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I am trying to reduce a 4 volt battery to a 3 volt source. I have used a 1 ohm 5 watt resistor, which drops the voltage to 2.9 volts, but the resistor is eating up the battreies current. I think the resistor was using up around 1 amp of current. Can someone help me design a voltage regulator circuit that will do the job and not use so much current?
Thank you.

What is the load?

Hi Russ,
If you mean how much current does it draw, I am guessing 1 amp or even lower. I am not real sure at this moment. I can measure it when I get home and let you know tonight.

Resistors do not use up current. The current through the resistor would be the same as that through the load. What the resistor does consume is battery capacity (units of mAh) which translates to reduced battery life. <p>To measure the current easily (without opening the circuit for a meter) measure the voltage across the resistor and the resistor value then divide I=V/R. Since you say you are dropping 1.1V across a 1 ohm resistor (assuming it is accurately 1 ohm) you have 1.1A flowing through your load. If you are using a 2200mAh capacity battery, this resistor will shorten useful charge time from 2 hours to 1 hour.<p>A resistor is never efficient and certainly will not regulate the output should the battery voltage drop when it starts to discharge. However, you cannot beat it for simplicity and low price.<p>If you require good regulation, you should probably go the route of a variable voltage regulator. With just a few resistors you should be able to get the voltage you want. The LM338has a minimum input voltage of 4.2V which is marginal for your application. There are probably other regulators with specs better matching your application.<p>When Russ asked what was the load I presume he was shortcutting a long list of questions like; Can it withstand overvoltage, undervoltage, does it require regulation and to what precision, is the current constant or variable and to what range etc. By simply telling us what it is, we can better make those assessments ourselves and maybe give alternate advice (or at least that's what I would have meant)<p>[edit: strike that last part, I just connected it with your other post about the Flourescent light]<p>Chris<p>[ September 17, 2003: Message edited by: haklesup ]</p>

By far the simplest way to do this would be to add one or two silicon diodes in series. Each would drop about 0.6 to 0.8 volts depending on current draw. Something like a 1N4001 to 1N4007. Or, if there is not much current draw, a red LED would do the trick.

What was I thinking, that calculation with the capacity was totally wrong.<p>If your dropping resistor is 1 ohm and the current is 1.1A and the voltage across the load is 2.9V that makes the load equivelent resistance = 2.6 ohms.<p>Obviously, increasing the series resistance will reduce the current in the loop thus using up the battery capacity more slowly and extending the useful time of the battery. <p>No matter what you use to drop the voltage it will consume 25% (1V out of the 4V) of the power thereby reducing battery life by that amount as compared to the ideal situation of having a load that uses all 4V or a 3V source. Since you have neither and the load probably does not need close regulation, the resistor should be sufficient.<p>The Diode is better than the resistor (for small currents)because the voltage drop is more constant (but not exactly constant) as the current through the diode changes. However, the voltage drop across a diode at 1A will be considerably more than the 0.7V threshold. The 1 ohm resistor has a lower resistance than the forward resistance of most diodes anyway.<p>FYI the battery capacity of a typical alkaline AA battery is about 1800mAh. Two in series is the same, two in parrallel will double it.<p>I find it hard to believe that you can light a fluorescent lamp from 3V without an inverter and that would take more than one xistor and require a transformer of some sort. I haven't seen a Flourescent lamp for use under 6V or 12V is common. <p>Search for "fluorescent lamp circuit" or "cold cathode driver" or some combination of these words.<p>I agree with Edd (other post) that you may have damaged the semiconductor by a temporary reversal of polarity. The 3 terminal device may not be a transistor but a regulator or other special IC. I would expect a lamp circuit (especially one designed to run on batteries) to tolerate a 1V overstress. Even a 2n2222 can handle 4V (C to E)but forward bias the B-E or B-C junction to 4V and put 1A through it and you will have popcorn. Battery operated equipment should tolerate reversal of the batteries though.<p>For a replacement, if you cannot read any markings, you may have to reverse engineer the circuit so that you (or we) can make an educated guess.<p>Is this one of those new camping lanterns Coleman Micro Lantern Most I have seen need 4 or 8 batteries. To get 6 hours out of that lamp it would need to consume 300mA. At 6V that uses 1.8W. The 1.1A you measure may be a result of the burned out transistor or the fact to get 1.8W out of a 3V source you need 600mA but on two AA batteries this would barely last over 3 hours. Does this agree with the performance of your lamp. at 1.1A the battery would be dead in just over an hour.

Hi. From the way I read your problem, you're concerned about the power being wasted by a series resistor. If you're using 1A from a 4V source, the current will pass through both your series resistor and the load. That means you will be wasting about 25% of the power output of the battery, and would like something more efficient.<p>Assuming you don't want to go with a higher regulator input voltage, a low dropout voltage regulator (such as the National Semiconductor LP3965ET-ADJ, available in unit quantities from Digi-Key for $3.21 ea.) would provide you with good voltage regulation without worrying about dropout voltage, like with the LM317CT. However, all the current going through the regulator will also be going through the load -- same situation from a wasted power standpoint, but much better regulation.<p>I would think that, if it's important enough (after all, it's only 25% -- even if you got 100% efficiency, your battery would only last 133% as long), you would look into a switching regulator setup. There are some newer switching regulators which are meant for battery use with low voltage dropout.<p>If this is really that important that you want to save most of the 25% of the battery power you're wasting on regulation, you might want to try a high frequency synchronous current mode adjustable switching regulator. These are used in battery-operated instruments and cell phones. They're not cheap in unit quantities, and you can't just cobble one up on a perfboard. You may want to use an etched PCB design, or get very creative with layout. One IC which will do the job is the Linear Tech LTC3411. It's a 10-pin MSOP, so a hobbyist might try "dead-bug" construction (glue it upside-down on a ground plane of copper-clad FR-4, and wire point-to-point with EXTREME care and thought), and you might get both the regulation of a LDO, and better efficiency.<p>Personally, I'd stick with the resistor or, if you need regulation, an LDO, and live with the wasted power. If this has to do with the previous post, your lights will require big current pulses, which will really mess up the input voltage if you're using a resistor -- that might not make the inverter circuitry for the fluorescent too happy, either. So, if this is for your fluorescent, I'd go with the LDO mentioned above -- it's easy, layout isn't too critical, and you can be done with it (once you fix the fluorescent driver). By the way, use a fairly large cap at the LDO output to help with the current pulses.<p>LP3965ET-ADJ pdf Datasheet
LTC3411 pdf Datasheet<p>Good luck.
Chris<p>[ September 17, 2003: Message edited by: Chris Foley ]</p>

bigger batteries

Two AAs :p

Sounds like the 1 Ohm resistor is parallel with the load, instead of in series. THAT will eat batteries!<p>Did you measure the current yet? And, what KIND of load?