Upgrading amps of a circuit to 12A

Hi,
I did motorize my kid's electric car with a 12v wiper motor that burns out any fuse lower than 12A because of it's high current at start. I think it's running amperage is about 6A.

I found this great circuit that fits almost exactly in what I want, except that it have a bit weak output amperage.
here is the circuit:
http://www.soloelectronica.net/dcdc_12v_a_18v.htm

I think this is also great because of the potentiometer for the output current that will be very useful.
If anyone with such knowledge in electronics could help me to replace a few components that help rising the output power would be perfect!!!
I also know that for this high current I should enlarge the pcb traces and place loads of solder on it.

best regards!

This circuit is odd that the motor is connected at the input, which means it rneans at a constant speed,Am I missing something here?

This circuit is a boost circuit, meaning it can only increase the output voltage, What you would need is a buck-boost, or a flyback. Or take what you have and add a buck stage to it (down conversion). The buck can be either before or after the boost, although it looks like the spec indicated 9V as the low operation point at which it cuts off, so likely put the buck after the boost and use it to control the output voltage to the motor (pot).

Or even better, modify this circuit (simple) to make it a buck, then can can go from about 11V to say 1V.

Oh, sorry the motor is for a fan motor. Never mind. Also the output is 18V and will draw much more current at that voltage. If it's a 12V auto motor, then 14.5V is the most it should ever see, not 18v. So no wonder it's blowing a fuse.

As mentioned this circuit won't do what you intend it to do. Need modified to a buck converter, not boost converter.

Hi Earl thank you for your answer!
Yes I want it to have 16v or 18v output. I want more juice on my wiper motor . but I also need it to have at least 12A output.

this board specs:
- Output current: 5A a 18V (10A peak)
- Input tension: between 10 y 15V CC
- Working frequency: 40kHz
- Efectiveness: 95%
- Output tension: between 12 a 26V
- Max output power: 100W
- Protections: Low tension 9,2V by CI, over power (10A) F1 on board, by polarity D2 and F1 on board.

There is one potentiometer to adjust output voltage from 12v to 26v (or less) as it can be seen in the pictures and here in this specs!

And yes it is a boost circuit. With output tension adjustable.
As I said, all specs are fine except the output current that is too low :S

regards!

Its fine to have 18v output since it have an pot for adjusting output tension that I would let it by 15v and test the motor behavior.

Or, perhaps, a slo-blo fuse?

It is rated input power as 100W. Output power is only really 95W or so. You could get perhaps 10 - 15W more out of the circuit by increasing the fuse t0 12A. That's basically the limit. It still may blow a fuse though because the inductor may saturate so a different inductor with a higher DC rating. Also the chip IC2 is really not a good FET driver. That could be replaced with a real driver, make the 10Ohm drive resistor 2-Ohms and increase the efficiency.

As is this is too close to it's design limits.

Dear Mr.Rafael Borowski,

Thank you for e-mail. I think the circuit has to be improved with appropriate components of suitable (better) 'derating factors'. This would definitely solve your problem.Pl.try this exercise and observe the output. .

Best of luck !!

With Regards,

KCPadhy,
For KICSPL,Bangalore-93, India,
Tel/Fax:+91-80-25241189 ,Mob:9916133982 .
E-mail: kicspl(_at_)qmsskc.org,
Url:www.qmsskc.org

Hi Earl,
I made this picture with a few possible mods.
What do you think?
Do you know any free software for drawing the circuit for printing?

My regards,
Rafael.

L1 - do not change the numver of turns to more. Parallel is ok but not more turns because that causes the amp-turns to go up and so does the DC flux, which absolutely could cause an issue. You can buy these pre-wound torroidal cores through a distributor. The thing to look for is the max DC load current you want, multiply that by 1.5, then look for the same inductance you have which is 65uH, which you need so not to cause any other things to go wrong (won't get into that).

IC2 should be replaced with an 8-pin MOSFET driver, TI makes them, many others, Vishay, Faitchild. Get those through distribution. They are real simple, +/- connections, in/out. You can get them in dual or single drivers. Either will work in your situation. There are single drivers with up to 6A of drive current available. The MOSFET drive resistors should be at least 1/4W (have not done the calculation but ball park idea of what it should be).

More output caps, I really don't think doule is needed. Go with the percentage increase in output current you want. So if you are going for 20% more output current, then up the caps by 20%.

I think the rest of the changes are probably OK.

Free Schematic Capture software with print capability:

* http://www.digikey.com/schemeit
* http://sourceforge.net/apps/mediawiki/tinycad/index.php?title=TinyCAD
* http://www.diptrace.com/download.php (the "freeware" links)
* http://www.cadsoftusa.com/download-eagle/
* http://www.sunstone.com/PCB123-CAD-Software.aspx
* http://www.expresspcb.com/expresspcbhtm/Free_schematic_software.htm
* http://www.olimex.com/pcb/dtools.html

Thanks a lot!

Hi,
I know it would take the circuit project part out of your project but I would recommend:

"Traxxas esc(traxxas esc)":http://www.amazon.com/Traxxas-Brushed-Waterproof-Electronic-Control/dp/B001ER092W/ref=sr_1_27?ie=UTF8&qid=1340648789&sr=8-27&keywords=traxxas+esc

To work with your current motor and V.

or

a BLDC motor and controller to replace everything, they can be found for less than $60.

http://www.youtube.com/watch?v=nzuCmBgE660&list=UU0moHebOrEjWKqYmOKjRL8g&index=3&feature=plcp
This link will redirect you to a test I made of amps draw by this wiper motors.
I am finishing the schematics and board (I've learned Eagle from Youtube yesterday) and hopefully will post here later today.
regards!

"sch file (Eagle Light)()":https://skydrive.live.com/redir?resid=EF237A0748651ACF!409&authkey;=!ACp-G5ZzyRFiOI0
"brd file (Eagle light)()":https://skydrive.live.com/redir?resid=EF237A0748651ACF!410&authkey;=!AEdjz0z2Ij1rT2w

Ok so this is better. Each FET should dissipate about 1.2W (could tripple if there is timing skew and RDS_on mismatch). Not the best way to drive these. Look up UC27518.

I don't know the parameters on your inductor other than it's a gaped ferrite. Some quick calculation and looking up the core data indicates saturation of a DC_inductor current of 5.25A. So you either need to change the inductor or it will saturate. You could reduce turns to get the inductance down to say 35uH and that would double you DC_inductor current but it would also double the p-p ripple current seen by both input & output caps.

What happens when an inductor saturates? It become just a coiled up air core inductor reducing the inductance to perhaps a few hundred nH. It would draw 1400A in 12uS which is your 50% cycle time. Can it dray 1400A, well no because you have a fuse. You need to get a different inductor, seriously or it won't work.

Here is a web site that has an inductor that would work.

http://www.coilcraft.com/dmt.cfm

You can actually order the part from their site or cal distributors like DigiKey, Mouser, etc. You have to dump you current inductor. Also perhaps you can get away with the MOSFET drive scheme if you use enough air from the fan. This circuit should be able run with no fan, if the MOSFET power dissipation is down. The diode losses are about 5W at 8A load current, which the heat sink could do at room temp.