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After some bad luck with cheap Chinese products, I'm looking to build myself a simple USB "pass-through" e-cigarette.

For anyone not familiar with e-cigarette, the idea is very simple : you put voltage across a thin wire acting as resistor, it heats up the liquid and that's it. Most e-cigarettes got some PCB inside to manage different voltages and more importantly hold a battery.

However what I want to build is much more simple as it doesn't include a battery : all I want is get the power from a usb connection (rated for 2A) and output a voltage around 4V.

As I got some basic knowledge about electricity I recalled the voltage divider circuit which is quite simple and easy to make :

schematic

simulate this circuit – Schematic created using CircuitLab

But it seems that V2 will change depending on the load (cf : https://en.wikipedia.org/wiki/Voltage_divider).

Another thing is I use 1.8 Ohm resistance as the heating element and I calculated that I would need R1 = 0.45 Ohm which I don't think is easily achievable.

So to wrap it up here is what I need a circuit for :

  • Input : 5v USB
  • Output : 4v across a 1.8 Ohm load
  • If possible the possibility to use a variable resistor to tweak the output voltage (about 3.5v to 5v)

Thanks a lot to anyone willing to help !

PS : CircuitLab is awesomely intuitive but why no MathJaX syntax ?

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    \$\begingroup\$ Why don't you think 0.45 Ohm is easily achievable? \$\endgroup\$
    – Finbarr
    Aug 13 '17 at 22:04
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    \$\begingroup\$ And why do you believe that a[ny] USB connection is rated for 2 A ? \$\endgroup\$ Aug 13 '17 at 22:10
  • \$\begingroup\$ And in any case, if you want your circuit to put 5V across your 1.8 Ohm load that's a current of 2.77A. You're going to need one of those magic Tesla generators I keep getting emails about. \$\endgroup\$
    – Finbarr
    Aug 13 '17 at 22:15
  • \$\begingroup\$ At these power levels, you want a switching regulator or to re-wind the coil for the input voltage. But your load exceeds USB specifications, so it's not clear how you intend to achieve your goal anyway. \$\endgroup\$ Aug 13 '17 at 22:30
  • \$\begingroup\$ Thanks for the feedback guys. AliChen the 2A I'm talking about is coming from a USB powering station (not the usb port of a computer). Finbarr I got some resistors and most of them are in the 100 to 100k ohms range so I don't know how I could reach .45 \$\endgroup\$
    – Furrane
    Aug 13 '17 at 23:26
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First thing to keep in mind is that a voltage divider is not a good circuit for delivering power. You've discovered the first reason why (it provides no voltage regulation as the load changes) but the second reason might be more significant in this case - it's not very power efficient.

To do what you describe, simply take 22 of your 10Ω resistors and connect them in parallel. That will give you 0.45Ω. Use that as R1 and use your load (heater) as R2. Provided your power supply can supply the 2.2A (given that it's only rated at 2A, this is doubtful), your load will see 4V.

Note however, that R1 will be dissipating 25% as much power as your heater! Provided that you use 1/4W resistors and they're well ventilated and are well within tolerance, then nothing should catch on fire at least, but it will be very hot.

Output voltage control will be difficult to add with this configuration as variable resistors are not commonly available in the range and power level you need.

The next step up in sophistication to do what you want to do is to put a Zener in parallel with the heater. You'll still need your chunky R1 but at least you'll get voltage regulation.

The next step would be a linear regulator, which is similar to the Zener solution but some regulators will allow you to add output voltage control (note however, it's very unlikely you'll be able to go above about 4.4V, because the regulator needs some voltage headroom to work).

And the next step up from there would be a switched-mode power supply (SMPS). That gives you voltage regulation, voltage control, and much better efficiency, but they are far more complex.

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