Electronic – Confusion in deriving ripple voltage for an unregulated power supply

power supplyregulationsripple

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Above is a full wave rectifier as an unregulated power supply

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(please left-click with your mouse to enlarge the image)

In many texts and tutorials the ripple voltage(|AX| in above figure) which is

deltaV = I / (2*f*C)

I think it comes from the fact that

C * dV/dt = I so deltaV = (I/C) * delta_t;

and according to texts since delta_t is T/2 = 1/2*f

then deltaV = I/(2 * f* C) where I is the load current.

I can understand that derivation but if you look at my image above, according to this equation the real delta_t is not T/2 but instead less than that. Hence the ripple is |AX| not |YC|. But the tutorials take delta_t as T/2 where it starts from the point A to point C. But actually delta_t starts at point A and ends at point B.

Im confused at this point. What is the real ripple Voltage here?

Best Answer

Here's a picture from a different web source that gives the same answer but, notably with small print at the bottom: -

enter image description here

So it's an approximation and it doesn't take account of diode drops either: -

enter image description here

What is the real ripple Voltage here?

This picture is better because it shows the forward volt drops in the diodes but it's still not exact - how far do you want to go getting to the real answer. Do you want to consider any of these: -

  • Non-linear resistance of the "real" load
  • Real load effects such as pulses of current it might take
  • Self-discharge of electrolytic
  • Transformer secondary leakage inductance and resistance
  • Volt drops in wires

It's just an approximation at the end of the day and at the point when the formula becomes grossly inaccurate it's probably a good idea to add more capacitance to the bridge output to reduce ripple voltage anyway!