voltage

This circuit is used when delaying the standing-up of the square wave signal.
Also, in the case having to do with the pulse of the input signal, it uses for the circuit as it changes the output voltage gradually by the occurrence frequency of the pulse, too. In the case, the resistor must be put in parallel with the capacitor to miss the electric charge which was stored up at the capacitor.

The use example : Door lock signal making circuit and voltage change circuit of Car speed detection device and so on.

The photograph above is the one of the following condition.
F : 50Hz
R : 10K-ohm
C : 0.22µF

The electric charge begins to store up in capacitor(C) when the voltage is applied to the input. The electric current which flows into capacitor as the electric charge is stored up decreases.
The electric current change which flows through the capacitor(C) and the resistor (R) is demanded by the following formula.

i = (V/R)e-(t/CR)

i : The electric current which changes in the time(A)
V : The applied voltage(V)
R : The resistance value(ohm)
C : The capacitor value(F)
e : The base of the natural logarithm(2.71828)
t : The elapsed time after the charging beginning(sec)
CR : The capacitive time constant (C x R)


The change of the voltage(Vc) which appears in the both edges of the capacitor(C) becomes the following formula.

Vc = V[1-e-(t/CR)]

It is as follows when showing the formula above by the graph.