Converting 9VAC 4A to 1.5VDC 15A - Page 1
Forum section: electronics / beginners

author: Wheres_my_chippy
author_rank: Newbie
timestamp: March 20, 2015, 06:16:36 pm
content: I have a 9V power supply that I want to convert to 1.5VDC 15A. My idea was to use a simple full wave rectifier to convert to DC but I'm not sure about how to get the ~8.5 VDC to 1.5 and the 4A to 15A.P.S. This is for creating a wired analog of a NIMH battery where the draw from the battery is 1.2VDC at around 15A.

author: Simon
author_rank: Unknown
timestamp: March 20, 2015, 06:18:00 pm
content: You need a switchmode power supply, you will be able to convert 70-95% of your input power

author: Anand
author_rank: Contributor
timestamp: March 20, 2015, 06:48:28 pm
content: https://en.wikipedia.org/wiki/Buck_converterStart with a circuit and do a simulation before you attempt to build the thing.

author: Wheres_my_chippy
author_rank: Newbie
timestamp: March 20, 2015, 07:47:48 pm
content: Just wondering, would a 7805 circuit not suffice? I know it wastes energy/ isn't that efficient, but I am, relatively, a beginner.

author: Cliff Matthews
author_rank: Unknown
timestamp: March 20, 2015, 08:01:58 pm
content: Pardon my asking, but isn't that cutting it close?You've only got 36watts in and expect to get 23watts out at low voltage. If 1.5v is required at the load, there's still at least one diode and some cable to pass through coming off the switch-mode transformer - so really you have to make about 2.1 v at 15amps to get any margin to work with.. this sounds like something that happens on multi-phase motherboard cpu regulators.

author: ConKbot
author_rank: Super Contributor
timestamp: March 20, 2015, 08:06:44 pm
content: You only have 4A available, so even if you could get a LM317 to drive a beefy array of output transistors to linearly regulate the output, you'd need a power supply that could do 15A out, which you don't have.What is your power supply? 4A is a tad large for a transfomer plug pack, though it might be. I'd consider re-winding the transformer, identify the primary and secondary, cut out the original secondary, replace it with 10 turns of wire, and measure your output voltage to figure out what the volts/turn is, and put the appropriate amount is. Shoot for ~1.35V open circuit after your bridge rectifier, and under load, hopefully it'll be in the 1.1-1.2 range, but that would really depend on the "regulation" of the transformer. (some are more ideal, where vin/vout = Npri/Nsec and the load doesnt make the output sag, some sag more under load)What is the application if you dont mind? I cant think of much that thats high power at that low of a voltage. I dont think Ive seen such a thing, even from the flashlight nutters over at candlepowerforums.If youre not keen on fiddling with the transformer, the LMZ22010 from TI looks interesting, 2 of them could share current, but a tad pricy at $16 each, and you'd have to make a PCB to go with it, or bodge it down on some protoboard, which since the switching is all internal could be doable.edit:Quote from: Cliff Matthews on March 20, 2015, 08:01:58 pmgood point, at 1.2v out, diode losses will be eating up a ton compared to the load, especially a bridge rectifier with 2 diode drops.

author: Simon
author_rank: Unknown
timestamp: March 20, 2015, 08:15:18 pm
content: 9V in -1.5V rectifier =7.5Vx4A = 30W 30Wx70% efficiency of a SMPS = 21W/1.5V=14A.Now that is not looking in detail at the efficiency of the SMPS, I expect that at lower output voltages the diode will cause more of a loss so a nice regulator that uses a mosfet switch for the off cycle will be needed and will need a nice low on resistance.The other option is what is this used for ? if for example its to power a heating element, or a light, or a motor or anything else that does not give much of a stuff about what type of power it gets so long as the average voltage is right then you could just pulse width modulate the 7.5V output at a 20% duty to get your equivalent of 1.5V. Much more efficient

author: Zero999
author_rank: Super Contributor
timestamp: March 20, 2015, 09:12:44 pm
content: Quote from: Cliff Matthews on March 20, 2015, 08:01:58 pmThere are ways to avoid diodes.It's possible to use the LT4320 with four MOSFETs to convert AC to DC at high efficiency.http://www.linear.com/product/LT4320The diode in the usual buck converter circuit can easily be replaced with a MOSFET.

author: mariush
author_rank: Super Contributor
timestamp: March 20, 2015, 11:34:35 pm
content: Wheres_my_chippy: you can not use a 7805 or other simple chips like LM317 because these are linear regulators.Linear regulators work by wasting the excess power as heat. So for example, if you have 9v DC at 4A available and you want 5v, then a linear regulator will only be able to give up to 4A of current but will waste ( 9v - 5v ) x 4 = 16w in the form of heat to achieve that. Linear regulators simply can not give more current than what the source power supply can provide.To reduce these losses in the form of heat and achieve more current at output, you need to use switching regulators chips which also need additional components like inductors, diodes, resistors and so on.Since you have a 9v AC power supply and almost all regulators work with DC, the first step is to convert your AC voltage to DC. This can be done using 4 diodes or bridge rectifiers (4 diodes in a convenient package)The rectification process converts your 9v AC into a wavy DC output with peaks of 9v x 1.414 = about 12.7v but there are some losses in the diodes in the bridge rectifier, usually about 1.5v. So in practice you end up with about 11-11.5v.The current doesn't remain the same, instead of 4A the actual current becomes about 0.62 x 4 = 2.48A ... so really, you're looking at best at 11.5v x 2.5A = 28.75 watts ... let's round it up to 30 watts.If you resort to chips like the already mentioned LT4320 plus the required 4 mosfets, then you would actually have about 12.5v x 2.5A = about 32 watts.Even the best dc-dc switching regulator chips will get you about 95% efficiency, but it's better to think 90%... if you want 1.5v x 14A or 21 watts with about 90% efficiency, then you need to have about 25 watts which turns out it's possible in your case.From here, it's just a question of finding a proper switching regulator or controller and using the datasheet to build a circuit with it. But judging by your posts, you're probably too much of a beginner to adventure into something like this, so I'd suggest at least buying a ready made adjustable dc-dc converter circuit.

author: Cliff Matthews
author_rank: Unknown
timestamp: March 21, 2015, 12:27:22 am
content: Looking at the original, Chippy said it's a 9volt supply. But it seems it was DC and assumed to be AC - that's why 8.5volts was said to be measured (likely unloaded after Chippy's rectifier?).To continue this and get the most from the transformer (a wall-wart?) pressure it slightly with a vise and use a sharp chisel or Exact o-knife to carefully break the two halves apart. Then use a full-wave schottky bridge and few thousand uF of low ESR caps to produce as much DC from this thing as possible, afterwards you may have a chance to get to your goal.

author: Flump
author_rank: Frequent Contributor
timestamp: March 21, 2015, 01:32:17 am
content: Quote from: mariush on March 20, 2015, 11:34:35 pmmariush that was a great post, thank you for sharing that