# How does a switch's series voltage affect discharge time in a relaxation circuit?

I made this relaxation oscillatory circuit in LTspice using a voltage controlled switch (sw) that switches off at $$\ 100V \$$ and back on at $$\ 50V \$$. To the left, the sw has $$\ V_{ser}= 0V \$$ series voltage and it takes the capacitor $$\ C_1 \$$ approximately $$\ \approx 0.7ms \$$ to hit the drop-out voltage at $$\ 50V \$$. The math confirms the discharge time I measured in the LTspice trace:

$$\ V_{d} = V_{c}* e^{-t \over R_{th}C } \$$

$$\ t = -\ln {V_d \over V_c} R_{th}C_1 = -\ln{50V \over 100V} * 10^6 * 10^{-9} = 0.693ms \approx 0.7ms\$$

where $$\ V_d \$$ is the drop-in voltage and $$\ V_c \$$ the cut-in voltage. $$\ R_{th} \$$ is the Thevenin resistance the capacitor sees - that is, $$\ R_2+(R_1||R_{on}) \approx R_2 \$$.

Now if I add a series voltage ($$\ V_{ser} = 20V \$$) to the switch sw (schematics on the right), I spotted no difference in the charge-up time but a significant increase in the discharge time ($$\ \approx 1ms \$$. I don't understand either case.

If the capacitor starts discharging at $$\ 100V \$$ and there's this series voltage of $$\ 20V \$$, the capacitor is effectively discharging a $$\ 100V - 20V = 80V \$$ of initial charges. The capacitor should then hit the drop-out voltage quicker than that without adding the series voltage in the switch.

$$\ t = -\ln {V_d \over V_c} R_{th}C_1 = -\ln{50V \over 80V} * 10^6 * 10^{-9} = 0.47ms \$$

But the simulation gives $$\ \approx 1ms \$$

How does the switch's series voltage affect the relaxation circuit - specifically, why does it increase discharge time, whereas the charge-up time seems not to be affected?

Due to the fact that in your circuit $$\V_∞ = 20V\$$ instead of $$\0V\$$.

You need to use this general formula for the capacitor charging/discharging phase:

$$V_C(t) = V∞ + (V_{start} - V∞) \times \left(e^{\frac{-t}{RC}}\right)$$

Where:

$$\V_{start}\$$ initial capacitor voltage.

$$\V∞\$$ steady-state final voltage.

Because now you have been using the simplified version that assumes $$\V_∞ = 0V\$$.

And if we solve it for the time we get this:

$$T = RC \times \ln \frac{V_{start} - V∞}{V_C - V∞} = 1\text{ms} \times \ln\frac{100V -20V}{50V - 20V} \approx 1\text{ms} \times 0.98 \approx 0.98\text{ms}$$