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We have the following simple circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

A capacitor is being charged by an resistor of 100 ohms from a 5V power supply. Let's say I want the capacitor to reach 2.15V in 1 milisecond. How do I calculate the capacitance I need?

I was looking online and came across this formula:

Formula taken from www.electronics-tutorials.ws

When trying to derive the capacitance from it (using photomath), the program spit out (to my eyes) this cursed thing:

Photomath result

v - Capacitor voltage (the one we are charging to), s - Power Supply Voltage, t - Charge Time, r - Resistance, c - Capacitance

I don't know what any of these greek letters symbolize, and how to interpret this.

So I am reaching out for help, is there some more traditional formula to calculate the capacitance given the charge time, supply voltage, charging resistor and the voltage we want to charge to in said time?

Thank You very much for your help.

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    \$\begingroup\$ You know the exponentiation goes away if you take logarithm of both sides? It's unknown what you tried to do in "photomath". Then solve for C and plug in the other values that are known. \$\endgroup\$ Commented Jul 17, 2023 at 10:28
  • \$\begingroup\$ Egads. Photomath has been downloaded, it says, by 220 million people. If it uses privileges and can abuse privacy and/or transmit personal info to some site then just the sheer number of downloads could make it crazy dangerous. That said, it doesn't look like it figured things out, either. Must be based on ChatGPT, which also cannot do math well. (But pretends at it.) Kuba, is this homework? \$\endgroup\$ Commented Jul 17, 2023 at 10:56

1 Answer 1

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The equation you found is actually the voltage on a capacitor that is discharging from \$V_S\$ toward zero. The voltage for a charging capacitor is actually

$$V_C = V_S \cdot (1 - e^{-t/RC})$$

Divide by \$V_S\$ and swap terms to isolate the exponentiation:

$$\frac{V_C}{V_S} = 1 - e^{-t/RC}$$

$$e^{-t/RC} = 1 - \frac{V_C}{V_S}$$

Take the logarithm of both sides:

$$-\frac{t}{RC} = \ln\left(1 - \frac{V_C}{V_S}\right)$$

Negate both sides:

$$\frac{t}{RC} = -\ln\left(1 - \frac{V_C}{V_S}\right)$$

Now you can either solve for \$C\$:

$$C = \frac{t}{-\ln\left(1 - \frac{V_C}{V_S}\right)\cdot R}$$

Or you can solve for \$t\$:

$$t = -\ln\left(1 - \frac{V_C}{V_S}\right)\cdot RC$$

Note that the latter equation gives you the classic formulas for the 555 timer. For astable operation, the threshold is reached when \$V_C\$ is half of \$V_S\$:

$$-\ln\left(1 - \frac{1}{2}\right) = 0.693$$

And for monostable operation, the threshold is reached when \$V_C\$ is two thirds of \$V_S\$:

$$-\ln\left(1 - \frac{2}{3}\right) = 1.1$$

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