From this it is clear that the current is closely linked to the change in voltage wrt time.
Set the differentiator op amp, i.e. in the inverting configuration, with a resistor between Vout and the virtual earth, and a capacitor between the virtual earth and Vin (I assume you are familiar with this set up)
You know that the voltage at the virtual earth is 0 volts therefore the current through the capacitor is solely due to the change in Vin (refer to the above equation). The current then goes through the resistor causing a drop across it which is equal to Vout. Remember the very high input impedance of the op amp means that it draws effectively no current, therefore Icapacitor = Iresistor.
If the rate of change in Vin is increased (dVin/dt), the current increases (remeber from i = C dv/dt, as dv/dt increases so I increases) and so the voltage across the resistor and hence Vout increases. So you say that output voltage is equal to the product of the resistance, the capacitance and the rate of change in input voltage:
Vout = RC dVin/dt
The negative sign comes from the fact that this is an inverting amplifier configuration, i.e. phase shifted by 180 degrees. So the final equation becomes:
Originally posted by Dave@May 17 2004, 03:13 PM So you say that output voltage is equal to the product of the resistance, the capacitance and the rate of change in input voltage:
I have perhaps not explained this very well. The idea of the product of the capacitance, resistance and rate of change of input voltage gives the output voltage is a contruct from Ohm's Law, Vout = iR where i = C dVin/dt.
