07-09-2017, 11:16 AM
By introducing the electric reactance into the feedback loops of the op-amp amplifier circuits, we can cause the output to respond to changes in the input voltage over time. By drawing their names from their respective calculation functions, the integrator produces a voltage output proportional to the product (multiplication) of the input voltage and time; and the differentiator (not to be confused with the differential) produces a voltage output proportional to the rate of change of the input voltage.
Capacitance can be defined as the measure of the opposition of a capacitor to changes in voltage. The higher the capacitance, the greater the opposition. The capacitors oppose the change of voltage creating current in the circuit: that is, they charge or discharge in response to a change in applied voltage. Therefore, the more capacitance a capacitor has, the greater its charge or discharge current will be for any given voltage change velocity across it. The equation for this is quite simple:
The fraction dv / dt is a calculation expression representing the rate of change of voltage in time. If the direct current supply in the above circuit was steadily increasing from a voltage of 15 volts to a voltage of 16 volts for a time interval of 1 hour, the current through the capacitor would probably be very small due to the very low voltage change rate (dv / dt = 1 volt / 3600 seconds). However, if we continuously increased DC supply from 15 volts to 16 volts in a shorter time span of 1 second, the rate of voltage change would be much higher, and therefore the charge current would be much higher (3600 times greater, to be exact). The same amount of change in voltage, but very different exchange rates, resulting in very different amounts of current in the circuit.
Capacitance can be defined as the measure of the opposition of a capacitor to changes in voltage. The higher the capacitance, the greater the opposition. The capacitors oppose the change of voltage creating current in the circuit: that is, they charge or discharge in response to a change in applied voltage. Therefore, the more capacitance a capacitor has, the greater its charge or discharge current will be for any given voltage change velocity across it. The equation for this is quite simple:
The fraction dv / dt is a calculation expression representing the rate of change of voltage in time. If the direct current supply in the above circuit was steadily increasing from a voltage of 15 volts to a voltage of 16 volts for a time interval of 1 hour, the current through the capacitor would probably be very small due to the very low voltage change rate (dv / dt = 1 volt / 3600 seconds). However, if we continuously increased DC supply from 15 volts to 16 volts in a shorter time span of 1 second, the rate of voltage change would be much higher, and therefore the charge current would be much higher (3600 times greater, to be exact). The same amount of change in voltage, but very different exchange rates, resulting in very different amounts of current in the circuit.