Q = C * V, C = capacitance in farads, Q = charge in coulombs, V = voltage potential
Q = I * T, I = current flow in amps, Q = charge in coulombs, T = time in seconds
F = 1 /T, F = frequency in hZ, T = period in seconds
Therefore,
I * T = C * V, or C = (I * T) /V
The delta “Δ” symbol indicates a change, such as a change in voltage or a change in time
C = I * ΔT / ΔV
In this case, ΔT = 1 /F e.g. the time period of a 50hZ waveform = 1 /50hZ = 0.02seconds
Therefore,
C = I / (ΔV * F)
Now including the 70% factor we get the final relationship:
C = 0.7 * I /(ΔV * F)
C = capacitance in farads, I = current in amps, ΔV = peak-to-peak ripple voltage, F = ripple freq in hZ
Note that ripple frequency in a full-wave rectifier is double line frequency. For half-wave rectification, the ripple frequency is the line frequency.
Solving for ΔV
ΔV = 0.7 * I /(C * F)
The 70% factor
This is an assumption—in cases where the capacitor is oversize, it will be perhaps 80%, but this is close enough for government work and not very critical.
Putting it to work, a practical example
Assume that we want to make a 9V, 500mA power supply using the LM7809 voltage regulator device, 12V transformer, bridge rectifier and filter capacitor. Line frequency is 50hZ. How large should we make the filter capacitor?
From the spec sheet, we learn that the dropout voltage of the LM7805 is 2.5V. Therefore, the valley of the peak-to-peak ripple should be 9V + 2.5V = 11.5V. Assuming that the average rectified DC voltage is 12VDC, the minimum negative peak of the ripple voltage = 12V – 11.5V = 0.5V. The peak-to-peak ripple voltage is double the peak or 1.0V. Plugging it into the formula:
C = 0.7 * I /(ΔV * F) = 0.7 * 0.5A /(1Vp-p * 100hZ) = 0.0035farads or 3500µf
Selecting the capacitor
Since 3500uf is not a standard value, simply scale up to 3900uf.