# Help Analyzing Circuit Using Ideal Diode Method

• Engineering
Hey guys, I see this website a lot when lot when looking for help so I made an account since I cannot find help with this problem anywhere.

## Homework Statement

Use the ideal diode model to analyze the circuit below (circuit (a) in the attachment). Start by assuming D1 is off and D2 is on.

## Homework Equations

Ohm's law, Voltage-division principle

## The Attempt at a Solution

The textbook has the answer (Electrical Engineering Principles and Applications 5th Edition, by Hambley, Example 10.5) but I do not understand it. The answer they have is the second attachment, and the circuits are in the first one.

I am having trouble with a few simple things. I understand that iD2 = 0.5mA from Ohms law, but cannot figure out why vD1 = +7V. With both those values, I am unsure why the first assumption (D1 off, D2 on), is incorrect, and why the fact that vD1 = +7V not consistent with the first assumption?

From there, how did they solve for vD2 = -3V with the second assumption (D1 on, D2 off), and why is that consistent with the assumption? I feel like this is a simple question but for some reason I find this confusing. Thanks for the help.

#### Attachments

• 35.9 KB Views: 583
• 92 KB Views: 446

## Answers and Replies

Related Engineering and Comp Sci Homework Help News on Phys.org
gneill
Mentor
Hey guys, I see this website a lot when lot when looking for help so I made an account since I cannot find help with this problem anywhere.

## Homework Statement

Use the ideal diode model to analyze the circuit below (circuit (a) in the attachment). Start by assuming D1 is off and D2 is on.

## Homework Equations

Ohm's law, Voltage-division principle

## The Attempt at a Solution

The textbook has the answer (Electrical Engineering Principles and Applications 5th Edition, by Hambley, Example 10.5) but I do not understand it. The answer they have is the second attachment, and the circuits are in the first one.

I am having trouble with a few simple things. I understand that iD2 = 0.5mA from Ohms law, but cannot figure out why vD1 = +7V. With both those values, I am unsure why the first assumption (D1 off, D2 on), is incorrect, and why the fact that vD1 = +7V not consistent with the first assumption?

From there, how did they solve for vD2 = -3V with the second assumption (D1 on, D2 off), and why is that consistent with the assumption? I feel like this is a simple question but for some reason I find this confusing. Thanks for the help.
Hi solomon684, welcome to PF.

An ideal diode conducts (ON) when it is forward biased, and behaves as an open circuit (OFF) when it is reverse biased. If the circuit wants to cause a current to flow in the direction of the diode's arrow, then that diode will turn on.

In figure B the potential at the top of the 6kΩ resistor (with respect to the bottom node) is 3V. The potential at the open end of the 4kΩ is the same as that of the 10V battery, or 10V. Thus the potential difference between the two points is 10V - 3V = 7V. The circuit would "like" to push current from the 10V point to the 3V point.

Since a 7V forward potential would want to make diode 1 turn on, the assumption that the diode is off is incorrect.

Ok got it, that makes much more sense, but then why is vD2 -3V? Wouldn't it be 3V - 10V = -7V?

And then what is it that makes that assumption correct? I cannot seem to figure this portion out. It seems like the negative voltage would make the current go left, turning on D2 and making the assumption incorrect. Obviously that is wrong but I am unsure why.

gneill
Mentor
Ok got it, that makes much more sense, but then why is vD2 -3V? Wouldn't it be 3V - 10V = -7V?
Referring to figure (C), the potential at the top of the 6kΩ resistor is determined by the voltage divider action of the 10V battery, the 4kΩ resistor, and that 6kΩ resistor. The result is that there is +6V at the top of that resistor. So the cathode of the diode has +6V and the anode +3V. Note the defined polarity of Vd is the potential from anode to cathode. That makes Vd = 3V - 6V = -3V.
[/QUOTE]
And then what is it that makes that assumption correct? I cannot seem to figure this portion out. It seems like the negative voltage would make the current go left, turning on D2 and making the assumption incorrect. Obviously that is wrong but I am unsure why.[/QUOTE]

The assumption is correct because it doesn't introduce any contradictions in the circuit behavior. Since the cathode of D2 is at a higher potential than its anode, it is reverse biased and no current will flow through it. D1 on the other hand has a forward bias provided by the 10V battery.

Referring to figure (C), the potential at the top of the 6kΩ resistor is determined by the voltage divider action of the 10V battery, the 4kΩ resistor, and that 6kΩ resistor. The result is that there is +6V at the top of that resistor. So the cathode of the diode has +6V and the anode +3V. Note the defined polarity of Vd is the potential from anode to cathode. That makes Vd = 3V - 6V = -3V.

The assumption is correct because it doesn't introduce any contradictions in the circuit behavior. Since the cathode of D2 is at a higher potential than its anode, it is reverse biased and no current will flow through it. D1 on the other hand has a forward bias provided by the 10V battery.
Ohhhhh okay I see now. In the first example (diagram (b)) the anode was 10V while the cathode was 3V, so it would be forward bias with current flowing through, so the diode would be on, meaning the assumption that it is off is incorrect.

Just a question though, is there ever a time when the value of the current would make an assumption incorrect, or does it only have to do with the voltage potential?

gneill
Mentor
Just a question though, is there ever a time when the value of the current would make an assumption incorrect, or does it only have to do with the voltage potential?
The open circuit potential across a diode location is usually diagnostic if the rest of the assumptions are made correctly. It's hard to say what effects a wrong assumption could have in a complex circuit. Current assumed flowing where it shouldn't can alter potentials here and there throughout the circuit. Incorrect currents can lead to incorrect potential calculations.

The open circuit potential across a diode location is usually diagnostic if the rest of the assumptions are made correctly. It's hard to say what effects a wrong assumption could have in a complex circuit. Current assumed flowing where it shouldn't can alter potentials here and there throughout the circuit. Incorrect currents can lead to incorrect potential calculations.
Alright cool, thank you very much for your help. I appreciate it.