# Realistic Diode-Resistor Series Voltage Drop

Discussion in 'Homework Help' started by Austin Clark, May 13, 2012.

1. ### Austin Clark Thread Starter Member

Dec 28, 2011
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How would you calculate the voltage drop of a diode in a diode-resistor series circuit? In real life, the voltage drop of a diode isn't always exactly it's bias voltage or whatever. The problem is, when I try to calculate the REAL voltage drop of a diode in the circuit, I get inter-dependencies and I get confused... Essentially, the voltage drop is based on the diodes resistance and the total circuit resistance:
VoltageDropOfDiode = Vin*(DiodeResistance/(DiodeResistance+ResistorResistance))
HOWEVER, the diodes resistance isn't constant, and is based on the voltage leading into it, which means it's voltage drop is based on its voltage drop. You need the variable you want to find to find itself!

I know it's not usual to think of a diodes resistance, but it's still technically correct.

How would you solve this problem?

Another question might be what would the I-V curve look like for a REALISTIC diode-resistor series circuit? and how would you find the total current through the diode-resistor circuit realistically?

2. ### bretm Member

Feb 6, 2012
152
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Using a datasheet's IV curve is pretty realistic. Real components don't always behave exactly like simple equations.

But if you want to solve it methematically, you'll need to use the "diode equation", ignore the "-1" part of it because it is negligible, and then use the Lambert W function to solve for voltage. Search for Lambert W and diode and you'll probably find an example somewhere.

3. ### Ron H AAC Fanatic!

Apr 14, 2005
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You don't need the Lambert W function (which I had never heard of) to solve the diode equation for forward voltage vs current. Simple natural logarithms will suffice.
The problem is, you generally don't know the value of Is (the saturation current), which is a function of, among other things, the area of the diode junction. This can be solved for if you have a single voltage and current measurement.
There is another term in the equation, η (eta). From the diode Wiki,

4. ### Austin Clark Thread Starter Member

Dec 28, 2011
409
44
Via a little experimentation (and a lot of excel) I've found that using
a*(e^(b*V))=I
is a good way to find a somewhat decent model. I literally just measured a bunch of states and checked and guessed the values for a and b until I got something that was pretty close to the measured values.

However, this alone won't help me solve my problem. My problem is finding the current with a resistor in series with it. I want to know where the circuit will "balance".

5. ### Ron H AAC Fanatic!

Apr 14, 2005
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I think the equation is transcendental, which means it can't be solved by standard algebraic manipulation. I missed your mention of the series resistor.
Bretm mentioned the Lambert W function, which I pooh-poohed. I see now what he was talking about.
A thousand apologies, bretm.

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6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I recollect that before we had programable calculators or PC's I used the load line graphical method to solve such problems. If you have the diode VI characteristic or the means of plotting it then a solution is within reach.
If you use goal seek in Excel [What-if Analysis] the problem is readily solved.

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• ###### Goal Seek - Series Resistor + Diode - spreadsheet details.png
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Last edited: May 14, 2012
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8. ### Sparky49 Active Member

Jul 16, 2011
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Quick question, is the 'e', the inverse of ln, or is it a placeholder?

9. ### MrChips Moderator

Oct 2, 2009
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11. ### Austin Clark Thread Starter Member

Dec 28, 2011
409
44
So, essentially, you're using excel to check-and-guess until a reasonable value is found? Sounds similar to what I was doing. I'll have to try and copy your sheet and experiment a bit with it. Thanks!

12. ### Austin Clark Thread Starter Member

Dec 28, 2011
409
44
What do you mean by drawing the resistor load line?
Could you show us an example?

13. ### MrChips Moderator

Oct 2, 2009
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Suppose your power supply voltage Vs is 10V and the series resistor is R = 200Ω.
The maximum current you can conduct through the diode would be Vs/R = 50mA. This is the point on the current scale (Y-axis).

The maximum voltage at 0mA will be Vs = 10V.
This is the point on the voltage scale (X-axis).

Draw a straight line between these two points. This is called the load line.
Where this line intersects with the diode V-I curve is your solution,
(Vd, Id).

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14. ### Austin Clark Thread Starter Member

Dec 28, 2011
409
44
Ah, ha! Thank you VERY much!!! Basically, you're finding the point where the amount of current through the resistor after the diode drops the voltage equals the current the diode will conduct with that dropped voltage. Super simple now that It's been shown to me! The current through each component must equal the total circuit current, and once you know that current, you just need to find out what voltage is necessary to conduct that much through the diode, and there you have it!

I'm gonna be staring at this chart for a good while now until I've observed it from every angle. Get my head fully around it

Such an elegant solution.

Will my previously mentioned equation
a*(e^(b*V))=I
by sufficient in helping my find a solution with this method? That is, is my equation a good model to use?

After pushing numbers around, I've come up with
ae^(xb)=(V-x)/R
where x is the voltage drop across the diode, V is the source voltage, and R is the resistor resistance. a and b are specific to the diode used.
I'm not sure how to, if it's even possible, set x equal to something. That is, solve this algebraically. It seems as though it IS transcendental after all.

Last edited: May 14, 2012
15. ### bretm Member

Feb 6, 2012
152
24
You need to get f(x)*exp(f(x)) on one side and then use Lambert W.

That equation is fine if you know what a and b are. If not, you can take two or three measurements (e.g. measure V at I=20mA and measure I at V=600mV) and then fit the values to the data.

Then if you want to solve it directly from a and b:

http://www.wolframalpha.com/input/?i=a+*+exp%28b+*+%28V-+x+*+R%29%29+%3D+x

(The "product log" function is the Lambert W function I mentioned.)

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16. ### MrChips Moderator

Oct 2, 2009
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Imagine a fixed load resistor R in series with a non-linear device with unknown resistance Rx, such as a diode.

The resistor load line (straight line) is the locus of all possible solutions for any value of Rx ranging from Rx = 0 to ∞.

i.e. when Rx = 0,

Vx = 0
Ix = Vs/R

When Rx = ∞
Vx = Vs
Ix = 0

Hence the solution (Vd, Id) must appear somewhere on the straight line.
Thus, the only possible solution given the diode V-I curve must be where the two lines intersect.

This is a common technique for finding the proper operation point (called the Q-point) using the transfer characteristic curves of tube and transistor amplifiers.

17. ### Austin Clark Thread Starter Member

Dec 28, 2011
409
44
So, would this solution be correct?
http://www.wolframalpha.com/input/?i=ae^%28xb%29%3D%28V-x%29%2FR

I would simply measure a few values and fit a and b to whatever diode I am using, and just plug in the rest?

also, I have one question about the "load line", it represents the current through the resistor given every possible resistance value of the diode, which makes sense, but would a better explanation be that the load line represents the current going through the resistor as V across it decreases and V across the diode increases (in other words, the diodes relative resistance in the circuit increases and thus the amount of voltage dropped across it increases). I guess my only problem is trying to understand why it's a straight line, and why it works. I'm trying to visualize and get an intuitive understanding of the concept, but I'm not quite there yet.

As for the Lambert W function... that's just gonna have to wait a bit. I'll figure that one out eventually

18. ### Ron H AAC Fanatic!

Apr 14, 2005
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The load line works because it is the locus of of current through the resistor for all possible values of voltage across it. If Vd is the voltage across the diode, then (Vs-Vd) is the voltage across the resistor. The current is identical for both curves at their intersection.

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19. ### Austin Clark Thread Starter Member

Dec 28, 2011
409
44
Welp, that actually made things click well for me, thank you very very much! You're basically finding the point where the current is the same for each, AND the voltages both add up to the source voltage.

What threw me off at first was that the X axis was labeled Voltage, so I was confused why the current through the resistor seemed to indicate that current went DOWN with voltage, which is false. It just threw everything off for me. BUT, now that it's been cleared up, it all makes perfect sense.

20. ### Ron H AAC Fanatic!

Apr 14, 2005
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You should be able to get the curve for your diode by taking voltage and current readings at two different currents. Using the diode equation, you can first solve for the ideality factor, η (take the ratio of I1/I2). Once you have done that, you can use one of the V-I data sets to solve for Is. Then you can plot the curve.