Solve Thevenin & Diodes Homework with Ideal Diode

In summary, The original poster is trying to solve a problem using Thevenin's method but is getting a different output voltage than expected. They are asking for help in identifying any mistakes. Another user responds, pointing out that the Thevenin equivalent resistance has been calculated incorrectly and explaining the correct method for calculating it.
  • #1
roeb
107
1

Homework Statement



http://img444.imageshack.us/img444/5601/scan0001my.th.jpg

Hey, I'm trying to solve this problem using thevenin's method. As you can see in method 1 I was able to solve for the correct output voltage of 7.5 volts. When I try Thevenin's method I get 4 volts. I have a feeling that I'm missing something simple. Does anyone see what I've done wrong? Note that this is an ideal diode so there isn't a voltage drop over the diode.

Thanks,
roeb

Homework Equations





The Attempt at a Solution

 
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  • #2
roeb,

You don't have the correct value for the Thevenin equivalent resistance. Remember that to calculate this, after removing the load, replace voltage sources with wires (and if there are any current sources, replace them with breaks).

With the 15V source in your problem replaced with a wire, what kind of network do the two resistors present? I suspect that, because the resistors are in series to form a potential divider, you have seen them as being in series for calculating the Thevenin resistance. Not so!
 
  • #3


Hi Roeb,

I see that you have correctly solved for the output voltage using the first method. However, when using Thevenin's method, you must take into account the effect of the ideal diode on the circuit. Since the diode has a voltage drop of 0 volts, it essentially acts as a short circuit when it is forward-biased. This means that the 10kΩ resistor in parallel with the diode will have no effect on the circuit.

To solve for the Thevenin equivalent circuit, you must replace the diode with a short circuit and calculate the voltage across the 10kΩ resistor. This will give you the Thevenin voltage, which in this case is 5 volts. Then, you can calculate the Thevenin resistance by considering the 20kΩ and 10kΩ resistors in parallel, giving a value of 6.67kΩ.

Using these values, you can then solve for the output voltage using the Thevenin equivalent circuit, which should give you the correct answer of 7.5 volts. I hope this helps and good luck with your homework!
 

FAQ: Solve Thevenin & Diodes Homework with Ideal Diode

1. What is Thevenin's theorem and how is it used to solve circuit problems?

Thevenin's theorem is a method used to simplify complex circuits into an equivalent circuit with a single voltage source and series resistance. This equivalent circuit, known as the Thevenin equivalent, can be used to analyze the behavior of the original circuit, making it easier to solve for voltages and currents.

2. How do ideal diodes differ from real diodes?

Ideal diodes are theoretical components that have zero resistance when forward biased and infinite resistance when reverse biased. In contrast, real diodes have some amount of resistance, known as the forward voltage drop, when conducting in the forward direction and some leakage current when reverse biased.

3. What is the purpose of using ideal diodes in circuit analysis?

Ideal diodes are used in circuit analysis to simplify calculations and assumptions. By assuming that a diode is ideal, it eliminates the need to account for the forward voltage drop and leakage current, making the analysis more straightforward and accurate.

4. How do you apply Thevenin's theorem to a circuit with ideal diodes?

The process of applying Thevenin's theorem to a circuit with ideal diodes is the same as for any other circuit. First, you need to find the Thevenin voltage and resistance of the circuit. Then, replace the circuit with its Thevenin equivalent, including ideal diodes. Finally, use the equivalent circuit to solve for voltages and currents.

5. Can ideal diodes be used in real circuits?

No, ideal diodes cannot be used in real circuits as they do not exist in physical form. However, they can be used in theoretical circuit analysis to simplify calculations and assumptions.

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