Circuit with Diodes

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  • Thread starter Tekneek
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  • #1
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I have to find the V and I in the following Circuit.

5fg2gh.jpg


I am not sure how to approach this problem. If someone can help me with this step by step it will be great.

Assume Forward voltage drop of 0.7 volts
 
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  • #2
CWatters
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There are two diodes. You could start by assuming that each diode is either conducting/on or off. That gives you four combinations. Redraw each of the four circuits replacing diodes that are on with a short circuit and off with an open circuit.

Roughly calculate or estimate the voltages on the various nodes to see if they are consistent with the diode being in the state for that circuit. Some of the four combinations will turn out to be inconsistent with your assumption about the state of the diode and can be discounted.

Once you figure out which diodes are conducting/not it should be easier.
 
  • #3
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There are two diodes. You could start by assuming that each diode is either conducting/on or off. That gives you four combinations. Redraw each of the four circuits replacing diodes that are on with a short circuit and off with an open circuit.

Roughly calculate or estimate the voltages on the various nodes to see if they are consistent with the diode being in the state for that circuit. Some of the four combinations will turn out to be inconsistent with your assumption about the state of the diode and can be discounted.

Once you figure out which diodes are conducting/not it should be easier.
I forgot to add that we are assuming the diodes have forward voltage drop of 0.7 volts, would that change anything?
 
  • #4
NascentOxygen
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I forgot to add that we are assuming the diodes have forward voltage drop of 0.7 volts, would that change anything?
The approach is exactly the same.
 
  • #5
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The approach is exactly the same.
But what does it mean by it by
consistent with the diode being in the state for that circuit. Some of the four combinations will turn out to be inconsistent with your assumption about the state of the diode
How would you account for the voltage drops because of diodes if I make them short circuit and open circuit?
 
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  • #6
NascentOxygen
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Well, if you assume that a diode is non-conducting and analyze the circuit voltages based on that assumption, you may find that the voltage across that diode is such that it is forward-biased. If it is forward-biased, then your assumption that it is non-conducting must be wrong.

A realistic diode model is where the diode is represented by a fixed 0.7v drop, when conducting.
 
  • #7
berkeman
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So now you have enough information to write some equations. Please show us your work.
 
  • #8
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Well, if you assume that a diode is non-conducting and analyze the circuit voltages based on that assumption, you may find that the voltage across that diode is such that it is forward-biased. If it is forward-biased, then your assumption that it is non-conducting must be wrong.

A realistic diode model is where the diode is represented by a fixed 0.7v drop, when conducting.
I think I get it, but do i really have to do this every single time. Like making assumptions about 4 combinations?
 
  • #9
berkeman
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I think I get it, but do i really have to do this every single time. Like making assumptions about 4 combinations?
In general, yes, unless you can use your experience and intuition to judge whether the individual diodes are probably on or not. In this circuit it gets easier...
 
  • #10
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So now you have enough information to write some equations. Please show us your work.
Well I assumed both of the diodes were conducting. As a result, I find that the voltage across 5k resistor (on the right side) is lower than voltage across 10k resistor (on the left side). Since current flows from high potential to lower potential, current cannot flow through the diode in the middle from right to left. Meaning, the diode in the middle is off.

And I used voltage divider to find the voltages.
 
  • #11
berkeman
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Well I assumed both of the diodes were conducting. As a result, I find that the voltage across 5k resistor (on the right side) is lower than voltage across 10k resistor (on the left side). Since current flows from high potential to lower potential, current cannot flow through the diode in the middle from right to left. Meaning, the diode in the middle is off.

And I used voltage divider to find the voltages.
Good. Sounds like you got it right.
 
  • #12
CWatters
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How would you account for the voltage drops because of diodes if I make them short circuit and open circuit?
Start by assuming they are ideal diodes with no forward voltage drop. With luck the voltages across the diodes will be such that it's obvious which ones are on or off.

If the voltages across the diodes turn out to be close to the forward voltage then it's not so clear if the diode is on or off and the problem gets quite a bit harder. In some cases it's easier to model the circuit in an simulator than to try solving the maths yourself.
 

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