Realistic Diode-Resistor Series Voltage Drop

Thread Starter

Austin Clark

Joined Dec 28, 2011
412
I tried finding values for the VI curve of a diode, but my mulitmeters resistance influences the current too much, and I can't get a curve to fit it, it's not a sharp enough curve. I guess I could try measuring the resistance of my meter, and try and figure out the real VI curve given that, but I dunno...


Also, bipolar junction transistors are supposed to "multiply current". Would this model be a better way to determine current through the Base/Emitter? Also, does the transistor still multiply current if I have a resistor between the collector restricting the multiplied current?
 

Ron H

Joined Apr 14, 2005
7,063
I tried finding values for the VI curve of a diode, but my mulitmeters resistance influences the current too much, and I can't get a curve to fit it, it's not a sharp enough curve. I guess I could try measuring the resistance of my meter, and try and figure out the real VI curve given that, but I dunno...


Also, bipolar junction transistors are supposed to "multiply current". Would this model be a better way to determine current through the Base/Emitter? Also, does the transistor still multiply current if I have a resistor between the collector restricting the multiplied current?
I think something else is wrong. A normal multimeter with 10Meg input resistance will only draw about 70 nanoamps when across the diode, when the diode has on the order of 1 to 10mA through it. This is insignificant.
A simple transistor doesn't work in a way that would be useful. You can convert current to voltage with an op amp, but you don't need to do that. Unless you have a high voltage supply or an adjustable current source (so that the voltage burden of your ammeter is insignificant when you insert and remove it), you will need two meters, one in series to monitor current and one across the diode to monitor voltage.

EDIT: How do you know the curve is not "sharp" enough?
 

t_n_k

Joined Mar 6, 2009
5,455
Alternatively, if you don't have much in the way of equipment this might be accomplished with one DVM / multimeter. Get a few 1% resistors (say 5 different values from 470Ω to 47kΩ) a DC source - a 9V battery would be OK. Construct a series circuit comprising the forward biased diode, resistor(s) and battery.

Swapping in / out series resistor values measure the resistor voltage drop and the diode forward voltage drop in each case - 5 resistors would give five measurement cases. The resistor voltage drop can be used to estimate the current [I=Vr/R] in each case. A 1% resistor tolerance should make the current estimate for each case reasonably good.

This gives 5 cases of diode forward bias voltage vs diode current from which you could make a reasonable estimate of the VI characteristic.
 

Ron H

Joined Apr 14, 2005
7,063
Alternatively, if you don't have much in the way of equipment this might be accomplished with one DVM / multimeter. Get a few 1% resistors (say 5 different values from 470Ω to 47kΩ) a DC source - a 9V battery would be OK. Construct a series circuit comprising the forward biased diode, resistor(s) and battery.

Swapping in / out series resistor values measure the resistor voltage drop and the diode forward voltage drop in each case - 5 resistors would give five measurement cases. The resistor voltage drop can be used to estimate the current [I=Vr/R] in each case. A 1% resistor tolerance should make the current estimate for each case reasonably good.

This gives 5 cases of diode forward bias voltage vs diode current from which you could make a reasonable estimate of the VI characteristic.
Good suggestion. Another alternative, if you have the specs on your multimeter, is to substitute its burden resistor when you remove it from the series circuit. For instance, my Fluke 87 manual says the voltage burden is 102μV/μA on the 5mA scale. This is, of course, equivalent to 102Ω. So, with my MM, if I measure current on the 5mA scale, then remove the meter and replace it with a 100Ω 5% resistor, the current will remain essentially the same while I use the meter to measure diode voltage.
 

Thread Starter

Austin Clark

Joined Dec 28, 2011
412
Those are some great suggestions. I'll try them out and post my results here sometime soon.


*EDIT: Here are some screenshots of my results. The Blue line is the actual recorded data, the Red line is the theoretical model, adjusted to fit the top and bottom values.
This isn't quite accurate enough for me, I was expecting at least within 5%, I was hoping for 1%. I used two meters, one measuring voltage and another current for this.
 

Attachments

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bretm

Joined Feb 6, 2012
152
To fit the curve at more than two points, you're going to have to use a model that has more than two degrees of freedom. For example, if you model the diode as having an equivalent series resistance and an equivalent shunt (parallel) resistance, that's two more degrees of freedom. You can calculate the two resistance values that will make your curve fit four of the measurements instead of just two.
 

Thread Starter

Austin Clark

Joined Dec 28, 2011
412
To fit the curve at more than two points, you're going to have to use a model that has more than two degrees of freedom. For example, if you model the diode as having an equivalent series resistance and an equivalent shunt (parallel) resistance, that's two more degrees of freedom. You can calculate the two resistance values that will make your curve fit four of the measurements instead of just two.
Yeah, I thought of that, but how would I try and calculate those values now? Check and guess? And even still, calculating the IV characteristic of the diode with series and parallel resistance requires knowledge of the "pure" or ideal IV characteristic.
 

Thread Starter

Austin Clark

Joined Dec 28, 2011
412
What is the part number of your diode?
I'm trying to learn theory, not develop a specific circuit, that all comes later. I like to have a 100% full working understanding before I begin projects. Of course, you learn a lot just by failing, but in this case there's not a lot to learn that way.
 

t_n_k

Joined Mar 6, 2009
5,455
I did a test with a 1N4003 diode and the assumed exponential curve fitting results using CurveExpert are shown.

The exponential fit works OK but it doesn't turn out to the the "best fit" for the range of data.

Volts Amps
0.625 0.00248
0.644 0.00381
0.668 0.00642
0.677 0.00772
0.687 0.00968
0.697 0.01228
Exponential Fit: y=ae^(bx)
Coefficient Data:
a = 2.52752891872E-009
b = 2.20752225433E+001
Interestingly the best fit for the range of data turned out to be a 4th order polynomial.
 

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Ron H

Joined Apr 14, 2005
7,063
I'm trying to learn theory, not develop a specific circuit, that all comes later. I like to have a 100% full working understanding before I begin projects. Of course, you learn a lot just by failing, but in this case there's not a lot to learn that way.
That's the longest part number I've ever seen.;)
What I meant was, what is the part number of the one you plotted?
You said "screenshots", but I only saw one.
 

MrChips

Joined Oct 2, 2009
30,712
There is no point attempting to fit the data to a 4th order polynomial since we know that the theoretical relationship is exponential.

Using the data you provided here are the results I obtained:

a = exp(-19.7641)

b = 22.0274

Using the standard diode formula:

\(I=I_0e^{( \frac {qV} {kT} )}\)

If we use the common value of

\(\frac{q}{kT} = 40\)


we can use:

\(I=I_0 e^{(\frac{40V}{n})}\)


where,

\(I_0 = e^{-19.7641}\)

and the ideality factor \(n = 1.816\)


Here is the relationship:


 
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Thread Starter

Austin Clark

Joined Dec 28, 2011
412
That's the longest part number I've ever seen.;)
What I meant was, what is the part number of the one you plotted?
You said "screenshots", but I only saw one.
AH, I misunderstood :)

I'm not sure the part number honestly, it was one from an old power supply I believe, but I highly doubt it's defective. I also tried with an LED, but it's IV curve was even MORE strange, it seemed to have a linear characteristic with a sudden curve at higher voltage. Also, half-way in making my measurements it seemed to change its current VS voltage, I literally had to go back and remeasure all the values I had just done moments ago because I was suddenly getting drastically different values... Maybe my equipment is just too shoddy, but it's all fairly new and I don't THINK it's too incredibly cheap. My thought was that it must have heated up or something, or maybe once you get it lit, going back down in voltage more current is allowed to keep flowing... no idea.
 

t_n_k

Joined Mar 6, 2009
5,455
The attachment is a lab write-up on the determination of an LED's V-I characteristic. I just found it doing a search. It's quite well done and comes up with a modified model of the purely exponential type under consideration in this thread.

As in all these matters a model doesn't always perfectly reflect the physical reality. One often has to tweak the model to match the physical reality - particularly over a large operating range of the defining parameters.
 

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t_n_k

Joined Mar 6, 2009
5,455
With the concept of applying some tweaking in mind I looked at the OP's data from post #25. Something similar was done by the author in the attachment included in my previous post.

A slight adjustment to the model equation might be of the form

\(I_{[mA]}=aV_{[mV]}+be^{\( cV_{[mV]} \)}\)

While not recommended I'm sticking with mA and mV variables which is what Austin Clark did in the spreadsheet analysis.

I can set up a user defined model in CurveExpert and look for a match with the data.

This is what the application came up with

User-Defined Model: y=a*x+b*exp(c*x)
Coefficient Data:
a = -4.027E-003
b = 9.882E-004
c = 1.850E-002
This result gives a better outcome than a purely exponential fit as follows

Exponential Fit: y=ae^(bx)
Coefficient Data:
a = 1.143E-004
b = 2.232E-002
The maximum error in the tweaked user defined fit was -4.9% at the 441mV value. At the same voltage the purely exponential form fit gave +18% error. I suspect the OP's data result for this value was an outlier anyway.

I think the user defined tweak I applied is effectively including a large parallel negative resistance (small conductance?) term with the purely exponential model term. It's "annoying" that the tweak turns out to be negative which makes a physical explanation difficult. This probably adds weight to my assertion that the first data value is an outlier and it skews the ideal exponential model inappropriately.
 
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gootee

Joined Apr 24, 2007
447
Sorry to dredge up such an old thread but I was just doing something similar (diode curve fitting) and thought that someone else might find the information to be useful, someday.

I was creating an Excel spreadsheet that does numerical solutions for power supply circuits, including a (scalable) transformer model with winding resistances and leakage inductances, and a bridge rectifier with non-ideal diodes, reservoir capacitance with ESR, optional wire and PCB trace ESRs and ESLs, and a constant current load. The spreadsheet usually uses a second-order differential equation for the circuit. But the diodes make it non-linear. I needed to model the diodes fairly closely, either with an equation or with a table-lookup, and was also planning to initially validate and verify the spreadsheet by comparing its results to a detailed LT-Spice simulation of the same circuit, that I already had.

I noticed that Excel (I'm using 2007) has a curve-fitting option (click in a plot then select Layout at the top).

And I found out that it's extremely easy to export the plot data from LT-Spice. I set up a simple simulation, just to generate the diode data: a simple swept voltage source and a diode; 0 to 1 volt in one second with .001 sec timesteps gave me more than enough data points (although using as many as you can seems to be an advantage).

Excel's curve-fitting leaves a lot to be desired. The only forms that would give decent fits were 6th-order polynomials and ax^b. But even using the Lab Fit software, which automatically tries hundreds of equation forms for you, the form that matches the diode equation never was in the top ten. (Don't necessarily reject the high-order polynomials, either. They are easy to differentiate and integrate! Also, Taylor Series, or something like them, are another often-overlooked way to get good equation models that are very easy to work with.)

The spice diode model that I was using was for the OnSemi MBR20100CT (10 A avg fwd current). I didn't exhaust myself trying different forms of equations. But I did spend at least a few hours at it. I was able to fit the I vs V data fairly well with Y=(A*EXP(B*X+C*X^0.5)+D*X, with A=0.16207E-11 B=-.25129E+02 C=0.56003E+02 D=0.97441E-01.

I also calculated the diode's resistance in a third Excel column and at one point I did a fit for R vs I, and Excel itself was able to get that one, with R = 0.50084*i^-0.79962 (which is nicely simple, in one way), with about 10% max error, although Lab Fit used Y=(A+B*X^C)/(D+X^C) and got 5% max error, and I didn't even try pushing it to get a better convergence.

NOTE that you might have to eliminate the data for extremely-small voltages, before trying to fit a curve that includes the diode resistance. It's also worthwhile to create another column for the equation's calculations, and one next to that in order to calculate the error (percent error is good but also check the absolute error, especially for extremely small data values). In Excel, it's so easy to copy and paste the original plot and then just edit the column letters to plot the error.

Interestingly, the pair that I had the most difficulty in curve-fitting was R vs V, which seemed like the most-reasonable one to use for a passive mathematical model. I think I ended up using Y = A*EXP(B*X^C)+D for R vs V.

Cheers,

Tom
 
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