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ultra low voltage converter?? (0.4v)

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huzefasi

New Member
I need 0.4v for an application which I cant share on public forum. What is the most efficient to get this? I dont want to use a linear converter because input source will be battery, I want to increase the current.
 

colin55

Well-Known Member
Don't help someone that thinks we are all IDIOTS.

The original poster needs to learn a few manners.
 

huzefasi

New Member
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.
 

huzefasi

New Member
Its for a "possible" commercial application. If it works out I will post more details.
 
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Dick Cappels

Active Member
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.


8820-BuckConverterSchematic.gif
 

KMoffett

Well-Known Member
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
 

huzefasi

New Member
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



No, Amp-sec is 1921970 . So for 1800 sec or .5hour i need 107amps.

EDIT: It is 192970.
 
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KMoffett

Well-Known Member
No, Amp-sec is 1921970 . So for 1800 sec or .5hour i need 107amps.
Where did you get "1921970"? As my math instructor (2nd grade) said frequently: "show your work!" ;)

ken
 

huzefasi

New Member
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.


8821-BuckConverterSchematic.gif

Thanks for diagram and advice. I will try to figure it out.
 

Mr RB

Well-Known Member
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...
 

bountyhunter

Well-Known Member
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.
So, you want a DC-DC converter to step down the 3.7V LI-Ion battery to 0.4V @ 107 Amps?

I would sure like to see that....:D
 

bountyhunter

Well-Known Member
i need 107amps.
Seriously? You do realize a power conveter which can output 107A is going to be at minimum the size of a shoe box?

To drive that kind of current you would prbably have to parallel about ten or fifteen FETs on the top side and the same on the bottom (rectifier) side. To get all that gate capacitance to switch fast will take drivers with a huge amount of current drive capability. IMHO, this is a seriously ambitious design.
 
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huzefasi

New Member
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...

Thats too bad. Thanks for your input though, I got to rethink this to see if there another way. My priority is light weight and really compact size.
 
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