg3 ?
Click to expand...
Right.


thaintrinh said:
Correct me if i'm wrong, here's my analysis:
+ The current source that represents XM1 is calculated as gm1*Vgs1 = gm1*Vid/2.
+ Hence, Vg3 = -gm1*((ro1 + XR2) || ro3 || (1/gm3))*Vid/2, since ro1, ro3 >> 1/gm3 => Vg3 = -gm1/gm3 * Vid/2 => XR2 has nothing to do with the voltage gain (Av) of this opamp circuit.
+ In my equivalent circuit, V1 is Vid through a input impedance Zin (which is infinite) => V1 = Vid ?
Click to expand...
You didn't yet explain neither V1 nor Vid.
And I must admit I don't feel quite fit with those equivalent circuits.

thaintrinh said:
+ In my simulation, as XR2 increases, Av increases as well ?
Click to expand...
Yes, about the same as the left branch current decreases:
erikl said:
a very little bit: in the order of XR2/(1/gm1)≈XR2/ro1
Click to expand...
... as I indicated above.
 
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