Does anyone know where the bottom equation is derived from?

Why is it 8V/9V?

description of calculation

Source: Mouser - Everything for Capacitive Power Supplies by TDK


1 Answer 1


Those are the minimum and maximum output voltages, given 1 Vp-p of permissible ripple.

The capacitor discharges exponentially, following the curve

$$V(t) = V_0 e^{-\frac{t}{RC}}$$

Solving for C gives:

$$\frac{V(t)}{V_0} = e^{-\frac{t}{RC}}$$

$$\ln\left(\frac{V(t)}{V_0}\right) = -\frac{t}{RC}$$

$$C = -\frac{t}{R\ln\left(\frac{V(t)}{V_0}\right)}$$

This isn't how I normally do this computation, however. A power supply is usually designed for a particular load current, not resistance, so there's simpler formula you can use to size the capacitor:

$$\Delta V = \frac{\Delta Q}{C} = \frac{I \Delta t}{C}$$

Solving for C:

$$C = \frac{I \Delta t}{\Delta V}$$

and given a load current of 9 V / 600 Ω = 15 mA, plugging in values:

$$C = \frac{15 mA \cdot 10 ms}{1 V} = 150 \mu F$$


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