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I have a 3.3v circuit with a device that indicates Power Good with a single open-drain pin. When there is a fault, PG is actively pulled low; when normally operating, it is high-z floating.

I'd like to use a single bi-color SMD LED, either independent or bi-directional. Something like Dialight's 5988621207F or similar.

Small footprint is generally optimal.

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  • \$\begingroup\$ How much current does the PG pin can sink and how much current you want to drive through the LEDs? We cannot know an optimal solution to a puzzle with unknown variables. \$\endgroup\$ Commented Feb 11 at 19:03
  • \$\begingroup\$ I looked on the sheet, and I thought it specified IpgMax, but didn't find it. I have two different devices, but they both have the same PG configuration. ti.com/lit/ds/symlink/tps25983.pdf ti.com/lit/ds/symlink/lmzm23601.pdf \$\endgroup\$ Commented Feb 11 at 19:29
  • \$\begingroup\$ One of the chips do define absolute maximum PG current. However, you never ever want to go near absolute maximum ratings. \$\endgroup\$ Commented Feb 11 at 19:36
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    \$\begingroup\$ There was a discussion recently - possibly here - in which a user recounted their struggles with a red/green status LED that made no sense to them. They were colourblind. So if this is going to be used by other people, take care. \$\endgroup\$ Commented Feb 12 at 9:55

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Add one small MOSFET (BSS123, or BSS138, etc) and two resistors. Note that the VGS threshold of the MOSFET must be less than the forward voltage of the LED.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Could you explain a bit how this works in the two states? I can't seem to get the simulation to work like I expected it would. I assume the FET on the left is the open-drain, right? \$\endgroup\$ Commented Feb 11 at 19:25
  • \$\begingroup\$ This seems right: snipboard.io/WLRdZC.jpg, using sunledusa.com/products/spec/XZMDKVG55W-4.pdf and assets.nexperia.com/documents/data-sheet/PMH600UNE.pdf. Can you double check me? \$\endgroup\$ Commented Feb 12 at 1:45
  • \$\begingroup\$ @Drise Your proposed combination of parts would seem to work. The PMH600UNE MOSFET has a threshold of 0.95V max and the LED forward voltages are around 2V depending on color. So, the FET will be able to turn on and so will the LEDs. \$\endgroup\$ Commented Feb 12 at 3:47
  • \$\begingroup\$ Are you sure the threshold is the critical parameter here? The FET needs to be significantly conducting at a lower voltage than the LED's forward voltage, not just conducting 10 μA or whatever this FET defines its threshold at. \$\endgroup\$ Commented Feb 12 at 3:56
  • \$\begingroup\$ The sheet says at 1.2V it's typically 1500mOhms RDSon, also mentioned at 1mA, which should be enough? \$\endgroup\$ Commented Feb 12 at 5:09
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Both chips should be able to sink around 5mA. That's plenty. Modern LEDs are extremely bright even at 1mA.

Depending on the LED, you may have different options. A red LED will have lower forward voltage than green LED. If you connect anodes of both LEDs to a common resistor, tie green LED cathode to ground, then you could simply drive the red LED cathode with the PG pin. When PG goes low, red LED has lower voltage over it, so green LED will not have enough voltage over it.

You could also put red LED and it's resistor to be driven by the open-collector output. And then have the output to pull green LED anode to 0V via green LED pull-up too.

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Drive one LED directly off the PG pin;

Drive the other with a BJT/FET inverter -- essentially making another open-drain output that is inverted.

If one of your indicators can be very-low current:

Drive one LED directly off the PG pin;

Drive the other, in the opposite polarity, from the pullup resistor.

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