I'm getting close to finishing my brand spanking new gate control design. I sincerely appreciate all the help I've received out here.

My latest challenge is to extinguish an LED when a circuit is grounded. To do so, I require a simple transistor inverter. However, I could not find anything that uses a ground signal. So I had to come up with one.

The parameters of the circuit are that the input signal has to be a ground, the circuit power has to be 12V, and I am trying to stick to transistors (I could probably use CMOS, but I'm trying to design the entire thing using transistor logic - it's been a great learning experience). This inverter is also powered by solar/battery, so current draw (particularly when active and LED is out) is important. The best I can do is 213.26 uA. Here is my design:


(Simulation link)

All 3 transistors seem to go into saturation mode when they are forward biased, and go into cutoff mode when they are not. I'd like to reduce the current draw, but I don't think it's possible and still maintain exactly 10 mA on the LED when it is lit. Is there a simpler design that I am missing? Am I overlooking the obvious (yet again)?

As always, any help is sincerely appreciated. I can't wait to finish this design and start building it.

  • \$\begingroup\$ tinyurl.com/y9lkrcnh \$\endgroup\$ – Rikki Jan 8 '19 at 2:25
  • \$\begingroup\$ There's probably a reason you didn't do this, but tinyurl.com/yc8tavfo ? \$\endgroup\$ – user253751 Jan 8 '19 at 2:40
  • \$\begingroup\$ Pulling 11.75 mA in off state. \$\endgroup\$ – Rikki Jan 8 '19 at 2:48
  • \$\begingroup\$ Although, thinking about it, seems like a lot of work to save the equivalent power consumed by a single LED. You have a point, and that's why I posted. \$\endgroup\$ – Rikki Jan 8 '19 at 2:56

I think there are two transistors too many. Most people will say that 70K is a bit high for the base resistor in this application, but we can leave that alone for now. Here is a more simple solution.


simulate this circuit – Schematic created using CircuitLab

Is there a reason you are not using a saturated switch to drive the LED? Also, why is exactly 10 mA LED current important?

  • \$\begingroup\$ Sorry, I don't know what a saturated switch is. I'll have to look it up. But in the meantime, you once again came up with an elegant (and INFINITELY better) solution. This is precisely what I was looking for. I tend to over complicate things. Your circuit is perfect (and uses less current). \$\endgroup\$ – Rikki Jan 8 '19 at 3:39
  • \$\begingroup\$ Oh sorry, forget to add.......I've seen circuits where the current used by different LED's is slightly different, and it drives me nuts because I notice it. I'm shooting for 10mA, and if I can just make sure I get close enough to not notice the difference, I will be happy. Right now I'm just simulating. -smiles- \$\endgroup\$ – Rikki Jan 8 '19 at 3:41
  • \$\begingroup\$ @Rikki "Saturated" means the transistor is "completely on", so that adding more base current won't turn it on any more. \$\endgroup\$ – user253751 Jan 8 '19 at 4:16
  • \$\begingroup\$ You mention "saturation" in your question. Same thing. When the intent of a transistor circuit is to go from full on to full off without operation in the linear active region between, the application is called a saturated switch. \$\endgroup\$ – AnalogKid Jan 8 '19 at 14:23
  • \$\begingroup\$ Got it. The circuit I provided went from cutoff to 30% of the way into the saturation region of operation when forward biased. I was confused by the term "saturation switch" since this is what my circuit was doing (and yours too). This is fun and I sincerely appreciate all your help. \$\endgroup\$ – Rikki Jan 8 '19 at 19:04

You are correct (and I was in error), your Q3 and my Q1 are operating in the saturation region, but barely. The rule of thumb (from the 1950's) is that the base current should be no less than 10% of the collector current for hard saturation. I think 5% (20:1) is a more reasonable number for today's parts. The problem with this is that your off-state circuit current is now higher.

To fix that, change Q1 to a small MOSFET such as the 2N7000 or 2N7002. Now the gate pullup resistor can be 100K or even 1 M, reducing the off-state current through the circuit. For the high resistor values I would add a small noise filter cap such as 100 nF or 1 uF from the gate to GND.

  • \$\begingroup\$ I see it as a different answer to the same question, albeit one that is very similar to the first answer. The FET can make for a significantly lower off-state current. \$\endgroup\$ – AnalogKid Jan 9 '19 at 4:10

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