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Current Reduction

Discussion in 'General Electronics Discussion' started by johnsonjp34, May 13, 2012.

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  1. johnsonjp34

    johnsonjp34

    5
    0
    Apr 10, 2012
    I'm working on a project that needs 18 V so I am using three 6V batteries in series. When I measure the current on my multimeter I get around 5.5 Amp. I need about half that current. So I used a 2 ohm 1/4 W resistor, but it burns out everytime. I don't understand why my other 1/4 W resistors say 470 ohm do not burn out. What type of resistor would I need for this to keep the resistor from burning out, maintaining ~18 V, and having around 2 Amp?
     
  2. CocaCola

    CocaCola

    3,635
    5
    Apr 7, 2012
    Please explain the application, most devices will only draw what they need from the batteries not the full amount... If you are working with LEDs there is likely a better solution...

    If the 1/4 watt resistor is burning up it's probably because you are exceeding it's 1/4 watt rating ;)
     
  3. johnsonjp34

    johnsonjp34

    5
    0
    Apr 10, 2012
    I'm building a geiger counter and I am attempting to increase the battery life by switching from 2- 9v's to 3 - 6 V batteries. The circuit involves a LM358 set up as a square wave generator that feeds a TIP31 to feed a transformer and voltage multiplier to produce 500 V. It works fine with the 2- 9V batteries in series (TIP31 does get a little hot) . When I connect three 6 Vs in series the TIP31 and LM358 fry. I think I need to step back the current... What watt rating resistor would be appropriate for this. The 2-9V batteries produce about 2.5 amp on my multimeter, so I would like to reduce the 3-6Vs to that. Using V * A = W wouldn't 18 * 2.5 = 45 W. I've never seen a 2 Ohm resistor in that watt rating... Any ideas?
     
  4. CocaCola

    CocaCola

    3,635
    5
    Apr 7, 2012
    Post up the circuit if you can...

    BTW they do make 50 watt 2 Ohm resistors (about $10 based on a quick Google search) but I doubt that is the best approach here...
     
    Last edited: May 13, 2012
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,418
    2,788
    Jan 21, 2010
    Try just using two 6V batteries in series.

    Chances are that the circuit relies (or copes) with the internal resistance of the batteries. You have replaced them with ones having a lower internal resistance and now a higher current flows.

    You might also have a fault.

    What voltage do you measure across the two 9V batteries when the device is running?

    Try two 6V batteries, then measure the input voltage and current, and the output voltage.

    I am surprised that the current draw is so high. The 500V supply has (or should have) almost zero current draw and I would be expecting a very low current from your batteries. I would expect maybe 20 to 30 mA once the output rail reaches 500V, and maybe significantly more (a couple of hundred mA max) as it brings the rail up.
     
  6. johnsonjp34

    johnsonjp34

    5
    0
    Apr 10, 2012
    The TIP31 may be what is draining my 9V's so much as it gets really hot. I've used a 2N2222 before but it lasts about 5 minutes. Is 18V sufficient to open the TIP31 correctly?
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,418
    2,788
    Jan 21, 2010
    Show us the circuit. 0.7 volts is all you need to operate the transistor. But "operate" is relative to what the circuit is supposed to do. It may not do anything useful until the voltage gets higher.

    It may also be that the circuit is poorly designed and is very wasteful of energy.
     
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,270
    Nov 28, 2011
    Sounds very much like a problem with the circuit.
    Two small 9V batteries can't supply enough current to do much damage, but I guess your three 6V batteries can. If the circuit is supposed to operate from 18V, then using three 6V batteries should not cause the semiconductors to lose their smoke!
    Also I have to ask, because this mistake has tripped people up in this forum in the past: You know that when you measure battery load current, you disconnect one wire from the battery and put the meter across the gap? You don't connect it straight across the battery.
    I ask this because you say that you "need about half that current" which is not a meaningful thing to say in this case. The circuit should draw the amount of current it needs; you shouldn't be adding resistors to change the current drawn by the circuit.
    We like to help, but we need MUCH more information.
    What frequency are you operating the circuit at? What are the details of the transformer?
    The most important thing for you to do is:
    POST THE CIRCUIT!
    If it's hand-drawn, scan it. That's fine, it doesn't need to be beautiful.
    Also a photograph could be useful.
     
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