# resistor combination circuit

simulate this circuit – Schematic created using CircuitLab

This is probably a very stupid question. I am new to electrical engineering in general, I was working with NOT transistor gates in order to make a simple SR Latch and came across this case. D2 and D1 in my circuit are both pulled to ground even though when simulated D2 should still be fine the only reason I decided to ask about this is because I couldn't find anything on the internet about it (maybe I was using the wrong terms?). If someone could take a moment of their time to tell me why this doesn't work that would be amazing. The desired result is to have D2 be ON while D1 is pulled to ground and OFF.

It seems I lack understanding in the topic and I'm trying to get a better grasp on it.

• You are shorting out D1 with the wire connection that is between R3 and D1 and in the text 'L' – Aaron Jan 12 at 22:31
• If you remove R2 and R3 and add an ordinary diode in series to one of LEDs, you can achieve the following: when connecting the other LED to ground, the first will be 'off'; otherwise, it will be 'on'. Does that work for you? – Circuit fantasist Jan 13 at 2:37

## 1 Answer

R1 and R3 form a voltage divider.

This limits the voltage available to R2 to only 2.5 V.

The forward voltage of your LED is 2 V.

The remaining 0.5 V is dropped over R2, but this yields us a current of only:

$$\frac{0.5\ V}{4.7\ kΩ} =1\ mA$$ Which might not be enough to see it light.

Note that this is an over simplified way of calculating the voltages and currents (it's actually worse than I've shown). I am ignoring certain effects, but the overall point is, you are drawing current over those resistances which is limiting the voltage and therefore current available to the upper LED.

If you could do away with R1, and increase R2 and R3 to compensate, this could work.

• Modern LEDs? You'll see it light at 1mA, indoors at least. Even 0.1mA :-) in dim light. – user_1818839 Jan 12 at 22:42
• He will actually get less than 1 mA...because as soon as the LED starts to conduct, even more voltage will be dropped across R1. – evildemonic Jan 12 at 22:43