12v and 5v power output equivalence - Page 1
Forum section: electronics / beginners

author: danbr
author_rank: Contributor
timestamp: October 07, 2015, 03:48:40 pm
content: I did a quick search and didn't find much.Pretty simple question, although its been racking my brain the past couple days.I'm pulling 12 volts (from a car) with a min fuse value of 5amps. I'm being cautious and saying 2.5amps will be the max current draw allowed by my system.Okay, so 12v X 2.5amps = 30 watts, no big deal.What I'm wondering is, my system runs off 5 volts that I am buck converting down from 12.Is it correct to say:30 watts / 5 volts = 6 amps @ 5 volts? I can safely pull those 6 amps at 5 volts. I am designing my little system around the power requirements/restrictions, so I need to know my power limit.Thanks!

author: SteveLy
author_rank: Regular Contributor
timestamp: October 07, 2015, 04:00:22 pm
content: You basically have the right idea. Power = Voltage x Current. You also need to allow for the less than 100% efficiency of the voltage converter.Suppose the efficiency is H, as in e.g., H = 0.85 for 85%. If you want to get P_out watts out then you need to put P_in/H watts in. For 30W out you need 30W/H in; e.g. H = 0.85, P_out = 30W, P_in = 30W/0.85 = 35.3W; let's say 36W to be on the safe side. Your V_in = 5V source must then supply P_in / V_in = 36W/5V = 7.2A. Equivalently, take the ideal 100% efficiency case input current of 6A you've calculated and divide it by the efficiency of the converter (expressed as a value between 0 and 1).

author: danbr
author_rank: Contributor
timestamp: October 07, 2015, 04:14:43 pm
content: Ah perfect, just the answer I was looking for.Erring on the side of caution should help me out here. The buck converting should be able to supply enough current at the 5 volts, just as long as the overall system doesn't draw more than its rated for.Thank you SteveLy!

author: Ian.M
author_rank: Super Contributor
timestamp: October 07, 2015, 04:54:03 pm
content: Its more complex than that. While the engine is running, your nominal 12V supply will typically be about 14V. During engine starts, it can drop to 10V while the starter motor is operating, for anything up to 30 seconds.There is also the issue of protecting your buck converter against spikes on its supply. Googe "load dump transient" to see what you are up against. If you are pulling 6A at 5V, you presumably will have quite a lot of valuable electronics on the 5V rail that wouldn't survive the buck converter series chopper transistor shorting out.

author: danbr
author_rank: Contributor
timestamp: October 07, 2015, 05:54:04 pm
content: After reading up some, I'm not worried about load dumps, the cars' battery will never be disconnected while the system is running. Cranking is another issue. Since I am not pulling directly from the battery, I'm fairly sure a voltage drop would be pretty crappy, but not the end of the world. The power schottky rectifier should take care of the input voltage dips/ripples, while the beefy 3 - 330uF caps should support the 5v for a couple seconds input voltage dips. I've tested this, and so far no ill effects. The car I'm testing on has an "eco start/stop" function, so while unfortunate as it maybe, the car has to restart constantly, ugh. I'll do more power testing/data verification.Thanks for the heads up, learning lots on this project.

author: Ian.M
author_rank: Super Contributor
timestamp: October 07, 2015, 06:10:19 pm
content: So you want to bet on the vehicle's battery wiring never developing a bad connection?At least its a modern vehicle with a smart engine management system, so it will have some load dump protection (high current TVS clamping) already fitted, probably at or in the alternator, so *IF* your switching regulator has a high enough input voltage rating, you *MAY* be OK.