> I am not as familiar as I want to be on this topic and I'm 
> looking for
> pointers. ;) My goal is to modify an electric cart to be 
> charged by solar
> panels mounted on the cart's roof. The cart charging 
> circuitry requires a 48
> volt input and the maximum output of the solar panels (in 
> series) is 34V and
> 7A.

> What I need then is a converter that can handle about a 
> max of 5A output at
> 48 volts. In my research, I'm drawn to a boost type 
> converter since I need a
> constant and higher voltage output than the input voltage. 
> A boost converter
> is also attractive because of its simplicity. Designing a 
> flyback type
> converter would be more difficult for me, but I want 
> something that works
> well!
>
> Can anyone give me pro/con about a boost topology 
> converter for this
> application?

If this is a one off then buy versus build needs to be 
considered.
That's possibly "buy and modify" but you may still find that 
it's cheaper and easier to do that.

Assuming build from scratch -

This is in the capabilities of a single switch simple boost 
converter, although commercial solutions at this level may 
often use some form of push pull topology.

You say the charging circuit needs 48V. Is that a 48V 
battery or a lower voltage battery with a charger that needs 
a higher voltage than the battery. That is an important 
question as it can severely affect the solution. If you want 
something that works with an existing 'black box' then 
optimum efficiency may not be the aim. if you want best 
energy use then more details of the whole system would be 
useful.

eg
xxx solar panel with N cells.
Vmax = 34V (68 cells?)(unusual)(measured?)
I max = 7A (measured)(short circuit?)(name plate?)...

Assuming that the 34V is Voc in full sun then Voptimum 
(Vmpp) will be about 34 * (0.75 to 0.8) = 25 to 27V

Assuming a 48V battery = 4 x 12V the required charge voltage 
will be more like 56 V.

So step up will be say 25 to 48 ~= 2:1

A boost converter with the inductor "standing" on the 25V 
rail will need to produce another 25V or so.
Efficiencies approaching 95% should be possible with care.

On the input side power in (first cut) = say 25*7 = 175 
Watts
Question: What resistance in the primary side will cause 1% 
losses?

1% = 1.75W say 2W
For 2W at 7A R = P/I^2 = 2/49 ~= 0.04 ohm.

So per 1% primary loss you can have 40 milliohm for wiring, 
inductor, MOSFET (hot) and more.

Secondary (boost mode losses) involve ~= 3.%A the wiring and 
7A pulses in the inductor, diode and 50V reservoir capacitor

Assuming Schottky diode and say 0.5V drop you get 7A x 0.5V 
x 50% = 1.75W say 2W losses or another 1%.

If you are aiming at 95% overall and allow ~= 50% in input 
and out put sides you are allowed a total input equivalent 
resistance of about 2.5(%) x 40 milliohm = 100 milliohm.

If the FET is half of this that's 50 milliohm. A FET with a 
real world 50 milliohm Rdson will need a spec sheet one of 
under 25 milliohm say 0.02 ohm as the spec sheets almost 
always quote pulsed values at very low duty cycles.

FET rating needs to be say 20A, 50V (abs min, more better), 
<= 20 mohm Rdson.
The IRFP064 is a good start. TO247 package.
$6 Digikey
Claimed 0.008 Rdson
55V 110A
A bit more voltage would be advised!

IXFH110N10P is a bit better and a bit dearer
http://ixdev.ixys.com/DataSheet/99212.pdf


http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail?name=IRFP064NPBF-ND
http://www.irf.com/product-info/datasheets/data/irfp064npbf.pdf


I'll stop there.
MUCH more can be said but lets see if we are going in the 
right direction for you.

This and friends could be useful
Note that Schottky diodes really really don't like over 
voltage - rate conservatively.

        http://www.vishay.com/docs/88941/v30100s.pdf



        Russell McMahon

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