Help Analyzing Circuit Using Ideal Diode Method

In summary, the conversation discusses a problem with understanding the ideal diode model in regards to a circuit with two diodes. The person has attempted a solution using Ohm's law and the voltage-division principle, but is having trouble understanding the answer provided in a textbook. They are specifically having trouble understanding the assumptions made about the circuit and how to solve for the voltage values of each diode. The conversation concludes with a clarification that the assumptions are correct and do not introduce any contradictions in the circuit behavior.
  • #1
solomon684
9
0
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.
 

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  • #2
solomon684 said:
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.
 
  • #3
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.
 
  • #4
solomon684 said:
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.
 
  • #5
gneill said:
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?
 
  • #6
solomon684 said:
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.
 
  • #7
gneill said:
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.
 

FAQ: Help Analyzing Circuit Using Ideal Diode Method

What is the Ideal Diode Method and how does it work?

The Ideal Diode Method is a simplified circuit analysis technique that assumes diodes behave like ideal switches. This means that the diode will either be fully conducting (ON state) or fully non-conducting (OFF state) depending on the voltage across it. This method simplifies the analysis of circuits containing diodes by treating them as open circuits when they are in the OFF state and as short circuits when they are in the ON state.

When is the Ideal Diode Method most commonly used in circuit analysis?

The Ideal Diode Method is most commonly used in circuit analysis when dealing with rectifier circuits, switching circuits, and power supply circuits. It is also useful in analyzing circuits with multiple diodes.

What are the limitations of the Ideal Diode Method?

The Ideal Diode Method is a simplified approach and does not take into account the non-ideal characteristics of real diodes. It assumes that the diode is either fully ON or fully OFF, which may not always be the case. Additionally, this method does not account for the voltage drop across the diode, which can affect the overall circuit performance.

How do I apply the Ideal Diode Method in circuit analysis?

To apply the Ideal Diode Method, you first need to identify the diodes in the circuit and determine their state (ON or OFF) based on the voltage across them. Then, you can replace the diodes with ideal switches and analyze the circuit using traditional circuit analysis techniques.

Are there any software tools available for using the Ideal Diode Method in circuit analysis?

Yes, there are various software tools available for circuit analysis that incorporate the Ideal Diode Method. These include LTspice, PSpice, and Multisim, among others. These tools allow you to simulate the circuit and analyze its performance with the Ideal Diode Method, making the analysis process more efficient and accurate.

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