How do I calculate the voltage of a diode in a simple electronics circuit?

[akaliuseheal]

Homework Statement

It's simple problem but I get stuck while calculating the V_D.

Currents and voltages in the circuit are required.
So V, V_D and I

Homework Equations

The Attempt at a Solution

Since V_A<V_K I would assume that diode is OFF and then the diode can be represented by break.
Upon testing that the diode is ON, I get i_D=(-5-3)/20k < 0 thus proving that it's OFF.
⇒I=i_D = 0, V_D < 0
Now, the problem for me is always finding the voltage of a diode since there is that break in the circuit.
English is not my native language so I hope I chose the right title of the topic and that I posted this in right thread.

 

[Merlin3189]

I have some difficulty here because V_A, V_K and V_D are not shown in the question (diagram), but let's try.

Since V_A<V_K I would assume that diode is OFF and then the diode can be represented by break.
Upon testing that the diode is ON, I get i_D=(-5-3)/20k < 0 thus proving that it's OFF.

I don't agree with your calculation, because you seem to be calculating the current in the opposite sense (direction) to the arrow. But that simply gives the wrong direction for the current. If current is <0 that simply means it is in the opposite direction to what you chose as the positive direction of current.

If you assume the diode is conducting, you get a current of (3 - -5)/20k = .4mA in the direction of the arrow.
That would mean the diode was conducting in the reverse direction. So either it is faulty or has a very high reverse leakage!
Or you could conclude that a reverse current is impossible in an ideal diode, so the assumption that it is conducting, must be wrong. In which case I must be 0 as you go on to say.

⇒I=i_D = 0, V_D < 0

Assuming V_D means the voltage across the diode in the direction of the arrow, Then the top is at +3V, the bottom is at -5V and the voltage across it is the difference. (I don't get it to be <0V)

Now, the problem for me is always finding the voltage of a diode since there is that break in the circuit.

Would you have a problem with any other circuit that did not have a diode in it - just the resistor and voltage sources? If the diode is off, just leave the break in the circuit and forget that it ever was a diode. ALL circuits contain breaks which cause no problems.

English is not my native language so I hope I chose the right title of the topic and that I posted this in right thread.

You did better than many English speakers! And certainly better than I could do in Russian or any of 100's of other languages!)

 

[donpacino]

Homework Statement

It's simple problem but I get stuck while calculating the V_D.

Currents and voltages in the circuit are required.
So V, V_D and I

Homework Equations

The Attempt at a Solution

Since V_A<V_K I would assume that diode is OFF and then the diode can be represented by break.
Upon testing that the diode is ON, I get i_D=(-5-3)/20k < 0 thus proving that it's OFF.
⇒I=i_D = 0, V_D < 0
Now, the problem for me is always finding the voltage of a diode since there is that break in the circuit.
English is not my native language so I hope I chose the right title of the topic and that I posted this in right thread.

I agree with you that the diode is off. now you need to find the voltage of the diode.

Well Vd will just be the voltage across it, so your voltage source (3v) - V (the voltage node in the middle).

You can find V by taking the -5V and the voltage drop across R1.

Hint (This is very easy): Like you said there is a break since the diode is off. Since there is a break, no current can flow through the resistor. so the voltage drop across the resistor is zero.

 

[akaliuseheal]

I have some difficulty here because V_A, V_K and V_D are not shown in the question (diagram), but let's try.

I don't agree with your calculation, because you seem to be calculating the current in the opposite sense (direction) to the arrow. But that simply gives the wrong direction for the current. If current is <0 that simply means it is in the opposite direction to what you chose as the positive direction of current.

Would you have a problem with any other circuit that did not have a diode in it - just the resistor and voltage sources? If the diode is off, just leave the break in the circuit and forget that it ever was a diode. ALL circuits contain breaks which cause no problems.

Usually not, but in electronics, I do sometimes get stuck.

 

[donpacino]

Usually not, but in electronics, I do sometimes get stuck.

id would have to be zero because it is a broken circuit. no current can flow through a broken circuit!

 

[akaliuseheal]

I get it to be 8V, but isn't the condition for diode to be off i_D=0, V_D<0 (ideally)
)

 

[akaliuseheal]

id would have to be zero because it is a broken circuit. no current can flow through a broken circuit!

Yes, that is the calculation that I used to check is it ON (i_D > 0 ?)
He wrote " I don't agree with your calculation "

 

[donpacino]

it depends on how you define vd. In your case I guess it would have to be -8V

 

[akaliuseheal]

it depends on how you define vd. In your case I guess it would have to be -8V

That was what was bothering me, that condition V_D< 0.

Thanks to you two for looking into this.

 

[Merlin3189]

I suppose I was being a bit pernickity. The I arrow shows the current flowing from +5V to -3V, so I think the PD is +5 -(-3) = +8V
But as I said, V_D wasn't specified. You could mark it on the circuit with its own arrow in the opposite direction.The remark about other circuits having breaks which cause no problem, was perhaps an obscure reference to the fact that all circuits have places where you could draw another wire, but you don't. And the wires that are not part of the circuit have no effect on it. If the diode is not there, then the circuit is just that much simpler.