Bridge rectifier VDC = VAC*1.414 only at no-load ?

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I have observed that it is always suggested that the secondary VAC of the transformer should be VDC/1.414 for a bridge rectifier with a capacitor filter. i.e an 18-0-18 for ±25VDC .
Now isn't the case only at no-load condition ? which is not quite what happens most of the time(no-load). There will always be a load - even minimal - like the chips drawing idle current. So the chips would really be operating at about ±18-20VDC - depending on whether the transformer is at full load or not and ofcourse the % regulation of the transformer. Am I thinking this right ? If so then why arn't people using something like 24-0-24 to get ±25VDC ?
Thanks for clearing this up for me.
Well, there are different things to consider, but the short answer is that when a xfrmr is described as "24VAC" that is what the manufacturer has rated the xfrmr as. The no-load voltage of the xfrmr secondaries will be higher. You can read up on rectification to understand where the value of 1.414 comes from.

Secondly, you are also ignoring the forward voltage drop of the rectifiers. I use VAC*sq rt 2 - 1.4 as a guideline to come up with VDC for a bridge.
The voltage will be VAC*sqrt(2) provided the following conditions are met:

-The diodes are ideal
-The capacitors are infinite and ideal
-The AC supply and transformer are ideal

Since neither is true, let's see what happens when you put a load on the unregulated AC->DC supply:

- the diode has a voltage drop approximately between .7 and 1.0 [for Si] depending on the current
- the diode has a resistance associated with it
- the capacitors are far from infinite and also have a resistance
- the transformer has a resistance

So on the first half cycle, the AC voltage rises to VAC*sqrt(2)-Vd [where Vd is the diode voltage drop] and charges the capacitor to this value.

If you add a load, the voltage will rise to VAC*sqrt(2)-RI-Vd where R is the resistance of all of the wires, transformer secondary, and the diode on-resistance leading to the load. I is the load current and the current to recharge the capacitor, initially a high value - only limited by the resistance of the source and the resistance of the capacitor.

So what you see if you watch the current through the diode bridge is that it is zero nearly all of the time and sees a large spike near the apex of the sine wave, which is why people use 100A bridges in devices that won't use nearly that much average current.

What you will also see unless the capacitors are infinitely large is a lot of ripple.

So on average the voltage will sag, but in reality there will also be an AC signal superpositioned on the DC.


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> VAC*sqrt(2)-RI-Vd where R is the resistance......

Yes, but remember that R is accounted for in the V spec.

But that spec is for a resistive load (not our spiky rectifiers).

A small transfomer has about 20% regulation. The "18VAC" is (normally) for full (resistive) load current. At no-load, it may be 18VAC*120%= 21.6VAC. Rectify that, and (at no-load) you are looking at 30.5V DC.

> There will always be a load

For very-very light load on a supply that can handle a heavy load, knock off a volt for rectifier and ripple. Say 29V.

I have a hefty 18-0-18VAC 3A transformer and a couple 4,700uFd caps as a utility supply. For loads from near-zero (I have 2K bleeders on it) to ~1ADC it gives 28VDC-24VDC. I think of it as "+/-25VDC".

Because of the spiked waveform (not to mention the 1.414 voltage transformation), the DC current probably should not reach half the rated AC current. Exact computation is a headache, and requires measuring exact transformer and rectifier parameters; in DIY a dummy-load or real-amp test is usually best. But if the raw rectfied and filtered DC voltage ever came down to the AC voltage, you would probably be badly overloading the iron or the caps. ACV * 1.414 * 0.8 is often a good guesstimate of safe loaded voltage, where "0.8" may be 0.75 to 0.9 if you use small or super-oversized filter caps.

For voltages over ~20V, rectifier drop usually vanished in the haze of overall uncertainty. For 5V supplies rectifier drop can bite you bad and you better allow for it.

> why arn't people using something like 24-0-24 to get ±25VDC?

Because for many practical parts, you get 24VAC*120%*1.414-1V= 40V at no-load. If you were expecting 25V, and picked your parts that way, you could burn-up a bunch of parts before you got your load hooked up. With 18-0-18, you get from 30V to 22V, more in line with "+/-25V" thinking.

That's all for unregulated supplies. If you regulate, at over about 10V, just set ACV = DCV(reg), 24VAC for a 24VDC result. The raw supply will be 40V-30V, +/-10% for utility voltage variation is 44V-27V, still leaving a few volts to drop in the regulator to arrive at 24V.
maybe I have a badly speced transformer (something I pulled out of my junk box for a *test*). or my dmm is not reading the ACV correctly. The spec says it measures "average rms". I believe the dcv is accurate.

Here are some of my measurements, let me know what you think could be the issue here -

Transformer spec = 8VAC 250ma, Bridge rectifier 50V 25A, 4700µf/35V filter cap.

No Load:
Secondary ACV = 11.16
Filtered DCV = 14.74

50ma Load(200ohm resistor):
Secondary ACV = 10.2
Filtered DCV = 11.27

160ma Load(50ohm resistor):
Secondary ACV = 8.48
Filtered DCV = 8.00
I'm just speculating but I think you would see better numbers with a "true RMS" multimeter. If you hook up an oscilloscope I'm willing to bet what you will see is a not-so-nice sine wave on the transformer secondary. Many small cheap transformers are inefficient and distort the signal pretty bad.

I'll do some testing later today with a 40VAC toroid and a cheap 6VAC radio shack transformer. I'll try to take some pictures if I can find my digital camera. (no expensive DSO for me yet)
this is what i noticed

using a 25 A, 200 v rectifier with metal housing

and a transformer of 0v- 12 v i measured 30 volts

with a small rectifier (the ones where the in and out are in line, i measured (with the smae transformer) 14.7 volts

what is the reason behind this?

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