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Running LED on 1.2V rail

I have a board I am working on, and 3 voltages i'm working with: 3.3V, 2.5V, 1.2V. I want to have "Power OK" LEDs for each voltage rail. A look on digikey shows me that 3.3V and 2.5V wont be a problem, but there are no LEDs with a voltage drop less than 1.2V... and if there is, theyre probably expensive. So what is a simple way to have a "Power OK" led for 1.2V rail? p.s. I'm using all 0603/surface mount stuff.

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whatsisface7 years ago
I'd use an OP-AMP comparator. Put a potential divider with sutiable resistor values to give VOUT equal to about 1.1 Volts between the 2.5V rail and 0V. Use this as the inverting input to the circuit and then use the 1.2V as the non-inverting input. Put the LED on the output of the OP-AMP and it'll light up when the 1.2V rail voltage is higher than the 1.1V fixed voltage from the potential divider.

The only problem I can see is that I'm not sure whether OP-AMP's are produced that can run from a single 3.3V rail.
samurai1200 (author)  whatsisface7 years ago
So I tried modeling this, but my output won't hit the negative supply rail, it gets stuck at ~1.5V. What am I doing wrong?
comparator-nothittingground.JPG
Ah yeah, that's the issue with op amps. They only ever reach +/- about 2V of the positive and negative rails. I think the transistor method below will be better.
samurai1200 (author)  whatsisface7 years ago
HAH yeah that was dumb of me... I forgot about rtr opamps...
Use an opamp with rail-to-rail output such as the Microchip MCP6041

http://ww1.microchip.com/downloads/en/DeviceDoc/21669c.pdf
øPossum7 years ago
Simple transistor circuit:
led1_2v.png
samurai1200 (author)  Ã¸Possum7 years ago
Here's my issue with this. Most BJTs dont turn on "hard" (saturation) at 1.2V. Do I have to calculate where in the linear area that this BJT is to figure out how much current is going through from C to E? Normally, this wouldn't be an issue, but in this battery-powered application, a few milliamps matter.
samurai1200 (author)  samurai12007 years ago
actually, i guess i could just simulate it instead of doing math =)
that sounds like the best idea, remember the base resistor
There is no base resistor, only an emitter resistor. It is a constant current source.
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