Understanding Diodes in a Circuit

In Summary, the circuit has two diodes with one facing the current and one facing the opposite. It measures the voltage across the two resistors and then calculates the current if the diodes were reversed.
  • #1
piisexactly3
36
0
I wanted to post a similar looking circuit but I couldn't find any, so I'll have to describe it unfortunately.

Imagine a circuit where you have two resistors and you want to measure the voltage across each of those resisistors so you put a voltmeter parallel to each one. Except, instead of voltmeters, imagine diodes with one facing the current and the other facing opposite.

The source pd is 3V with negligable internal resistance, the forward facing diode is parallel to a 10Kohm resistor and the reverse facing diode is parallel to a 5k resistor.

We want to know the pd and current for each resistor and diode, and then what the pds and currents would be if the diodes were reversed

Homework Equations



V= IR, Kirchoff 1st and 2nd law, forward diodes have pd of 0.6V...

The Attempt at a Solution



the pd is 3V for each diode and resistor? (incorrect)
the pd is 0.6V for the forward facing diode (correct) so the pd for the parallel 10K resistor is 2.4V (incorrect)

I know that voltmeters have a very high resistance so that they can read a pd as close to what the resitor's pd is and so the diode with its very high resistance in the reverse would read the same pd as the parallel 5k resistor. However I still can't work that out as I don't know the current for the circuit, And I can't can't work out the current because I don't what the total resistance is because I don't know what the resistance of the forward diode is.

I've tried a lot of algebra but can't get anywhere.
 
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  • #2
Does the circuit look as that in the picture?

You can take that the resistance of the forward diode is zero, and the resistance of the reverse-biased diode is infinite.

ehild
 

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  • #3
The answer that my textbook gives for the current of the forward diode is 420 microA, so I calculate the resistance as 1429 ohms.
Hmm I think my textbook has made a mistake in giving me too many unknowns with not enough knowns.
 
  • #4
Oh and yes the circuit does look like that except the diodes are facing each other.
 
  • #5
piisexactly3 said:
The answer that my textbook gives for the current of the forward diode is 420 microA, so I calculate the resistance as 1429 ohms.
Hmm I think my textbook has made a mistake in giving me too many unknowns with not enough knowns.
The textbook is okay, there is nothing wrong with the question.

What characteristics are you going to assume about the diodes?

Does any current flow, for the arrangement you describe? Trace the current path.
 
  • #6
Ok I've got it.

I want to show my calculations because I'm not sure I did it the most efficient way.

Firstly I can see from Kirchoff's laws that the pds will be 0.6V for the forward diode and parallel resistor.
Therefore, the pd for the resistor and reverse diode will be 2.4V

Now the tricky part was getting the current. The current for the 10K resistor was simple as 0.6/10,000 = 60 microAmps. I don't know the current for the forward diode because I don't know its resistance. However I do know that the current leaving the junction is equal to the current leaving the junction for the other diode-resistor pair. Because the other reverse diode has no current, the current leaving the junction is simply 2.4/5,000 = 0.48milliA which also what the current through the parallel resitor is.

I can then use this equation V1/R1 = V2/R2 and plug in 0.48mA = 0.6/R2 . This then gives the value 1250 ohms.
Becuase I now know the total resistance leaving the junction, I can use 1/1250 = (1/10,000) + (1 Rdiode) . The resistance comes out as 1429 ohms for the forward diode which then gives me the current 0.42mA if use the 0.6V information.

it was fun getting there but I'm still not sure if I took the simplest route...
 
  • #7
Good, that's one way. But you didn't have to determne an equivalent resistance for the diode. 480uA - 60uA would have got it.
 
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  • #8
Doh! But thanks
 

Related to Understanding Diodes in a Circuit

1. What is a diode and what is its purpose in a circuit?

A diode is a two-terminal electronic component that allows current to flow in only one direction. Its main purpose in a circuit is to control the flow of electricity and to prevent current from flowing in the wrong direction.

2. How does a diode work?

A diode works by utilizing a semiconductor material, usually silicon, to create a one-way flow of current. When the positive terminal of a battery is connected to the anode (positive) side of a diode and the negative terminal is connected to the cathode (negative) side, the diode allows current to flow through. However, if the polarity is reversed, the diode blocks the flow of current.

3. What are the different types of diodes and how do they differ?

There are several types of diodes, including rectifier diodes, Zener diodes, Schottky diodes, and light-emitting diodes (LEDs). These differ based on their specific design and materials used, but they all serve the same purpose of controlling the flow of current in a circuit.

4. How are diodes used in practical applications?

Diodes have a wide range of uses in various electronic devices. They are commonly used in power supplies, voltage regulators, and signal processing circuits. They are also used in solar panels to convert sunlight into electricity and in LEDs to produce light.

5. Can a diode be damaged or fail in a circuit?

Yes, diodes can be damaged or fail in a circuit. This can be caused by excessive voltage or current, overheating, or manufacturing defects. It is important to choose the correct type of diode for a circuit and to ensure that it is not being subjected to conditions that could cause failure.

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