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I have a LiPo that is 1800mAh,25C,11.1V. it can supply a current of 1.8*25 = 45A. I need a current of 3.5A to drive a LOAD(a Nichrome wire of length 5-6cm) and 7.4V. How do i proceed with this ?

I have tried looking into voltage regulators and Voltage Dividers if you have a solution or a different method that is of great help as well.

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  • \$\begingroup\$ Sounds like you just need a DC-DC buck converter that can supply 7.4V, and is capable of delivering at least 3.5A. What is the load? \$\endgroup\$ – mkeith Oct 18 '17 at 6:19
  • \$\begingroup\$ You do realize that your LiPo system can only supply your load for about a half hour, or so? Is that okay? (\$80\%\frac{11.1\:\textrm{V}\cdot 1800\:\textrm{mA-hr}}{7.4\:\textrm{V}\cdot 3.5\:\textrm{A}}\approx 37\:\textrm{m}\$, and probably less than that.) In any case, you might look on ebay: ebay.com/itm/192264665783 (that's not a recommendation... just a pointer.) \$\endgroup\$ – jonk Oct 18 '17 at 6:19
  • \$\begingroup\$ yes i just need the LOAD for a couple of seconds \$\endgroup\$ – Karan Motiramani Oct 18 '17 at 6:24
  • \$\begingroup\$ @KaranMotiramani Those converters are not designed for high speed pulsing nor do they work well with cases where you run them with the load completely removed. How do you plan on turning it on and off? You actually may be better off with a linear drop, if this is only momentary. What is the load, as well? \$\endgroup\$ – jonk Oct 18 '17 at 6:28
  • \$\begingroup\$ @jonk i didnt get your point my load is a Ni Wire that i need to heat up..its 5-6cm in length how do you suggest i proceed ? \$\endgroup\$ – Karan Motiramani Oct 18 '17 at 6:33
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I need a current of 3.5A to drive a LOAD(a Nichrome wire of length 5-6cm) and 7.4V. How do i proceed with this ?

Let's get started with a couple of equations that we're going to need.

Sloppy Ohm's law: \$V=A×Ω\$

Sloppy power equation: \$W = V×I = \frac{V^2}{R}\$

By saying sloppy I mean that I'm using the units rather than their common variables that holds their values (U for voltage, I for ampere etc..).


The power dissipated in the nichrome wire is \$7.4V×3.5A=25.9W\$

So let's see how much current we would only require if we want \$25.9W\$ and we're using \$11.1V\$ instead.

\$W=V×A <=> A = \frac{W}{V} = \frac{25.9W}{11.1V}\approx 2.3A\$

Now how are you going to get 2.3A instead of 3.5A? Either use a buck converter that is 100% efficient, or do something else that's much easier and 100% efficient (no step down loss).

Just make the nichrome wire longer, that way you will linearly reduce the power dissipated into the nichrome wire. It is only linearly reduced because the voltage is fixed and I'm treating your battery as an ideal voltage source.

So we know that \$W=\frac{V^2}{R}\$, we know that you want \$25.9W\$, let's calculate how much you're getting with the "5-6" cm you got now.

The resistance of the nichrome wire is \$\frac{7.4V}{3.5A}\approx 2.1Ω\$

\$W = \frac{11.1V^2}{2.1Ω} \approx 58W\$, this means that you need a wire that is \$\frac{58W}{25.9W}=2.25\$ times longer. So if the wire is 5 cm, then you can replace it with \$11.25\$ cm wire and get \$25.9W\$ dissipated in your new longer nichrome wire.

And there's no transistor involved anywhere, no step down losses, the only real losses is that first nichrome wire you got that you might have to throw away.

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