240v to 12v transformer, how do you calculate current?

happyhero

[FONT=&quot]Hi I have a partial electrical background and I'm sure I knew this but I just cant work it out now. I'm retired now maybe I'm forgetting this stuff after not using it much.[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]Anyway, I have a power supply for some LED tape, they all came together and I have one of those 2A round pin sockets to plug it into. I felt like checking everything was correct, i.e that it does not draw more than 2A.[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]The power supply says Input 100V-240V 50/60htz 1.6A so ok for 2A socket and the ouput says 12V 6A but how do you get from 1.6A to 6A, i.e. how do you calculate this?[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]I know the equations V=IR and W=IV so for example 240/1.6A (V/I=W) = 150watts but 12/6 =2watts so is there a calculations to work out what the Amps are on the other side if you know one side, i.e if you knew it was 1.6A on the 240 side could you calculate what Amps it would be on the 12V side or is it all down to the number of windings on the power supply/transformer?[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]I remember someone used to say to me in the past that with the 12V MR16 lamps that if the lamp was 20 watts it would draw 20watts more or less on the 240 side as well, is that correct?[/FONT]

Undervalued

happyhero wrote: »

[FONT=&quot]Hi I have a partial electrical background and I'm sure I knew this but I just cant work it out now. I'm retired now maybe I'm forgetting this stuff after not using it much.[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]Anyway, I have a power supply for some LED tape, they all came together and I have one of those 2A round pin sockets to plug it into. I felt like checking everything was correct, i.e that it does not draw more than 2A.[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]The power supply says Input 100V-240V 50/60htz 1.6A so ok for 2A socket and the ouput says 12V 6A but how do you get from 1.6A to 6A, i.e. how do you calculate this?[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]I know the equations V=IR and W=IV so for example 240/1.6A (V/I=W) = 150watts but 12/6 =2watts so is there a calculations to work out what the Amps are on the other side if you know one side, i.e if you knew it was 1.6A on the 240 side could you calculate what Amps it would be on the 12V side or is it all down to the number of windings on the power supply/transformer?[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]I remember someone used to say to me in the past that with the 12V MR16 lamps that if the lamp was 20 watts it would draw 20watts more or less on the 240 side as well, is that correct?[/FONT]

Apart from any inefficiencies in the transformer (energy lost as heat or noise!) the power (watts) is the same at either end.

So if you draw 10 amps at 12 volts that is 120 watts. At the 240 volt end that is only half an amp (0.5 x 240 also equals 120).

So if you assume the transformer is 80% efficient (and it should be better than that) then allow that safety margin.

onomatopoeia99

happyhero wrote: »

[FONT=&quot]I know the equations V=IR and W=IV so for example 240/1.6A (V/I=W) = 150watts but 12/6 =2watts so is there a calculations to work out what the Amps are on the other side if you know one side, i.e if you knew it was 1.6A on the 240 side could you calculate what Amps it would be on the 12V side or is it all down to the number of windings on the power supply/transformer?[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]I remember someone used to say to me in the past that with the 12V MR16 lamps that if the lamp was 20 watts it would draw 20watts more or less on the 240 side as well, is that correct?[/FONT]

If you work on the basis that transformers are 100% efficient to make the sums easier and ignore internal resistance ...

240V at 1 amp is 240 watts
12V at 20 amps is 240 watts

That should be enough information to get you going.

DoaM

The error in post #1 is that you give the formula correctly (W = V x I) but then actually use it incorrectly (W = V / I).

(R = V / I)

EssexExile

Surely P not W, or is that another thing they've changed since I left school?

kwikbreaks

EssexExile wrote: »

Surely P not W, or is that another thing they've changed since I left school?

From memory when I was taught this stuff in the sixties W for watts was used.

Apart from the calculation error there is a concept error too.

The figures on a PSU do not tell you how much current it will use and how much it will supply they are maximum figures. On the input side there may be a current surge as the thing starts up so the current mentioned is the maximum drawn. On the output side the current stated is the maximum it can supply - the actual current will depend on the load.

Most PSUs are at least the 80% efficiency mentioned - put your finger on one and see how warm it gets to find the poor ones.

Heedtheadvice

There you have it then.
Depends upon if your unit is a transformer or a power supply. If you LEDs do not have a regulator in them to give them the correct voltage supply then it it likely that the unit will be a power supply to regulate the voltage provided to the lamps (jailbreaks (????)post applies) unless of course it is just a cheap and nasty transformer (with a very basic power supply to convert to DC, that does no regulation)! Otherwise the approximate general calculations provided above should suffice.

And apologies to kwik who seems to have got out of jail!!

DoaM

EssexExile wrote: »

Surely P not W, or is that another thing they've changed since I left school?

I've always used P (for Power), but I didn't want to confuse the OP.

were

ok so the transformer give 12v AC, Won't you need DC? Also that 12V is rms value, so the peak voltage is 1.414 times higher once it is smoothed

Heedtheadvice

Sorry were, but the peak dc voltage depends upon the type of rectification used, you are correct for one case at least..... and we should not forget the diode voltage losses too......

were

Heedtheadvice wrote: »

Sorry were, but the peak dc voltage depends upon the type of rectification used, you are correct for one case at least..... and we should not forget the diode voltage losses too......

No, still the peak of AC even is still 1.414 times higher than the RMS value, even if you do not rectify it.

If you convert AC to DC then you have pulsed DC and your diode(s) will drop voltage.

Though to correctly power LEDs for maximum brightness, you should use constant current source, not voltage.