For the above circuit, I calculate that the equivalent resistance of the charging circuit is 2.668 kΩ, and the equivalent capacitance is 2.055 µF. Therefore, the time constant would be 2.668*2.055 = 5.48 ms as the amount of time required to charge.
I simulated the circuit and got the following time constant:
It is almost twice as doubled, while it should be approximately the same. Can anyone explain to me the difference?
Thank you.
Best Answer
As drawn R4 and R3 are in series making the equivalent charging resistor
\$ R = \frac{1.62 \cdot 1.78}{1.62 + 1.78} + 1.82 + 2.43 = 5.098 k\Omega\$
and capacitor
\$ C = \frac{1.4 \cdot 2.2}{1.4 + 2.2} + 1.2 = 2.055 \mu F\$
The time constant is therefore
\$\tau = C \cdot R = 10.47 \text{ ms}\$
As pointed out in comments you have neglected R4. If R4 is only supposed to effect the discharge time it should be connected between the switch and earth.