# Homework Help: Which Diode will turn on?

1. Feb 15, 2016

### Marcin H

1. The problem statement, all variables and given/known data

2. Relevant equations
V=IR

3. The attempt at a solution
I'm not sure how to figure this problem out. At first I was thinking that the diode on the left would turn on because that segment of wire has less resistance than the wire with the second diode. Current flows where there is least resistance, so it will flow to the left one, but I don't think that is right. I know that after the 1Kohm resistor I will have 5mA of current flowing and splitting into the 2 wires, but I'm not sure if that helps.

2. Feb 15, 2016

### Staff: Mentor

Your instincts are right, but the full answer to the problem depends on what the datasheet characteristics of the diode are and what is meant by "ON" in the question.

In general with problems like this, you can start by assuming that one diode is ON, and work out the voltages and currents. Then assume the other diode is on, and do the same. Then assume both are ON, then assume both are OFF, etc.

The problem here is that even if the left diode is ON, the right diode may still be OFF or ON, depending on the datasheet curves for the V-I characteristic, and what is considered ON versus OFF on that curve...

3. Feb 15, 2016

### Marcin H

IT doesn't say, but I think I can use the offset ideal model for a silicon diode where Von = 0.7V. That's what I used for other problems. I'm still not sure how I could use that to check if they would be on or off though.

4. Feb 15, 2016

### Staff: Mentor

Well if the left diode is ON with 0.7V forward drop, and the right one is ON, then the right one will have less than 0.7V across it, so it will be OFF. But then both diodes have 0.7V across them, so the right one is ON, but then...

5. Feb 15, 2016

### Marcin H

So are you saying that only one of the diodes can be? Also, why will the right one have less than .7V across it? Don't the resistors have an effect on the diodes?

6. Feb 15, 2016

### Staff: Mentor

In a real circuit with real diodes, both would be ON at some level. Use a real diode datasheet like for a1N4148 to calculate how "ON" each diode is (what is Vf and If for each?). Then look at the schoolwork problem and ask what do they give you for the equivalent V-I curves for the diodes in this problem? If they suggest that you use ideal diode models, then the problem is poorly constructed, IMO.

7. Feb 15, 2016

### Marcin H

I have one example in my notes, but that one doesn't make much sense either. I'm only given Von in this as well.

Red LEDs: Left LED on and right 3 are off. I=(6-2)/100=40 mA
I'm not sure how this makes sense or what the current really tells you...

8. Feb 15, 2016

### Staff: Mentor

What is Von for the left LED? If you put that voltage across the 3 right side LEDs, what will each of their Vf values be? Will that be enough to turn them on? This example is very different from your original post -- can you say why?

9. Feb 15, 2016

### Marcin H

All the LED's are at 2V in this example. I think the reason was that the left one will be on because there are less diodes to power. The 3 diodes on the right add up to 6V, so the voltage source which is also at 6v would not be able to power all those, so it will only power the one on the left.

10. Feb 15, 2016

### Staff: Mentor

A better way to think about it is that if the left LED is on, that will clamp the voltage across the right 3-LED string to 2V. That only gives 2V/3 across the right LEDs, which is not enough to turn them on.

11. Feb 15, 2016

### Marcin H

Hmm. Then how can we look at the original problem? Both LED's are at the same voltage, so would only one LED be able to run at a time?

12. Feb 16, 2016

### Harrison G

If Von on the right diodes is 2 volts, that means you are going to need 6 volts at the top of the right diode circuit. That upon the fact you have a resistor before them and that u have anotuer diode means that tje right leds wont light up

Last edited by a moderator: Feb 17, 2016
13. Feb 16, 2016

### Harrison G

By the way sorry for the
Imagine it like this:Each diode needs 2v to turn on. You have 3 diodes which means you need 6v to power all of them. But you have a resistor. Now, imaginr that in the verry first moment the diodes didnt see the resistor and they give some current. When that current reaches the resistor, the electrons are slow down. Through coulombs force these electrons reppel those behind them and they are also slowed down. In that point between the diodes and the resistor act two forces. One of atraction(that of the power supply attracting electrons) and the other is the other force is coused by the electrons who arr slowed down by the resistor and they reppel the electrons attracted by the power supply. The two forces are opposite of one another. Litteraly the voltage at that point fall down below 6 volts and there is a lack of voltage that is needed to pull any further current from the diodes. Sorry if its hard to understand, its a bit weird :-D, but that is how i explain stufff

Last edited by a moderator: Feb 17, 2016
14. Feb 16, 2016

### cnh1995

I believe you mean diodes.. There are no LEDs in the OP..
If the turn-on voltage of both the diodes is same,only one diode (the middle one) will be turned on.

For the other diode to be turned on, its turn on voltage should be less than that of the middle one.
For the exact analysis, diode equation is needed, which will increase the complexity of the analysis.

15. Feb 16, 2016

### Marcin H

So do you use KVL to figure that out? The turn on voltage is the same for both, I think, at .7V. Using KVL I noticed that in the loop of the left box you would have a voltage drop of 4.3V across the 1kohm resistor. So you would only have .7V travelling towards the junction after the 1kohm resistor and since the left diode has less resistance than the right one, it will light. The diode on the right has a resistor in series with it and current flows where there is least resistance. So left one will turn on. Is that the right way to think about this?

16. Feb 16, 2016

### cnh1995

For the rightmost diode, 0.7V will be the total voltage across the branch. For the diode to turn on, you need 0.7V across it alone.

17. Feb 16, 2016

### cnh1995

For the rightmost diode, 0.7V will be the total voltage across the branch. For the diode to turn on, you need 0.7V across it alone.

18. Feb 17, 2016

### Marcin H

Right so you would have enough voltage to power only one of them? and that .7V after the 1kohm resistor goes to the left diode because there is less resistance there?

19. Feb 17, 2016

### cnh1995

Yes. If there weren't that 50 ohm resistor, both the diodes would be turned on.

20. Feb 17, 2016

### cnh1995

But this is true only when the diode voltage is "just enough" to turn it on, i.e. below that voltage, the diode will be off. It can be predicted only by using the diode equation. In this case, you have assumed the diode has "just turned on" at 0.7V. If the turn on voltage of the diodes were say 0.6V and the actual voltage across the middle diode were 0.7V, then both the diodes would be turned on.

21. Feb 18, 2016

### Staff: Mentor

In my opinion, any analysis of this circuit is flawed if it boldly concludes that “only one diode will be ON”. The resistor joining the pair is only 50 Ω.

22. Feb 24, 2016

### Merlin3189

May it be that this question is there to make you realise that diodes aren't all or nothing on/off devices, as Berkeman pointed out early on?

For any real diode, if both are the same, both must be on. Once one diode (the LH) conducts, then the pd across the other must be enough to make it conduct. The only way the pd across the second diode (RH) can be less than the pd across the first diode is if it conducts! Otherwise there is no voltage drop across the 50Ω resistor. (This is just restating Berkeman.)

If you had a single diode, you could draw a 1k load line (0V,5mA to 5V, 0mA) on the I-V graph for the diode to find the operating point.
The second diode with the 50Ω resistor will have an I-V graph offset by IR (ie 50xI) on the voltage axis.
Since the current is then potentially shared between them, you can draw a load line for the average current (0V, 2.5mA to 5V, 0mA equivalent to 2k for each of the parallel diodes.) The operating point is then the voltage where the LH diode current is the same ΔI above the load line as the RH diode is below it. A little messier than a simple load line, but possible if you can construct the graphs.