slew rate calculation for a two stage op amp

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anhnha

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Could you explain a bit why slew rate is calculated by that formula along with that assumption?

I know that slew rate is defined as the maximum rate of change of output voltage per unit of time and is expressed as volt per second.
However, the lecture below give that formula without any explanation is a bit confusing.

 

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For example if M2 is fully ON and M1 is OFF, the current trough M3 is 0, so current trough M4 is 0. This means that the voltage (charge) over (on) Cc will change, in this case depending on the current source which is realized with M5.

Simplified, you could write:
I=C*V/t => I/C=V/t

Hope this helps.

BR Jerry
 
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    anhnha

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In most cases a current of output stage is much higher than in first one, while Cc is comparable with Cload. So time for (dis)charging loading capacitor by output stage current is much smaller than time needed by tail source current to charge/discharge compensation capacitor.
 
Thank you everyone. I think I misunderstood a bit. I thought that the maximum current flowing capacitors will be I5 when M1 is on and M2 is off.
M1 is on and entire current I5 will flow through M1. This current will be copied to M4 and therefore I4 = I5. In turn, this current will flow through two series capacitors.
What is wrong with this thinking?

Here is what I thought. I wrote that before reading your posts. Please comment.

 

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