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?

LTSpice: Vser in a relaxation circuit


1 Answer 1


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:

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


\$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:

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


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