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I have a task to drive four LEDs from a 4-20mA loop. I have 24V for the LED side so I need to convert the current into something that can drive the LEDs.

I know I can use a 250\$\Omega\$ resistor as the converter in the loop which will give me 1-5V from 4-20mA respectively, now I need to get this voltage to drive the LEDs.

Do I simply connect a transistor base to the 1-5V point, ground the emitter and drive the LEDs from the 24V via dropping resistors via the collector??

It needs to be a variable drive as the brightness of the LEDs must vary with the 4-20mA signal.

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  • \$\begingroup\$ Thanks all so far, i will do some tests today on the bench and see what plays best. I have the LDR and leds plus some resitors arriving today so will be able to try the whole setup. Interesting about the 4mA current load to extinguish the leds. Setup tested and works well, i can get a 5k - 40k swing by padding the led with 375R to keep it dim at 4mA, that is plenty enough swing for the application and i have moved on to building the unit now. I'm amazed that it can be done with only a reseistor, an led and an LDR, very useful stuff. Thanks to all who helped, this is a great site. \$\endgroup\$ – user3056 Feb 17 '11 at 7:14
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I believe you can simply connect all four LED's in series (observe correct polarity) with the 250 ohm resistor. The 4-20mA transmitter will take care adjusting the current/voltage for you. Just be sure the LED's rated average current is around 20mA, that way the LED's are fully lit at 20mA and dim at 4mA.

*EDIT - The 4-20mA transmitter should have a specification called "load range", your resistor should somewhere in this range.

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  • \$\begingroup\$ Wow, didn't know that! So no dropping resistors, just the load resistor across the loop? That's pretty simple. The LEDs I have are 30 ma but maybe that won't make much difference, worth trying anyway. What effect does the resistor have then? If say I just connected the LEDs in series across the loop without it the resistor, what would that do?? Or do loops HAVE to have a certain resistance?? \$\endgroup\$ – user3047 Feb 16 '11 at 21:07
  • \$\begingroup\$ No sorry, everything needs to be in series in the loop including the resistor. \$\endgroup\$ – Prof. Meow Meow Feb 16 '11 at 21:19
  • \$\begingroup\$ @davek, test your LEDs with a battery and resistor (or a variable-current supply if you have one) to see if their brightness is sufficient for you. \$\endgroup\$ – markrages Feb 17 '11 at 3:29
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    \$\begingroup\$ If so, then you will need to add a resistor across the LEDs to take 4 mA current at something less than their forward voltage. But this will steal some current at full brightness too, leaving maybe 15 mA for the LED. \$\endgroup\$ – markrages Feb 17 '11 at 3:30
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    \$\begingroup\$ You don't need the series resistor. In a voltage driven system its function would be to limit the LED current, but here the current is the given, and the voltage is just a consequence of the current passing through the LEDs. \$\endgroup\$ – stevenvh Jun 13 '11 at 10:38
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If we assume 2 V red LEDs, then 4 of them in series (if that's what you want) is 8 V. You don't want the 4 mA through the LEDs as they would always light (though faintly). So use a 8 V/4 mA = 2 kΩ resistor in parallel with the LEDs. This will bypass the 4 mA. Only if the current gets higher than 4 mA the voltage drop will be large enough to pass current through the LEDs as well. The resistor will still draw 4 mA, so the LEDs will have 0 to 16 mA.

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