[SOLVED] Second pole location with compensated opamp

[petelee]

Hi Analog experts,

I'm seeing this material in this link people.seas.harvard.edu/~jones/es154/lectures/lecture_6/pdfs/lecture37.pdf.

In page 8, there's this below description. I don't understand physically how it can be diode-connected. If it's diode-connected at a certain frequency, the drain and gate of M5 in page 3 should be in same potential, but I can't imagine how they can be same. Could anybody enlighten me? I'm trying to understand with the scenario that M5's gate is high frequency sine wave and trying to imagine what the M5 drain would be.

Interpretation:
At frequencies around ω2 (>> ω1), the impedance Zc= (1 /jω2Cc) is small enough that M5 can be considered diode-connected
Load capacitance sees a Thévenin resistance of 1 / gm5-->
ω2 is set by the load capacitance in parallel with 1 / gm5

Regards,
Pete

 

[nitishn5]

At high frequencies the compensation capacitor acts as a short for AC signals only i.e it is a short only for those high frequency signals.

Let us assume that the gate of M5 is at V_gs,M5 DC potential.
and that the drain of M5 is at V_ds,M5 DC potential.

Now when we add a high frequency voltage at the gate, V_gs,M5 + a₀sin(ωt).
Now the capacitor is only a short for this high frequency signal and hence, at the drain we would see, V_ds,M5 + a₀sin(ωt).
The DC voltages would remain the same.

Similarly the impedance seeing from the drain of M5 the impedance at high frequencies would by 1/g_m5

 

[petelee]

Thanks a lot Nitishn5~~

Now I can clearly understand why it has diode connection in ac signal domain assuming the Cc acts a short circuit.

I have another question. What is the impedance of CL when Cc becomes virtually a short connection? Isn't it much lower impedance in the same high frequency range where we assume Cc is a short connection?
My question would be stupid. If you educate me, it'd be really appreciative.

 

[nitishn5]

What if I ask you 'So what if C_L is also a short?"

Think of it like this,

The resistance at the drain node of M5 is 1/g_m5 as explained in the paper. Rest are very big by comparison.
The capacitance at that node is C_L and some form of C_c in parallel.
(I say 'some form of' because C_c is not a capacitor to ground like C_L and the both cannot be taken in parallel directly. Use the Miller Multiplication method to know how much the effect of C_c would be at the output node.)

The effective capacitance would be just C_L since it would usually be large.

So the effective second pole location would be ω₂ = g_m5 / C_L

 

[petelee]

Thanks for detailed explanation.
I'd like to re-phrase my question in other way.
Basically, I don't still get how physically the gate and drain of M5 at a certain high frequency have same voltage. But, if I just believe blindly that the gate and drain of M5(two nodes of Cc) have the same voltage like a0sin(wt) at a high frequency, it doesn't have any current across Cc. And if M5 drain voltage is a0sin(wt), then I expect some current flowing between Vo and gnd thru the output resistance of Vo node and CL. It doesn't meet Kirhoff's law. I'm trying to understand with real waveform of sine wave.
Could you help me on understanding this?

 

[petelee]

Hi Nitishn5,

I think you don't need to answer my question. I opened the old text book of circuit theory to recall the basics of rc circuits. Based on your comments and RC circuit theory, now I think I can understand why it meets KCL. I forgot the basics.
Thanks for your help.