Powering LED from AC line

DECKY999

Member
2007-07-18 7:06 am
OK I know it can be done but I did something wrong. I have a LED with 3V f_voltage and 30mA current with an idea to power it directlu from the AC trafo supply (not the mains 240V).

I pinched the supply from the transformer AC line (~25V), on the junction with the rectifier bridge. From there the scematics goes like this:

~ - 1N4004 diode - cap 100uF 25V + leg - 380R resistor -
AC supply |
~ ------------------ -- cap - leg- LED cathode - o - LED anode

It worked for a while and then the cap blew up.

The idea is that with the half bridge rectification the LED can see only half of the supplied AC voltage. Therefore I thought that the 25v capacitor should be enough. 380R resistor is to provide desired voltage drop from 12.5 to 3V.

Am I doing something wrong here?
 
DECKY999 said:
The idea is that with the half bridge rectification the LED can see only half of the supplied AC voltage.

No, half-wave rectification doesn't give you half the voltage. You get the same voltage as you would with a full-wave rectifier, except it only sees that voltage every 1/60th of second instead of 1/120th of a second for a full wave bridge.

If your transformer's outputting 25 volts RMS, then the cap will see 25 x 1.414 or 35.35 volts. Your cap was only rated for 25 volts which is why it blew up.

As for the LED's 3 volt forward voltage, that doesn't mean you have to drive it with 3 volts. That's just the voltage across the LED when it's conducting. So you don't have to drop your voltage down to 3 volts, you just have to limit the current.

So, the first order of business is to use a higher voltage cap. I'd recommend 50 volts.

Next, if that 25 volts is from the transformer's datasheet and not what you've actually measured, then you should do some measuring. The specs rate the voltage for a given load current. Since you're not going to be drawing but a few milliamps through the diode, you'll likely end up with a higher voltage.

So build the circuit with just the diode and the cap and measure the voltage across the cap to get a starting point.

Although the LED's rated at 30mA, you don't need to run that much current through it. Just a few milliamps is sufficient.

So your LED circuit will be a resistor in series with the LED and those two wired across the cap.

The value of the resistor will be the voltage you measured across the cap, minus the LED's forward voltage, divided by the amount of current you want through the LED.

So if for example the voltage across the cap is 35 volts. Subtract 3 volts leaving you with 32 volts. Divide that by 0.003 and that gives you 10,666. Use a 10k resistor in series with the LED and you'll end up with about 3mA through the LED.

se
 

DECKY999

Member
2007-07-18 7:06 am
Re: Re: Powering LED from AC line

Steve Eddy said:


No, half-wave rectification doesn't give you half the voltage. You get the same voltage as you would with a full-wave rectifier, except it only sees that voltage every 1/60th of second instead of 1/120th of a second for a full wave bridge.

If your transformer's outputting 25 volts RMS, then the cap will see 25 x 1.414 or 35.35 volts. Your cap was only rated for 25 volts which is why it blew up.

As for the LED's 3 volt forward voltage, that doesn't mean you have to drive it with 3 volts. That's just the voltage across the LED when it's conducting. So you don't have to drop your voltage down to 3 volts, you just have to limit the current.

So, the first order of business is to use a higher voltage cap. I'd recommend 50 volts.

Next, if that 25 volts is from the transformer's datasheet and not what you've actually measured, then you should do some measuring. The specs rate the voltage for a given load current. Since you're not going to be drawing but a few milliamps through the diode, you'll likely end up with a higher voltage.

So build the circuit with just the diode and the cap and measure the voltage across the cap to get a starting point.

Although the LED's rated at 30mA, you don't need to run that much current through it. Just a few milliamps is sufficient.

So your LED circuit will be a resistor in series with the LED and those two wired across the cap.

The value of the resistor will be the voltage you measured across the cap, minus the LED's forward voltage, divided by the amount of current you want through the LED.

So if for example the voltage across the cap is 35 volts. Subtract 3 volts leaving you with 32 volts. Divide that by 0.003 and that gives you 10,666. Use a 10k resistor in series with the LED and you'll end up with about 3mA through the LED.

se

Thanks - I understand my stupidity now.