Luxeons, resistors and current measurement

[Streak]

I have read many CPF articles about the LS and how to prevent frying it. I have managed to get hold of one to play with and have some questions.

One CPF member sais that he fed his LS at 3.4v and 870mA. How is this achievable? I fed about 5 volts into mine with no resistor and only got about 300mA. Others have achieved 300mA with 3.2v. How can I be exceeding the suggested Vf of 3.42v with no resistor and not achieving 350mA?
Am I missing something here, forgot how Ohm's law works or am I miss reading my Fluke 77?? Set on 300mA scale and in series with the positive. I know the 300mA scale has a shunt in circuit which could give false readings. Should I be using the 10A scale instead?

The application sheet for the LS says that with a 6v battery pack you should use either an 11 or 8 ohm (depending on ranking) resistor to keep currents down to 350mA. Applying Ohm's law that should mean 5.5 ohms at say 3 volts and 8.25 ohms at 4.5 volts.

However at 3 volts being below Vf and current below 350mA there should be no need for the series resistor.

I have been using a mini PWM circuit driven by 6 NiCd and a 20 ohm resistor in series with the LS, to control brightness. I guess I could just as well use 3 NiCds and a 10 ohm resistor to get the same results??

Lots of questions, lots of rambling. Help me understand.

Thanks.

 

[Klaus]

Sreak - try using the 10A scale - my Fluke too is way off at the lower scale.

I had good results in using 3 Nimhs cells with the LS and no resistor (just for a short experiment)

Klaus

 

[remuen]

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Streak:
I have read many CPF articles about the LS and how to prevent frying it. I have managed to get hold of one to play with and have some questions.

One CPF member sais that he fed his LS at 3.4v and 870mA. How is this achievable? I fed about 5 volts into mine with no resistor and only got about 300mA. Others have achieved 300mA with 3.2v. How can I be exceeding the suggested Vf of 3.42v with no resistor and not achieving 350mA?
Am I missing something here, forgot how Ohm's law works or am I miss reading my Fluke 77?? Set on 300mA scale and in series with the positive. I know the 300mA scale has a shunt in circuit which could give false readings. Should I be using the 10A scale instead?
<HR></BLOCKQUOTE>

It seems you have missed the most important posts

.

If you put the amp meter in series with the LS (which is basically right) you won't get a correct result. This is because the internal resistance of the amp meter is in series with the LS. And it seems that the internal resistance of your DMM is much too high. That's why you will never get 350 mA at 3.x volts.

You should use a 0.1 ohm resistor in series with the LS, measure the voltage across it and then calculate the current with the formula I = U / R.

You will find some more informations about this in the 'Zetex-Thread' eg. on page 9 ( http://www.candlepowerforums.com/cgi-bin/ultimatebb.cgi?ubb=get_topic&f=3&t=000983&p=9 ) and on page 23 ( http://www.candlepowerforums.com/cgi-bin/ultimatebb.cgi?ubb=get_topic&f=3&t=000983&p=23 ) where MrAl has explained it.

 

[remuen]

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Klaus:
Sreak - try using the 10A scale - my Fluke too is way off at the lower scale.
<HR></BLOCKQUOTE>

Klaus, sometimes thinking about something is faster than fast doing something

- oh sorry, I forgot that you like smoke and fire

 

[bikeNomad]

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Streak:
I have read many CPF articles about the LS and how to prevent frying it. I have managed to get hold of one to play with and have some questions.

One CPF member sais that he fed his LS at 3.4v and 870mA. How is this achievable? I fed about 5 volts into mine with no resistor and only got about 300mA. Others have achieved 300mA with 3.2v. How can I be exceeding the suggested Vf of 3.42v with no resistor and not achieving 350mA?
Am I missing something here, forgot how Ohm's law works or am I miss reading my Fluke 77?? Set on 300mA scale and in series with the positive. I know the 300mA scale has a shunt in circuit which could give false readings. Should I be using the 10A scale instead?

The application sheet for the LS says that with a 6v battery pack you should use either an 11 or 8 ohm (depending on ranking) resistor to keep currents down to 350mA. Applying Ohm's law that should mean 5.5 ohms at say 3 volts and 8.25 ohms at 4.5 volts.

However at 3 volts being below Vf and current below 350mA there should be no need for the series resistor.

I have been using a mini PWM circuit driven by 6 NiCd and a 20 ohm resistor in series with the LS, to control brightness. I guess I could just as well use 3 NiCds and a 10 ohm resistor to get the same results??

Lots of questions, lots of rambling. Help me understand.

Thanks.<HR></BLOCKQUOTE>


You need resistance in series with the LS. In your circuit, that's coming from the meter's shunt resistance and the internal resistance of the batteries.

You're risking damage to the LS if you don't use a resistor and put your meter on the 10A scale.

I don't know how you got your projected resistor values ("using ohm's law..."), unless you believe that the LS has resistance, which it doesn't.

The 8 ohm figure for 6V supply came from:

(6-3.2)/0.350

So with a 4.5V supply you'd use a

(4.5-3.2)/0.350 = 3.7 ohm resistor

But the problem is that at 350mA the LS could show from 2.9 to 3.8V forward voltage. Using a resistor without measuring the LS forward voltage at desired run current can result in your current being too high.

 

[Jonathan]

It is very rare that I disagree with bikeNomad, but this is one of those rare moments.

Of course a Luxeon has resistance. There is a potential difference across the device, current is flowing through the device, and power is being dissipated in the device. However what the Luxeon is not is a resistor.

Ohm's law states very simply that the voltage across the device is equal to the current through the device times the resistance of the device. It doesn't say anything about any of the terms being constant.

A resistor is a device which in the ideal case has a _constant_ resistance. Most other devices have resistance which changes with the applied voltage, and even real world resistors show some amount of resistance change as the applied voltage changes. While you can use resistance to describe a component at a particular applied voltage, what you really need to do in the case of LEDs is look at a graph of current versus voltage for the particular device.

For each point on the current versus voltage curve, you can take the ratio and calculate the resistance _at that particular voltage_. Additionally, you can look at the slope of this graph, and calculate the 'incremental resistance', which tells you how much the current will change when the applied voltage changes. An ideal resistor has a graph which is a straight line through the origin (0 amps at 0 volts).

Many LEDs can be described by a current versus voltage graph that holds at zero current until some threshold voltage, and then shoots up very rapidly. Up to the threshold voltage, the LED essentially has infinite resistance; above the threshold voltage the incremental resistance is essentially zero.An approximate model of an LED is a fixed voltage drop in series with a low value resistance.

It is from the above model that bikeNomad produced the equations in his post. Think of your system as a series connection of the battery as a fixed voltage source, the LED as a fixed voltage drop, and the resistor in series. You simply subtract the LED threshold voltage from the battery voltage, and you are left with the voltage that the resistor must drop. You use Ohm's law for the calculation of this component.

Now, real LEDs actually have a finite incremental resistance after their threshold voltage, and batteries have internal resistance as well. That is why people can get away with connecting batteries directly to LEDs, if they are lucky. As bikeNomad has pointed out, there is actually quite a bit of variation the the threshold voltage of the Luxeon devices, which means that some Luxeons will match some batteries, without any resistor at all, and that if you are designing a system with a resistor, there will be a pretty wide variation in resistor values to use.

Regards,
Jon

 

[Streak]

Many thanks for all of your comments. I now understand and can continue experimenting. In the absence of a step down regulator it seems the best design to minimise resistor heat is to use as low a voltage as possible. Instead of the 7.2 v I have been using with a 10 ohm resistor that needs to handle in excess of 1.2W I should rather use 4.5v with about a 3.5 ohm resistor at .35w.

Thanks again

Streak

 

[Jonathan]

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Streak:
Instead of the 7.2 v I have been using with a 10 ohm resistor that needs to handle in excess of 1.2W I should rather use 4.5v with about a 3.5 ohm resistor at .35w.

Thanks again

Streak<HR></BLOCKQUOTE>

This is correct from the point of view of minimizing losses in the resistor. However remember the point about different LEDs having different threshold voltages...the lower the 'headroom' between the battery voltage and the LED voltage, the greater the variation in voltage across the resistor as the LED voltage changes, and the greater the resulting variation in current as LED voltage changes.

For example, with a 7.2V supply and a 2.8V Luxeon, your resistor needs to drop 4.6V, and for 350mA you would need a 13 ohm resistor. If it were a 3.5V Luxeon, then the drop would be 3.7V and you would need a 10.5 ohm resistor. But if you used the same 13 ohm resistor, then the current would have dropped to 285mA, which would be fine for many applications.

Now do the same calculation for possible threshold voltages, but start with 4.5V. 2.8V Luxeon -> 1.7V drop -> 5 ohm resistor. 3.5V Luxeon -> 1V drop -> 3 Ohm resistor. If you use the higher value resistor and you get a high voltage Luxeon, then your current will be quite low.

In general, with simple drop resistors and LEDs, you get higher efficiency with lower battery voltages as compared to LED string voltages, and you get better regulation of current with higher battery voltages as compared to LED string voltages.

Good Luck!

Jonathan Edelson

 

[bikeNomad]

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Jonathan:
It is very rare that I disagree with bikeNomad, but this is one of those rare moments.

Of course a Luxeon has resistance. There is a potential difference across the device, current is flowing through the device, and power is being dissipated in the device. However what the Luxeon is not is a resistor.
<HR></BLOCKQUOTE>

Jon is right, of course. I was just trying to point out that you can't pretend that the LS is a fixed-value resistor and use Ohm's law. The resistance is non-linear, which makes the model of a voltage drop in series with a resistor the right one to use.

Remember too that once you're into the conductive range of the LS (say >20mA or so), the equivalent resistor value is about 1 ohm (that is, imagine a 3V drop and a 1 ohm resistor in series). So for every 0.1V of voltage increase you get 0.1A of current increase.