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Unread 13-01-2008, 22:58
Ronald_raygun Ronald_raygun is offline
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Circuit Design

NOTE: This has nothing to do with the FIRST competition.

I'll be participating in another Fuel cell car race this year with the NorthEast Regional Science Bowl.

Last year, my car was the fastest primarily because I was able to use a capacitor to store the fuel cell's electricity.

Championshiop Race

First Race

I expect that next year, my opponents will be using capacitors of their own cars against me, so I need to step up my own car. My target last year was 8MPH, and I hit it exactly. This year, I wanna try and reach 20MPH. A little higher, but I think very possible.
  • Hydrogen Capacity was increased
  • Oxygen Intake and filter developed
  • New starting system which uses a push-push (on-off) set of switches.
  • Multiple capacitors in a series-parallel chain.

It's the capacitor problem which I'm having trouble getting through.

I'm chaining two 2.7v 10F capacitors in series, and getting two of those chains in parallel, for a total of 5.4V and 10F.

How do I wire those capacitors to a DPDT switch so that when the fuel cell is charging them, they are individually charging in parallel, and when armed for a race, the capacitors are in the series-parallel arrangement described above.

Any thoughts?

(For schematic-drawing software go to www.expresspcb.com and download their software. )
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Last edited by Ronald_raygun : 13-01-2008 at 22:58. Reason: Missing bracket
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Unread 14-01-2008, 08:16
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Re: Circuit Design

Jose,
It is quite easy. Just wire two two pairs of two capacitors in parallel. The pairs then act as one single cap. Then it is a simple matter of switching one of the pairs to be in series with the other for normal operation and in parallel for charging. The second pole of the switch is used to switch the low side of the upper pair between the negative side of the charging circuit and the positive side of lower pair.
If you need a schematic, let me know.
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Unread 14-01-2008, 17:42
Ronald_raygun Ronald_raygun is offline
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Re: Circuit Design

Is this what you meant? (Sorry for the resolution)

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Unread 14-01-2008, 19:27
EricVanWyk EricVanWyk is offline
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Re: Circuit Design

Please see attached. The switch is in its "discharge" mode. And yes, I did that in Altium

What you are looking for is a standard "switched cap boost" circuit. Usually, they do this to create a supply that is (nearly) double the input voltage. This is done with a few more diodes and FETs, and fewer switches. Also, they flip the "switch" at about 100kHz (or more).

Be sure that the Fuel cell doesn't mind getting double its output voltage shoved in its face.

Tie this circuit in parallel to your battery.

Good luck!
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Unread 14-01-2008, 22:17
Ronald_raygun Ronald_raygun is offline
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Re: Circuit Design

The fuel cell can take up to 4V. I'm sure we can put a voltage limiter if needed.

The way I envisioned this system was to have two power sources: One from the fuel cell at ~2 volts, and one from the capacitor at ~5 volts.

When I put two batteries with uneven voltages in parallel, doesn't the load draw more current from the weaker battery?
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Unread 15-01-2008, 07:52
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Al Skierkiewicz Al Skierkiewicz is offline
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Re: Circuit Design

Jose,
This is what I had in mind. I thought you were looking for a way to charge all capacitors in parallel and then switch them to be in series for the double voltage. When the switch is up the caps are in series and when down they are in parallel.
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Unread 16-01-2008, 12:48
SteveJanesch SteveJanesch is offline
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Re: Circuit Design

Jose,

When you put two batteries with unequal voltages in parallel, the lower voltage battery acts as large load on the higher voltage battery. The stronger battery tries to recharge the weaker one in addition to drive whatever load is placed on them. The net output voltage is somewhere between the two voltages, depending on the relative strengths of the batteries.

When the fuel cell charges the capacitors, the caps' voltage will eventually equal the maximum voltage the fuel cell can produce. If you then stack the caps in series, that will nearly double their output voltage.

Question 1: What's the output voltage and current capability of your fuel cell?

Question 2: What happens when you put the stacked (double voltage) caps across the output of the fuel cell? I think that at best the output current of the fuel cell will drop to zero, since the higher voltage of the capacitor stack will dominate the cap/fuel cell combination. You'll be driving your motors off the capacitors only, which may be okay. Worse could be that the fuel cell draws current out of the caps, away from the motors, and you get no net benefit. Worse yet is that somehow the high voltage on the output prevents the fuel cell chemical reaction from happening, hydrogen gets vented from the fuel cell near something hot (from the overvoltage at the output), and manages to ignite outside the fuel cell. I'm dangerously ignorant about the fuel cell you're using, but I've seen a few things catch fire when too much voltage is applied.

I'd suggest you first check your data sheets for what happens when your fuel cell is subject to too high a voltage on the output. Next, if explosion is unlikely, you might try to mimic the situation you're proposing with a power supply or batteries. Good luck.

Steve
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Unread 19-01-2008, 06:43
Ronald_raygun Ronald_raygun is offline
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Re: Circuit Design

Excellent points stj.

The Fuel Cell requires 4V to charge, and has a measured maximum discharge voltage of 2.1v. (Syringe pushing hydrogen through.) It runs well, if that's what you mean by capability. 30cc of hydrogen at room pressure can power the fuel cell for 5 min under minimal load before it chokes on the water buildup.

I don't know what happens when you introduce 5.4V into the fuel cell. It'll probably be closer to 6v. I suspect however that at that voltage, the PEM will break.

The Fuel Cell I'm using is a double-reversible fuel cell. That means it can convert electricity into hydrogen and back. You can find it here.

I don't have any data sheets for this particular fuel cell except for the specs on the website:

Power Fuel Cell: 1W
Power Electrolyzer: 2W
Active Area: 2x3 cm2
Power Supply: 3.2-4.0 V DC
HxWxL: 54 x 60x 42 mm
Weight: 70g

Attached is a schematic of the current setup. Both switches are in the Armed/race position, and both switches need to be flipped if the fuel cell is charging the capacitors.

Just curious, would I be able to isolate the fuel cell from the capacitors with some series of diodes? Because if the fuel cell creates a large load, I may need to find another way to get more power to the motor.



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Last edited by Ronald_raygun : 19-01-2008 at 06:48. Reason: Forgot to put the picture.
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Unread 22-01-2008, 13:07
SteveJanesch SteveJanesch is offline
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Re: Circuit Design

By "current capability" I meant how much current the fuel cell will provide. No worries.

Judging by the website you provided, it looks like the fuel cell will provide up to half an amp at 2 Volts (depending on hydrogen flow, I imagine) and will electrolyze water into H2 and O2 when you apply 4 Volts, and draws about half an amp while doing so. Charging up the caps to 2 Volts and stacking them will put 4V on the fuel cell - no more than that. I'm not familiar with the cell, but it sounds like if you stack the caps the stored energy might to into hydrolyzing the water that built up while you were generating the electricity in the first place. Whether you wind up hydrolyzing more water or not, the motor will be running off the stored energy in the caps and not the fuel cell.

Since I don't know what the spirit of this competition is supposed to be, or what the rules are (particularly on stored energy), I don't know if what you're proposing is going to be legal.
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Unread 22-01-2008, 15:03
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Re: Circuit Design

Jose,
Sorry I didn't get back to this thread. I thought you were planning one of two things. Either put the charged capacitors in series with the cell for about 6 volts or you were going to start the race with the caps and then switch over to the fuel cell after a few seconds. Remind me what you did last year.
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Unread 22-01-2008, 23:56
Ronald_raygun Ronald_raygun is offline
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Re: Circuit Design

Quote:
Originally Posted by stj_1533 View Post
By "current capability" I meant how much current the fuel cell will provide. No worries.

Judging by the website you provided, it looks like the fuel cell will provide up to half an amp at 2 Volts (depending on hydrogen flow, I imagine) and will electrolyze water into H2 and O2 when you apply 4 Volts, and draws about half an amp while doing so. Charging up the caps to 2 Volts and stacking them will put 4V on the fuel cell - no more than that. I'm not familiar with the cell, but it sounds like if you stack the caps the stored energy might to into hydrolyzing the water that built up while you were generating the electricity in the first place. Whether you wind up hydrolyzing more water or not, the motor will be running off the stored energy in the caps and not the fuel cell.

Since I don't know what the spirit of this competition is supposed to be, or what the rules are (particularly on stored energy), I don't know if what you're proposing is going to be legal.

Based on that explanation, it sounds like the capacitors when in discharge mode will both charge the fuel cell and run the motor. Not really disastrous in a distance competition, but I'm in a speed competition. I'm expecting the caps to discharge themselves completely at the 4-5meter mark. I could use the momentum to finish the race, but I don't want to take that risk.

The rules define the fuel cell as the source of energy. Source is defined by the organizer as something that converts energy from one form to another. A capacitor doesn't do that so it's legal. In real life, I'd imagine that this method could be used in cars to help with acceleration.


Al,

What I did last year was run the capacitor in series with the fuel cell. It worked, until the semi-finals when the cap wasn't charged fully, so I ran the championship race in parallel.



This year I guess what I'm trying to do is get two separate sources of electricity powering the motor at separate times. The capacitor's job is to get the car up to speed. The fuel cell's job is to maintain the speed.


I remember reading something about the TRIMET which would use the solar panel to charge a set of capacitors, and when the capacitors reached a certain voltage, discharge them. Perhaps a similar system could be applied?

TRIMET
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Unread 23-01-2008, 07:36
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Re: Circuit Design

Jose,
The current density on small solar cell arrays is very small so your device would need to be in sun for quite a while to charge the caps. You could use either a relay or an electronic switch that would change over to the fuel cell after the caps discharge. You could make the control using a timer or a comparator. You need to see how long the car can run on just the caps and then a 555 timer might work just fine. I am just brainstorming here so it might not work but you get the idea. A 5 volt relay across the series combo of the caps and fuel cell would drop out at 3-4 volts. Just be sure to get a low current coil on the relay.
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Unread 23-01-2008, 23:39
Ronald_raygun Ronald_raygun is offline
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Re: Circuit Design

I was thinking more on the mechanism behind charging and discharging the capacitors on the TRIMET.

As I understand it, the TRIMET works by using the solar cells to charge the capacitor, and when the capacitor is at a certain voltage, the capacitor discharges into the motor.

I'm thinking of using a similar method, but in reverse. When the capacitor decreases to a certain voltage, it is sort of disconnected from the system, and the fuel cell is allowed to dominate.

You explained to me last year something about the drop out voltage. If I had that wired so that if the capacitor's voltage dropped below, the relay then connects the fuel cell to the motor?
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Unread 24-01-2008, 07:26
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Re: Circuit Design

Yes, think of the relay taking over for the switch in your diagrams. They come in the same flavors of DPDT, SPDT, etc.
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