Talkin' PSUs

My last post was necessary to understand the manipulation of the smoother capacitance. It is also essential to know where the charge is "held."

If current is the flow of charges around a circuit, then that current must return to where the charge is kept (Kirchhoff conservation laws).

The main charge is kept in the smoother. We can use one capacitor or use any number in parallel to make up the value determined by the formula.

To ensure the voltage to the 7824 remains within its input range for all grid supply voltages, we determined a maximum ripple voltage of 1-volt.

C = IT/V

= 1(amp) x 0.01(sec) / 1(volt) = 0.01 Farad (which is 10,000 uF)

But what if we don't have a 10,000uF capacitor? Say we have two 4700uF capacitors we can put in parallel, making 9400uF. Will that suffice?

Using the rearranged formula

V = IT/C

= 1(amp) x 0.01(sec) / 0.0094(Farads) = 1.064 volts

Will 0.064 volts make that much a difference?

Let me answer that question by stating that electrolytic can capacitors (which are the only type having the capacitance we require) have tolerance of +/-20%.

It is possible (although not very likely) that a 10,000uF capacitor could be 8,000uF.

Let's run that through our formula.

V= IT/C

= 1(amp) x 0.01(sec) / 0.008(Farads) = 1.25 volts

Likewise, 9400uF would become 7520uF, and the ripple voltage 1.33V.

Based on discussions so far, we might see sufficient latitude and trust in hope by using two 4700uF capacitors that both can't be -20%, or we could wish to reach perfection. Being pragmatic, will we need precisely 1 amp output? And if not, we can risk a third of a volt on a "brown-out" voltage of 207V on a 220V supply grid. If the grid voltage is 240V, there is no need to worry.

The non-technical layman understands electricity due to it being a greater part of modern living.

He understands Voltage that, in most cases, it appears between two points. The live and neutral of a mains plug and the pointed end and the blunt end of a battery are two examples.

In a mains derived DC power supply, both types of electricity exist. The input is from the live and neutral of a mains plug. The output mimics that of a battery.

A battery contains charge, and in use, the charge flows in the circuit it powers, and that flow is called current.

As the charge is consumed, the current flow diminishes, and eventually, all the charge has gone.


The transformed and rectified waveform is like a series of upward going loops. The top of each loop is the peak value of the AC voltage.

VDC = √2 x Vac

The charge is used up in current flow in the circuit powered by the power supply.

The smoother capacitor voltage decays, only to be met by the next loop. These loops are 0.01 seconds apart.

Rather than the DC voltage being a straight line over time, it ventures up and down by an amount called ripple voltage.

The ripple voltage is directly related to the current being drawn from the power supply. If we don't draw any current, then this DC voltage is a straight line over time, but as soon as we draw current, the Voltage starts to ripple.

The letter used for charge is Q, and it stands for coulombs. Charge is related to Capacitance (the storage of charge) and Voltage.

Q = CV

Charge is used up by current draw over time, so

Q = IT  (I stands for current in amps, T stands for time in seconds)

Therefore Q is related to both, so

Q = CV = IT

If we want to find Capacitance and know what ripple voltage is acceptable and the current the load draws, we also understand that the rectified waveform repeats every 0.01 seconds. We can use

CV = IT

Rearranging it to obtain C on its own involves dividing both sides of the equation by V.

The V's of CV/V cancel to become C. On the other side, we have IT/V.

C = IT/V

If we want to know the ripple voltage, then we divide both sides by C

V = IT/C

So now, even the layman should have a better idea of designing power supplies.

A diode is NOT a stupid idea. ()


Edited by Graham Slee - 22 Jan 2021 at 10:20am