Better Transistor Inverter Using Ground Signal?

Question

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.

Answer 1

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?

Answer 2

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.