# Current through secodary of a step down transformer

Discussion in 'General Electronics Chat' started by IcedFruits, Apr 9, 2015.

1. ### IcedFruits Thread Starter Member

Jan 15, 2014
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hi, lets say there is a step down transformer with secondary output rated as 6-0-6 500mA.

then would it mean after converting it to DC with bridge rectifier will yield an o/p of 12v 500mA maximum (not considering diode voltage drops)?

i know, due to ac wave peaks, the actual o/p without load will be higher voltage, but under load conditions that would be the limit , right ?

(p.s. i am not actually trying to draw maximum rated current from a transformer, just trying to understand the figures)

Jul 18, 2013
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The current limit is whatever the AC secondary is rated at (Va).
Max.

3. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
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If you are planing to use a bridge rectifier across the entire secondary, yes, the unloaded dc output will (assuming nominal conditions) equal 12*(2^.5) (i.e. ~17v) post (capacitive) filtration... Whereas the 'tapped secondary' topology will render half this value (all else being equal)...

Note, however, that, inasmuch as the continuous current rating is based upon power handling, it is important that you understand the parameter format (i.e. 12v*500ma vs. 6v*500ma [6VA vs. 3VA respectively) note also that, strictly speaking, 1VA should be regarded as equivalent to (2^.5)/2 (i.e. ~.707) watts...

Best regards
HP

Last edited: Apr 9, 2015
4. ### IcedFruits Thread Starter Member

Jan 15, 2014
84
2
thanks.
thanks, actually i was more like trying to know about how much current can be drawn from the secondary.....

Last edited: Apr 9, 2015
5. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
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No problem! I've just edited my post to address that point!

6. ### dl324 Distinguished Member

Mar 30, 2015
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Transformer secondary is specified in RMS. You multiply by 1.414 to get peak. Under full load, secondary voltage drops approx 10%; that's how current rating of unmarked transformers is sometimes determined.

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7. ### crutschow Expert

Mar 14, 2008
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That's for a sine output. For a peak rectified output using a bridge the average DC output current should be derated to about 50% of that (due to the high peak current RMS power dissipation in the winding resistance).

8. ### IcedFruits Thread Starter Member

Jan 15, 2014
84
2
now i this is where i get confused, where that 50% number comes from ? the manufacturers consider these parameters and rate with the actual figures, no ?

for understanding sake lets say i put a regulator of 0.7 power factor, then from a 6VA secondary will i be getting a maximum of 6x0.7 = 4.2 watts ?

9. ### crutschow Expert

Mar 14, 2008
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Where did you get a 0.7 power factor?
The derating is an approximation based upon the peak current drawn by a typical rectifier-capacitor DC supply.
The power dissipated in the transformer winding resistance is proportional to I^2 * R. This means that, for a given average current out, the power dissipation increases for the short, high-current pulses drawn by a peak rectifier circuit.
Typically you thus have to reduce the average DC rectifier output current by about 50-60% as compared to a full-wave continuous sine current (which is how the transformer is rated), to keep the transformer resistance losses the same.
This is affected by the size of the filter capacitor and the rectifier output. The larger the capacitor, the higher the peak current, and the more you have to derate the transformer.
If you want an exact derating factor you can simulate your specific circuit and observe the dissipation in the transformer winding resistance (equivalent primary and secondary resistance).

10. ### #12 Expert

Nov 30, 2010
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This tells a lot.

• ###### Transformer-rectifier-output-current-ratios.pdf
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Jul 18, 2013
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Not always realized is that increasing the size of the capacitors after rectification in order to reduce % ripple can increase the calculated Va required when ran at the full rated current.
Max.

12. ### The Electrician AAC Fanatic!

Oct 9, 2007
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13. ### crutschow Expert

Mar 14, 2008
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Below is an LTspice simulation to show the difference in the transformer power dissipated between a sinewave load and a bridge rectifier capacitor supply with the same DC load current .

R_trans1 and R_trans2 represent the 1 ohm equivalent winding resistance of two identical transformers.

The first has a DC load of 1A from the rectifier-filter (whose capacitor value gives a reasonable 1V ripple at the capacitor filter).

The second has a sinusoidal AC load current of 1Arms. This generates an average (real) power dissipation in its 1 ohm resistance of 1W

As can be seen, the current through R_trans1 is a series of high current pulses which generates an average (real) power dissipation of 2.44W in the 1 ohm winding resistance (small Waveform window).

Thus the power dissipated in the transformer winding resistance with the rectified capacitor load is 2.44 times the dissipation of the transformer with the pure sinusoidal load.

This ratio difference would be reduced for higher values of transformer resistance at a given current since the higher resistance reduces the peak current value in the rectifier capacitor filter.

• ###### Transformer diode-cap derating.asc
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Last edited: Apr 10, 2015
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14. ### #12 Expert

Nov 30, 2010
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@crutschow Can you present a formula for this based on uf per ampere?
If this isn't a hijack...

Jul 18, 2013
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This is something I transcribed from a empirical lab testing done at U of Maryland.
I think it is all there.
Max.

• ###### RippleCalc.pdf
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16. ### crutschow Expert

Mar 14, 2008
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It is, sort of.
I don't have a formula but Max appears to have covered it.

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17. ### IcedFruits Thread Starter Member

Jan 15, 2014
84
2
^ thanks everybody for all those details, graphics & attachments, i will read them all up soon, and will try to put up some thoughtful questions if i can.

at a few minutes glance, it appears higher the filter capacitor means less time the secondary gets to conduct & the coil current peak goes higher during conducting period. this is keeping the filer o/p current keeping constant right ?

one more thing, during this non conducting period of secondary coil, what happens at the primary, current should minimize there too, right ? how this nonconducting time is registered in energy meters for billing ?

18. ### crutschow Expert

Mar 14, 2008
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The peak secondary current will appear in the primary current, reduced by the turns ratio.
The energy meter should respond to the peak current and only bill for the actual energy used by these peaks.

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19. ### #12 Expert

Nov 30, 2010
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All moments of, "no load" on the secondary result in "no load" on the primary. The voltage keeps changing, but the current approaches zero until the voltage gets high enough to kick some current through the diodes.

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20. ### The Electrician AAC Fanatic!

Oct 9, 2007
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This result is theoretically true, but just doesn't happen in the real world. The problem is that the "underlying AC waveform" is shown to remain a perfect sine wave no matter how large the filter capacitance gets. This could only be true if the transformer had no winding resistance and leakage inductance, the diodes had no resistance, the capacitor had zero ESR, etc., and the primary were driven from a perfect voltage source with no internal impedance.

With a real transformer, this is what happens: