Need help with "Ohm's Law"

robk

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Need help with \"Ohm\'s Law\"

I've been making quite a few ZLT+ boards for use in 2D Mags for buddies of mine. They work great, although they are not terribly efficient - but with D cells, it doesn't really pose a problem. Here's my question: I use a 0.1 ohm / 1% current sense resistor in series with the LED to "tune" each board for about 500mA (or 50.0 mV across the sense resistor). If I read , say, 490 mA, what is the true current when I remove the current sense resistor and package the whole thing up? It's got to be 10 - 50 mA higher, doesn't it? It's becoming more important now, as I am etching the emitter resistor on the board instead of using a thin wire. Thanks for any assistance!
Rob
 

CM

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Re: Need help with

robk,

I think you'd have to know the I/V characteristic of the LED to answer that question. So let's say you increase the voltage to the LED by 50mV in your example. Knowing the I/V characteristic let's you know how much the current increases when the forward voltage is increased by 50mV. Unfortunately, this is not a trivial measurement to do. My guess (beware) is that this curve is roughly the same shape for LED's from different dies but it's location varies in the X-axis (voltage axis) depending on the Vf of the LED in question. Lumiled has this curve on one of their data sheets. Unfortunately again, there isn't a whole lot of resolution in their graph. You may be able to blow this up and then estimate the dI/dV. Here's the link to the data sheet. It's Figure 3a

http://www.lumiled.com/pdfs/protected/DS23.PDF

Good luck

CM
 

robk

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Re: Need help with

That's what I was afraid of! There doesn't seem to be an easy answer, but does it really matter? As long as I am a bit conservative on setting the current draw with the current sense resistor in place, the final circuit current should be just a bit over what's measured.
Thanks CM!
Rob
 

Doug Owen

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Re: Need help with

Fifty milivolts out of thirty six hundred or so? Not to worry. Less than two percent?

If you're really concerned, measure the battery current before and after, but this should be a trivial increase, far less than heat and other variables.

Doug Owen
 

CM

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Re: Need help with

robk,

Just looked at the data sheet again and did some calculations for you. 50mV correspond to approximately 42mA. At the nominal drive of 350mA, this is a 12% increase, which IMO is not trivial. So as a first order estimate, whatever current you read with the sense resistor, figure on 10-12% increase when you remove it. As Doug mentioned, you should measure the battery current before and after to see how it compares with the "expected" increase which we derived from the data sheet. Let us know what you find.

CM
 

MikeEvans

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Re: Need help with \"Ohm\'s Law\"

Here's the mathematical answer, it's a problem I have to deal with often. You have a 3 element series circuit, with R1 (current test resistor), R2 (all the rest of the electronics) and the LED. My luxeons showed a voltage drop of 2.9 plus 2.0 times the current I applied, (I graphed it out it was pretty linear from 0 to over 300 ma), using that in the calculation, ohms law, simple algebra...Current flow is I1, remove resistor R1 (make it 0), new current is I2... ratio new current to old: (I2/I1) is equal to 1 + (R1 / (R2 +2)) *The 2 came from my current test that show the LED for this equation acts as a 2 ohm resistor. So in worst case if R2 were 0 the current increase by removing .1ohm resistor would be 5% but if R2 were acting as 8 ohms the current increase would be only 1%. If the 'rest of the electronics' is active circuitry it may not be a fixed value but for the tiny current change it won't likely affect the numbers very much. The math part is HERE
 

Doug Owen

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Re: Need help with

[ QUOTE ]
CM said:
robk,

Just looked at the data sheet again and did some calculations for you. 50mV correspond to approximately 42mA. At the nominal drive of 350mA, this is a 12% increase, which IMO is not trivial. So as a first order estimate, whatever current you read with the sense resistor, figure on 10-12% increase when you remove it. As Doug mentioned, you should measure the battery current before and after to see how it compares with the "expected" increase which we derived from the data sheet. Let us know what you find.

CM

[/ QUOTE ]

While this calculation would be the case for a proper voltage regulator, it's not correct here. The ZLT is not a traditional regulator in that it doesn't monitor output but rather monitors charging current in the inductor. A controlled amount of energy is stored in the inductor and then released. The output current and voltage go where the load tells them to.

Consider that when the resistor is there it drops less than 2% of the output Voltage, the rest is in the LED itself. This means less than two percent of the total energy developed. I maintain that this is the only energy available to raise the current in the LED. Just where did you expect the 'extra' 10 plus percent to come from?

(No 'magic' parts here....).

Like I said, before and after current reading will tell the story, but my moneys on conservation of energy......

Doug Owen
 

eluminator

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Re: Need help with

I'd use the empirical approach. That way I don't have to think, and thinking gives me a headache.

You could vary the resistance and see how that affects the current. Putting two .1 ohm resisters in parallel would cut the resistance in half. Putting two .1 ohm resisters in series would double the resistance.

With luck you might find the current doesn't change much and could probably assume it won't change much if you remove the resistor.

This is a good lesson for those unfamiliar with how a digital multimeter works. They are probably wondering why you don't use one to measure the current. As you know, but some don't, a DMM actually only measures voltage. To get a current reading, it puts a small resistor in series with the circuit and measures the voltage across it, just like you are doing.
 

Doug Owen

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Re: Need help with

If you use a high enough Amps range, you actually end up using a smaller resistor. For instance here, a 10 Amp range typically uses a .01 ohm resistor, dropping 100 mV at 10 Amps (this is still one Watt!). Bummer is you typically have many times that in lead resistance.

Doug Owen
 

CM

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Re: Need help with

Doug,

Just looked at the ZLT circuit. You are right. It does monitor the inductor current (not the load current like the regulators I'm used to) so a different method must be used for calculating. That kinda sucks since too many other factors come into play. Guess that's the price for keeping it simple. Oh, yes I do know about the laws of thermodynamics (blah blah blah) so you don't have to rub it in. It was simply a misunderstanding of the mode of operation of the part /ubbthreads/images/graemlins/grin.gif

I also noticed the reference voltage on this part varies from 19mV +/- 25%. That's a huge swing. Though I like it's simplicity, I think I prefer other monolithic solutions from Linear Tech/TI/National. Not a slam to Zetex, since they have a solution many flashaholics can work with easily.

CM
 

Doug Owen

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Re: Need help with

quote]CM said:
Oh, yes I do know about the laws of thermodynamics (blah blah blah) so you don't have to rub it in. It was simply a misunderstanding of the mode of operation of the part

I also noticed the reference voltage on this part varies from 19mV +/- 25%. That's a huge swing. Though I like it's simplicity, I think I prefer other monolithic solutions from Linear Tech/TI/National. Not a slam to Zetex, since they have a solution many flashaholics can work with easily.


[/ QUOTE ]

I'm sorry, my intent was not to 'rub it in', but to cover all the bases. I don't know you from Adam's off ox, and have to make guesses about your skills based on what I read. The point was you were making some very authoritarian statements condemning my statements and were in fact yourself in error. Members were in danger of taking your (bad) advice as Gospel. Having no real way to be sure how well I was getting my point across, I used a couple of 'arguments'. Remember, in this format you and I aren't the only ones in the conversation, right? You're not the only one I need to convince.

Again, my intent was not to 'dis' you. I hope if you remove personalities from this you can see that. To those of us who deal with the nuts and bolts of the real world, 'right and wrong' are different from 'correct and mistaken'.

I agree with you that the ZLT device is not the best for many uses, but it does offer many outstanding advantages to those who might want them. This is, of course, the true beauty of the Capitalist system, the customer gets to decide (not you and I) what's a good product. Based on their success, some see advantages.

Doug Owen
 

robk

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Re: Need help with

Thanks guys. A lot of good ideas brought up. I'll try measuring battery input current (just with a DMM on 20A range) with and without the current sense resistor in series with the LED, see how much it changes. Also I like the idea of adding another 0.1 ohm and see if the current to the LED drops - so simple, and would give a rough indication of what happens when I remove all resistance.
Rob
 

CM

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Re: Need help with

Doug,

No problem, my apologies as well for my first post. If you notice when I wrote it, it was very late and I think I was getting delirious (been playing with a new toy). You're right, I should have read the details of the Zetex part to get my facts straight before making my post.

CM
 
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