I am intending to make an indicator led circuit which shows the negative 9 V supply is coming to the circuit and led becomes on, but if there is zero volts, the led will be off. I have done this circuit with pnp BJT however i know that if there is 0 volts at the input, led will still on. Do you have any idea about how i can reach my aim? Thank you so much.

This is my circuit;

enter image description here

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    \$\begingroup\$ Put the LED and series resistor directly across the negative supply? \$\endgroup\$ – AaronD Jul 31 '17 at 18:36
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    \$\begingroup\$ Are you looking for a specific indication? Do you need a full-on LED only at a specific voltage threshold (and hysteresis?) Or is just a varying brightness as the negative supply rail "comes up" good enough? \$\endgroup\$ – jonk Jul 31 '17 at 18:38
  • \$\begingroup\$ I just need to see that there is some negative voltage without a treshhold actually. I will not work too much about magnitude of the input. But at least, if there is not a connection and zero volts at the input, it can distinguish this situation and negative voltage. Thats my purpose. \$\endgroup\$ – layout789 Jul 31 '17 at 18:42

I am intending to make an indicator led circuit which shows the negative 9 V supply is coming to the circuit and led becomes on, but if there is zero volts, the led will be off.

I am understanding that you simply want an indicator to show that there is a negative voltage available. The LED configuration below will accomplish this.

When the source V1 reaches the Forward Voltage of the LED it will illuminate, and when V1 is 0V the LED will be off.


simulate this circuit – Schematic created using CircuitLab

Let's look at an implementation example:

In the circuit below the +5V is DC and can be supplied from any source, i.e. a bench supply or wall wart (but an Arduino could have trouble sourcing enough current.)

The Murata MEV1S0509SC is an isolated DC-DC Converter. It is a switching converter that will convert +5VDC to +9VDC. Because it is isolated it is completely acceptable to connect the positive output to ground and use the negative output as shown.

enter image description here

Now for the LED. I picked an HLMP-3301 which has a forward voltage of 1.9V typical. A diodes forward voltage is the voltage required for current to flow. Sparkfun has a good artical about it here: Sparkfun - Diodes.

enter image description here

Operation: When -Vout is "ON" (less than 1.9V below ground) the LED will have a forward voltage applied (0v - -Vout) and the LED will illuminate. The resistor is there to limit the current through the LED so it doesn't burn up. In this example, I picked a 10mA current at -9V Vout.

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    \$\begingroup\$ Except the LED is backwards.... \$\endgroup\$ – John D Jul 31 '17 at 19:18
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    \$\begingroup\$ Oh my... You are right. I will fix it now. \$\endgroup\$ – Scott Geier Jul 31 '17 at 19:32
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    \$\begingroup\$ There we go. It should be correct now. Sorry about that, Monday and all that. Thank you for the comment. \$\endgroup\$ – Scott Geier Jul 31 '17 at 19:36
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    \$\begingroup\$ Messing up the LED polarity is exactly the kind of error that ordering the power voltages in a schematic + to - top to bottom helps avoid. Instead of flipping the LED from the original, you should have flipped everything else. Ground should be at top, -V1 at bottom. \$\endgroup\$ – Olin Lathrop Jul 31 '17 at 20:41
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    \$\begingroup\$ @layout789 That looks like it will work. One note: You are correct when your supply is at -9V, the current flowing through your LED will be (9V - Forward Voltage)/270 = 22mA. However, you may be able to reduce that current (seldom do I run my LEDs at their rated current value) it just depends on the brightness you want. This will give you a place to start, and adjust accordingly. Good luck. \$\endgroup\$ – Scott Geier Aug 1 '17 at 13:47

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