Slew rate is basically a degradation effects on the high frequency response of active amplifier due to compensating capacitor. Slew rate is due to the maximum supplied current available for charging up the compensating capacitor. It can be found using a formula
slewrate = I(max)/C
So, if you have a total output current of 1 ma and if the compensating capacitor is 1000 pF, the slew rate is
SR = 1ma/1000 pF = 1 V/us
Also, you want to make sure that slew rate > 2*pi*f*v(peak)
V(peak) is the largest possible amplitude of the input signal to avoid distortion due to slewing.
So, if your slew rate is 0.5 and frequency is 20000 Hz, then
Thank for your reply.
Can you tell me why SR>2*pi*f*V(peak)?
Let us turn to this problem. If we look the schematic as a two stage opamp, the first stage SR is Ibias/C, while the second is Itail/Cout. How can he calculate the formula 1?
Can someone tell me ? thanks !
The SR of the first diff amp stage is Ibias/C and it is gained by Gm of the second stage which is Itail/Von. It is like output voltage SR of an opamp can be gained with output stage, if the output stage has gain more than 1.
The SR of a sine wave is 2*pi*f*V(peak), so you need SR of the amplifier more than SR>2*pi*f*V(peak) in order to avoid large distortion (sine wave would be triangular if SR not big enough).
The SR of the first diff amp stage is Ibias/C and it is gained by Gm of the second stage which is Itail/Von. It is like output voltage SR of an opamp can be gained with output stage, if the output stage has gain more than 1.
The SR of a sine wave is 2*pi*f*V(peak), so you need SR of the amplifier more than SR>2*pi*f*V(peak) in order to avoid large distortion (sine wave would be triangular if SR not big enough).