Find thevenin equivalent resistance with diode.

[bnosam]

Homework Statement

I have a circuit with a diode in it. I have to find the thevenin equivalent in order to reduce the circuit to a DC source, single resistor and single diode. How do I account for a diode when trying to find Rth and Vth? Do I just remove it and have it act as a short circuit or do I open circuit it?

 

[NascentOxygen]

Hi bnosam, between which 2 points do you wish to model the circuit as its Thévenin equivalent?

Is the 2.5k part of the circuit to be modeled, or is it an external load? Quite possibly you are needing to model only what is to the left side of the diode, and not including the diode itself?

 

[gneill]

What is the load? Is the 2.5 kΩ resistor part of the circuit to be "Thevenized", or is it the load?

You can always work stepwise through a circuit to convert to a Thevenin equivalent. Start by breaking the circuit before the diode and convert that part first:

Edit: Whoops! NascentO got there first!

 

[bnosam]

Hi bnosam, between which 2 points do you wish to model the circuit as its Thévenin equivalent?

Is the 2.5k part of the circuit to be modeled, or is it an external load?

I'm assuming the section near the battery. I need to reduce the whole thing to a equivalent circuit with just a battery, resistor and diode. So whichever would get me to that the easiest would be best, so I can apply the real properties on a simplified model of the diode's circuit.

What is the load? Is the 2.5 kΩ resistor part of the circuit to be "Thevenized", or is it the load?

You can always work stepwise through a circuit to convert to a Thevenin equivalent. Start by breaking the circuit before the diode and convert that part first:
View attachment 106443Edit: Whoops! NascentO got there first!

So if I broke it off before the diode. I would have the 10V source and a simplified 10K resistor.

 

[gneill]

Consider the simplification I suggested first. After that you should be able to see the way forward.

 

[bnosam]

Consider the simplification I suggested first. After that you should be able to see the way forward.

Are you meaning like this?

 

[gneill]

Are you meaning like this?

No, but that might work also depending upon what your end goal is.

I was suggesting cutting the circuit as per the red sectioning arrows that I drew and simplifying everything to the left first. I didn't know if the 2.5 k resistor was important to what you intend to accomplish; You haven't told us any details of what you are trying to find.

 

[bnosam]

No, but that might work also depending upon what your end goal is.

I was suggesting cutting the circuit as per the red sectioning arrows that I drew and simplifying everything to the left first. I didn't know if the 2.5 k resistor was important to what you intend to accomplish; You haven't told us any details of what you are trying to find.

Then everything on the left would simplify to the 10V source and just a single 10K resistor, no? The reason I want to simplify it is to apply the real diode model's properties to find current the compute output voltage of the diode using iterative analysis.

 

[gneill]

Then everything on the left would simplify to the 10V source and just a single 10K resistor, no?

No. You should "cut" the circuit where indicated and do the analysis to obtain the Thevenin model. The two 5k resistors are not in series. Hint: It's a voltage divider. Here's the circuit re-drawn to make it more obvious:

You should convince yourself that this is in fact the same circuit as the given drawing, only the voltage source's connection to the ground (common) node is made explicit.

The reason I want to simplify it is to apply the real diode model's properties to find current the compute output voltage of the diode using iterative analysis.

By "output voltage of the diode" do you mean the potential drop across the diode, or the potential drop across the 2.5k load resistor?

If the former then perhaps your idea of finding the Thevenin model at the open terminals of the diode position would be appropriate. But you can always get there by doing the simpler voltage divider simplification first. As I mentioned before, you can do Thevenin simplification in stages, one part of the circuit at a time.

 

[bnosam]

No. You should "cut" the circuit where indicated and do the analysis to obtain the Thevenin model. The two 5k resistors are not in series. Hint: It's a voltage divider. Here's the circuit re-drawn to make it more obvious:
View attachment 106448
You should convince yourself that this is in fact the same circuit as the given drawing, only the voltage source's connection to the ground (common) node is made explicit.

By "output voltage of the diode" do you mean the potential drop across the diode, or the potential drop across the 2.5k load resistor?

If the former then perhaps your idea of finding the Thevenin model at the open terminals of the diode position would be appropriate. But you can always get there by doing the simpler voltage divider simplification first. As I mentioned before, you can do Thevenin simplification in stages, one part of the circuit at a time.

Ohh ok. I'm not used to drawing circuits in the manner that I presented it in, that's just the form my book uses. I struggle a lot with circuits in general. It makes sense that it is a voltage divider, though. So the voltage divider where R1 = R2 the output voltage is half of the input voltage. 10 * 1/2 = output from those two voltages would be 5 volts?

The potential drop across the 2.5K load resistor would be the output voltage that I am looking for.

 

[gneill]

Ohh ok. I'm not used to drawing circuits in the manner that I presented it in, that's just the form my book uses. I struggle a lot with circuits in general. It makes sense that it is a voltage divider, though. So the voltage divider where R1 = R2 the output voltage is half of the input voltage. 10 * 1/2 = output from those two voltages would be 5 volts?

Yes, the Thevenin voltage for the voltage divider would be 5 V. You still need to determine the Thevenin resistance.

The potential drop across the 2.5K load resistor would be the output voltage that I am looking for.

Ah. So then leave the 2.5 k resistor in place. You'll have a Thevenin model consisting of a voltage source in series with a resistance (R_th) in series with a diode and 2.5 kΩ resistor.

 

[bnosam]

Yes, the Thevenin voltage for the voltage divider would be 5 V. You still need to determine the Thevenin resistance.

Ah. So then leave the 2.5 k resistor in place. You'll have a Thevenin model consisting of a voltage source in series with a resistance (R_th) in series with a diode and 2.5 kΩ resistor.

So I would remove the voltage source and then with that voltage source set to short circuit, the two 5K resistors are in parallel and the Rth = 2.5k Ohms. Is that correct?

 

[gneill]

So I would remove the voltage source and then with that voltage source set to short circuit, the two 5K resistors are in parallel and the Rth = 2.5k Ohms. Is that correct?

Yes.

 

[bnosam]

Yes.

Then I have the source, diode and two resistors in series in the circuit?

So I can add the two resistors to make a 5k ohm resistor in series with the source and diode now?

 

[gneill]

Sure.

 

[bnosam]

Sure.

Thank you very much for you assistance!