array wizard

[abnormal]

Hi fellow LED'ers,

Cool site!

Maybe someone can help with a question I have.

I've been using the handy array wizard (very nice tool btw ) and wondered if I was getting correct results.

With 4V source, 2V LED V drop, 15 mA current, and 2 LEDs the wizard gives me:
a string of 2 LED's and 1 ohm R.
How can any current flow when the source V = the LED V drop?
If you change the current to 20mA or 5mA or...., it still gives a 1 ohm R.

I think this may confuse noobies or am I doing something wrong?

Cheers!

[chopper]

Continue playing with the wizard entering different source voltages and have Your forward voltages combined equal the source and it will tell You to use a 1 ohm 1/4 watt every time(I think) That is a good thing since little power is consumed by the resistor! Yes it is correct although a little confusing till You build the array or give it some thought. One of the bigguns will post another reply telling You what law this pertains to and that it was named after some person that was born back when moby dick was a minnow. Carl

Oh...I nearly forgot, Welcome to the boards!!

[justDIY]

One of the bigguns will post another reply telling You what law this pertains to and that it was named after some person that was born back when moby dick was a minnow. Carl

Oh...I nearly forgot, Welcome to the boards!!

LOL - yea, what he said 

the 1 ohm resistor is one of two things, mathmaticaly ohms law would say you need zero ohms ... but LEDs without any resistors is asking for trouble ... so rounding up, the next logical size is 1 ohm.   This way, if you get for some reason, 4.2 volts, you'll have a little life insurance for your LEDs.

perhaps the Big Chief himself will come along and give us a few paragraphs on "why one ohm!"

and last but certainly not least, welcome to the forums.

[SurJector]

the next logical size is 1 ohm.   This way, if you get for some reason, 4.2 volts, you'll have a little life insurance for your LEDs.

The 1 ohm resistor is certainly a little life insurance, but not a very good one: if your voltage drop is 4.0V and 4.1V comes from your generator, you'll get 100mA through your LEDs and (at least) one will fry in short order.It sounds more reasonable to have a 10ohm resistor, even though you will have less intensity and thus less light.

[Rob]

"why one ohm?"

To take a step back and deal with why use a resistor at all: The resistor is intended to prevent changes in the current-vs-voltage curve from damaging the LED's junction (the working part).

The exact current that flows for a given applied voltage is an exponential function of a few terms, one of which is temperature. So with nothing to limit current, the LED warms as it operates, and will tend to draw more current from a voltage source. Assuming your voltage source is ideal, the voltage will remain perfectly constant and things go wrong rapidly--the junctiuon heats up more, and draws more currnet, leading to a thing called thermal runaway.

So, is one ohm enough to do the job? Every milliamp of additional current will cause a millivolt of voltage drop. I'm not sure what the basis of SurJector's calculation is, or where the 100 mA figure comes from. Here's how I'd figure it (put your math goggles on  )...

We'll use the Shockley equation to figure the diode current at a given voltage:
 i = I_S(exp(vq/kT) - 1), where i is the forward junction current and v the forward junction bias.

I'm going to do the usual shortcut with the constants and the temperature and take kT/q = 26mV, which corresponds to room temperature. (Check my math if you like, k is the Boltzmann constant, and q is the protonic charge--set T = 300K and you should get my value).

We need a value for I_S, the reverse saturation current. Since we have one point on the operating curve, we can solve the Shockley equation for I_S. I'm going to use 2V @ 15mA from your original post.

I_S= i / (exp(vq/kT) - 1) = 0.015A / (exp(2V/0.026V) - 1) = 5.87 × 10^-36A

Now, Kirchoff's voltage law tells us that the voltage drop on the diode plus the voltage drop in the resistor must equal the source voltage. We need to rearrange that Shockley equation to get the voltage interms of the current, but we now have a "constant" value for I_S.

V = 0.026V * (ln( 1 + i/I_S)), where I_S = 5.87 × 10^-36A

So what we'd like is current in the diode as a function of source voltage, for a 1 ohm resistor. We know that the voltage across the diode has to be the source voltage minus the voltage drop across the resistor (i*R). So that gives us a handy equation:

i - IS * (exp ((VS-i*1ohm)/0.026V) - 1) = 0

anyhows, for R = 1 ohm, and VS = 2V (I'm just doing the single LED case to avoid doubling the Shockley term) I get between 10 and 11 mA for the current. If I drop in 10 ohms as SurJector suggests, then I get between 3 and 4mA.

If I bump that VS up to 2.1V, R = 1ohm, then I get between 62 and 63 mA.
For VS = 2.1V, R = 10 ohms, then between 10 and 11 mA.

All the above assume the junction remains at room temp, which may not be a terrific assumption. I'd expect real observed currents to be higher.

*** math goggles off now ***

So, SurJector's point about "good insurance" is pretty much correct. Bigger resistors are better insurance.

I had to make a design decision about what the lowest "useful" value resistor would be. I still get a lot of mail asking "one ohm? can't I just skip it?" I'll have to consider all the ramifications, but it seems like 10 ohms would probably be a better bottom of the range based on this calculation.

I've attached the Excel worksheet that I used to solve for current, in case you guys want to fool with it, or see the effects of changing the model's inputs.

[Rob]

Forget about the spreadsheet, I am pretty close to done with an online calculator for the Shockley model:
http://led.linear1.org/shockley.wiz

I'm fixing some presentation issues right now, but the calculations are working. I've been meaning to write this for a while now.

[Rob]

I removed the temperature as an input, becuase I couldn't figure out a way to model the carrier concentration dependency on temperature. This raises an interesting discussion:

If you followed along thrugh the math writeup above, you probably wanted to post and ask "Rob, you say that thermal runaway happens when things get hot, but the exponential term is related to the inverse of temperature, what's up with that? Are you crazy, or just lying to us? Perhaps both?"

The answer there is that not just the exponential term, but also the reverse saturation current I_S depends on temp. Specifically, the mechanism is that when things get hot, I_S goes up due to the increase in majority carrier concentration. This varies proportionally with exp(-E_A/kT), where E_A is an activation energy.

Since I'm backing into that term using the operating point from (probably) the datasheet, I'm limited to that temperature, as far as I can see. I locked that at 25C to be consistent with most datasheets I've seen.

[Rob]

anyhows, for R = 1 ohm, and VS = 2V (I'm just doing the single LED case to avoid doubling the Shockley term) I get between 10 and 11 mA for the current. If I drop in 10 ohms as SurJector suggests, then I get between 3 and 4mA.

If I bump that VS up to 2.1V, R = 1ohm, then I get between 62 and 63 mA.
For VS = 2.1V, R = 10 ohms, then between 10 and 11 mA.

All the above assume the junction remains at room temp, which may not be a terrific assumption. I'd expect real observed currents to be higher.

You can now run all these cases with my Shockley calc. I linked the quote accordingly.

Differences are because I'm using kT/q at 25C instead of 0.026 mV like in the excel workbook.

[abnormal]

Thanks for the help and kind welcome guys.

Rob,  wow... you have reached the panicle of best replies I’ve ever received on the net!
Thanks for the great explanation and work on the new Shockley diode modeler.

This is in part why constant current sources are nice for LED array applications.
Current holds fast through source voltage and LED temperature changes.

I still have my googles on   
Thanks,
Tom

[SurJector]

I'm not sure what the basis of SurJector's calculation is, or where the 100 mA figure comes from.

I've just made the usual approximation: Vf=const=2V and perfect voltage source. If you add .1V to the voltage source, you add .1V to the voltage across the 1ohm resistor which translates into a .1A=100mA current increase. There are various inaccuracies in this model, but I've given an idea of the current increase. Your computation is obviously more precise !

[xmastree]

Hi, I joined up just to say this...
About the array wizard and the 1 ohm thing. We all know that 1 ohm is no good, as any variation in supply voltage will cause a huge current difference in the LED.

For example, I tested it with sixteen 3V@20mA LEDs from a 12V source, and it put four in series with 1ohm. If I were actually making that, I'd put three in series, leaving more voltage available for the resistor.

I see it's been discussed a little, but to me the answer is simple. I think the wizard ought to factor in a minimum resistor voltage, depending on the application. For computer lighting, where the 12V is pretty stable, 3V would probably be ok. For automotive applications, where the 12V is usually 13.8V, I think 5 or 6V would be better.

Assuming perfect (same voltage at any current) LEDs, that 1.8V shift (from 12 to 13. would cause a 30% increase in current with a nominal 6V across the resistor. With 1 ohm it doesn't bear thinking about... 

[Rob]

Welcome to the boards. As noted above, I'm considering how to make the improvement.

[justDIY]

when we start talking about minimum voltages, and such, that really introduces a lot of variables, into an otherwise easy to use calculator.

Personally, I like to keep the voltage across the resistor to a minimum ... 3v is way too high, you have to consider the power dissipation, and I like to dissipate my power into LEDs instead of into resistive heaters.

Situations where the 1 ohm resistor generally comes into play arise out of erronious input, eg the current tail-light fad.   Inputting 12 volts as the supply voltage is inaccurate, the only time you'll see that voltage is with a damaged charging system or a dead battery, or for the brief instant the starter motor is engaged.  Inputting 13.7 volts yields the proper response.

I suppose, using your example, in the situation of a case-mod application where the regulated 12v rail will actually be held at 12v, we would need a different solution.   Perhaps a recommendation from the wizard that the supply voltage is too close to the forward voltage to ensure proper current limiting, in the case of the series only wizard.  In the case of the series/parallel wizard, a resistance value of some arbitary number (10?) should be set as the minimum, and if R < 10 then subtract a diode from the series string, and/or start a new string.

[xmastree]

Personally, I like to keep the voltage across the resistor to a minimum ... 3v is way too high, you have to consider the power dissipation, and I like to dissipate my power into LEDs instead of into resistive heaters.

If you're going for ultimate efficiency, ok. But the lower the resistor, the more sensitive the whole thing is to voltage fluctuations.

Three 3V@20mA LEDs and one resistor, also dropping 3V, on a 12V supply.
240mW total power, 60mW, or 25% of the total is wasted in the resistor.
The important thing is it would still light at 11.5V, where 4 LEDs is series may have difficulty.

Inputting 12 volts as the supply voltage is inaccurate, the only time you'll see that voltage is with a damaged charging system or a dead battery,

Or stopped, with the headlights and heated rear window on. You roll down the window to speak to the cop and his partner sees your tail lights go out as the extra load causes the voltage to dip.

In theory, the charging system should maintain 13.7V, but in the real world people do have less than perfect systems.

[Rob]

But the lower the resistor, the more sensitive the whole thing is to voltage fluctuations.

You have it right.

I'm trying to figure out how to accomodate high-current situations in particular. I suppose I could have a certan floor resistance as long as current was below some threshold, and a different lower floor above that threshold.

The wiz currently just finds (Vs-Vf)/Id for each series string, then calls the rounding routine on that value.