3

I'm looking to power nichrome wire from a battery. The problem I ran into is that desired length/width of the wire has very small resistance of ~ 8 Ω.

Several kinds of batteries I've tested have internal resistance much higher than that, which leads to the wire staying cold and the battery getting hot.

Is there any way to increase the current drawn from the battery? Some kind of amplifier maybe?

Edit:

  • 1.2 V (rechargable) and 1.5 V one, while not getting hot itself did not provide enough power to heat the wire.
  • 3 V did not provide enough power, even stacking three of them in series, only provided about 1 V on the wire, and not enough current to heat the wire.
  • 9 V battery did provide enough current to heat the wire, but at the same time got very hot, very quickly, itself.

PS. I was wondering if a joule thief might work for this purpose?

anilberg
  • 397
  • 3
  • 12
Marcin Raczkowski
  • 152
  • 1
  • 1
  • 8
  • What kind of batteries (chemistry, voltage, capacity) have you tried? A typical AA alkaline battery (1.5Volts) has an internal resistance well below 1 ohm. – JRE Oct 20 '15 at 14:35
  • Ok. There have been several tries, and I didn't measure internal resistance directly - I assume that's the reason because of low voltage on the wire. 1.5 AA battery did not provide enough current to heat up the wire. 3V Coin Cell provided ~ 0.5V on the wire itself. 9V did provide enough current to heat up the wire, but battery iself also got very hot very fast. – Marcin Raczkowski Oct 20 '15 at 14:38
  • 1.5V Alkaline or 1.2V NiMh Rechargeable? Your comment says one but your edit the other. – Passerby Oct 20 '15 at 14:47
  • More voltage! Put a bunch of AA's in series. You'll need several watts of power, (a guess) to get it hot. – George Herold Oct 20 '15 at 14:48
  • re: 1.5 and 1.2 - actualy both, More voltage doesn't work. Tried stacking 3x3V and it did nothing, while 9V battery heated up wire alright. – Marcin Raczkowski Oct 20 '15 at 15:09
  • Lead acid batteries have very low internal resistance. A 12V car battery will heat your wire very effectively! – Icy Oct 20 '15 at 15:19
  • Yea. I was thinking about creating something i could carry in a pocket, so Lead-Acid is kinda oversized for the usage :) – Marcin Raczkowski Oct 20 '15 at 15:32
  • @George Herold: More voltage means more current, and series cells provide the same current. Stacking batteries in series is only going to make things worse if the OP can't get enough juice out of a single one. – Mister Mystère Oct 20 '15 at 15:33

4 Answers4

3

The issues you are facing are pure Ohm's Law issues.

First, the coin cells. They have a high ESR, Equivalent Series Resistance (or Internal Resistance). 20 ~ 40 Ω, depending on the load or battery state. Stacking them increases the ESR as well. So from a 3V coin cell like the cr2032 (250 mAh average), and a 8 Ω nichrome wire, we solve for I. I = V / R where R = R-Wire + R-ESR.

0.108 A = 3 V / (20 Ω + 8 Ω)

And since we know the current through the two resistors (current is the same when in series) and the resistance, we can see how much voltage is actually on each, with more Ohm's Law. V = I * R.

2.16 V = 0.108 A * 20 Ω and 0.86 V = 0.108 A * 8 Ω.

2.16 Volts of the 3V available, are wasted inside the coin cell. Only V * I = P .86 V * 0.108 A = 0.09 Watts or 90 MilliWatts of Power are going through the Nichrome Wire. Coin Cells are useless in your application.

Alkaline Primary Batteries, like your typical AA, C, D and 9V have much lower ESR, and can be considered negligible for this purpose. So pure Ohm's Law applies.

0.1875 A = 1.5 V / 8 Ω and 1.125 A = 9 V / 8 Ω

A 1.5V AA battery, at a sixth of a 9V battery's voltage, with the same load resistance, will produce a sixth of the current. But the Power difference is much greater. In Power, that's 0.28 Watts and 10 Watts. While the ESR of the 9V is negligible, it still exists, at likely <= 1 Ω, and while reducing the power going through the wire some, the amount of Power drawn by the load is enough to heat the battery up. 9V batteries are designed for low current draws. Drawing 1+ Amps through it is not ideal.

AA, C, and D can handle this better. Each has less ESR than the smaller, and a higher capacity. Due to the lower voltage, you need to combine a few in series to get the desired current and power draw through the Nichrome Wire.

See How do i find nichrome temperature for specifics on heating and Pulse-powering heavy loads with a coin cell on battery loads.

Community
  • 1
Passerby
  • 72,580
  • 7
  • 90
  • 202
  • In relevant quotes, *Hey McFly, you bojo, those batteries don't work on wires!* **Unless you've got *POWER!*** – Passerby Oct 20 '15 at 15:48
  • I kinda specified that I know the problem is with high internal resistance of the batteries. I was wondering if there's any way to work around load resistance being lower then the internal resistance. Adding a Shottky Diode was suggested in another answer and I'm going to give it a try. From my calculations ESR of 9Volt battery (one that I have) is between 1-2 Ohm, and is enough to cause a baterry to dangerously heat up when connected to 8Ohm wire. – Marcin Raczkowski Oct 20 '15 at 15:52
2

If it were me, I would obtain a few lithium ion batteries, either "18650" type of "21700" type. They can be ordered for less than 5 dollars each and there are simple charging modules for $5 or $10 on line. Many household devices use packs of multiple lithium ion "18650" or "21700" batteries today. For example, cordless drill or vacuums.

The Benefits of switching to lithium ion batteries are that they are similar in size to a typical AA or 9 volt battery, they are rechargeable, they will actually make your nichrome red hot, and they will last a lot longer than alkaline.

P.S. T try getting a "used but functional Cordless drill with a proportional speed capability". In this case, you will also receive a "proportional trigger switch" which you can use to control the power supply to your nichrome wire.

Said another way, get a cordless electric drill, remove the electric motor and connect the 2 wires that were connected to the drill motor to your nichrome wire, one at each end. This should create a well controlled and plentifully powered system for heating your nichrome wire.

; )

Robert Bobby
  • 29
  • 2
0
  1. At a given ambient temperature, it takes a set amount of energy to heat the nichrome wire to fusion temperature, and then another amount of energy to melt it. Calculate the power required across your nichrome wire depending on the power losses and the required time-to-melt.
  2. Deduce what the required voltage and the corresponding current across the nichrome wire are
  3. Rule of thumb: choose a battery technology that has a capacity (in Ah) equal to or greater than the required current (in A) (10x greater for sustained use) if possible (otherwise step 4 is mandatory). Stack as many cells in series as required to make up the required voltage

AA batteries are usually quite good in terms of resistance (~0.2Ohm for alkaline) and current capability (up to 1 or 2A, not for long though). They wildly vary in specs depending on the manufacturer though, therefore if yours didn't work, you should try with a better quality one before trying the C and finally D type batteries.

  1. If the best stack you can afford (in terms of space, mass or whatever) can't provide the current or gets too hot, here is a potential solution: insert a Schottky diode that can handle the voltage and current in series with the stack (note that the stack voltage will be reduced by 0.2 to 0.5V), and duplicate that assembly in parallel as many times as required to reduce the current each stack has to provide. The diodes will prevent batteries fighting for the net voltage.

Note that this technique (step 4) is normally used to switch between power supplies, not increase the current capability. Only one diode will conduct at any time*, meaning that here the entire current will be supplied by one source before its voltage droops sufficiently to make another one conduct, and so on. This could work okay for this application though.

*: considering an ideal diode

Mister Mystère
  • 9,477
  • 6
  • 51
  • 82
  • So something like this: http://imgur.com/Peqwq3d ? – Marcin Raczkowski Oct 20 '15 at 15:18
  • Yes, although I'd recommend using Schottky diodes instead of regular diodes to avoid losing too much voltage. – Mister Mystère Oct 20 '15 at 15:22
  • I couldn't find symbol for Shottky in circuitlab, that's why I used standard one. Thanks for the answer. I'm trying to figure out how the diode will help with the battery heating, and also why only one diode conduts at the given time – Marcin Raczkowski Oct 20 '15 at 15:30
  • When you do that, if you have the time please post a picture of what the voltages across the diodes look like, I'm curious as to how stable this solution is. – Mister Mystère Oct 20 '15 at 15:36
  • Ok. I think I finally figured out how (4) helps. It splits the load between all the power sources in parallel so each unit heats up less :) – Marcin Raczkowski Oct 20 '15 at 16:14
  • Yes, but as I say the load can't be evenly split between the sources since that would mean all diodes conduct, which never happens in practice with real voltage sources. Instead, I think the power sources will take turn for the full current, and I don't know how frequent the switches will be. – Mister Mystère Oct 20 '15 at 16:38
  • Note that this is if you really want to use batteries that can't handle the current. I'd go straight to good AA's, C's and D's personally to avoid the trouble, but perhaps you don't have the space for them. – Mister Mystère Oct 20 '15 at 17:20
  • Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/30498/discussion-between-marcin-raczkowski-and-mister-mystere). – Marcin Raczkowski Oct 20 '15 at 17:22
0

Your problem is defined by the amount of power you can extract from the battery. Once you are operating at the limit the best you can do with fancy circuitry is trade off current for voltage. But the power won't increase.

For example, (disregarding efficiency effects), if the 9v battery gave you 100mA, you could change that to 1.5v at 600mA. The resistance changed from 90 ohms to 2.5 ohms, but the power is still less than 1 watt.

Your best bet is to determine which cells present the lowest internal resistance, and then, if necessary, connect several in parallel to reduce the resistance further.

gbarry
  • 8,615
  • 21
  • 31
  • As I understood the NiChrome temperature scales mainly with current not voltage. While thanks to ohms law increasing voltage will increase current on the same resistance, the problem I'm facing is that resistance of load is similar (or sometimes smaller) then resistance of internal resistance of batery, and I was lookign for a way around it. – Marcin Raczkowski Oct 20 '15 at 17:22
  • What you say is correct. What you have to realize is that you can't change just one parameter. The rest of them change with it. In short, there's no free lunch. But, get a battery that's capable of the power level, and you can make it work (maybe with a converter). – gbarry Oct 20 '15 at 17:29