Voltage concerns on a bridge rectifier-capacitor circuit

I'm thinking...ripple voltage.

D from BC

Your rectifier does not charge your capacitor to the measured AC voltage, but to the peak ac voltage(minus 2 diode voltages, ~1.4V). So the change from 6 to 8.5 is about what you would expect.

If this arrangement works, perhaps you should stop there.

An AV voltage of 7.9 volts RMS, (as read with a volt meter) means that this AC voltage will heat a resistor the same as a 7.9 volt DC voltage would. But twice a cycle, AC sine wave voltage is zero volts. At the two peaks of the sine wave, it is 7.9*1.414=11.2 volts.

When you use a bridge rectifier on this waveform, you get

11.2 volt peak ripples minus the 2 volts or so of rectifier drop. A little of the voltage also drops across the resistance of the transformer windings as you bring the load current up. So without the capacitor, you had a rippling voltage that had an average value of about 6 volts. That means the average of the zero volt periods and the peaks and everything in between was about 6 volts. Obviously, some parts of that rippling voltage had to be higher than 6 volts.

When you added the capacitor, it tended to charge up to the peak voltage of the rippling waveform, and supply the motor current during the low parts, with the rectifier supplying current only just before and during the peaks of the waveform, instead of all the time. So the new voltage waveform ahas peaks almost the same as before, but the valleys are almost filled in by the capacitor, so the average voltage must go up. At no load, it should go up to almost the 11.2 volts peak of the AC minus about 1.2 volts of unloaded rectifiers or about 9 volts. Under load, the diode drops increase to about 2 volts, the capacitor voltage sags a bit between the peaks, and the transformer drops more voltage from the large capacitor charging current peaks, so the voltage is less than 9 volts.

Agreed.

I see nothing strange. you have just demonstrated the properties of a capacitor filter on rectified sine wave AC.

See:

"ChrisPikula"

** DMMs set to read DC volts will read the *average* value of a wave connected to them.

Fine, if it is steady DC - but can be misleading if it is not. In your example, the wave is a sine wave that has been rectified so each half wave has the same polarity.

The average value is less than the peak value seen at the top of each sine wave crest - by a factor of 0.637 ( or 2/pi). So, when the DMM read 6 volts DC, the peaks were 6 / 0.637 = 9.42 volts.

Adding that filter capacitor raised the average DC value UP to almost meet the peak value - however adding 20 fans pulled the voltage down due to resistance losses in the transformer's copper wires and the diodes in the bridge.

Nothing strange is happening.

........ Phil

Okay, wow, that was *much* quicker than any other response I have ever had on Usenet. By an order of magnitude. I am *impressed*.

Okay, it was ripple current, neat. I was thinking it could be something like that, but, alas, I didn't understand why, and didn't think it would be that significant. But thanks to you, John and Phil, I see why. It's simple, sure, but as this was my first time ever making a powersupply, I didn't even think that an ideal AC-DC converter would have a voltage increase. Thanks to you both for the help, and the equations, they are greatly appreciated.

Cheers! Chris

It's is obviously a centre-tapped secondary transformer with wire 2 as the C.T. If you bridge wire 1 to wire 3 and connect the bridge rectifier input to wires 1 and 2 you will get the same output voltage but double the current capability. This way half the load current is supplied by each of the output windings instead of all of the load current supplied from one winding as you now have it.

enough,

If you measure the unloaded AC voltage from the transformer this will be the RMS value, Vrms, which is = 0.707 x Vpk. When you connect a capacitor across the dc output (no load) the capacitor charges to the peak value of the waveform (Vpk = 1.414 x Vrms). The diodes in your bridge rectifier will drop a total of around 1.8V on each half cycle thus leaving you around 6Vrms to supply the load. Depending upon the capacitor value and the load current the voltage across the capacitor will try to maintain as near the peak voltage value as possible. In your case Vpk = 6 x 1.414 = 8.5V which is what you read.

If the load current increases for a given capacitor value the voltage across the load will sag a little and the ripple voltage will increase and vice versa. Your 15000uF cap is quite large and will easily maintain near peak voltage with your fan load.

That is what we would expect.

It seems that you may have connected both caps in series and then onnected the combination across the load - correct? If this is so then you effectively halved the value of the capacitor appearing across the load and this would result in greater ripple current for a given load current. If you connect the second cap in parallel with the first then you double the effective capacitance thus reducing ripple voltage and improving the ability of the circuit to maintain output voltage near to Vpk for a given load.

The

Another thing to consider is the current specification of the transformer - When the voltage goes up as a result of filtering ~1.4 times, the max current you can safely take from the transformer is the reciprocal or ~.707 times.

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