# oscilloscope time charging different from the math

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.

• Did you forget R4? – Brian Drummond Apr 8 '14 at 4:57
• @BrianDrummond I thought R4 was for when the circuit is discharging. – Vu Chau Apr 8 '14 at 5:05
• You appear to have R4 in the wrong place. I think you intended it be in the circuit only when the switch is in the down position. Instead, it's in series when switch is in the up position, and produces a confusing result. – gwideman Apr 8 '14 at 5:20

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.