# slew rate calculation for a two stage op amp

Discussion in 'General Electronics Chat' started by anhnha, May 12, 2014.

1. ### anhnha Thread Starter Active Member

Apr 19, 2012
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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.
For a capacitor: i = C*dt/dt and therefore dv/dt = i/C.

• ###### Slew rate.PNG
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2. ### anhnha Thread Starter Active Member

Apr 19, 2012
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Please look at this and tell me if it is correct or not. Thank you.

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3. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Slew rate occur when op amp input stage is overdrive.
So simply assume M1 is full on and M2 is OFF. So in this case I4 = I5. And notice that this is the only current that is available to change the voltage at Vout, and Cc limits just how rapidly this can be done.

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4. ### anhnha Thread Starter Active Member

Apr 19, 2012
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Hi,

I am confused now. Why I6 isn't included in the formula?

5. ### MrAl Well-Known Member

Jun 17, 2014
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Hi,

They probably assume that the magnitude of I6 is the same as I7 except only one is flowing at any given time, so it is only a matter of whether we are charging the cap or discharging the cap, which depends on the state of the inputs.

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6. ### anhnha Thread Starter Active Member

Apr 19, 2012
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Hi, MrAl.

Do you mean that I6 is charging and I7 is discharging so the net voltage across the cap is remained?
Slew rate is calculated for the opamp with cap CL, right?
Let's call V_CL is the voltage across load capacitance.
Slew rate = max(dV_CL/dt )= max (current flowing through CL/CL).
The maximum current flowing through CL is I5 so, we have the result above.
Is that right?

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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You don't need to include load capacitance if you want to find slew rate.
Slew rate occur when op amp input stage is overdrive. So if we assume M1 is full ON and M2 is OFF. Then Q6 will also be OFF. And I4 = I5 and this current will charge the capacitor.
But if we flip the overdrive then M1 is OFF and M2 is full ON.
So now capacitor is discharged in this path:
Vdd--->M6 -->Cc---M2--->M5 and again the discharge current is equal to I5

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8. ### MrAl Well-Known Member

Jun 17, 2014
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Hi,

I was just saying that I6 is the same as I7, it just depends on what (binary) state the inputs are at. Remember the slew happens whether the output is ramping up OR ramping down. If they just state one of these two conditions, that still gets you the slew rate.