I have an inverter in my vehicle for my tablet. I that know it uses 5v and my vehicle is a 12v system. My inverter has a usb jack and also 2 3-pronged US sytle plugs. Its a fairly beefy one rated at 750 watts continious. Its good for power tools on a job site without a good outlet. I also use it to power my tablet on camping trips.
I bought a cheap Chinese volt/ammeter to moniter my draw. Turning on my inverter costs around .5 amps. Plugging in my tablet costs another .25 amps (on a 12v scale). So, most of my draw is from the inverter doing it converting.
I am wondering what the base draw would be to just have a dc to dc voltage drop? I can seee on Ebay that they are fairly inexpensive but I wonder how they work (guessing they do the switch on and off thing really quick to immitate ac so you could use a transfomer? And of course you could just put a resister of the right size to drop it down, but I would guess that would be the least inefficient.) My main question is though, if is it would be less of a draw than than the inverter. And, if so how much?
Best Answer
You are drawing very little power from an inverter capable of much more so it's not surprising the efficiency would be abysmal.
The kind of adapters (12V->USB) you mention are typically non-isolated DC-DC converters. They would likely have pretty good efficiency. Many newer vehicles have USB outlets built in to the dash.
One caveat is that cheap 12V-USB converters might well have no overvoltage protection, so a single-point failure could put 13.8V (vehicle supply voltage of a '12V' car with motor running) right on the USB jack of your tablet, almost surely destroying the tablet. That's a lot less likely with an inverter- usually if they fail the voltage goes to zero.
Here's a typical buck converter circuit capable of 500mA with 25V in (maybe not as much with 12V in):
In operation Q1 switches on and off at perhaps 50-100kHz, staying 'on' and 'off' just long enough to maintain the proper output current through L1 at the proper output voltage. That's controlled by a feedback circuit that maintains a certain 'on' time and adjusts the off time to maintain 1.25V at the output of the voltage divider R1/R2 (meaning the output voltages is nominally 1.25 * (R1+R2)/R1 = 5.0V). Because the inductor (ideally) has no loss and (ideally) the switch Q1 and diode have no losses the losses are all due to 2nd order effects- switch/diode voltage drop and inductor resistance and core loss, mostly. So efficiency is ideally 100%. In practice, maybe 70-85% at moderate loads. A linear 5V regulator has an ideal efficiency of 36% with 13.8V in.
Should Q1 fail short (perhaps due to a transient on the 12V bus with poor protection in the converter) or there be a bad solder joint or crack on R2 then the output will rise to much over 5V.
There are better more modern DC-DC converter circuits, this is just one that is sometimes used in inexpensive converters.
Personally, for an expensive tablet, I would seek out a device sanctioned by the tablet manufacturer (for example, one sold in the Apple store for an iPad). That way it's most likely acceptable quality and you have some recourse if things go badly.