# Passing 10 A through a nichrome wire (2.3 Ω) with a 4.2 V Li-ion battery source

I have a certain electrical problem with me that I'm unable to figure it out. I need to pass 10 A current through a 2.3 Ω nichrome wire to heat it up using a 4.2V Li-ion battery with a 70-150 C discharge rate. I currently have the following MOSFETs with me:

IRLB8721 (NMOS)
IRF9530 (PMOS)
IRF9640 (PMOS)
IRF9540 (PMOS)

The circuit I'm trying to emulate with a current source (and working) is attached.

10 A current source with an NMOS acting as a switch

I tried many approaches like using a current mirror to pass the 10 A through the load, or simply replacing the current source with a PMOS to bias it at 10 A and allow it to pass through the load (all of these simulations are done in LTspice) and none of these approaches seems to work.

Also my concern is that 10 A passing through 0.5 Ω resistance will need 23 V while I'm limited to a 4.2 V source.

How can this limitation be overcome? Through the use of current amplifiers or any other thing, and what would the circuit look like?

Edit: I made a mistake in calculating the actual resistance of nichrome wire. It's actually 0.5-0.8 Ω. The internal resistance of the meter was calculated to be around 2 Ω. Also the battery I'm using is from this website.

I also tried a boost converter that I had lying around. I calculated the open-circuit voltage to be around 12 V and then when I used it as a power source, it immediately dropped to 5-6 V and was hardly drawing 1 A of current through the load of 0.5 Ω. I even bumped the open-circuit voltage up to 24 V and the result was still the same.

The objective is to heat the nichrome wire as fast as possible. I've used the 10 A current source from the power supply (the circuit I've already shown). The wire heats up fast with that. So all I'm trying to do is replace that current source with an equivalent circuit. Now I'm not sure exactly what I'm missing here or what I'm doing wrong here. I had the power supply giving me 10 A and 3.7 V.

• You will need to step up your voltage, there is no way around it. There are boost converters that can help. Commented Mar 14, 2023 at 18:09
• start by connecting the nichrome wire directly across the battery ... that way you have no semiconductor components causing a distraction in your head Commented Mar 14, 2023 at 18:57
• You've provided the rated discharge rate of the battery, but that's not very useful without knowing the battery capacity. Please edit the question to include that information, and then flag this comment as "No longer needed". Commented Mar 14, 2023 at 19:02
• Also, this question is a rather typical example of an XY problem. You think the problem is "which mosfet should I use", but that's not a problem at all, and you won't be needing any discrete mosfets at all... Commented Mar 14, 2023 at 19:11
• Cut that nichrome wire into four equal pieces, connect them in parallel, and that will be 2.3Ω/4 = 0.575Ω. 3.6V/0.575Ω = 6.26A, 6.26A*3.6V=22.54W. Search for those MOSFET datasheets and see what their Vgsth or gate threshold voltage is - they might need 6V or more at the gate to even start turning "on." Commented Mar 14, 2023 at 19:36

There is nothing to figure out, because it is not possible.

A 2.3 ohm load on a 4.2 V battery will draw a current of 1.8 A.

And the battery is almost never that full to have 4.2 V, typical battery has a nominal voltage of 3.6, so it can be less too.

You would need to have multiple batteries in series or a boost converter to provide larger voltage, or multiple smaller resistances in parallel as load.

In any way, the battery would have to be rated to provide the required 10A without damaging or overheating, because that is a lot of current. Just make sure your battery can handle it.

To use a boost converter, providing 10A at 23V into a 2.3 ohm load means 230 watts, and so the battery would have to provide about 55 amps at 4.2V.

• Probably need to account for the internal resistance, as well. Commented Mar 14, 2023 at 18:48
• @DianBasit: Please don't burn down your house / building by pulling too much current from your battery. Pulling even more power at the same voltage means even more current. Your battery will need to be pretty large to handle a 230W load, or more since the boost converter won't be perfectly efficient. I worry that since you didn't realize that 4.2V can't push 10A through 2.3 ohms, you might be missing other simple calculations that could lead to safety problems if you hook up something that can pull a lot more current from your battery. Test outdoors if possible, battery fire is serious. Commented Mar 15, 2023 at 8:21
• @DianBasit whatever problem you're trying to solve here by putting 10 A through your 2.3 ohm nichrome wire, taking that power from a single lithium battery: I'm almost certain your missing something. As multiple engineers have told you now, you're requiring your battery to provide 250 W in power. That doesn't usually go well with lithium batteries as long as you do not add temperature control and you probably need to change your approach very drastically to make this even theoretically possible. Commented Mar 15, 2023 at 9:26
• @PeterCordes Yes I'm taking super precaution in this and definitely not peforming this indoor or at home. Thank you for your concern. Appreciate it. Commented Mar 15, 2023 at 14:56
• @user253751 see my answer. The battery product page must be bogus, or dangerous, or both. That's a 2A rated connector on that battery, and the whole device weighs 20 g. I have my serious doubts about the 140 C rating. Commented Mar 15, 2023 at 20:13

That's a 1/4kW heater, and the 2.3Ω resistance on the nichrome wire is the cold resistance. We must assume the wire will likely be quite a bit warmer though, as will be the wire that supply it, so its resistance when warm can be let's say 3..4Ω. So we need not a 3.7V supply, but something that can push up to 10A into the load, at a voltage of 24V (cold) to 36-40V (hot), to have ample margin.

The step-up/boost source should be configured for constant power operation, not for constant 10A current. As the wire gets hotter, it'll be outputting the same thermal power at several amps less than the cold current of 10A.

In any case, you're looking at a rather stout boost constant-current converter, and an additional control loop that adjusts the current to be inversely proportional to voltage, to maintain the constant power - since that's presumably what you want: a constant thermal power.

Most any boost converter will have an on-off control logic input, so you don't need to worry about discrete mosfets for that.

If you want/need something else, please edit the question to provide what the exact application is this nichrome wire used in.

• Thank you for a detailed explanation! I'll try this boost converter approach. Turns out my requirement were unrealistic. I was wondering if the requirement could be matched based on mosfet. I'll try this approach! Commented Mar 14, 2023 at 21:25
• (@DianBasit You didn't spell out your requirement - please clarify in the question: is it to dissipate 230 W?) Commented Mar 15, 2023 at 4:22
• @greybeard I just mentioned the update in the original post above. No, the idea is not to dissipate 230W but to heat the nichrome wire as fast as possible. Just like how vape technology heats up the coil as fast as possible. Commented Mar 15, 2023 at 14:51
• @DianBasit then, where does the number 10A come from? You want as many amps as possible, why limit yourself to 10? Commented Mar 15, 2023 at 19:23
• @user253751 You're right I can do as many amps as I can. It's just the power supply I was using had a limit of 10A. But if I can do more than 10A with circuitry I would definitely go for that. That means the wire would heat up faster which works for my application. Commented Mar 15, 2023 at 19:53

Dian, this all makes no sense, the 10A have nothing to do with your objective! if you directly connect the 0.5Ω wire to your battery, you'll probably get more current than 10A (at least for a short duration), before your resistance starts to increase due to the wire heating up. Do you really want to "heat the nichrome wire as fast as possible", or does "heat it up to a specific temperature within a specific time" suffice?

These are different things, and "as fast as possible" is a hard optimization problem, which needs your DC/DC converter to change its voltage with wire temperature.

So, I'll go with "if it gets really hot really fast, I'm happy". Let's optimize for that!

If the maximum burst discharge rating is correct (and I have doubts when it comes to websites selling batteries without extensive datasheets), then 140 C means you can discharge your battery in 1/140 of an hour, so that's a maximum current of 650 mA · 140 = 91 A.

Assuming that under that load, the battery still supplies 2 V, that means it has an effective internal resistance of 17 mΩ (and not the 1.7 mΩ that the description falsely claims).

Therefore, a 17 mΩ load would maximize the power you draw from the battery. That would be, roughly, 1/25 of your current nichrome wire resistance of 0.5 Ω.

That's nice, because it means all you have to do is cut up your wire into 5 equally long parts, and put them in parallel (with a very good contact).

Then you'll be converting roughly 200 W (Because of 2 V that the battery still can do under load, and the 90 A it can supply, and because exact numbers are hard here when a milliohm makes a large difference) to heat in the wire.

Note that this can't go on for long. When the internal battery resistance is as high as the external resistor, you're achieving maximum power transfer, but you're also converting roughly the same amount of power to heat inside the battery. Also, the wires on that product photo do not at all look like they are capable of 90A. Seriously.

So, I really can't trust that product page. Even 17 mΩ seems pretty unlikely, considering a mass of barely 20 g, and these cables.

Even going with the more conservative 70 C, that's still 45 A. The JST-PH-2.0 connector that this battery pack has is rated for 2A! It'll melt, quickly, under 45 A load. At 90A load, it'll possibly catch fire. (again, won't happen, your wires are far too tiny and have multiple mΩ of resistance themselves.)

4.2v ÷ 0.5ohms = 8.4A = 35W. That will be enough to heat your wire in a few seconds

Mosfet N channel IRLB3034 in lowside configuration, like your draw