Ir Led Driver Circuit
A while ago I bought a 10W 12V LED on line to see if I could build a small projector. Then somewhere, I hear you need to build a driver for it, otherwise you WILL blow the thing. So I start googling and tinkering and find a few basic formulas, P=IV, > P/V=I > 10w/12V=0.833A
So I need to get a current of 833mA at 12v to run this light, next step.
According to most sites, the cheapest way of doing this is with a simple LM317-T Regulator.
The ADJ voltage on this is 1.25v, so, V=IR > V/I=R > 1.25v/0.833A= 1.5 Ohm.
So I pick up a few regulators and a fist full of resistors at my local electronics shop, dust off an old soldering iron and thow it all together.
Then I plugged it on to a 12v 1.0A Power Adapter. I'd heard the 317 had a Vdrop of about 3 volts, so I took some readings to make sure everything was ok. Got 620mA between the Power/LED, the LED/LM317, and LM317 to the Power, all is good.
Voltage was 12.13v across the Power, 9.20v across the LED, and 2.91 across the LM317v (0.95 over the resistor). All the numbers looked fine.
Here's where it got tricky. Knowing the LED was rated for 12v, and the drop over the LM317 was ~3v, I dug up a 15.2v 1.2A Power adapter, and plugged that in instead.
When I took the readings however, my current was up to 810mA, but the voltage was still down at 9.56v. The Power adapter was at 15.20v, and the 317 was dropping 5.67v.
This is the bit I don't understand.
How can I get my volts back?
I know the LM317-T regulates its current via the resistor, but I'd assumed the voltage was dependent on whatever you plugged in, minus a constant 3v drawn by the regulator. This assumption is obviously wrong.
Is there a way I can get the voltage back?
The first circuit will provide around 17ma, the second is variable up to 0.5A for high power LEDs. If you have a large number of LEDs you can replace the two diodes and the 1KΩ resistor with a LM317 without any biasing resistors, where it is a 1.25V regulator. The IC LM317 promises countless interesting and useful power enhancement circuits, applicable in numerous fields where well dimensioned power supplies become crucially important. Some assorted useful lm317 application circuits from National Semiconductor pdf datasheet has been discretely explained in this chapter through the relevant circuit illustrations. The compilation includes.
Lm317 Application Circuits
Lm317 Circuit Examples
A while ago I bought a 10W 12V LED on line to see if I could build a small projector. Then somewhere, I hear you need to build a driver for it, otherwise you WILL blow the thing. So I start googling and tinkering and find a few basic formulas, P=IV, > P/V=I > 10w/12V=0.833A
So I need to get a current of 833mA at 12v to run this light, next step.
According to most sites, the cheapest way of doing this is with a simple LM317-T Regulator.
The ADJ voltage on this is 1.25v, so, V=IR > V/I=R > 1.25v/0.833A= 1.5 Ohm.
So I pick up a few regulators and a fist full of resistors at my local electronics shop, dust off an old soldering iron and thow it all together.
Then I plugged it on to a 12v 1.0A Power Adapter. I'd heard the 317 had a Vdrop of about 3 volts, so I took some readings to make sure everything was ok. Got 620mA between the Power/LED, the LED/LM317, and LM317 to the Power, all is good.
Voltage was 12.13v across the Power, 9.20v across the LED, and 2.91 across the LM317v (0.95 over the resistor). All the numbers looked fine.
Here's where it got tricky. Knowing the LED was rated for 12v, and the drop over the LM317 was ~3v, I dug up a 15.2v 1.2A Power adapter, and plugged that in instead.
When I took the readings however, my current was up to 810mA, but the voltage was still down at 9.56v. The Power adapter was at 15.20v, and the 317 was dropping 5.67v.
This is the bit I don't understand.
How can I get my volts back?
I know the LM317-T regulates its current via the resistor, but I'd assumed the voltage was dependent on whatever you plugged in, minus a constant 3v drawn by the regulator. This assumption is obviously wrong.
Is there a way I can get the voltage back?