Which diode will turn on in this circuit?

[Marcin H]

Homework Statement

Homework Equations

V=IR

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.

 

[berkeman]

Homework Statement

View attachment 95920

Homework Equations

V=IR

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.

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...

 

[Marcin H]

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...

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.

 

[berkeman]

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.

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...

 

[Marcin H]

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...

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?

 

[berkeman]

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?

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.

 

[Marcin H]

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.

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.

The answer is:
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...

 

[berkeman]

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.
View attachment 95928
The answer is:
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...

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?

 

[Marcin H]

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?

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.

 

[berkeman]

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.

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.

 

[Marcin H]

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.

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?

 

[Harrison G]

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.
View attachment 95928
The answer is:
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...

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 won't light up

 

[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 won't light up

By the way sorry for the

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?

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

 

[cnh1995]

Hmm.Both LED's are at the same voltage, so would only one LED be able to run at a time?

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.

 

[Marcin H]

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.

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 traveling 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?

 

[cnh1995]

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?

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.

 

[cnh1995]

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?

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.

 

[Marcin H]

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.

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?

 

[cnh1995]

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?

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

 

[cnh1995]

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

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.

 

[NascentOxygen]

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.

 

[Merlin3189]

May it be that this question is there to make you realize 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.