Relation between slew rate and gain-bandwidth product?

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Slew rate related to the actual voltage and maximum frequency at that voltage and only losely to the GBWP. Given a sine wave of frequency wt, the voltage, V = Vp*sinwt. The slew rate is the first derivitive of the voltage: dV/dt = Vp*w*coswt. w is omega = 2*pi*f, and the maximum value of the cosine = 1.

Therefore the slew rate, p = dV/dt = Vp*2*pi*f

Example: if the peak voltage, Vp = 10 volts, the frequency = 1 MHz, p =

The slew rate of amplifiers and other circuits is determinded by the current available to charge and discharge the capacitances in the circuit. dv = (1/C)int *i *dt. In most amplifiers, there is a dominant capacitace usually used for compensation that the available current must charge and discharge. That usually defines the slews rate as well as the GBWP. But the relation is not simple because the available voltage must be taken into account.

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If it helps - Bw = .35 x rise time

Only if you reciprocate something.

good catch, thx make that bw = .35 x 1/t bg