With the lessons on the transformer, diodes and capacitor this should be fairly self explanatory as to how this works, but I'll go through step by step.

What we need is to take a 240 volt supply an turn this into a 12 v supply for a circuit.

So we start with a transformer.

our mains voltage is 240, out ideal output voltage is 12

240/12 = 20, our transformer needs to be 20:1

After this we need to stop the voltage being an AC wave and rectify it.

We'll use the full wave rectifier discussed earlier. this gives us that two humps per cycle (as the negative waveform becomes positive).

But this wave form isn't all that useful, you see each time it dips to zero it's effectively turning off.

in order to smooth out the humps in the power supply we add a capacitor across the power rails, this acts like a battery storing power from the peaks of the humps and discharging, (and thus keeping the rail voltage high) as the hump drops to zero and whilst it's on it's way back up again.

**Ripple Voltage**

the series of humps has now been transformed into a straight DC line with a ripple in it.

That ripple can be very unwanted, consider what you can hear (20hz - 14Khz).

mains voltage is 50Hz, if you connected a speaker to the mains you would hear a low hum, (just before the fire started).

well, if the supply voltage in the circuit has a 50Hz ripple on it, that's also going to appear as hum at the output.

the ripple voltage is calculated as

Vpp = i/2fc where a full wave rectifier is used (as above)

Where a half wave rectifier is use it is Vpp = i/fc

So, where do we get the values from.

well,

vpp is the peak to peak value for the ripple voltage

I is the current in the circuits,

C is the value of the capacitor

and F is the frequency of the AC power

for the example above lets say that the load of the circuit is 1200Ohms.

we know that the votlage applied to that circuit is 12 volts

so the current is

12/1200 = 0.01A

The capacitor value is 0.001F (1 milifarfad)

and mains voltage is 50Hz

so the ripple voltage here is

vpp = 0.01/(50 * 0.001) = 0.2

so in this case the voltage goes between 11.8 and 12 volts, the ripple is 0.2 volts

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