#### Hugo Holden

##### Veteran Member

There are two common Full wave rectifier circuits you will see in analog power supplies.

One uses a center tapped transformer and essentially two diodes (unless a negative supply is required as shown with the blue diodes) the other uses a bridge rectifier (4 diodes).

They are called full wave because the filter capacitor gets charged on peaks on the sine wave from the transformer at twice the rate of the line frequency. So in a 50Hz line power system they get charged every 10mS and in a 60Hz system every 8.3 mS.

If you look at the voltage across the filter capacitors with the scope, with the power supply normally loaded, you will see a ripple voltage at these frequencies where the capacitors get charged on peaks and between those times the voltage is falling, and the rate that it falls depends on how high the current drain is and is very roughly inversely proportional to the uF value of the capacitor. So if the capacitor loses uF value over the years, the voltage troughs down more quickly between peaks. If it troughs down too low, ripple will appear in the output of the voltage regulators and foul up the operation of the computer. The first thing you would notice as it started to happen is a ripple in the drive signals to the VDU. On the PET because the vertical scan stage is very sensitive to this , it makes the picture bounce up and down vertically, if the large 23,000uF capacitor is losing capacity.

Generally for most supplies, the 0V connection is also connected somewhere to the chassis and the chassis is generally connected to the Line power Earth connection. In the PET they have the 23,000uF capacitor off the board away from where the rectifiers are so it makes the arrangements of its connections less obvious. It also creates the possibilities of poor connections on the connectors.

As you can see from the circuit configurations they are essentially the same except for the use of a center tap on the transformer secondary winding and not in the other case. Both have advantages and disadvantages, the center tap design is useful where a + and - supply is required. There are less rectifier losses as only two diodes, but more I^2R copper losses in the transformer has two windings with thinner wire and a higher overall resistance are required.

It is quite easy to work out what the ripple voltage amplitude is (or should be) if you know or measure the current consumption I and know the capacitor's uF value. A linear model is ok because a ripple voltage is low in magnitude (normally compared to the total voltage and over that ripple voltage range the current drain from the load is fairly constant)

From the basic equation for the capacitor charge Q in Coulombs = Voltage x Capacitance in F in Farads:

Q= CV, then over a time increment dt

dQ/dt = CdV/dt

dQ/dt is rate of change of charge with time which is current I , therefore:

I = CdV/dt

And we are interesting in the change of voltage with time dV , so rearranging:

dV = ( I/C)dt

Then lets say C = 10,000uF , I = 1 Amp, and in a full wave 50Hz system the capacitance is charged every 10 mS, then

dV = (1/10,000 x 10E-6) x 10 x 10E-3

dV = 1 Volt.

So a handy rule of thumb is that if the filter capacitor is 10,000uF and the current 1 Amp, the ripple will be 1V (and a little less than that in a 60Hz full wave system as the cap gets charged every 8.3 mS rather than 10 mS). Knowing this you can scale it in your mind, for example 2 Amps would have twice the ripple amplitude, or a 20,000uF capacitor would have half a volt ripple at one amp, or a volt ripple at 2A etc.

One uses a center tapped transformer and essentially two diodes (unless a negative supply is required as shown with the blue diodes) the other uses a bridge rectifier (4 diodes).

They are called full wave because the filter capacitor gets charged on peaks on the sine wave from the transformer at twice the rate of the line frequency. So in a 50Hz line power system they get charged every 10mS and in a 60Hz system every 8.3 mS.

If you look at the voltage across the filter capacitors with the scope, with the power supply normally loaded, you will see a ripple voltage at these frequencies where the capacitors get charged on peaks and between those times the voltage is falling, and the rate that it falls depends on how high the current drain is and is very roughly inversely proportional to the uF value of the capacitor. So if the capacitor loses uF value over the years, the voltage troughs down more quickly between peaks. If it troughs down too low, ripple will appear in the output of the voltage regulators and foul up the operation of the computer. The first thing you would notice as it started to happen is a ripple in the drive signals to the VDU. On the PET because the vertical scan stage is very sensitive to this , it makes the picture bounce up and down vertically, if the large 23,000uF capacitor is losing capacity.

Generally for most supplies, the 0V connection is also connected somewhere to the chassis and the chassis is generally connected to the Line power Earth connection. In the PET they have the 23,000uF capacitor off the board away from where the rectifiers are so it makes the arrangements of its connections less obvious. It also creates the possibilities of poor connections on the connectors.

As you can see from the circuit configurations they are essentially the same except for the use of a center tap on the transformer secondary winding and not in the other case. Both have advantages and disadvantages, the center tap design is useful where a + and - supply is required. There are less rectifier losses as only two diodes, but more I^2R copper losses in the transformer has two windings with thinner wire and a higher overall resistance are required.

It is quite easy to work out what the ripple voltage amplitude is (or should be) if you know or measure the current consumption I and know the capacitor's uF value. A linear model is ok because a ripple voltage is low in magnitude (normally compared to the total voltage and over that ripple voltage range the current drain from the load is fairly constant)

From the basic equation for the capacitor charge Q in Coulombs = Voltage x Capacitance in F in Farads:

Q= CV, then over a time increment dt

dQ/dt = CdV/dt

dQ/dt is rate of change of charge with time which is current I , therefore:

I = CdV/dt

And we are interesting in the change of voltage with time dV , so rearranging:

dV = ( I/C)dt

Then lets say C = 10,000uF , I = 1 Amp, and in a full wave 50Hz system the capacitance is charged every 10 mS, then

dV = (1/10,000 x 10E-6) x 10 x 10E-3

dV = 1 Volt.

So a handy rule of thumb is that if the filter capacitor is 10,000uF and the current 1 Amp, the ripple will be 1V (and a little less than that in a 60Hz full wave system as the cap gets charged every 8.3 mS rather than 10 mS). Knowing this you can scale it in your mind, for example 2 Amps would have twice the ripple amplitude, or a 20,000uF capacitor would have half a volt ripple at one amp, or a volt ripple at 2A etc.

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