Doing the math - simple battery regulator

[burpee]

How about discussing the exact values for the most optimal way to use a 8xAA NimH batteries for a 7.4V Li Ion battery pack.

In this case I need the 8xAAS battery back {9.6V?}to deliver no more than .85A at 7.4V.

My calc. resulted in a 7.5V 1.3watt diode and a 2.4k resistor, 1 watt.

Any comments. There must have been a thread on this that spells this out, but I can't crap on these forums, or else I find too much.

 

[Mr Happy]

Does this help at all?

Poorman Mutli-Lux setup method

It's not exactly the question you asked, but perhaps it could offer the solution you are looking for.

 

[Selectron]

Your calculation for the resistor value is a long way off-target. You didn't say what the load consists of, but for now that doesn't matter. We know that with an applied voltage of 7.5 volts it will draw a maximum current of 0.85 amps, and so from R=V/I we can calculate that its effective resistance is 7.5V / 0.85A = 8.8 ohms.

Forget about the zener diode for a moment, and just consider what happens when you place the proposed 2.4K resistor in series with the load - total series resistance is now 2,400 + 8.8 = 2,408.8 ohms. Now let's apply 7.5 volts and see what the current flow will be: I = V/R = 7.5 / 2,408.8 = 0.003 amps (3mA).

So your load won't ever receive the 850mA which it requires, because the 2.4K resistor will limit the maximum current to a mere 3 milliamps.

I can go through the zener diode resistor calculation if you want, but just out of curiosity, what type of device will you be feeding with the 7.5 volts? With a current flow as high as nearly one amp, a simple resistor/zener combination is unlikely to be the best solution.

 

[burpee]

So your load won't ever receive the 850mA which it requires, because the 2.4K resistor will limit the maximum current to a mere 3 milliamps.

Thanks Selectron, I knew I had something wrong. I inadvertently entered "85" instead of 850 mA in the web calculator. I now find that they suggest a three ohm resistor.

However, (in the meantime) I discovered my learning curve hampered from using English. I did not know what the hell was meant by "buck circuit."

Now that I know what I want, I guess I can learn something.

I want a DC to DC circuit that will tolerate the variances associated with eight fully charged AA NiMH cells in series as a power source and deliver or "step" this voltage to a light that is meant to run in between 4.5 and 5.8 watts. Usually at power level as high as 7.4 volts at 850mA and as low as 6.5V at 650mA. The light appears to have a "cut out" and powers down to 10mA when it can no longer run at 4.5 watts or more.

It is my understanding that I can build such a circuit using as little as two transistors and two resistors. And because of the small voltage being regulated it should be rather efficient as well.

If I can't to that, how well would the single diode, single resistor solution work? At least I get to use up my stock AA NiMH, right?

 

[Mr Happy]

If I understand correctly, you have a light that usually runs on two lithium ion cells in series and you wish to run it on NiMH cells instead?

The answer is to use 6 NiMH cells in series and don't try to insert any kind of regulator in the way. Using 8 NiMH cells will give you no advantage in run time or power unless you use a very sophisticated bucking voltage regulator. There is almost no point in trying to do this.

In short, the simplest and easiest way to adjust the desired output voltage of a battery pack is to add or remove cells until you get the right voltage. Six fully charged NiMH cells will have a voltage of about 6 x 1.4 = 8.4 V, and six depleted NiMH cells will have a voltage of about 6 x 1.1 = 6.6 V. These voltages almost exactly match the charged and depleted voltages of a two cell lithium ion battery.

 

[burpee]

The answer is to use 6 NiMH cells in series and don't try to insert any kind of regulator in the way. Using 8 NiMH cells will give you no advantage in run time or power unless you use a very sophisticated bucking voltage regulator. There is almost no point in trying to do this.

Finally, now we are getting somewhere. Well Mr. Happy, that arrangement was tried and resistance was such that the light drew too much current and went into thermal shutdown. (that's a guess - the light quit)

At the time the light quit -it was drawing 8.1V at .86A. Hence my assumption regarding its own circuitry. Besides, part of the rational for the eight cell pack over six cells was extended run time. Using six cells will deliver a little over two hours of power - not enough to warrant the modding.

By the way a 6AASx2P was tried with similar "lights out" results. I added a couple of 1ohm resistors and fried them - but the light stayed on longer, and I don't have 5watt resistors right now. Hence questions about getting the right stuff the first time.

I've already ruined one really nice light, so I'd like to get this "right" the first time if I can. My custom branded Li Ion packs are both over a year old and being able to rotate in another AA battery source while their run times deteriorate is of some use. Of course by next year I'll chuck out more bucks for a new Li Ion pack.

Like I said I know of two "web builds" one using a single diode and resistor, the other using two transistors and two resistors. I guess should try the "dumb circuit" and see how much efficiency is lost. For all I know, there is probably a single IC that could handle these parameters. 9.6 to 7.4 isn't a big stretch.

Thanks for your attention, any comments are appreciated. I figure there are two dozen builders on these forums who have already applied and or studied these kinds of circuits.

On the other maybe I should buy a Fenix TK40 and forget it. Oh wait, I just remembered my wife said something that.

 

[Mr Happy]

If you can be specific about the light you are trying to run it may help with providing suitable options. Your earlier posts suggested some kind of regulated light (low voltage "cut out"), but now you suggest it is a direct drive unit that can be fried.

Whatever the case, using 8 cells with a resistor will not provide any more run time than using 6 cells. The run time depends on the current drawn from the battery, and since the current is the same in both cases the run time will also be the same. Maybe you just need to use 6 cells in series and drop an extra 0.6 V or so with a diode?

 

[burpee]

Maybe you just need to use 6 cells in series and drop an extra 0.6 V or so with a diode?

I thought this was a very typical modification with dozens of forum users doing this type of thing. I've seen other threads that go to considerable lengths describing circuitry that produces some fairly inconsequential results.

Thanks anyway. No use screwing up a good light trying to make NiMH work with a light head made for Li Ion, obviously its impossible!