Some quick ideal diode analysis

[STEMucator]

Homework Statement

Find the currents and voltages indicated for the circuits given:

Homework Equations

The Attempt at a Solution

I needed some practice, and I want to ensure I understand for the harder problems later.

(a) The diode is going to behave like a short circuit, so the current can be found from $I = \frac{5V - (-5V)}{10k} = 1 mA$. The voltage will be entirely across the $10k$ resistor, and so $V = 0 V$.

(b) The diode is going to behave like an open circuit, hence the current in the branch is $I = 0 A$. None of the voltage will be across the resistor since there is no current, and so $V = 5V - (-5V) = 10V$.

(c) The diode is going to behave like a short circuit again, so the current can be found from $I = \frac{5V - (-5V)}{10k} = 1 mA$. The voltage $V$ will be entirely across the resistor, so $V = 5V - (-5V) = 10V$.

(d) The diode will behave as an open circuit, so $I = 0 A$. The voltage is a bit confusing to me though, would it be $10V$ again? I'm finding it difficult to reason it out.

 

[milesyoung]

(a) If the diode is represented as a short, then V and the bottom node will be one and the same, so what is V?

b, c and d have the same problem.

 

[phinds]

You are totally ignoring any possible effects of whatever is hooked to the node labeled V but your "answers" do not reflect that.

 

[STEMucator]

Okay, so:

(a) $I = 1mA, V = -5V$.

(b) $I = 0, V = -5V$.

(c) $I = 1mA, V = 5V$.

(d) $I = 0, V = 5V$.

Weird, the harder problems in the book are easier than these were. I had a dull moment looking at these.

 

[phinds]

I have no objection to your ignoring the rest of any possible circuit if that's what you want to do, but you need to SAY that that's what you are doing (otherwise it just looks like a serious oversight on your part) and you should give some thought as to whether or not that's a good idea in terms of learning how circuits work.

 

[STEMucator]

I have no objection to your ignoring the rest of any possible circuit if that's what you want to do, but you need to SAY that that's what you are doing (otherwise it just looks like a serious oversight on your part) and you should give some thought as to whether or not that's a good idea in terms of learning how circuits work.

Miles helped me realize it was all the same node in the cases of (a) and (c) when I visualized the short circuit. In the cases of (b) and (d), the open circuit looked a little funny at first, but I know I can't ignore it.

 

[milesyoung]

Okay, so:

(a) $I = 1mA, V = -5V$.

(b) $I = 0, V = -5V$.

(c) $I = 1mA, V = 5V$.

(d) $I = 0, V = 5V$.

Weird, the harder problems in the book are easier than these were. I had a dull moment looking at these.

(b) If there's no current, then there's no voltage drop across the resistor. The top node and V are then at the same potential, so what is V?

Same thing goes for (d).

 

[STEMucator]

(b) If there's no current, then there's no voltage drop across the resistor. The top node and V are then at the same potential, so what is V?

Same thing goes for (d).

Okay, so literally every detail needs to be accounted for at this point, including the fact the resistor is is short circuited due to zero current. Then I could say $V = 5V$ and not $-5V$ for (b).

Similarly, for (d), the open circuit would cause $V = -5 V$, due to zero current.

 

[milesyoung]

Okay, so literally every detail needs to be accounted for at this point, including the fact the resistor is is short circuited due to zero current.

I wouldn't say it's shorted, it just doesn't have any voltage across it.

Then I could say $V = 5V$ and not $-5V$ for (b).

Similarly, for (d), the open circuit would cause $V = -5 V$, due to zero current.

Yes. :)

 

[STEMucator]

I'm wondering if you could help me with a curiosity now that I have the question sorted out. Using the exact same diagrams and problem statement, I want to find the voltages and currents using the constant $0.7V$ voltage drop model.

(a) The diode behaves like a short circuit. I believe $I = \frac{V_{DD} - V_D}{R} = \frac{4.3V}{10k} = 0.43 mA$ and $V = 0.7V$.

(b) The diode behaves like an open circuit. Hence $I = 0$ and $V = 5V$.

(c) I'm a bit confused by this one. The diode behaves like a short circuit. Would it be $V = 4.3V$ and $I = \frac{9.3V}{10k} = 0.93 mA$?

(d) The diode behaves like an open circuit. Hence $I= 0$ and $V = -5V$.

Do these seem reasonable given the different conditions?

 

[milesyoung]

(a) The diode drops 0.7 V, i.e. V must be 0.7 V higher than -5 V. What is V then?

That doesn't make any difference for the current, though. For the ideal case, the resistor had 10 V across it. Now it has 0.7 V less, which must be the same for both (a) and (c), i.e. the current must be the same.

(c) V must be 0.7 V lower than 5 V, so V = 4.3 V is correct.

Edit:
This is a minor thing, but a short circuit in circuit analysis typically means a connection with an ideal wire, which has no voltage across it by definition. It would probably be more correct to say the diode is either forward or reverse biased.

 

[STEMucator]

(a) The diode drops 0.7 V, i.e. V must be 0.7 V higher than -5 V. What is V then?

That doesn't make any difference for the current, though. For the ideal case, the resistor had 10 V across it. Now it has 0.7 V less, which must be the same for both (a) and (c), i.e. the current must be the same.

(c) V must be 0.7 V lower than 5 V, so V = 4.3 V is correct.

Edit:
This is a minor thing, but a short circuit in circuit analysis typically means a connection with an ideal wire, which has no voltage across it by definition. It would probably be more correct to say the diode is either forward or reverse biased.

Back from more problems. I think this one is clear now.

(a) I see now that there should be a $9.3 V$ drop across the resistor $R$ just looking at it... The diode drops the other $0.7 V$. Hence the voltage $V = -4.3 V$. The current is then given by $I = \frac{5V - (- 4.3 V)}{10 k} = \frac{9.3 V}{10 k} = 0.93 mA$.

(b) The diode is not conducting, so my answer has not changed, $I = 0$ and $V = 5V$.

(c) I actually understand this fully now. Since the diode is conducting a reasonable amount of current, $V$ is $0.7V$ less than $5V$ and so $V = 4.3V$. Hence the current $I = 0.93 mA$.

(d) Same answer as before. Open circuit, $I = 0$, $V = -5V$.

I think after doing so many problems I've pieced this diode thing together. I actually get the voltage drop idea at this point, so I should be okay.

Back to work, thanks for all your help.

 

[milesyoung]

(a) I see now that there should be a $9.3 V$ drop across the resistor $R$ just looking at it... The diode drops the other $0.7 V$. Hence the voltage $V = -4.3 V$. The current is then given by $I = \frac{5V - (- 4.3 V)}{10 k} = \frac{9.3 V}{10 k} = 0.93 mA$.

That's true. I took the view that $V$ must be at some value that's higher than -5 V, since the diode has a voltage drop, i.e. what value of $V$ would allow you to drop 0.7 V and end up at -5 V? That has to be -4.3 V, as you wrote.

Since the diode is conducting a reasonable amount of current ...

I assumed you were using the "ideal diode with a twist"-model, where the diode drops 0.7 V at any current if forward biased, or drops any voltage necessary to keep a current from flowing if reverse biased.

Back to work, thanks for all your help.

You're very welcome.