as a start, primary volt-amperes x 0.7, so that if your traffo has a 100VA primary rating, then you can safely use 70watts as secondary power draw...
a voltage doubler gives a dc output unloaded of Vsec x 2.8, so that if you have 100volts secondary voltage, that becomes 280 volts unloaded, then current draw can be 10/280 or 0.25A, actual will be lower since under load you can expect that 280 volts drop to something line 230 volts in some cases....
a voltage doubler gives a dc output unloaded of Vsec x 2.8, so that if you have 100volts secondary voltage, that becomes 280 volts unloaded, then current draw can be 10/280 or 0.25A, actual will be lower since under load you can expect that 280 volts drop to something line 230 volts in some cases....
your transformer is rated for a certain output (or input) power - 125Vac, 1.25A => 156VA. this rating is based on (among other things) copper loss in the windings and some maximum allowable temperature rise.
if you pull more power out of the transformer, it gets hotter: doubling the output power doubles the current, and the copper loss quadruples (P_Cu = I*I*R).
the temperature rise is the power (copper) loss times the "thermal resistance" of the transformer, so if you quadruple the copper loss then the temperature rise also quadruples.
But its worse than that - as it gets hotter the winding resistance increases - Copper has a positive temperature coefficient. so when you double the power and hence the current, the copper loss and therefore temeprature rise more than quadruples. This will ultimately lead to thermal runaway (fried transformer).
So your output power is limited to pretty much the transformer rated power - in your case 156VA. So if you use a voltage doubler, the maximum output current will be about 625mA.
But thats assuming a lossless voltage doubler. in practice the available output power (and hence current) will be less than that.
Edit: This carefully ignores power factor, which Tony helpfully includes (primary VA*0.7). Tony picks 100VA & 100V as it makes the maths easy.
if you pull more power out of the transformer, it gets hotter: doubling the output power doubles the current, and the copper loss quadruples (P_Cu = I*I*R).
the temperature rise is the power (copper) loss times the "thermal resistance" of the transformer, so if you quadruple the copper loss then the temperature rise also quadruples.
But its worse than that - as it gets hotter the winding resistance increases - Copper has a positive temperature coefficient. so when you double the power and hence the current, the copper loss and therefore temeprature rise more than quadruples. This will ultimately lead to thermal runaway (fried transformer).
So your output power is limited to pretty much the transformer rated power - in your case 156VA. So if you use a voltage doubler, the maximum output current will be about 625mA.
But thats assuming a lossless voltage doubler. in practice the available output power (and hence current) will be less than that.
Edit: This carefully ignores power factor, which Tony helpfully includes (primary VA*0.7). Tony picks 100VA & 100V as it makes the maths easy.
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Sorry my tx have exactly 125VAC 100W, hey guys still not clear how much power using voltage doubler & how can I calculate?
100W x 0.7 = 70W @ 250VDC?
125VDC x 2.8 (voltage doubler) = 350V unloaded, then current draw can be 10 (where take this value?)/350 or 0.35A, actual will be lower since under load you can expect that 350 volts drop to something line 300 volts in some cases....
100W x 0.7 = 70W @ 250VDC?
125VDC x 2.8 (voltage doubler) = 350V unloaded, then current draw can be 10 (where take this value?)/350 or 0.35A, actual will be lower since under load you can expect that 350 volts drop to something line 300 volts in some cases....
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your secondary is 125v ac at 1.25A.....so that is about 156 watts available....
your dc unloaded voltage then becomes 125vac x 2.8 = 350volts dc unloaded.....
available current then assuming steady 350volts then becomes 156watts/350volts or 445mA
these are available to you, notice that we did not take into account the dc resistances your of your traffo windings, what are you building anyway?
your dc unloaded voltage then becomes 125vac x 2.8 = 350volts dc unloaded.....
available current then assuming steady 350volts then becomes 156watts/350volts or 445mA
these are available to you, notice that we did not take into account the dc resistances your of your traffo windings, what are you building anyway?
you must apply the manufacturers de-rating when feeding a capacitor input filter.
If the output VA is 156 when feeding a resistor and the manufacturer tells you the de-rating factor for a capacitor input filter is 0.7 then the maximum continuous output VA becomes 156 * 0.7 = 109VA..
If the DC output voltage is 2.8times the AC secondary voltage then the current available is 156VA*0.7 / secondary AC / 2.8 = 109/125/2.8 = 312mAdc.
Expect a voltage doubler output to sag quite badly with this loading.
I'd suggest that the doubler output be reduced to <30% (~100mAdc) of the maximum rating for reasonable voltage performance under load.
If the output VA is 156 when feeding a resistor and the manufacturer tells you the de-rating factor for a capacitor input filter is 0.7 then the maximum continuous output VA becomes 156 * 0.7 = 109VA..
If the DC output voltage is 2.8times the AC secondary voltage then the current available is 156VA*0.7 / secondary AC / 2.8 = 109/125/2.8 = 312mAdc.
Expect a voltage doubler output to sag quite badly with this loading.
I'd suggest that the doubler output be reduced to <30% (~100mAdc) of the maximum rating for reasonable voltage performance under load.
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