Replacement resistor value - help please

Discussion in 'General Electronics Chat' started by Moejoe, Sep 3, 2013.

  1. Moejoe

    Thread Starter New Member

    May 6, 2013
    11
    0
    Hi, Sorry this is really simple and boring but a large resistor has burnt out on a very very simple circuit for charging a battery within a torch.

    From a 6v DC (300ma) plug, the positive terminal goes through the large (probably 3 or 5W) burnt out resistor then through a 1N4007 Rectifier Diode which then splits off to the battery and the lamp. The battery requires: Cyclue use = 4.82-4.99V, Standby use = 4.5-4.6V and Max charging current of 0.84A.

    The resistor is burnt beyond recognition, so I'm hoping someone wouldn't mind telling me the value of which i can replace it with (and remind me why). Also wether it would be better to go with a higher or lower value than this depending on what is available to me.

    I greatly appreciate your time and help, I have attached a picture of said circuit and a (poor) diagram of the layout.

    Thanks
     
  2. #12

    Expert

    Nov 30, 2010
    16,298
    6,808
    It's very simple. Maybe that's why it committed suicide. No safety margin.
    It's hard to guess why somebody would design a charger this badly, and even harder to guess what size resistor to install in a bad design. I suspect the battery went flat and the charger couldn't survive that. Then again, how does a 1.8 watt charger melt a resistor that size?

    Try 22 ohms, 3 watts
     
    Last edited: Sep 3, 2013
  3. adam555

    Active Member

    Aug 17, 2013
    858
    39
    I guess that if you want the 840mA through that resistor, and taking into account the diode, the resistor should be around 6Ω. But, considering that the 6V power supply is limited to 300mA the resistor should be around 17.5Ω. And all that without even considering how many amps the lamp needs... sorry, tough call. :(
     
  4. wayneh

    Expert

    Sep 9, 2010
    12,123
    3,048
    I agree with starting high, like 22Ω (or higher), and seeing how it goes. That is, check the voltage across it and calculate the current. The only negative to a higher value is that it will limit the maximum charging current. That might be a good thing. The peak voltage charging the battery will still approach 5.3V, which is fine, it will just get there more slowly with a higher resistor value.
     
  5. #12

    Expert

    Nov 30, 2010
    16,298
    6,808
    I figured the 22 ohms to limit the current to the maximum ability of the 6V, 300ma supply, even if the battery is replaced with a dead short. 330 ohms will probably work in the LED part, even if the supply voltage increases when the battery has finished charging and there is very little load on the supply.
     
  6. PackratKing

    Well-Known Member

    Jul 13, 2008
    850
    215

    Biting Tongue :D
     
    #12 likes this.
  7. Moejoe

    Thread Starter New Member

    May 6, 2013
    11
    0
    Thanks for the replys from everyone i greatly appreciate it. It is not an expensive torch but even so I was rather surprised to find so few parts within it. It is used on a fishing boat so judging by what is left of the board it has gotten wet at some point but for the sake of a resistor it's not worth throwing in the trash.

    You have allready answered my question excellently, but just for my curiosity's sake would someone mind giving me the basic function of the diode within this circuit? I wasn't sure if this would affect the choice of resistor I was looking for.

    As an ex-physics student I understand all the ohm's law etc but I have next to no practical experiance. If my memory serves me correctly then isn't the purpose of this diode to restrict the size and direction (one way) of the current and to only allow DC, or 'rectify' any AC to DC flowing through. If so the power supply is allready DC and has a much smaller current rating than that of he battery hence I struggle to see its purpose here or is it just filtering out any stray AC?

    Sorry if that sounds silly, but its been a while :)

    Again many thanks to everyone
     
  8. #12

    Expert

    Nov 30, 2010
    16,298
    6,808
    The diode prevents smoke if you connect anything backwards.
     
  9. wayneh

    Expert

    Sep 9, 2010
    12,123
    3,048
    I think they are also taking advantage of the voltage drop of the diode, to help match the supply to the battery.
     
  10. Moejoe

    Thread Starter New Member

    May 6, 2013
    11
    0
    That makes sense. Thank you for the help :)
     
  11. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    Just to be clear you should get a 5 watt 22ohm resistor.

    When you mount it, keep it 5mm clear of the board, to assist airflow and thus cooling. Power resistors should not normally be mounted flush like other types.
     
  12. #12

    Expert

    Nov 30, 2010
    16,298
    6,808
    I did skimp by 10% on the power rating in a certain way of looking at the problem.

    I did that because the resistor will only suffer the effects of full power if the battery is ACTUALLY replaced with less than 2 ohms or a zero impedance voltage less than .256 volts. Not likely to happen and not likely to last for weeks or months if it does happen. At the 3 watts that I specified, the resistor is rated at 183% of the highest possible power you can get out of 6.0 volts. You, studiot, are talking about the tradition of using resistors rated at 200% of the actual highest power they will be subjected to on a chronic basis. This is a method to improve reliability, and it not only works, it is necessary. Not putting in a 100% safety factor results in crystallized solder joints in only a matter of a year or two.

    Now that I have explained my approach, I leave it to Moejoe to decide which he will choose, 3 watts or 5 watts.
     
Loading...