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Voltage sag from a power supply.

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MikeT1982

Mon Aug 13 2007, 07:54AM

Registered Member #621 Joined: Sun Apr 01 2007, 12:37AM
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Posts: 119

I am curious about this. If you have an imaginary power supply consisting of say a 120VAC to 12.6VAC step down transformer rated at 3 amps, a full wave bridge rectifier, and a smoothing capacitor bank of sufficient size to have less than a 10% ripple with 3 amps draw, and totally OMIT the voltage regulator. And I am talking about a smoothing cap bank that was designed with the CV=It equation you guys showed me to yeild between a 5% and 10% ripple on the peak DC voltage of 17.64 volts.

I am guessing the capacitor bank will charge to the 12.6 volts x 1.4 yeilding 17.64 volts, giving an output voltage of 17.64 volts.

Now my question of curiosity is, as you place a load on it, gradually increasing and stop at 3 amps load that the transformer is rated, is the power supply actually still pushing close to 17.64 volts, or is there a point at which the capacitors will "dump and stay dumped" and you will be running at your RMS 12.6 volts? I'm wondering if the caps will actually keep the voltage higher than the 12.6 RMS all the way untill the transformer falls flat on its face (i'm guessing it would at) at 3 amps.

Thanks!

Mike

Bored Chemist

Mon Aug 13 2007, 09:03AM

Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022

Most transformers produce a little more than the rated voltage when not loaded. Ignore the rectifier + cap for the moment. Plug the transformer into the mains and it will give something like 13V (it will depend on the transformer's design).
If you put a 4.2 Ohm resistor across it you will get 3 Amps lowing and the voltage will fall to 12.6V
Now, to go back to the original idea with diodes and a cap, the first thing you need to consider is the volage drop across the diodes (typically a volt or so each but current and temperature dependant).
Then there's the fact that the mains is not all that stable. Calculating to 4 digits really isn't apropriate for this sort of thing. A load drawing more than 3A will not only drop the voltage more but will overheat the trasnsformer slightly ( you can get away with this for a while). A much bigger load than 3A will cause problems more quickly but there's no particular "cut off" point until something burns out or the fuse blows.

Sulaiman

Mon Aug 13 2007, 09:49AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

Just to add to the above;
For a 37.8 VA (12.6V x 3A) transformer the 'regulation' will typically be about 15%
so with no load I'd expect 14.5 Vac, so your capacitors need to be rated for 20.5V dc minimum.
As above, you should get a minimum dc voltage 12.6 x SQRT(2) - (2 x Diode Voltage) - Ripple voltage = 14V minimum
Using schottky diodes (I do because I've got stock) that would be about 15V minimum.

With a full-wave rectifier (dc current) = (ac current) x 0.62
so I'd expect 1.86 Adc to heat the transformer within specifications.

So your 37.8VA transformer will actually give 26 Watts dc minimum.

As you increase the dc current the dc voltage will fall due to
- Resistance of the transformer windings
- Increasing resistance of the transformer windings due to heating
- Increased voltage drop across the diodes/rectifier
- increased voltage ripple across the capacitor

The limit to how much current you can continuously draw will be determined by heating of the transformer and diodes.

MikeT1982

Tue Aug 14 2007, 09:03PM

Registered Member #621 Joined: Sun Apr 01 2007, 12:37AM
Location:
Posts: 119

Guys thanks for all the detailed info, it's greatly appreicated. I don't want to start a new post but maybe someone will see this. I was thinking about how in a big capacitor charging step-up circuit you should have a current limiting resistor between the transformer and the cap bank unless it is a current limited transformer like an NST....

In these little, low voltage step-down power supplies though is the "short circuit" seen by the step-down transformer at first "energize time" irrelevant (maybe due to small capacitance of smoothing bank in comparison?) that we don't need current limiting resistors?

Steve Conner

Wed Aug 15 2007, 10:17AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Yes, in small power supplies, no current limiting is needed, because the resistance of the transformer windings limits the current fine.

In larger ones though, the surge at startup would blow the fuse, trip a breaker, or blow the rectifier diodes, so you have to do something about it. The usual way is to put a large power resistor in series with the AC line, that gets shorted by a time-delay relay a second or two after turning the equipment on.

You usually see this soft-start circuit starting to appear in DC power supplies around the 500VA-1kVA level, especially with toroidal transformers which have a vicious inrush current of their own. I last used it in an audio amp with two 160VA toroids and 40,000uF @ 50V of filter capacitance.

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