Circuit design question - Running a Cree/Seoul off of 2 CR123s

[stoven]

My e-fundies are a little rusty and am trying to figure out a cheap and easy way to run a Cree or Seoul off of two CR123 batteries.

Is it possible to use a zener diode to reduce the input voltage from 6 volts down to something closer like 3.3 volts? I know zener diodes regulate voltage but not current. If I use a zener diode to output a slighter higher voltage than the emitter needs and put a resistor in series with the emitter to limit current to 1A would it work?

Any insight would be appreciated.

Thanks,

Steve

 

[Curious_character]

You don't want or need to regulate the LED voltage, you want to regulate the current. The cheapest and most effective way to do this is simply with a series resistor. Since the forward voltage of the LED is roughly 3.7 volts and the nominal cell voltage is about 3.0 volts, you need a resistor that drops about 6 - 3.7 = 2.3 volts at the current you want. For example, if you want to drive the LED at 1 amp, you'll need a 2.3 ohm (use 2.2 or 2.4, which are standard values) resistor. It'll have to dissipate I^2 * R watts, or over 2 watts in this case. This power will be lost, so you'll be throwing away about half the battery's energy into heat. The resistor will provide much better current regulation than a zener, for the same loss in power.

The only way to get better efficiency is a switching regulator, such as the Downboy from the Sandwich Shoppe. But that doesn't fit your requirement of being inexpensive.

(This post was edited. The original post was appropriate for Li-Ion cells, and it was edited to apply to lithium primary cells, which is the subject of the question.)

c_c

 

[stoven]

I don't want to waste half the battery power through the resistor so that's why I wanted to use a Zener. So the only more efficient way than a resistor is using a downboy buck converter? I was hoping that using both a zener and a resistor would work.

 

[Curious_character]

I don't want to waste half the battery power through the resistor so that's why I wanted to use a Zener. So the only more efficient way than a resistor is using a downboy buck converter? I was hoping that using both a zener and a resistor would work.

Sorry, the zener will waste exactly the same amount of power as a resistor with equal voltage drop. Power = I * V, and it doesn't matter whether the V is across a zener or resistor. The advantage of a single resistor over a zener or zener-resistor in this application is that it will maintain a much more constant current as the battery and LED voltages change. Yes, the only more efficient way is a switching converter like the Downboy.

As Robert Heinlein observed, TANSTAAFL. There Ain't No Such Thing As A Free Lunch.

c_c (I've done electronic circuit design for a living for over 30 years.)

 

[stoven]

One last question -

As the batteries drain their voltage decreases right? So as the voltage decreases the change in voltage across the resistor decreases causing a reduction in current right?

 

[AZ Sky]

One last question -

As the batteries drain their voltage decreases right? So as the voltage decreases the change in voltage across the resistor decreases causing a reduction in current right?

That is correct.

 

[Curious_character]

One last question -

As the batteries drain their voltage decreases right? So as the voltage decreases the change in voltage across the resistor decreases causing a reduction in current right?

Yes, that's correct. Now, let's see just how much it'll decrease. All you need to know is Ohm's law to follow it.

For simplicity, I'm going to consider the LED to have a voltage that doesn't change with current. It does, of course, but not much. At around 1 amp, a 200 mA current change causes only about a 0.1 volt change in LED voltage. So we'll consider it constant at 3.7 volts for this analysis.

Your fresh cells will have a voltage of around 3.0 volts at one amp, although they won't stay at that level very long. Suppose you chose a 2.2 ohm series resistor. The initial current would be (6.0 - 3.7) / 2.2 = 1.05 amp. In a short time, the voltage will drop to around 2.5 volt per cell, and stay close to that for most of the discharge period. Then your current will be (5.0 - 3.7) / 2.2 = 590 mA. When the cells are just about dead, their voltage will be about 2.0 volts, so the current wil be (4.0 - 3.7) / 2.2 = 140 mA. That drop will occur, though, only very near the end of the discharge period. Also, at that low current the LED voltage will be significantly lower, so the current drop won't be this bad. If you had used Li-Ion cells with voltage of around 4.0 - 3.5 volts per cell during discharge and chosen a 4.3 ohm resistor, you'd have (8 - 3.7) / 4.3 = 1.0 amp at 4.0 volts/cell and (7 - 3.7) / 4.3 = 770 mA at 3.5 volts/cell. The higher the resistor value, the more constant the current. But the efficiency is lower with the Li-Ion cells and higher resistor value.

You won't find a very good zener with a drop around 2 volts, but let's say you did, and you used it and 0.33 ohms instead of just the resistor. Then when the battery voltage is 6.0 volts (3.0 volt/cell), the current will be (6.0 (battery) - 2.0 (zener) - 3.7 (LED)) / .33 = 910 mA. It'll decrease from 910 mA to zero as the cell voltage drops from 3.0 volts to 2.85 volts, which will take only minutes.

In practice, the internal resistance of the cell and other stray resistances will alter things a bit, but this will give you an idea of what's happening. You can improve the current regulation with a linear current regulator circuit, but it'll have the same or less efficiency than the resistor.

c_c

 

[AZ Sky]

I agree with Curious.

The linear current regulator is probably the best way to go for an inefficiant current regulator (albeit just as power wasting as a resistor).
A switching regulator is really the only way to get an efficiant driver.

I'm thinking that the original poster was thinking of placing the zener in parallel with the LED in the erroneous thought that it would mantain a constant voltage over the led. This would be a useless endeavor as the led is a forward biased diode and would overide the zener. (I might be wrong in the OP's thoughts)

 

[Curious_character]

I'm thinking that the original poster was thinking of placing the zener in parallel with the LED in the erroneous thought that it would mantain a constant voltage over the led. This would be a useless endeavor as the led is a forward biased diode and would overide the zener. (I might be wrong in the OP's thoughts)

Ah, that connection didn't occur to me, I guess because it wouldn't work well at all. You could find a zener (which is a reverse biased diode running at breakdown) with either a higher or lower voltage than an LED which is, as you say, a forward biased diode. The LED has a much higher forward voltage than a conventional silicon diode. But in that connection, whichever has the lower voltage will get virtually all the current. And both an LED and low voltage zener have negative temperature coefficients. So whichever initially gets the current will get warmer, dropping its voltage, so it'll hog the current even more.

c_c

 

[stoven]

Thanks for the help guys. You cleared up a lot for me.