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

No, it's completely right.

Without the capacitor there, what you had was the average voltage of a rectified sine wave. When you add the capacitor, it charges up to the *peak* voltage of the rectified sine wave.

All quite normal.

Graham

Actually no. That's something else.

Graham

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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Because of the crest factor, it'll actually be rather less than that.

Graham

More like 0.55

Best regards, Spehro Pefhany

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That seems extreme. A simple linear transformer fed, sine wave, capacitor input filter?

Educate me. How do you get .55?

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I'll take a shot. When you produce an average DC current with narrow pulese from the transformer, the heating effect of those pulses is higher than the average DC current would have. The larger the capacitor filter and the lower the transformer leakage inductance (and other impedances in series with the transformer, like line filters), the narrower and higher the current pulses. The rule of thumb for converting DC output current to transformer RMS current, with a capacitor input filter with a full wave rectifier is to assume the transformer RMS current is 1.8 times the DC output current. The reciprocal of 1.8 is about 0.56.

The actual multiplier depends a lot on the details (and the math is messy, meaning nonlinear) so if you want to cut things more closely, you need either make measurements or build a good spice model of the whole system.

OK, I'll try too. ;-)

Say you've got a 12V, 2A transformer. These are RMS. That's

24 volt-amps. But, when you full-wave rectify it, you get 12 * 1.4 V = 16.8V (assuming perfect diodes), which at 2A, is 33.2 watts, which is more than 24, so the current has to be derated.

But that's just a factor of 0.7. Is the .15 difference just fudge factor? Or did the diodes do it? ;-)

Thanks, Rich

What "did it" was the contraction of the current to a narrow pulse near the peak of the voltage wave (with a blast of power dumped there), with the copper doing nothing the rest of the time. The heating of that copper goes up instantaneously as the square of the instantaneous current.

So if you contract the pulse duration to half, but keep the same average current, the peak current must double. That doubling increases the instantaneous IR losses (I^2*R) to

4X. Average that 4X power peak by half duration pulse over the same half cycle period and the average power dumped into the winding resistance, doubles.

BTW: A good rule of thumb for all transformers feeding rectifiers is that the VA rating of the transformer should be more than 1.5 times the load side watts. If you find your calculated values far from that point, you've made a mistake.

That's only the half of it.

Yes, you do also lose real watts as heat dissipated in the rectifiers (let's change that 0.7 to say 0.65) but that's still not the whole story.

The waveform of the current drawn from the transformer in a capacitor input filtered PSU is highly non-sinusoidal in shape. This results in the rms value of the current being considerably higher than its average value. See 'crest factor'.

It's this rms value that causes a greater heating effect in the windings than the average value that requires the transformer to be derated further.

Graham

The conduction period of full-wave rectified 50Hz supplies of this type is typically 2.5 ms compared to the 10ms half-period of the input. This means the current during the conduction period must be *at least* four times (in practice rather more) greater than the average current.

Graham

Ok Thanks

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Got it. Thanks. Peak currents push I^R losses higher than steady RMS current delivered to the load would.

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