AC power VS DC power

[yoosefheidari]

hi
have a question about ac and dc power rating of my transformer
i have 200VA transformer which output voltage is 36V and can deliver up to 5.5 amps current.
question is here : this 200VA refer to AC power of transformer?what about DC power?
if i attach a 6/5 ohm resistor to output it consupt about 5/5amps AC and power loss on resistor is about 200watts.
but if i use a bridge rectifire and output capacitor the output DC voltage is about 50V and if attach same 6/5 ohm resistor then it consumpt about 7/7amps DC and power loss on resistor is about 385 watts
what happend here about AC and DC power?200VA AC power equal to 385VA DC power?

 

[KlausST]

Hi,

A transformer is for AC only, thus it is not specified for DC.
It is specified for current RMS.
Now it depends on the connected circuit how much RMS current it draws from the transformer.

A bridge rectifier with a bulk capacitor maybe draws 2..4 times higher RMS input current than average DC output current.
(Not calculated)
With series inductor the factor will be less.

Klaus

 

[FvM]

As already stated transformer VA rating is the product of rated output voltage and current, both are rms values. The ratio of rectifier DC output current and input rms current is often specified as "form factor". A typical value for a bridge rectifier with medium size filter capacitor is 1.6. The exact value depends on capacitor value and transformer winding resistance and leakage inductance.

Form factor 1.6 results in 3.4 A maximum DC output current of your transformer with rectifier.

 

[yoosefheidari]

for better understanding i did a test with a 12VA transformer.no-load voltage of trnsformer was 12.7v
i use a 19.6 ohm resistor as a load
first i connet resistor directly to transformer's output and voltage droped to 11.3 and cunsumped 0.573amps current
then i use a bridge rectifire+3300uf bulk capacitor.no-load DC voltage was 16.5v
i connect same resistor to capacitor's terminals and DC voltage droped to 11.3v and consumped 0.573amps current.output voltage of transformer was 10.7v and current was 0.834amps
(all measurments done by a true rms DMM)
now:
why dc voltage droped so much?it should have been about 14volts!
why AC power at first test is eqeal to DC power of second test?
and why AC power in second test is so high about 9 watts?

 

[KlausST]

Hi,

RMS voltage measurement in a DC supply makes no sense (in most cases).

RMS voltage drop on a rectifier load doesn't tell much because the waveform will be distorted.

Use a scope to check current waveform and voltage waveform.

Klaus

 

[RCinFLA]

Rough approximation way to do this is you had 12.7v no load to 11.3v loaded with 0.576 amps. This is roughly equivalent to a series resistance of 1.4v/0.576A = 2.43 ohms. Model transformer as 12.7v source with 2.43 ohm source resistance.

You have diode drops at capacitor peak charging current. If a bridge you will have two diode drops. Diode drop could be a lot higher then you think due small conduction angle on charging capacitor. Charging current is quite high peak value so the drop across the source resistance of 2.43 ohm can be high. You need to look up diode forward drop at the peak cap charge current.

All depends on average DC load you put on output. It determines conduction angle on cap charging and how much ripple voltage you will end up with.

 

[Easy peasy]

For a transformer with sine AC o/p if the voltage is 15V rms, the rectified average is 15/1.111 = 13.5V, this is what you get with an LC smoothing filter, for just a cap after the bridge rectifier - you get the peak on the cap ( 15 * sqrt2 ) = 21.2Vpk and then falling until the next AC peak - the ave volts now depends on the size of the cap and the load - there can be a fair bit of ripple if the cap is too small, and/or the load to large in proportion.
The transformer is rated in VA, since the Vout is fixed this is really a thermal limit on the amps in the sec. This is given in amps rms as this is the equivalent to the DC value for wire heating, so if you have 200VA / 36V = 5.55 amps rms, so you could draw a square wave of current at 5.55A pk, or a sine wave of current with 5.55A rms ( 7.85Apk ). But ...
your cap input filter has a much peakier current with a typical factor of ave/rms = 0.65, thus for 5.55 A rms allowed - you can draw 3.6A ave on the DC - which will be 5.55 rms due to the high peaks of current in the transformer - you can model this in LTspice to see for your self.
BTW, the LC choke can draw a near square wave from the mains for large enough L & C and thus allow the full 5.55 amps to be drawn ....