# Resistor and diode in parallel

Discussion in 'Homework Help' started by afaik, Nov 2, 2008.

1. ### afaik Thread Starter Member

Nov 2, 2008
22
0
Hi, sorry to have to be asking such a simple question, but I've spent too much time on it and need to move on so I can finish this lab. What I have is a circuit with 5v connected to a 2.2KΩ resistor. To the 2.2kΩ resistor is connected a Si diode and a 1k resistor in parallel. What I need to find is Vo, Vr1 and Id.

What I do know is that when you have two parallel diodes, the diode with the lower turn on voltage would be the one with all the current flowing through it. If it's a Si and Ge in parallel, the Germanium would be turned on and the Si would be considered open. But now that we replace one of the diodes with a resistor, I'm not sure where to go. My book isn't much help and I'm having difficulty looking for information online. I did find this forum though so if someone can help me that would be great.

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2. ### Ron H AAC Fanatic!

Apr 14, 2005
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Are you to assume that the diode has fixed forward voltage drop Vf (e.g., 0.7V), or do you have a more sophisticated model? If your model has fixed Vf, then calculate Vo without the diode. If it is more than Vf, then you can assume that the diode is conducting, and Vo is equal to the Vf of the diode. Subtraction will give you Vr1, and Ohm's law will give you Id (Id=Ir1-Ir2).
If your diode model is more sophisticated, tell us what it is and we can take it from there.

3. ### afaik Thread Starter Member

Nov 2, 2008
22
0
I should have wrote 'what I am familiar with so far" instead of "What I do know", I really don't know much yet. The model in the attachment is the one I am working with. Simple to the experienced, daunting to those who aren't.

I wired up the diagram and have found some numbers to help me.
Like you said, my diode is conducting and Vo is equal to the voltage drop across Vd but I'm still not sure how I can get that by doing calculations only.

When you say to subtract to get Vr1, you mean Vr1 = E-Vd = 5 -.6 = 4.4V.

And to find Id you mean (4.4/2200) - (.6/1000) = .002 - .0006 = .0014A Id = 1.4mA

If this is right, I think I've got most of it down.

Last edited: Nov 2, 2008
4. ### afaik Thread Starter Member

Nov 2, 2008
22
0
Unless it's as simple as using KVL which in this case would yield:

E - Vr1 - Vd = 0
5v - .6v = Vr1 = 4.4V
Vr1 = 4.4V

E - Vr1 - Vd = 0
Vd = Vo = E - Vr1
5 - 4.4 = .6V
Vo = Vd = .6V

A bit redundant but it works.

5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
I think you've got it.

6. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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Yep. A forward diode makes a lousy zener...but for math purposes, it works.

eric

7. ### afaik Thread Starter Member

Nov 2, 2008
22
0
I was looking at a similar problem recently and came back to this thread to clarify my confusion, but I realized, while I know now what happens, I still don't know why they diode and the resistor sync up. For example, if you have in parallel an Si diode and a resistor and apply a voltage to the circuit, why would the voltage across be the voltage of the diode and not the applied voltage?

8. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,393
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The voltage in a parallel circuit must be the same.

The forward biased diode (conducting) controls the voltage in the parallel circuit. If the diode were reverse biased (not conducting), then the resistor would control the voltage in the parallel circuit. See the attachment.

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9. ### afaik Thread Starter Member

Nov 2, 2008
22
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Alright, I think I'm almost there, but I need to pick at what you're saying for a sec.

What I need to understand from this statement is why the conducting diode controls the voltage in parallel. I understand that a reverse biased diode is the equivalent of an open circuit, and so according to your diagram, in the second example there is no parallel circuit, just R3 and R4 and series. But if we short R1 and have 5 volts across both the diode and the resistor, why does the diode force not only the voltage across the parallel elements to ~.7V, but the whole power supply to drop down from whatever voltage it is switched to, to ~.7?

I know I seem to be dragging this on so if the answer is right in front of me and I still don't see it, I'll drop the issue.

10. ### Distort10n Active Member

Dec 25, 2006
429
1

Because a practical lab supply is designed to source or sink a maximum current. If you think about an ideal diode model, there is no voltage drop and the diode is an ideal short when conducting. What happens when you short out a supply? The voltage will drop to 0 but the current tends toward infinity.
Practically speaking, there will be a very small series resistance in the supply and a diode drop to deal with so the current will not be infinity but will be high, perhaps pulling the maximum current from the supply.

11. ### Ron H AAC Fanatic!

Apr 14, 2005
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Either the power supply's maximum output current is less than the maximum current the diode can handle, or the diode goes up in smoke.

12. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Maybe this will clarify it for you ...

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13. ### afaik Thread Starter Member

Nov 2, 2008
22
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I think I got it. Because once the diode reaches .7ish, it's considered a short circuit and -correct me if I'm wrong- very little or no current flows through the resistor. So although the resistor has a voltage across, it does not have current going through it and turns to a short, the reason the parallel voltage dictates the voltage. Hopefully I've finally got it, so thank you all in advance for helping me jump over a hurdle that's been holding me back. If the concept isn't right though, I'll keep at it...

14. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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If a resistor has voltage across it, it will have current flowing through it. A diode does not magically invalidate Ohm's Law.
Look at the attachment and see if it helps.

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