# Find the V1 value interval (V1min, V1max) for which D2 emits light?

I know I'm supposed to use Thevenin equivalent circuit, but can someone please guide me through the steps of solving this problem?

Edit: I have simplified resistors R1, R5 and R3 to a resistor R2 of 2K ohms. And I know that a minimum voltage of 0.65 needs to go through D2 and a minimum voltage of 6.2 needs to go through D1. I also now that I need to find the Thevenin circuit which I found to have a resistor with a resistance of 5K ohms in series with D2 and D1. Is that correct? What's the next step?

$$(R1+R5)//R3=\frac{1}{\frac{1}{3K+1K}+\frac{1}{4K}}=2K=R2$$
$$R_{TH} = R2//R4=\frac{1}{\frac{1}{2K} + \frac{1}{2K}} = 1K$$
$$V_{TH} = V*\frac{2K}{2K+2K}=\frac{V}{2}$$

When I reach here, I don't know what to do:

$$\frac{V}{2} - 0.65 - 6.2 - 2000i_0 = 0$$ This is the equation I get for the Thevenin circuit. Does having the $$i_0>0$$ make diode D2 emit light? Does negative voltage not work in this case since D2 would be in reverse-bias?

• I have simplified resistors R1, R5 and R3 to a resistor R2 of 2K ohms. And I know that a minimum voltage of 0.65 needs to go through D2 and a minimum voltage of 6.2 needs to go through D1. I also now that I need to find the Thevenin circuit which i found to have a resisotr with a resistance of 5K ohms in series with D2 and D1. Is that correct? – Kyron Calliste Jan 30 at 4:42
• Sorry to pint this out, but you are already using a simulator, test them by simulating. As it would take longer for some one to do the mathematics and check your solution. – jDAQ Jan 30 at 5:03
• I want to understand how to solve it analytically though, which is why I'm asking for help =) – Kyron Calliste Jan 30 at 5:14
• Ok, ok. first keep R4 where it is. Also, the Thevenin equivalent resistance seems way to high, even if you combined R4 into it. Redo the calculations, in fact, you can edit the mathematical steps in your post using latex, that way people can check if your steps are correct without having to do it from scratch. – jDAQ Jan 30 at 5:35
• @Kyron: Your question title should give a clue as to what the question is about. Can you edit it? Note that on SE you don't thank in advance, you thank afterwards by accepting and upvoting answers. – Transistor Jan 30 at 15:04

This looks like a homework question. I will offer some suggestions and pointers but I won't solve it for you. Modify your question as you work out the details and we will help you reach the answer.

1) Look closely at the resistors R1, R3, R5. Simplify those to a single resistor.

2) Observe that neither D1 or D2 will conduct at below a certain input (V1) voltage. Work out what that voltage is - that value is important.

Modify your question with those calculations.

The circuit is a bit messy and you might have missed that:

• $$\ R_1\$$ is in series with $$\ R_5\$$.
• $$\ R_3\$$ is in parallel to the series equivalent of $$\ R_1\$$ and $$\ R_5\$$.
• You could also incorporate $$\R_6\$$ into your Thevenin model.
• Finally you would have a Thevenin equivalent and the branch with $$\D_2\$$,$$\R_4\$$ and $$\ D_1 \$$

You are correct that LEDs only emit light if they are forwardly biased, so you must have $$i_0>0$$ for it to start conducting.
Regarding the equivalent resistance measured from the R4 terminal, consider that, if the diodes are conducting, the "resistance measured" would be just $$\ R_{th} \$$, even though it might not be as simple to measure such resistance (an ohmmeter wouldn't work in this situation), but you could empirically find that value by varying R4 and calculating how much $$\ R_{th} \$$ is.