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Random Quote: When the tide goes out, we find out who has been swimming naked (Warren Buffet)
Does the exact current requirements make that big of a difference? Can't we assume that the circuit should work up to 15 amps, and any thing below that?
I think it does. If you design for 15 amps, and only need 1 amp, then in my opinion it is way over designed. The size of a holding capacitor for 15 amps would be very large indeed.
The way to determine the correct size cap is fairly simple. C = I * (dt/dv)
C(in farads) ='s the load current in amps * (how long the period of time is in seconds / how much voltage change you can allow over the period of time).
so for example, setting the cranking interval to 3 seconds, 15 amps for the load current, and allowing only a 1/2 volt drop during that 3 second interval.
C = 15 * (3 / .5) = 45 farads. Thats really really big, swimming pool big. As you can see, it will be a trade off between cranking time, voltage drop, and how big of swimming pool you can afford.
I forgot to add, but due in part to our butt thumping friends of the car audio world you can actually buy caps this big, or at least batteries masquerading as caps. LOL http://cgi.ebay.com/ws/eBayISAPI.dll...category=14932
Ok, I only need 2.5 amps (Chartplotter/radar) for a max of 1 second(HPDI's start instantly). Not sure exactly what my volt drop is, what would be about normal?
My unit draws 5.5 amps, and I need to maintain that current with with only a 1/2 voltage drop for 3 seconds.... I would need a capacitor equal to C = 5.5 * (3/.5) = 33 farads.
And with you circuit, the capacitor should be hooked up in parallel to my units power, with tha anode of the Cap connected to the positive wire and the cathode connected to the negative wire. And between my fuse and the capacitor I should wire in series a Schottky diode with the anode connected to load side of my fuse and cathode connected to load (+) side of the capacitor.
These you can hunt for on ebay. In fact I have seen them at walmart too.
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The diode is dependant upon the caps series resistance, as the cap will be the big current load for the diode as it is charging up. Hmmm this leads me back to the original circuit that was proposed. Maybe a combination of the 2 circuits is better for this application (sorry my brain is in design mode at the moment).
The problem I just noticed is the large current requirements to charge the cap initially. This would drive a larger wire gauge, and fuse rating. The cap looks like a short circuit when power is initially applied to it.
So lets do this. Lets take the original circuit with the diode and resistor on one side between the cap and the load, and my diode on the other side between the fuse and the load. What this does, is allow the capacitor to charge up with lower current as the resistor will limit the current flow to it, and my additional diode can serve as the isolation from the battery as it's voltage is pulled down while cranking.
A Dual diode package with the cathodes tied together, for instance digikey.com part number 220CMQ030-ND, looks OK but expensive. The reason I like is it will accept terminal lugs rather than soldering to it. But any approch is fine.
The first is a pic of the part, and the second is the schematic. The difference of the two approaches is the new diode between the fuse and the load.
Sea_Dad, Thanks for sharing your knowledge. Thanks also to Gil Morgan.
I just have one more question, when buying capacitors should I be concerned with its voltage rating? I am looking at a cap rated for 50 farads and a voltage rating of 2.3V. Will this work in the above circuit?
Wow, what the hell happened over the last few days?? Come back and see lots of new stuff on capacitor/diode clamp circuits. Some good ideas, some not so good. Need some time to look things over.
Sea-Dad, you may want to change that 10 ohm 2W resistor to at least 20W. You will be dissipating at least 14.4 watts initially.
Wayne, those 2.3 volt capacitors will not work. Putting capacitors in series to get higher voltage capability is no good. Unless they are perfectly matched, one will tend to take more of the voltage than the other, causing it to break down.
More later.
I'm confused because the pkg you have has 2 stripes.
Most diodes that I have worked with only have 1 stripe which is the Cathode (-neg) side.
We also must keep in mind that different mfg's equipment have different input voltage specs.
My 2006C will operate from 8-32V. I'm working on a design to supply at least 9-10V not 12V.
I'm leaning towards a small 12V 1.2 to 4 Ah (general purpose lead acid, Yusa part number NP1.2-12)wired with 3 diodes to prevent it powering anything but my GPS. This size battery is overkill($15-20) but I have room for it.
My mormal 12v system will also keep this battery charged.
I wish I could draw a schematic and submit it.
Thanks, I was concerned with spiking the electronics but your diagram helps a lot.
There is just one more thing. I was wondering what function the resiitor has in the loop. Is that to prevent the spike upon discharge, or does it slow the charge rate of the capacitor.
I wasn't able to find a 10ohm 2w resitor. Can I put 2 -10ohm 1watt resitors and get the same effect.
Sea Dad,
I think you will find that "batteries" and their "longer dis-charge rates" would work better for this application.
Your design you posted on 11-3 is fine but I would use a battery instead of the cap.
I hope this not turn into "which" battery is best.
I need to purchase few materials, construct a proto-type and bench test.
I'll post my results.
Like BW23 points out, a battery will most likely be the better solution for these current demands. The cap is just too big in value to be economical.
Sean, the 4700uf cap would drop below 10 volts in less than 8 milliseconds by using this circuit. (the resistor is to allow current to flow around the diode so the cap can charge.)
Maybe BW23 can supply a link for the battery he is going with. It really does sound like the better approach for current draws larger than 1/2 amp or so.
To use a battery just wire a single diode in series with the fuse. Anode to the fuse, and cathode to the load & battery. Just wire the battery in parallel with the load.
Maintain Voltage Drop Capacitors ??? 
