So I have two same DC/DC step down converters and I'd like to have a green LED driven with the converters' open drain PGOOD signals. The signal is active-high and the LED should be on, when the power is okay and off, when not.

So I came up with this design:

enter image description here

In the converter's datasheet they use a 100k pull-up for PGOOD. I only used a smaller one because I was simulating this circuit in LTSpice and got too small current through the LED.

So when the power isn't okay, the PGOOD is connected to GND and the converters should sink current and the NPN won't conduct - the LED should be off.

And when the power is okay, the PGOOD is Hi-Z so the NPN will conduct and the LED should be on.

Will this work as expected? And if yes, is the design done optimally?

  • \$\begingroup\$ Based on my personal preference a failure LED is more helpful (usually you see if the device is working, but don't know where the error is, if it doesn't work), which could be implemented even easier. \$\endgroup\$
    – Arsenal
    Sep 5, 2018 at 16:30
  • \$\begingroup\$ Yes, it should work. You could connect 12V to 3K3 to LED to GND, then PGOODs to the LED anode. When either power isn't good, the LED anode voltage will be too low to glow. \$\endgroup\$
    – TonyM
    Sep 5, 2018 at 16:33
  • \$\begingroup\$ Thanks for the suggestions. You can post schematics in an answer, I'll upvote. \$\endgroup\$
    – Luka
    Sep 6, 2018 at 10:07

2 Answers 2


It will work, but not well. Your PGOOD pullup resistor is way too large.

Let's start with BJTs and gain. A BJT (Bipolar Junction Transistor) will display current gain in your configuration. In some circuits, you want a collector voltage somewhere near the middle of the supply voltage (6 volts here). This is called "linear mode" and is what you use to make an amplifier. In this case, you might well get a gain (collector current divided by base current) of 100 or so.

But. That's not what you want to here. You want to turn on the transistor hard, so that as little voltage as possible appears at the collector. This is called being in saturation. Basically, this term applies when the collector-emitter voltage is less than the base-emitter voltage, usually about 0.6 volts.

In saturation, the gain of the transistor is much less than when active, and a standard ballpark figure for gain is 10. In your circuit, since you have a 1k LED resistor, 12 volt supply, and an LED voltage drop of 2-3 volts, you clearly want about 10 mA in the LED. This means that you want about a 1 mA base current.

Base current will be set by the PGOOD resistor, which will see about 2.6 volts (3.3 minus the 0.6 of the base-emitter junction). So you want something like a 2.6 kohm resistor. A standard 2.7 kohm would be an excellent choice.

  • \$\begingroup\$ Thank you very much for the extensive explanation! What I don't quite understand is how you make the transistor go into saturation? By choosing the base and collector resistors of the right size? \$\endgroup\$
    – Luka
    Sep 6, 2018 at 10:15
  • 1
    \$\begingroup\$ @Luka - The collector current is determined by the supply voltage and the load resistance, since if the transistor is in saturation the emitter-collector voltage will be very small (less than .5 volts). Then you set the base resistor to give a base current about 1/10 of the collector current. Note that this is not a precision calculation, since the 1/10 is, as I stated, "a ballpark figure", and 20% either way will certainly work fine. \$\endgroup\$ Sep 6, 2018 at 13:51
  • \$\begingroup\$ Yes, I understood your saturation gain estimate, I just didn't get when transistor goes in saturation. But it's getting clear now. Thanks! It depends on voltages, V_b > V_c > V_e, right? And then you set the I_be to be 1/10 of I_ce \$\endgroup\$
    – Luka
    Sep 6, 2018 at 15:04
  • 1
    \$\begingroup\$ @Luka - Don't get carried away with worrying about saturation. For switching purposes you want Vce to be as small as possible so that the power dissipated in the transistor (as opposed to the load) is minimized. At a gain of 10, most transistors will have a Vce of 0.2 to 0.4 volts. Unless the current is too large for the transistor, which is why you should read the data sheet. Otherwise, don't worry about what mode you're in. \$\endgroup\$ Sep 6, 2018 at 15:52

The base current for your 2N5550 is only 27uA with a 100k Ohm pullup to 3.3V.
This probably will work, but you have no idea what the leakage current of the PGood signal is, or the actual Hfe of the transistor, so I'd be hesitant. For example you might find it does not work when cold when the transistor Hfe is at a minimum.

If you want to run your LED from 12V then you could do this:


simulate this circuit – Schematic created using CircuitLab

The PGood signal is able to be used up to 16V so you should have no problem at 12V. Almost any generic Nchan FET would do providing it's VGS will withstand 12V, I showed the 2N700 here, which will withstand +/-20V.

  • \$\begingroup\$ Thanks! I accepted WhatRoughBeast's answer because of a better explanation. Could you please explain why you chose a FET instead of a BJT? Is it because of higher voltages? \$\endgroup\$
    – Luka
    Sep 6, 2018 at 10:02

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