How Does an Ideal Diode Affect Voltage in a Circuit?

[Mutaja]

Homework Statement

1a) Use an ideal diode model and find V in both cases. Show your work. (I would assume an ideal diode model means with a "perfect" u-i characteristics?)

A diode circuit is as shown:

1b) If V₁ = 2.8sin(t), sketch to periods of V₁ and V₀ in the same graph. Use an idea diode model.

1c) Repeat exercise d with a constant voltage drop model. (here I assume it's a typo, and I should repeat exercise b).

Homework Equations

The Attempt at a Solution

First things first, 1a).

So, the way I understand a diode, current either goes though, or it doesn't. I'm going to assume, in the first circuit (a), that the current goes through both diodes. What happens then at the point b?

Am I going to use the voltage and resistors to calculate currents I and I_D2, then the voltage potential "between" ground and D2? I'm not familiar with this type of layout of circuits.

Is there a negative potential difference between the point B and past the 5k resistor?

Any input on this will be greatly appreciated.

 

[gneill]

Hi Mutaja. If the diodes are ideal and you assume that they are both conducting, then you can "replace" them with a short (a wire). If they are not ideal but have a fixed potential drop then you can replace them with voltage sources with that potential.

The trick is always determining whether or not a given diode will be conducting or not. One way is to remove that diode from the circuit and see if the resulting potential across the open terminals where it was connected would forward bias the diode.

Note that quite often when diodes are "replaced" with a wire or a fixed voltage source and you're looking for a voltage at some location, the diodes will act as voltage clamps, fixing a potential at a certain value (take a look at D1 in figure (a). If D1 is conducting, what MUST be the potential at node B?).

 

[Mutaja]

Hi Mutaja. If the diodes are ideal and you assume that they are both conducting, then you can "replace" them with a short (a wire). If they are not ideal but have a fixed potential drop then you can replace them with voltage sources with that potential.

The trick is always determining whether or not a given diode will be conducting or not. One way is to remove that diode from the circuit and see if the resulting potential across the open terminals where it was connected would forward bias the diode.

Note that quite often when diodes are "replaced" with a wire or a fixed voltage source and you're looking for a voltage at some location, the diodes will act as voltage clamps, fixing a potential at a certain value (take a look at D1 in figure (a). If D1 is conducting, what MUST be the potential at node B?).

I'm not sure I completely understand what you mean in your second paragraph here, but that might be because, as I said, I'm not very familiar with the layout on the drawing. There's no voltage source, and it's not given which way the current flows. As for voltage, or potential, itself, the numerous grounding symbols confuse me.

As for your question, I would say 0 volts, because node B would be the same point as the grounding. But I really don't have much clue.

 

[gneill]

I'm not sure I completely understand what you mean in your second paragraph here, but that might be because, as I said, I'm not very familiar with the layout on the drawing. There's no voltage source, and it's not given which way the current flows. As for voltage, or potential, itself, the numerous grounding symbols confuse me.

The arrows labeled +10 V and - 10 V indicate that they lead to voltage sources of those values. You can take it that they both have those potentials with respect to ground.

The ground symbols indicate that all those wires are connected to a common reference node. You could replace them with a wire connecting all the places where you see a ground symbol, but that would make the diagram messy.

As for your question, I would say 0 volts, because node B would be the same point as the grounding. But I really don't have much clue.

Yes, that's correct. The diode would effectively clamp node B to ground if it conducts.

 

[Mutaja]

The arrows labeled +10 V and - 10 V indicate that they lead to voltage sources of those values. You can take it that they both have those potentials with respect to ground.

The ground symbols indicate that all those wires are connected to a common reference node. You could replace them with a wire connecting all the places where you see a ground symbol, but that would make the diagram messy.

Yes, that's correct. The diode would effectively clamp node B to ground if it conducts.

Ah, I understand! So basically, in 1a) figure (a), what they're asking for is the voltage across the diode D2?

If we assume that D1 is conducting, then V is in reality V_D2. The current flowing through the 10k resistor and D2 will be 1mA, since the resistance is 10k ohm, and the voltage is 10volts across the 10k resistor and the diode. But that wouldn't make much sense, because looking at the voltage divider, all the voltage will be across the resistor, and none across the diode?

So I understand more, but not everything

 

[gneill]

But that wouldn't make much sense, because looking at the voltage divider, all the voltage will be across the resistor, and none across the diode?

No, it makes sense. It just means that the output voltage will be zero. It's effectively clamped to the reference node via the two diodes, if both are conducting.

 

[Mutaja]

No, it makes sense. It just means that the output voltage will be zero. It's effectively clamped to the reference node via the two diodes, if both are conducting.

Alright, then I'm confused as to how I can see/compute if they're conducting or not, if they're on or off.

Can I assume that both the diodes are conducting because of the potential difference on the end of the two resistors? Or at least one of them have to be conducting. Does it affect the circuit at all if the diode D1 isn't conducting?

 

[gneill]

Alright, then I'm confused as to how I can see/compute if they're conducting or not, if they're on or off.

Can I assume that both the diodes are conducting because of the potential difference on the end of the two resistors? Or at least one of them have to be conducting. Does it affect the circuit at all if the diode D1 isn't conducting?

See my middle paragraph in post #2.

Another method is to test all combinations of diodes ON/OFF until you find one that doesn't produce a contradiction to the assumptions. Such as finding a current running "backwards" through a diode that you assumed was forward biased.

 

[Mutaja]

I've tried to understand that, but I don't...

So, I've removed both diodes in figure (a) (should I check one and one?)...

Trying your other method, as I have only two diodes.

If D1 doesn't conduct, D2 doesn't conduct: No current will flow through the circuit?
If D1 does conduct, D2 doesn't conduct: 2mA will flow through the D1? (I would assume, since the potential is -10, the current will flow against the diode in the direction it doesn't conduct...) Am I overthinking this?
If D1 doesn't conduct, D2 does conduct: 1mA will flow from the -10 to the -10v, in the direction which the diode won't let any current pass through. This is because -10V and 5k ohm creates 2mA, and 10v and 10k ohm creates 1mA.
If D1 does conduct, D2 does conduct: The current will split at the B point in which case no current can flow through D2 since it's faceing the wrong way, and 2mA will flow through the 5k ohm resistor.

Now, I know most things here are wrong, but this is how I'm thinking right now. What am I missing here? What is my mind not comprehending...

 

[gneill]

I've tried to understand that, but I don't...

So, I've removed both diodes in figure (a) (should I check one and one?)...

Yes, do them one at a time.

Trying your other method, as I have only two diodes.

If D1 doesn't conduct, D2 doesn't conduct: No current will flow through the circuit?

Yes, but there's a contradiction since the anode of D2 will be at +10V and the cathode at -10V, so that would force D2 to conduct, right? So that case can't be true. (Also D2 would be forced to conduct since its cathode would be -10V lower than its anode, so double contradiction).

If D1 does conduct, D2 doesn't conduct: 2mA will flow through the D1? (I would assume, since the potential is -10, the current will flow against the diode in the direction it doesn't conduct...) Am I overthinking this?

I don't know about overthinking, but the direction of current though D1 would be from ground to the -10V supply. So that's okay. But then node B would be at ground potential while the anode of D2 would be at +10V (assuming no current through D2). That would be a forward bias condition, forcing D2 to conduct. So the assumption that D2 could be off if D1 is conducting is clearly false.

If D1 doesn't conduct, D2 does conduct: 1mA will flow from the -10 to the -10v, in the direction which the diode won't let any current pass through. This is because -10V and 5k ohm creates 2mA, and 10v and 10k ohm creates 1mA.

Check your current calculation. If D2 is conducting and D1 not conducting, then the path for current goes from +10V to -10V as you say, but that means a total of 20V potential difference from one end of the path to the other. And there's two resistors in that path, 10k and 5k. So what's the current? What potential would that put at node B?

Note that it's often convenient to use nodal analysis in these situations to quickly find the potential at a node. Write:
$$\frac{V_B - (-10V)}{5 k} + \frac{V_b - 10V}{10 k} = 0$$
Solve for $V_B$.
Will that potential at B allow D1 to remain off?

If D1 does conduct, D2 does conduct: The current will split at the B point in which case no current can flow through D2 since it's faceing the wrong way, and 2mA will flow through the 5k ohm resistor.

No. Current will flow from the ground connection down through D1 towards the -10V source. That's the correct direction for the diode. Current will also flow from the +10V source down through D2 towards the -10 V source.

Having D1 conducting will pin node B at ground potential. So you can work out the individual currents easily enough.

 

[Mutaja]

Now I finally understand the logic of figuring out if the diode conducts or not.

They're both conducting, since Vb is supposed to be 0 in your equation, but I end up getting - (10/3).

Anyhow. I end up calculating that the current through D2 is 10v/10k ohm = 1mA from +10V to the -10V.

The current through D1 will be 2mA from ground through the 5k resistor and towards the -10V.

The voltage drop across the 10k resistor will be 10V, therefore V will equal 0?

Stay with me on this... haha

 

[Mutaja]

This is how I can explain it right now.

We can see that both the diodes has to conduct to satisfy the properties of the circuit. What happens then, is point B is grounded and since the current flows from +10V to -10V, there will be no voltage across D2.

Question: If D2 had faced the other way, would there we a potential difference of -10V across it?

Edit: I just realized, that totally neglects the resistors in the circuit, and both the figures, (a) and (b) will be identical to me...

 

[NascentOxygen]

Are you saying that you think the resistor values make no difference here? How could you prove or disprove this?

 

[Mutaja]

Are you saying that you think the resistor values make no difference here? How could you prove or disprove this?

I'm saying using my method in post #12 neglects the resistors, the way I see it. And my logic tells me that's completely wrong.

I'm assuming the resistors makes a difference because they affect the current flow, and that would also affect the voltage across the resistors and diodes.

What I was thinking, however, is that if current flows through the diode, and it's conducting, then there's no voltage across it, and if it's not conducting, there will be voltage across it equal to the negative amount of what the voltage between the grounding point and source.

Like this:

 

[gneill]

Yes. That's correct.

If you wish you can think of an ideal diode behaving as a switch, either open or closed.

 

[Mutaja]

Yes. That's correct.

If you wish you can think of an ideal diode behaving as a switch, either open or closed.

What about my statement in post #12 then? How will the resistors affect the circuit? Because I get the same answer for both circuits in exercise 1a).

 

[gneill]

What about my statement in post #12 then? How will the resistors affect the circuit? Because I get the same answer for both circuits in exercise 1a).

You'll need to go through the exercise of checking to see which diodes are on or off. Remember at one point checking to see what the voltage at node B would be if D1 didn't conduct but D2 did? And then finding that the resulting potential across D1 would actually forward bias it, so that case had to be rejected?

Well, things may be different this time with the resistor values swapped...

 

[Mutaja]

You'll need to go through the exercise of checking to see which diodes are on or off. Remember at one point checking to see what the voltage at node B would be if D1 didn't conduct but D2 did? And then finding that the resulting potential across D1 would actually forward bias it, so that case had to be rejected?

Well, things may be different this time with the resistor values swapped...

I checked the voltage V_B through voltage divider rule. The resulting potential that I found was negative 10/3 voltage. Is it the negative part that forward bias the diode?

So when I calculate the voltage across the diode D1 in the second example with D1 off and D2 on, I get a positive voltage (still 10/3V), which allows the diode to stay off?

 

[gneill]

So when I calculate the voltage across the diode D1 in the second example with D1 off and D2 on, I get a positive voltage (still 10/3V), which allows the diode to stay off?

Yup. D1's cathode will be more positive than its anode, so that's a reverse bias.

 

[Mutaja]

Yup. D1's cathode will be more positive than its anode, so that's a reverse bias.

Awesome, thanks. I will do more of these exercises to get some training in recognizing whether or not the diode is conducting.

For my exercise 1b), the diode will work as a short circuit for the positive values of the sin wave, and as an open circuit for the negative values.

Sketching two periods of the voltages V₁ and V₂ in the same graph, will leave a regular sin wave (top point 2.8, one period = 2pi) for two periods, while the positive values will be sketched twice because that's how V₂ will look?

Like this, drawn in excel (not sure how to only get positive values)

And there would be a double line following the positive values, like this:

 

[gneill]

Not quite. When the diode is reverse biased it effectively "disappears" from the circuit, leaving the voltage divider to scale the signal. So the negative half of the input signal is going to be affected by the voltage divider, but it still gets though.

 

[Mutaja]

Not quite. When the diode is reverse biased it effectively "disappears" from the circuit, leaving the voltage divider to scale the signal. So the negative half of the input signal is going to be affected by the voltage divider, but it still gets though.

Will half of it get through? The period of V1 will be as my first picture, and V2 will have positive top values of 2.8, and negative "top" values of 1.4?

My thinking behind this is the voltage drop across the first 1k resistance. Since they're both equal, the voltage drop will be half.

 

[gneill]

Yes, that it. The voltage divider scales the signal by half when the diode is not conducting.