Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Voltage reducer 12v to 9v

  1. Sep 28, 2009 #1
    Any EEs out there who can help me with a circuitry question?

    I purchased an Estes rocket for my kids that uses a small chamber to run DC current through water and produce hydrogen which is, in turn, the fuel to launch the rocket. It's great when it works but the problem is that it runs on 6 "D" cells and quickly drains them.

    I was able to find one post online of a person who figured out a way to hook his launch pad to a 12 volt garden tractor battery--he stated that he had measured the peak current load at approximately 6 amps. However, he didn't elaborate (but I assume he didn't simply use the 12 volt battery or else he would have fried the ignitor filament in the hydrogen chamber that glows to ignite the H2 when you push the launch button).

    I purchased a portable 17 Ah power pack (they type you can use to jump a battery) that has a 12v outlet on it. I was hoping to come up with a way to reduce the voltage from 12v to 9v but maintain a 6 amp (peak) draw. However, everything/anything I can find online that reduces the DC voltage from 12 to 9 volts is only rated at 2 amps or less.

    Can someone who is familiar with circuits outline what size and rating resistors would be required (and whether in series or parallel) to be able to accomplish this? Obviously, I am hoping to come up with something relatively inexpensive and that I could place in some sort of small enclosure. At one point I thought perhaps I could use an old power supply from a PC, but any that I've found that have a 9v lead only have a minimal amperage rating.

    Thanks for your help
     
  2. jcsd
  3. Sep 28, 2009 #2
    3 volts at 6 amps is 0.5 ohms. I2R is 18 watts. Try two of these in parallel:
    http://www.radioshack.com/product/index.jsp?productId=2062290
    Your 17-AH battery-booster power pack might put out something closer to 14 volts to boost a car battery, so perhaps you should get two more 1-ohm resistors so you would have one ohm total with 4 resistors. Still, the zero current voltage output would be equal to the battery or booster voltage.
    Bob S
     
    Last edited: Sep 28, 2009
  4. Sep 28, 2009 #3
    Are you serious; an over the counter O2 + H2 rocket? Doesn't it go off all at once?

    You might risk a few small charges to see if the ingiter can ignite the charge at 12V.
     
  5. Sep 28, 2009 #4
    The 3 volts being the drop that I'm looking for--right? Would putting the 2 of these in parallel be the equivalent of a "zener diode"? Somewhere I read that a zener diode would do it, but it seemed as though none of the parts they listed on that site would handle the watts needed?

    Will these resistors get pretty hot with 18 watts? That doesn't seem like much but I guess even a 20 watt bulb can be too hot to touch.

    Thanks
     
  6. Sep 28, 2009 #5
    Phrak,

    It takes about 5 minutes to bubble (has a solution of citric acid in the chamber and a catalyst which I think I read somewhere is made of zinc), then goes into countdown mode and you push the launch button which basically energizes the filament to make it glow and ignites the small amount of H2. Wonderful when it works, but I was frustrated to only get a couple of launches out of a new set of "D" cells. Maybe the real problem isn't the batteries at all, but I figured it was just sucking too much out of them too quickly because I wasn't getting reliable H2 generation and also wasn't getting reliable filament glow.

    I bought about 4 of these units when they were selling them out at the Shack for $10 each (last year). On one I already burned out the filament when I wired it directly to a 12 volt car battery, so figured the components can only handle the 9 volts.
     
  7. Sep 28, 2009 #6
    Guess I didn't think about the fact that during the phase of H2 generation when it's not pulling 6 amps, the equation will no longer be valid and my voltage will not be constant, so perhaps this will cause a problem elsewhere?
     
  8. Sep 29, 2009 #7
    There's not going to be any really simple, cheap, efficient way of dropping 3V. Especially using passive components.

    You could use a high-powered buffer to replicate a 9V input, and use the +12 as a supply voltage. That would cost around $20.

    You could buy a pre-made 9V 6A power supply, such as http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=VOF-65-9-ND which would cost around $28.

    You could make a AC-DC supply yourself for around $15.

    Believe it or not, DC-DC converters are quite complex in general compared to AC-DC or AC-AC converters.
     
  9. Sep 29, 2009 #8

    vk6kro

    User Avatar
    Science Advisor

    One way of doing this would be to put 5 high current Silicon diodes in series.

    Each will drop at least 0.6 volts, so you get a drop of 3 Volts, even at small currents

    But you do need to know the currents.
    It would be good if you could measure the current from the 9 volt supply when producing your gas mixture and also when firing the explosion.
    These diodes would get hot, so they would need a heatsink and would need to be electrically isolated from the heatsink.

    I have seen a Hydrogen explosion and I'm surprised that they get away with doing this safely.
    Are you sure they really use a hot wire to fire the explosion? I would have expected a spark.
    The website for Estes rockets showed one Hydrogen model that could get to 250 ft. That would be amazing.
     
  10. Sep 29, 2009 #9
    Thanks for the link--I did a lot of web searches but hadn't found anything like this. When you mention the high-powered buffer, would that still be a completely DC system? Guess if possible I was hoping that I could keep this all DC so that we can take it out in the middle of a field, etc., without a cord (although we usually launch it close enough to an outdoor outlet that we could go that route if needed). If it would be all DC, could you elaborate?

    Also, on the ac to dc 9v 6a supply link you supplied, I note that it is rated just over 6 amps. Is my understanding of the loads correct--the load will only pull the amount of amps it needs (ie, even if the battery is capable of more) but my experience was that the higher voltage (of going directly from a 12v battery) is what burned out my filament?

    Thanks
     
  11. Sep 29, 2009 #10
    The filament is actually more like a glow plug--looks sort of like a small spring stretched between 2 terminals suspended in the middle of the gas accumulation chamber. I have read of a few people who managed to "explode" their chambers if they ran the unit through several of the gas accumulation cycles without effecting a discharge--but my experience has been that, if anything, it's hard to get enough gas to accumulate just to have a decent launch.

    As for measuring the currents, the multitester I own isn't rated for currents in that range. I've looked at a couple that can test up to 10 amps, but I have noticed that they say that if you measure anything that high for over 30 seconds it will damage the tester. But I guess if it's under the 10 I should be OK. I'll try to pick one up and test a cycle with fresh "D" cells.

    Thanks
     
  12. Sep 29, 2009 #11
    Here in thumbnail is a regulator that produces a relatively constant 3 volt drop for different loads and different input voltages. The 2N3055 is rated at 10 amps collector current, and needs a sufficient heatsink for 20 watts.
    Bob S
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook