Problem with voltmeters and a lamp with an internal resistance

Thread Starter

EobardThawne

Joined Apr 30, 2017
8
In this problem's solution, I can't understand why the lamp's resistance is used to calculate the voltage of the diode: I assumed it would be the other way around: Since the current is the same in the whole circuit, because the connection is in series, then I could apply Ohm's Law on the lamp, with the lamp's resistance and the current of the diode, to obtain the voltage of the lamp. What am I not getting here? I upload the letter of the exercise, along with the solution, in a file. Thanks in advance.
 

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wayneh

Joined Sep 9, 2010
16,229
So I'm right, or my interpretation is wrong also?
No, I think you've got it. The current through the known resistive load gives you ∆V across the bulb directly by Ohm's law. The remainder of the ∆V has to be across the diode, which is not an ohmic device.

I don't know if it was intentional, but whoever set up the solution made a mistake, in my opinion, by using two different sets of subscripts on the voltages. It's far too easy to make a mistake if you bury yourself in unnecessary complexity. That may have been the intention, to see of you could cut through it. It's actually a very easy problem.
 

WBahn

Joined Mar 31, 2012
25,220
If an instructor provided this solution, you need to RUN away from that course and probably the entire school!

Ohm's Law relates the current THROUGH a RESISTOR to the voltage ACROSS THAT resistor.

First, the person is trying to use Ohm's Law on a diode, which is a highly non-ohmic device.

Then they are throwing the current through one component (the diode) and using the resistance of some other component (the lamp) to calculate a voltage.

Wrong, wrong, WRONG!

Then, after getting 10.5 V as the voltage across a forward-biased diode, they didn't even bother to think if that was in any way, shape, or form even conceivably a reasonable answer.

Then they couldn't be bothered to check if their answers agreed with the problem. They claim that the voltage across the lamp is 1.5 V and with a 30 Ω resistance that means that the current is 50 mA, not the 350 mA stated by the problem.

Your approach is correct. But consider whether the result you get is reasonable. What voltage do you get across the diode? Does it match the model you are using for how a diode behaves?
 

wayneh

Joined Sep 9, 2010
16,229
Then, after getting 10.5 V as the voltage across a forward-biased diode, they didn't even bother to think if that was in any way, shape, or form even conceivably a reasonable answer.
That was the tip-off to me, before I made any effort to look at the math. 10.5V across a diode in series with a bulb with 1.5V across it? I don't think so. A brief inspection of the math exposed the problem.
 

WBahn

Joined Mar 31, 2012
25,220
That was the tip-off to me, before I made any effort to look at the math. 10.5V across a diode in series with a bulb with 1.5V across it? I don't think so. A brief inspection of the math exposed the problem.
Me too.

I just did an interesting little calculation.

If I have a diode that obeys the diode equation ideally (i.e., no equivalent series resistance and no heating effects) that conducts 1 mA of current with 0.7 V across it at room temperature, what is the voltage across it when every electron in the known universe is flowing through it every second?

I think I'll start a thread over in the Math forum to see if there are any takers.
 

Thread Starter

EobardThawne

Joined Apr 30, 2017
8
Thank you very much for your answers!!! I was getting really puzzled with this exercise. And yes, the solution was given by my instructor. I guess he messed up with the subscripts, like Wayneh said.

Then, after getting 10.5 V as the voltage across a forward-biased diode, they didn't even bother to think if that was in any way, shape, or form even conceivably a reasonable answer.

[...]

consider whether the result you get is reasonable. What voltage do you get across the diode? Does it match the model you are using for how a diode behaves?
The result I got was those two same numbers, but interchanged. How were you able to tell 10.5 V was unreasonable for a voltage across a forward-biased diode?
 

wayneh

Joined Sep 9, 2010
16,229
A diode conducts very well, with 'low' resistance, whenever the ∆V across it rises over 0.65V or so. (∆V is a function of current, so the exact voltage depends on the exact current you want to call 'conducting'.) If you find 10.5V across a diode, that means it should be passing a current high enough to destroy any diode I can imagine. I think @WBahn is looking into whether it's even possible.

Usually, you can assume ~0.7V across any conducting diode when you look over a circuit. So even the 1.5V in your problem bothers me as fake.
 

WBahn

Joined Mar 31, 2012
25,220
Thank you very much for your answers!!! I was getting really puzzled with this exercise. And yes, the solution was given by my instructor. I guess he messed up with the subscripts, like Wayneh said.
We all mess up. He set himself up to make a stupid mistake because he insisted on using two sets of subscripts. But blindly claiming that the diode had 10.5 V (or even 1.5 V) across it and not batting an eye is pretty unforgivable.

The result I got was those two same numbers, but interchanged. How were you able to tell 10.5 V was unreasonable for a voltage across a forward-biased diode?
Have you gotten to the point where you have been introduced to the diode equation:

\(
i_d \; = \; I_s e^{\( \frac{v_d}{V_t}\)}
\)

If not, then you are probably using a constant voltage model that says that the voltage across a forward-biased diode is a constant 0.7 V (or something close to that). Whether the notion of 1.5 V across a diode sets of warning bells is a function of how strongly you've been beaten in the head with that model.

Look at the thread I started over in the Math forum:

https://forum.allaboutcircuits.com/threads/the-power-of-exponentials.135424/#post-1132377
 

Thread Starter

EobardThawne

Joined Apr 30, 2017
8
Thank you both again!

We all mess up. He set himself up to make a stupid mistake because he insisted on using two sets of subscripts. But blindly claiming that the diode had 10.5 V (or even 1.5 V) across it and not batting an eye is pretty unforgivable.



Have you gotten to the point where you have been introduced to the diode equation:

\(
i_d \; = \; I_s e^{\( \frac{v_d}{V_t}\)}
\)

If not, then you are probably using a constant voltage model that says that the voltage across a forward-biased diode is a constant 0.7 V (or something close to that). Whether the notion of 1.5 V across a diode sets of warning bells is a function of how strongly you've been beaten in the head with that model.

Look at the thread I started over in the Math forum:

https://forum.allaboutcircuits.com/threads/the-power-of-exponentials.135424/#post-1132377
You're right, we're using the constant voltage model
 

MrAl

Joined Jun 17, 2014
7,100
Hi,

It seems a little obvious that the author made a small mistake in the writing.
He/she exchanged VB for VD and vice versa.

Try swapping them and see the answer pop out. The voltage might be a little high for a diode though, so maybe they are calling an LED a diode which is sometimes also done and then 350ma would be a typical current for a high power 1watt LED.
 
Last edited:

Thread Starter

EobardThawne

Joined Apr 30, 2017
8
Hi,

It seems a little obvious that the author made a small mistake in the writing.
He/she exchanged VB for VD and vice versa.

Try swapping them and see the answer pop out. The voltage might be a little high for a diode though, so maybe they are calling an LED a diode which is sometimes also done and then 350ma would be a typical current for a high power 1wall LED.
Thank you! That could be it. He usually represents LEDs with the two arrows pointing upwards, but he has used different notations for some components before.
 

WBahn

Joined Mar 31, 2012
25,220
Go to your instructor and ask them if they are sure that there is 10.5 V across the diode. If they say yes, then ask them if this is actually possible with a real typical diode. If they catch their mistake, then ask them if 1.5 V across a diode is realistic. Ask him whether it might not be more reasonable to assume that the voltage across the diode is about 0.7 V making the voltage across the lamp about 11.3 V making the current in the entire circuit about 377 mA. How they answer these questions will tell you a lot about how much you are likely to learn from him the rest of the term.
 
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