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question about slew rate of opamp

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hayowazzup

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If an opamp has a slew rate of 0.8V/microsecond, the gain of opamp is 4, and
at t = 0, Vin = 1V, and takes 5 microsecond to constantly increase to 2V.

Will Vo take (4*2)/0.8 * 5 = 50microS to constantly increase from 4V to 8V at t=0?

thanks in advance
 

No. Both the input and output changes occur concurrently, and the output slew rate is decently faster (by a factor of 4) than the input slew rate. So I think -- in first approximation -- you could add the input and output rise times quadratically (root of summed squares), i.e
total output rise time ≈ √(52 + 52) ≈ 7µs .
 

erikl said:
-- you could add the input and output rise times quadratically (root of summed squares), i.e
total output rise time ≈ √(52 + 52) ≈ 7µs .
Click to expand...

I don`t think so.
The internal opamp slew rate is 0.8V/us. The input signal rises with 1V/5us=0.2V/us.
Because a gain of 4 the required output change is 4*0.2V/us=0.8V/us.
Because this is just the capability limit of the opamp the opamp output should be able to follow the input signal - that means: 4 volts in 5 us.
 

There are 2 parts to the total rise time delay namely, slew rate limited one which is totally decided by the output stage current, and bandwidth limited delay or ringing before settling, which is decided by gain of op-amp. In this case the slew limited delay is (8-4)/0.8=5us and the delay through op-amp due to gain is same as 5u, hence it should rise in 5us(max between both). Hope this helps.
 

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