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

[happyhero]

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.

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.

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?

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?

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?

 

[SFK]

HappyHero,
You wrote the equation correctly and then changed it so it is wrong when you said: "W=IV so for example W=240V/1.6A (W=V/I) = 150watts, but W=12V/6A =2Watts"

It should be
Input W = VI = 240V*1.6A = 384W (at 240V)
and
Output W = VI = 12V * 6A = 72W (at 12V)
This means that at the Input side it will draw a maximum of 384W and at the output side it will provide a maximum of 72W. This seems uncommonly inefficient (ie it suggests there is 312W of losses), but it might just be them covering their bases for the 100 to 240V operation, or it being able to produce more than 72W at 12Volts.

For you the output side is important with regards to your LED tape.
Most LED tape (but you should check yours) is about 5W per meter length.
So this means you can drive a maximum length of (72W / 5W/m =) 14meters of 5W/m LED tape.
I normally leave a 10 to 20% margin to minimize possible overheating, so 14Meters x 85% = maximum of 12meters of LED table with a 15% margin.

sfk

 

[bernardgreen]

The ratio of currents is the inverse of the ratio of voltages

Primary 240 volt , secondary 12 volt Ratio = 240 over 12 = 20 to 1

So Vp/Vs = 20 ( which is also the turns ration of the transformer )

Hence current ratio Ip over Is = 1 to 20

Ip/Is = 1/20

Secondary load current 6 amps

Primary current = 6 * 1/20 = 0.3 amps

Check 0.3 * 240 = 72 = 6 * 12

This is true only for inductive wound transformers. When dealing with "electronic" transformers there is no fixed turns ratio

 

[endecotp]

The power supply says Input 100V-240V 50/60htz 1.6A so ok for 2A socket

That's all you need to know - the max input current is 1.6A, which is fine for your 2A supply.

and the ouput says 12V 6A but how do you get from 1.6A to 6A, i.e. how do you calculate this?

You don't. Those numbers aren't really comparable, because they are limits. 1.6A is probably the maximum input current it will ever take, and since it's rated for 100 - 240 V operation, that is likely to be when it's connected to a 100V (i.e. North-American) supply. 6A may be the minimum that the output is guaranteed to be able to supply. It will never be both consuming 1.6A at the input (at 240V) and supplying 6A at the output at the same time, as that would imply an efficiency of less than 20%.

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?

Yes, with the difference being the inefficiency of the transformer, which you can feel as heat.

 

[aptsys]

That's all you need to know - the max input current is 1.6A, which is fine for your 2A supply.




You don't. Those numbers aren't really comparable, because they are limits. 1.6A is probably the maximum input current it will ever take, and since it's rated for 100 - 240 V operation, that is likely to be when it's connected to a 100V (i.e. North-American) supply. 6A may be the minimum that the output is guaranteed to be able to supply. It will never be both consuming 1.6A at the input (at 240V) and supplying 6A at the output at the same time, as that would imply an efficiency of less than 20%.



Yes, with the difference being the inefficiency of the transformer, which you can feel as heat.

Correct.

 

[ericmark]

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.

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.

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?

100 x 1.6 = 160W and 12 x 6 = 72W however the 1.6A is the peak not the average current, and when you switch on a switch mode power supply it has to charge the capacitors so if you plug the power supply into an energy meter likely after first few seconds it will be drawing around 0.35 amp.

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?

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?

What you need to remember is an electronic power supply with a input voltage of 100 ~ 240 volt will draw more amps at 100 volt than it does at 240 volt. I was asked to PAT test a school laptop trolley, it had 20 laptops all plugged into sockets which in turn connected to a single 230 volt outlet, if you simply added up the amps of each power supply you were looking at around the 30 amp mark, however in real terms, I know because I measured, on first plug in it peaked to around 12A for a few seconds, then dropped to around 7 amp.

The switched mode power supply turns the 230 Vac into DC, then it turns it back to AC at a much higher frequency, as a result the transformer if used is a lot smaller, it samples the output and changes the mark/space ratio of the bit which produces the AC at high frequency, so 50/50 is about max output and minimum output is likely something like 5/95 it can't go to 0/100 as just can't switch that fast, so as well as a maximum output there is also a minimum output. Also the capacitors are charged when the voltage is above the existing voltage so it does not draw even through the cycle, the current varies much more than it would do with a simple sine wave.

 

[aptsys]

so 50/50 is about max output and minimum output is likely something like 5/95 it can't go to 0/100 as just can't switch that fast, so as well as a maximum output there is also a minimum output. Also the capacitors are charged when the voltage is above the existing voltage so it does not draw even through the cycle, the current varies much more than it would do with a simple sine wave.

There's no such limit. Maximum duty cycle is merely limited by saturation of the transformer core. There is no theoretical minimum duty cycle - if the load is so small that the duty cycle is minimal, most will just go into pulse skipping mode as this is far more efficient. Minimum duty cycle may be limited if you use a DSP for the controller, but almost all general purpose switcher ICs don't use digital PWM generators because the loop gain is typically not as good and more difficult to get sufficient phase margin on the feedback.

 

[ericmark]

There's no such limit. Maximum duty cycle is merely limited by saturation of the transformer core. There is no theoretical minimum duty cycle - if the load is so small that the duty cycle is minimal, most will just go into pulse skipping mode as this is far more efficient. Minimum duty cycle may be limited if you use a DSP for the controller, but almost all general purpose switcher ICs don't use digital PWM generators because the loop gain is typically not as good and more difficult to get sufficient phase margin on the feedback.

I think you are likely spot on, however I was trying to keep it reasonably simple.

 

[FrodoOne]

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.

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.

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?

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?

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?

The Power Supply is a Switch Mode Power Supply, where a Bridge Rectifier charges an electrolytic capacitor (to somewhere near the peak of the AC voltage waveform).
For 240 V AC this would be about 340 V, so the capacitor concerned is likely to be rated at 400 V.

The electronics in the device "switch" the voltage from the input capacitor on and off through the primary of a transformer (at a frequency of between 50 kHz and 1 MHz) and the secondary voltage from this transformer is rectified, applied to a second electrolytic capacitor and this voltage is monitored to control the "on" period of the switching concerned.
For lower input voltages, the "on" period would be longer than for higher input voltages, and hence the average primary current would be higher with lower input voltages.
However, this device could be operated on 100 V AC (in Japan).
1.6 A would be the peak current at switch on.

If the device is supplying 12 V at 6 A this is 72 VA.
If the device was 100% efficient, the primary current at 240 V AC would be 0.3 A
Even if its efficiency is only 75% (and a good design could have efficiency of 95%), the primary current at 240 V AC would be 0.4 A

(I note that much the same has been said by others while I was composing this - and doing other things.)

 

[aptsys]

The Power Supply is a Switch Mode Power Supply, where a Bridge Rectifier charges an electrolytic capacitor (to somewhere near the peak of the AC voltage waveform)

This may be the case for some small devices, but it leads to terrible power factor and harmonics. Normally the active PFC stage has the input is fed into a boost regulator which shapes the input current to match the incoming AC waveform.