The latest project I'm working on is making adapter to basically use the mains for an ecig rather than usb or an battery. For anyone whos familiar with ECIGS they basically use around 3.7-6volts. It's my understanding that the resistance in the coil a ecig determines the required amps, for example:

  • 3.7 volts with coil resistance of 3 ohm you will use around 1.233 amps. This is 4.56 watts.

  • 3.7 volts with coil resistance of 4.5 ohm you will use around 1.233 amps. This is 4.56 watts.

  • 6.0 volts with coil resistance of 3 ohm you will use around 2 amps. This is 12 watts.

  • 6.0 volts with coil resistance of 4.5 ohm you will use around 1.333 amps. This is 8 watts.

The step down I'm using is rated as so:

Input voltage range:4-40V
Adjustable output voltage range:1.25-37V
Operating current:2A

The ac/dc adapter I'm using is rated:

INPUT: 220-240v AC 50Hz 350mA
OUTPUT: 12v DC 750mA

Question: If I step down an input of 12v with 750mA to say 6v does the 750mA max remain the same or does this change because the voltage is lower? if so how much would it increase by and what is the forumula when down stepping.

... Apologies if this question has been asked time and time again, I'm sure it has but searching through all the step down questions I couldn't find an appropriate answer or question, only that of increasing voltage.

  • \$\begingroup\$ PS. I haven't plugged in anything as of yet and wanted to learn before doing so... Don't want to harm myself or my mains adapter ;) \$\endgroup\$ Commented Aug 23, 2014 at 17:01

1 Answer 1


If the "step down" is a linear regulator you'll get slightly less than 750mA out of it, at most.

If the "step down" is a switching regulator, you'll get somewhat less than the input power, say at least 75%. Since your input power is 12V * 750mA = 9W, you could probably count on getting maybe 7W out of it. So you could power the first two indefinitely (if the specs are correct) but not the last two.

Of course the output voltage has to be set appropriately for the load resistance before connecting the load.

Edit: so that means that with an output voltage of 6V your switching regulator would be able to supply 1.5A minus the losses in the regulator. Even if you reduce the voltage below 4.5V you should not attempt to draw more than the 2A limit of the regulator.

  • \$\begingroup\$ It's a switching regulator and doing a meter test its more like 90%. I'm not sure if you understood the question, but I mean does the AMP on an AC/DC at 12v 750mA increase when decreasing. For example is 12v 750mA converted to 6v then become 1500mA without counting for the 75%-90% loss of course. \$\endgroup\$ Commented Aug 23, 2014 at 17:31
  • \$\begingroup\$ Power is volts * amps, so (ignoring switching regulator losses) the available current would be 1.5A at 6V. In practice there are losses that will vary with input and output voltage and current, but they should be in the 10-25% range. \$\endgroup\$ Commented Aug 23, 2014 at 17:36
  • \$\begingroup\$ That's absolutely fanstatic. Do you mind editing your question and add regarding the 1.5a at 6v and I'll mark the question as accepted because thats exactly the information I wanted with the bonus of learning about the loss in volt/watts too. \$\endgroup\$ Commented Aug 23, 2014 at 17:39

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