# Problem with diodes and voltage

#### an1996

Joined Nov 20, 2020
4
Hello, I am new in this forum.
I had this problem in my last online exam, it's a circuit with a diode and a LED and the exercise ask me for the tension in Vo.

The specs are the following:
R1 = 84 [Ω]
R2 = 109 [Ω]
Vs=3,8 [V]
Vu = 0,7 [V] (threshold voltage for D1)
Vux = 3,8 [V] (threshold voltage for the led D2)

In my opinion, Vo should be closer to Vs (3,8 [V]), but the correct answer is 0. Why? Thanks in advance. #### crutschow

Joined Mar 14, 2008
32,041
In my opinion, Vo should be closer to Vs (3,8 [V]), but the correct answer is 0.
Yes, I also think the voltage will be above 0V.
I would expect it to be approximately Vs (3.8v) minus the threshold value of D1.

• an1996

#### an1996

Joined Nov 20, 2020
4
Yes, I also think the voltage will be above 0V.
I would expect it to be approximately Vs (3.8v) minus the threshold value of D1.
I made a simulation with LTSPICE (with other values of course, but conceptually similar) and it shows what you said, there is a small voltage drop in D1, due to a very small (almost zero) current flow.
But according to some classmates and the professor Vo should be zero, because the current is zero; this doesn't make sense to me.

#### crutschow

Joined Mar 14, 2008
32,041
Vo should be zero, because the current is zero; this doesn't make sense to me.
Depend how you model the diodes.
You can have a voltage with essentially zero current if the diodes are viewed as having a voltage threshold independent of current down to near zero (which is not specified in the problem).
All in all it's a rather ambiguous idealized problem, which would not operate that way in a real circuit.
Have the professor built you the circuit to show that if he's so sure it's zero. • an1996

#### MrAl

Joined Jun 17, 2014
10,075
Hello, I am new in this forum.
I had this problem in my last online exam, it's a circuit with a diode and a LED and the exercise ask me for the tension in Vo.

The specs are the following:
R1 = 84 [Ω]
R2 = 109 [Ω]
Vs=3,8 [V]
Vu = 0,7 [V] (threshold voltage for D1)
Vux = 3,8 [V] (threshold voltage for the led D2)

In my opinion, Vo should be closer to Vs (3,8 [V]), but the correct answer is 0. Why? Thanks in advance.

View attachment 222876

The question sounds ambiguous because there seems to be some information left out. From their 'correct' answer as you state it they must be viewing the LED as having a much lower dynamic resistance than the diode D1.

Here is something to think about though while you are figuring this out...
If you replace the LED with another diode EXACTLY the same as D1, and make R2 equal to R1, the theoretical output will be one half of Vs no matter what voltage Vs was set at.
Now since the LED forward voltage is stated as 3.8v if you replace that with five diodes in series each with a forward voltage of 0.76 volts (which we assume th diode has too) that means the drop across the 'new' LED would be 5 times that of the diode for any Vs greater than 0 volts.

So they might be making a mistake when they say the output is zero volts. They may be using the diode/voltage source analogy where the diode voltage is a constant voltage source and they set it to zero when it is not conducting instead of an infinite resistance.

So there are a lot of angles we could look at here. If you like you can go over several different possibilities so that you learn to understand each approach and why it might come to be that way and what the importance of each technique is. That will give you a much better understanding that would go beyond this simple exercise which would prepare you for a lot of different circuits you may encounter in the future.

• an1996

#### an1996

Joined Nov 20, 2020
4
Now since the LED forward voltage is stated as 3.8v if you replace that with five diodes in series each with a forward voltage of 0.76 volts (which we assume the diode has too) that means the drop across the 'new' LED would be 5 times that of the diode for any Vs greater than 0 volts.
Exactly, the voltage drop across the LED is much higher than in D1 in all the simulations I made

So they might be making a mistake when they say the output is zero volts. They may be using the diode/voltage source analogy where the diode voltage is a constant voltage source and they set it to zero when it is not conducting instead of an infinite resistance.
The professor says that the diode, in this case, can be modelled as an infinite resistance or a switch. In my opinion, this simplified model doesn't make sense here since the results we obtain using it are very different from reality, I think the Shockley equation should be used here

So there are a lot of angles we could look at here. If you like you can go over several different possibilities so that you learn to understand each approach and why it might come to be that way and what the importance of each technique is. That will give you a much better understanding that would go beyond this simple exercise which would prepare you for a lot of different circuits you may encounter in the future.
Sure, but if I have trouble with such simple exercises my hopes are not very high.

And then there is the fact that it is very hard for a professor to admit his mistakes...

#### MrAl

Joined Jun 17, 2014
10,075
Exactly, the voltage drop across the LED is much higher than in D1 in all the simulations I made

The professor says that the diode, in this case, can be modelled as an infinite resistance or a switch. In my opinion, this simplified model doesn't make sense here since the results we obtain using it are very different from reality, I think the Shockley equation should be used here

Sure, but if I have trouble with such simple exercises my hopes are not very high.

And then there is the fact that it is very hard for a professor to admit his mistakes...

Well some of these exercises are not always specified correctly and as i said if you really want to understand this for yourself and not necessarily for your coursework you may want to go over different versions of this circuit with different diode and LED specifications, which yes may include the real diode equation.

When i make the diode have infinite resistance i cant ignore the LED resistance which also would have to be either the same as the diode (replacing it with a diode to gain more insight) or making it fives times as great. When i make them the same i get Vs/2 and when i make the LED have a resistance 5 times the resistance of the diode and then let the diode resistance go to infinity i get Vs*5/6 and that is due to the voltage divider effect.

What you can do to clear this up is ask the professor to show you the math behind his view and how he developed the equations. That will show either the error or a possible silly assumption.

• an1996