mr. RB,
Input voltage is 3.7v from li-ion battery. Duty cycle is inversely proportional to current. For 1800S or 30min duty cycle, current required is about 107amps.
colin, I wouldnt be posting here for help if I thought you all our idiots. I am not sure what part of my post you found so offensive.
I'm assuming that you are multiplying current times time to get "107amps". Current times time is AH (amp-hours). This relates to battery capacity So if I divide your 107/0.5Hrs I get 214 AH. If I divide your 107 by 1800 I get ~0.06A. Is that the amount of current you need out of your circuit.
Ken
Where did you get "1921970"? As my math instructor (2nd grade) said frequently: "show your work!"No, Amp-sec is 1921970 . So for 1800 sec or .5hour i need 107amps.
What is the LiON batteries current rating?
Use a buck converter, like the one below, except use a second FET instead of a diode. Search the web for more details, but a word of warning: You are talking about a bunch of amps at a low voltage, and in such applications, subtle variations in the design, such as the type of wire or foil used in the choke, can have a huge affect on efficiency. You might find it best to hire a consultant for this.
**broken link removed**
That depends on how much I need to convert to 107amp.
So, you want a DC-DC converter to step down the 3.7V LI-Ion battery to 0.4V @ 107 Amps?mr. RB,
Input voltage is 3.7v from li-ion battery. Duty cycle is inversely proportional to current. For 1800S or 30min duty cycle, current required is about 107amps.
Seriously? You do realize a power conveter which can output 107A is going to be at minimum the size of a shoe box?i need 107amps.
0.4v 107A output from 3.7v input is NOT going to be efficient. You need to generate higher voltages to get good FET turn-on so that is complexity and power wasted.
Even with synchonous recification 107A will still drop a heap of millivolts in the rect FET and the inductor coil and PCB connections. Ripple voltage in the inductor and cap will be a lot of millivolts too so it needs a lot of capacitance. Even a good toroid with windings that will do 107A average and enough size to keep the ripple voltage down will be large enough to have a nasty coil resistance maybe in the tens of milliohms. I wouldn't be surprised with losing 0.5v to 1v just in the IR losses in the output loop of the buck ie FET/inductor/cap/PCB. That puts it at maybe 30% efficiency just with output losses.
So it's going to be large (brick sized), complex, have a very expensive inductor, and MAYBE get you 30-40% efficiency if you really know what you're doing...