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I've got a bi-color LED in a push-button that is red when powered in one direction and green in the other direction. I would like to control it from a Raspberry Pi GPIO pin, e.g. High=Red, Low=Green.

I thought I could use some kind of H-Bridge to reverse the polarity, something like this perhaps?

enter image description here

The idea is that when input is High the BC547 is open and FET B and C are open providing polarity in one direction. When the input is Low the BC547 is closed and the FETs A and B are open providing polarity in the other direction.

I want to power it from Raspberry Pi's 5V pins however the GPIO is only 3.3V (that I can level shift up to 5V if needed).

The LED doesn't need a resistor, the button has an internal one.

I've put 4x 2N7000 (N-channel MOSFET) in the schematic as I've got a bag of them but happy to replace them with BC547 / BC558 or anything else.

  • Is the overall idea correct?
  • Will the gate levels be high enough to open the FETs?
  • Will the voltage drops on the FETs will still allow the LEDs to operate?

Thanks!

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    \$\begingroup\$ Wouldn't you be better just getting a bi-coloured LED with a common cathode and just driving the LEDs individually, rather than needing all that extra circuitry. You could also just use a single NOT gate if you want to control them off one GPIO. \$\endgroup\$ – HandyHowie Oct 15 at 6:56
  • \$\begingroup\$ @HandyHowie The LED is built in the push button, I can't change it unfortunately. \$\endgroup\$ – KeepLearning Oct 15 at 9:35
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H bridges are for driving high current coils or motors. You have no current limiting R.

CMOS (74HC family ) has about 50 OHms driver R so you can include this with your current limiting R.

Since Iv intensity in xxxx mcd is common now , the curent can reduced to a few mA for indicators with a single Rs = 470 to 1k. schematic

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  • \$\begingroup\$ Note that you can do this with 2 x 5V inverters in my answer vs 1x 5V inverters in the accepted answer. But if you had 5V uC then only 1 inverter needed/ \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 16 at 22:52
  • \$\begingroup\$ Also why do you put an extra resistor for the red LED? I tried without, with just one shared one like the other answer shows and it worked fine. Using 74HC04. Thanks! \$\endgroup\$ – KeepLearning Oct 16 at 22:52
  • \$\begingroup\$ Since R/G has a different Vf the extra R adds the 1.3V+/-0.2 difference typ 2.1 to 3.4 but reality is choice of LED Iv from 100mcd to >10,000 mcd at 20mA means you can tune equal brightness by R or choice of LED but the RED will draw more current, then operate at lower current like 5mA.. But Iv efficacy differences can make it look equal \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 16 at 22:55
  • \$\begingroup\$ I see, that makes sense. Thanks for a quick reply! \$\endgroup\$ – KeepLearning Oct 16 at 22:57
  • \$\begingroup\$ So single R is adequate then choose brightest cheap LEDs that are roughly equal and run at low mA lens reducing angle 50% makes almost 2x brighter \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 16 at 22:59
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Here is how I drive bi-color LEDs:

schematic

simulate this circuit – Schematic created using CircuitLab

Use a logic family that can source or sink 20ma or whatever you need to drive the LEDs. I actually don't use inverters (that solution was closest to your circuit), but a serial to parallel shift register (like a 74xx591) to drive 4 bi-color LEDs using the 8 outputs. For two output pins you end up with 00 -> off, 01 -> red, 10 -> green, 11 = off.

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  • \$\begingroup\$ That's brilliant, thanks! Works like a charm with 74HC04 :) \$\endgroup\$ – KeepLearning Oct 16 at 22:48
  • \$\begingroup\$ Unfortunately I can't upvote with my reputation yet :( \$\endgroup\$ – KeepLearning Oct 16 at 23:06

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