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For my project, I need to heat up a nichrome wire (which acts as a heating element) and provide a power of 100mW to it. I have already done computation and I need current of approximately 22mA to generate that power, also the voltage that will be generated across the heating element will be 4.51V. I am planning on using a NPN BJT as a current driver to limit the current at a constant rate and because my current requirement is lower than 1A, I think NPN BJT will be a better option than using MOSFET. I am using MCU connected to this nichrome wire through current driving circuit.

  1. Since, I'll have to provide a source voltage for the driving circuit, I plan on using a 3.6 V battery. The problem that I am confused about is, will my circuit work properly with 3.6V at source? Does 4.51V generated across heating element have any significance in regards to this?
  2. I researched about load switches which use P-channel MOSFET, because of it's lots of advantages, I am confused if it will be a better choice or not, if someone can suggest.
  3. Also, I read on a website that while using MCU, it's better to use a high side switch, since MCU always needs to be ground. So, on using an NPN BJT, wouldn't it work properly?
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    \$\begingroup\$ You're missing a key issue --- As the nichrome gets hot, its resistance will increase. Your calculation for required current and required voltage only is accurate at one specific temperature of the wire. Is that the 'cold' (room temp) resistance? As it heats, the resistance will go up, the current will go down, and your power will change. You will have to play/experiment a bit to dial in a 100mW output in your application. Calculating is a great way to start, but you'll likely have to tweak it, so make sure your circuit allows for that - Don't lock yourself in. \$\endgroup\$ – Kyle B Oct 5 '20 at 6:52
  • \$\begingroup\$ Can you split the wire in two and supply each half with 22mA and 2.25V? That would be a simple way to allow you to use the 3.6V battery. \$\endgroup\$ – user_1818839 Oct 5 '20 at 11:56
  • \$\begingroup\$ @Kyle B, The temperature coefficient of resistance of Nichrome is only 0.00013 / ° C. \$\endgroup\$ – vu2nan Oct 5 '20 at 14:13
  • \$\begingroup\$ @vu2nan Thanks for keeping me honest. I had googled it and found -0.0004 BTW (4x yours). Anyhow, probably depends on the exact alloy. Besides, I have to make a correction - I didn't notice (and maybe you either?) that it's a negative coefficient - i.e. the resistance DROPS as temp goes up. I never have used it myself, and I assumed it would have positive coefficient like almost everything else! My bad (not too proud to admit when I'm wrong ;) ) There's also going to be some temperature effect of the mechanical junction (you can't solder to NiChrome, usually it's clamped). \$\endgroup\$ – Kyle B Oct 5 '20 at 17:40
  • \$\begingroup\$ My point really was just that, no matter how precise his calculations, they weren't going to be "right" and OP will not want to solder together his solution w/o dialing it in on the bench first. It's pretty clear he's not an EE by trade, so he may be thinking we are more precise than we really are ;) (It's interesting to see my mechanical colleagues faces when I tell them a capacitor may have +-50% tolerances...) If he wants "100mW", it'll have to be dialed in. If he wants 100mW +- 20%, that's a different story.... \$\endgroup\$ – Kyle B Oct 5 '20 at 17:42
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Lets fill a few gaps. The wire is just a component like any other, it has a resistance and depending on the voltage through it, the current will change. If you want to have 22mA and 4.51V, this means that the wire's resistance is 205 Ohms. This of course depends on the length of the wire and changing the length will change the resistance. Here the wire must be very short, because with 22mA, it will not heat a lot, and the bigger it gets, the less will it heat.

  1. Driving from a 3.6V battery, means you have to decrease the resistance, or the current will not be the same. Of course when you decrease the length of the wire, and the power for it, you will have even less heat.

  2. A MOS FET would be a better choice. The reasons are that the MOS will allow most of the voltage(from the 3.6V battery) to be applied over the wire and the current will depend only on the resistance. On the other hand a BJT can be used, but it will be controlling the current through the wire, depending on the base current, the resistance of the wire and the voltage source connected to the collector and emitter. If you set it to 22mA, but the resistance is 205 Ohms, you will get about 17mA, because the source is not 4.51V. It can be used to fine control the current. Its normally a good idea, but it will eat a part of your voltage, which will be applied on the transistor itself(collector-emitter junction) and the power over the wire will drop even more. A MOS has very low Rds[on], which should be better for your purposes.

  3. This is only valid when you are powering something from the MCU's pin. If the power comes from the source instead of the MCU, this will allow you to power more things and just ground them through the MCU's pin. A whole MCU port of 8 pins can sometimes supply only 10mA as a whole, despite that an idividual pin is also capable of supplying 10mA, but if 1pin takes all the current, the rest will have nothing.

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  • \$\begingroup\$ thanks for the explanation. So, what I understand is that in order to get 22mA current, I need to provide source greater than 4.51V whether its a MOS or BJT. Am I right? \$\endgroup\$ – Gagan Batra Oct 5 '20 at 17:42
  • \$\begingroup\$ ^^^ Yes. Or use two pieces of NiChrome in parallel \$\endgroup\$ – Kyle B Oct 5 '20 at 17:50
  • \$\begingroup\$ So, if I use a larger source voltage such as 9V, then will MOS provide any advantage? Also, if I use MOS, I would need to use PWM to control the Vg which is generally around 3V, which I don't need to do if I use BJT. \$\endgroup\$ – Gagan Batra Oct 5 '20 at 18:49
  • \$\begingroup\$ You don't need a pwm to control the MOS. This is only dome when you have high currents, the switching allows the MOS to dissipate the heat. Yes you have a voltage drop over the transistor and wire. The parallel solution works. \$\endgroup\$ – CFCBazar com Oct 5 '20 at 20:37
  • \$\begingroup\$ If you use BJT, you'll loose a little more power (as heat in the BJT, but minimal). Because fully "on", the BJT will still present a 0.2V drop. A MOSFET on the other hand, it's voltage drop is dependent on its resistance. With a modern MOSFET that can be fractional ohms, so it's almost not in the circuit. You can PWM either type of transistor w/o issue. (There are differences that would become apparent in a more advanced application, but those shouldn't be of any concern to you here) \$\endgroup\$ – Kyle B Oct 5 '20 at 20:46

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