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Transformer voltage drop

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IamSmooth

Sat Dec 29 2007, 06:25PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

Is there a simple way to estimate the voltage drop of a transformer when under a specific load? Let's say I have a transformer with a secondary open voltage of 10.5v and a maximum current capablility of 4A, and I want to know what the voltage will be when there is a 3 ohm load across the output. Do I have to calculate the reactive resistance of the secondary and consider the circuit as two resistances in series, and this becomes a voltage divider? Any suggestions? Thanks

Steve Conner

Sat Dec 29 2007, 08:26PM

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

The manufacturer should specify the percentage regulation.

If you don't know this, but you have the transformer, it's time to do a short-circuit test on the transformer to measure its impedance. You short the secondary and energize the primary with a low variable voltage, and measure how much voltage is needed to make the transformer's rated current flow.

If you don't know the regulation and you don't have the transformer, you might as well go and do something else.

Sulaiman

Sun Dec 30 2007, 06:26PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

I newer thought of that way to measure regulation .. thanks Steve

IamSmooth

Mon Dec 31 2007, 10:10PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Ok. I did this. The open circuit secondary voltage is 10.7v. I shorted the secondary with a current meter and increased the primary to about 15v when i read 3.2A on the secondary. What does this tell me and how will this predict the voltage drop under various loads?

Sulaiman

Tue Jan 01 2008, 05:35PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

I can give you some 'Rules-of thumb' that I use;

For transformer heating with a full-wave rectifier and large capacitor equal to ac-rated heating, Idc = Iac x 0.62
So a 4A ac transformer should give about 2.5 Amps dc
I'd probably use 1N5820 schottky diodes for about 600 mV forward voltage drop at 10A peaks

15Vac for 3.2 A ac implies 18.74 V ac for the rated 4A ac.
So if using a 230Vac supply the output would be 10.7 V ac x (230 - 18.74)/230 = 9.83 V ac at 4 A ac load.
So the rectified output should be about 9.83 x 1.414 - 2 x 0.6 = 12.55 V dc. - ripple voltage.
I'd say 12 V @ 2.5 A dc = 30 W dc power supply.
Since you've probably got a 50 VA transformer, 30 W may seems low, but that's how it goes.
On no load the dc voltage should rise to almost 10.7 x 1.414 = 15 V dc no-load.

If you're using 110 Vac then it's worse - 10.7 x (110 - 18.74)/110 = 8.88 V @ 4 A ac.
or about 11 V @ 2.5 A dc with ripple. 27.5 W dc.

To predict the output under various load conditions is very tedious by hand, try to use a circuit simulator.
Consider the transformer as a 10.7 V rms sinewave source with 0.2175 ohms in series ... (10.7 - 9.83)/4
OR 10.7 V rms with 0.455 ohms in series if you're using 110 Vac.

IamSmooth

Sat Jan 05 2008, 12:43AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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I just finished rewinding my transformer to get the right voltage for the load. For future reference, what needs to be done to improve the voltage regulation? Thicker wire? Better ferrite material or more cross-sectional area for the expected flux?

Steve Conner

Sat Jan 05 2008, 01:32PM

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

For better regulation, just make everything bigger

ConKbot of Doom

Sat Jan 05 2008, 10:05PM

Registered Member #509 Joined: Sat Feb 10 2007, 07:02AM
Location:
Posts: 329

Steve Conner wrote ...

For better regulation, just make everything bigger

I was under the assumption that primary leakage inductance was the main culprit in poor regulation. Obviousley larger wires and cores could help with that...

Bjørn

Sat Jan 05 2008, 10:25PM

Registered Member #27 Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058

You short the secondary and energize the primary with a low variable voltage

At this point it is easy to get a bit careless because the voltage is low. Remember to turn the voltage all the way down before you disconnect. It is not possible to explain by words the kick you can get from a large transformer when doing this completely wrong.

kell

Fri Jan 11 2008, 02:02PM

Registered Member #142 Joined: Sat Feb 11 2006, 01:19PM
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Posts: 102

Ever hear of Thevenin? You can model your transformer as an ideal voltage source in series with a resistance. It doesn't account for inductive effects, but you can predict the voltage drop due to resistance in the secondary for any (resistive) load. The load and the transformer's secondary resistance form a voltage divider.
If your meter can't measure low resistance accurately (cheap ones can't), then load the output and measure the voltage. Knowing the open circuit voltage, the loaded voltage, and the load resistance, you can calculate the secondary resistance (remember, voltage divider).
Calculations are left as an exercise for the reader ;)

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