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I am new here so I would like to say hi to everyone. I just recently jut into electronics and I have a question on a project I am working on.

I am trying to power a 20W heating element off of a 3.7v 18650 lithium battery. I wanted to know if it is possible to design a boost converter that can take the 3.7v and boost it to 20W? I know in order to calculate watts it is voltage x amps= wattage. the element I am working with says 5v and 4-amps to make 20W. I have googled this and I keep confusing myself.

Summary: I want to know if I can design a boost converter that can has an input of 3.7 and can output 5v @ 4-amps, or any combination there/of to achieve 20W?

Kevin Reid
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Brandon Mimiaga
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    The usual way to do this is to use a heater with the correct resistance for the voltage you want to use. And: any reason to use electrical power from an accumulator for heating instead of a chemical reaction? – Janka Feb 01 '17 at 23:58
  • @janka what do you mean when you say using chemical reaction? – User323693 Feb 02 '17 at 00:43
  • Yes, this is well within the realm of possibility. Basically you just need a 5V boost converter capable of putting out 4 Amps. Your battery will not last very long. Maybe 15 minutes, but it depends on the battery. – user57037 Feb 02 '17 at 03:03
  • @Janka, since it is a battery, the output power would vary as the battery discharges. But it would be easier, I think, to use a buck regulator rather than a boost. So a lower resistance heat element would be a good idea. – user57037 Feb 02 '17 at 03:11
  • @Umar: electrical energy stored in accumulators is a valueable ressource. There have to be reasons why someone want to waste it in a heater instead of using another means of producing heat. For example there's easy no way to get rid of combustion gases (that's the reason for modern kitchen appliances), or you are operating the device in a flammable atmosphere. – Janka Feb 02 '17 at 10:34
  • @mkeith: Lithium accumulators have a flat dicharge characteristic. As soon the voltage has sunken under 90% of the tableau voltage, the accumulator is empty and you should stop discharging it further immediately. That's within the characteristic of a heater. No need to provide a more stable voltage to a heater. – Janka Feb 02 '17 at 10:37
  • Simpler : use it as an 11W heater directly connected to the battery (via a battery protector to cut off the battery when it's discharged) –  Feb 02 '17 at 11:23
  • @ mkeithI am using a thin-film heating element that needs to get to 204C/399F. I was unable to find one off the shelf that fit my size requirements. I found a company that designed me one and it uses 20 watts. – Brandon Mimiaga Feb 02 '17 at 16:05
  • @Janka, I work with lithium batteries all the time. "Flat" is not a precise term. The voltage does decrease during discharge as I am sure you know. Whether the amount of decrease is acceptable in a given application is application dependent. If the heater is modeled as a resistor, the power will vary with V^2, so even though the voltage range is limited, the power range will be greater on a percent basis. – user57037 Feb 02 '17 at 17:56
  • The application is a heater. 0.9^2 is still only 0.81. I doubt anyone will notice the difference between a 20W and a 16W heater. – Janka Feb 02 '17 at 21:03

2 Answers2

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It is possible to boost 3.7V to 5.0V. You cannot boost power, you can boost voltage. If you need 4A at 5V, and boost convertor has efficiency say 90%, than your battery must be able to provide 6 amperes. If you have such battery, than you can do it.

Chupacabras
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  • I can supply the 6 amps with the 18650 cell, however battery run-time becomes a problem. Can I boost the 3.7v to 24v then just have the battery supply the ~0.83 amps. – Brandon Mimiaga Feb 02 '17 at 16:06
  • There is no getting around power limitations of a battery. The only option is to add more batteries. You could use two or three in series to increase run time. In this case you would use a buck converter rather than a boost converter. – user57037 Feb 02 '17 at 17:50
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    Brandon, you are wrong. If you need 20W then it does not matter what voltage you are boosting to, you will still draw 6 amps from battery! Battery voltage X battery current = Boosted voltage X current from booster – Chupacabras Feb 02 '17 at 18:13
  • mkeith, I understand what your saying, however I would rather just run two or three in parallel to increase run time. the 18650 cell can provide up to 20A contunious. From what I have read running the 18650 cell in series can be dangerous. – Brandon Mimiaga Feb 02 '17 at 18:47
  • @Chupacabras Now I'm starting to understand what is going on so thank you. One more question I had that is related. I've been looking at booster converters online and I keep running across ones that say POL or point of load. Here is an example. http://www.powerstream.com/avn20-3v3s05.htm . I've googled the term but no one can define it for the laymen. – Brandon Mimiaga Feb 02 '17 at 19:01
  • Of course it is bad idea to power heater from single battery cell. But question was if it is possible to make booster, answer was yes. If you want to ask what battery configuration is best for this you can create new question for that. – Chupacabras Feb 03 '17 at 06:13
  • @BrandonMimiaga, you can do parallel, too. The approximate formula for run time is 0.8 * battery energy/20W. For example, one battery might have 2.5Ah * 3.6V = 9 Wh. So for two batteries, you get 18 Wh. That will give you roughly 0.8 * 18 / 20 = 0.72 Hours. That is around 43 minutes. – user57037 Feb 04 '17 at 07:27
  • For your purposes, just pretend "point of load" is meaningless. http://electronics.stackexchange.com/questions/231325/what-is-a-point-of-load-converter – user57037 Feb 04 '17 at 07:32
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Yes, you can build such a boost converter. However,in your case, problems appear:

1.You said you're a beginner. Boost converters are complex designs which require fine biasing and a careful selection of components. I would suggest you buy one instead

2.Even if you could actually build it and make it work, the battery would probably stand in the way of your goal unless it can output the necessary current. For the boost topology, it must be able to supply more than 4A. See if you can check this on its datasheet. You'll pobably want to constantly check its temperature even more if you can't do that.

As a general idea, in order to heat things up (eg. water) you need powerful batteries. It's easier to do this if the heating element requires less to heat up.

Daniel Tork
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