Power Supply Help

Discussion in 'General Electronics Chat' started by Hobbles, Nov 26, 2009.

  1. Hobbles

    Thread Starter New Member

    Jan 11, 2009
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    Hi I have a 9v 2A wall adapter and I need to bring down the current to about 150ma and my multimeter just broke so I really need help on this one. It seems like an easy thing to do and I cant find any other wall warts with a lower current around.
     
  2. russ_hensel

    Well-Known Member

    Jan 11, 2009
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    Power supplies are generally constant voltage, not constant current. You should not have to do anyting unless you are driving a constant current device ( like a LED with not current limiting resistor ). The current listed is the max current for the device.
     
  3. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Perhaps if you were to explain a bit further you might get more effective help, as you seem to have some misconceptions.

    How did your meter break?

    The current drawn from a power supply is determined by the load , not the power supply so your 2amp unit will happily supply 150 milliamps if that is all that is genuinely required. The 2amps is the maximum draw.
     
  4. Hobbles

    Thread Starter New Member

    Jan 11, 2009
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    Designing a battery charger and it needs to run at 150ma for 1.5h or 45ma for 5h also my little brother smashed it and broke the needle.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    An LM317 voltage regulator can be used as a current regulator.

    It will drop a minimum of 3v across itself though, so with a 9v adapter, you wouldn't be able to charge a battery that needs more than 6v for charging.

    You just need the LM317 and 1 resistor.
    Connect +9v to the IN terminal of the LM317.
    Connect one end of the resistor to the OUT terminal.
    Connect the other end of the resistor to the ADJ terminal
    Connect the ADJ terminal to the + battery terminal.
    Connect the - battery terminal to the 9v return (-9v wire)
    To calculate the value for the resistor:
    R = 1.25/DesiredCurrent
    If you want 45mA, then:
    R = 1.25/0.045 = 27.77 Ohms.
    You also need to calculate the resistor wattage. 1.25 x 0.045 = 56.25. Double it for safety; 112.5mW. You can use a 1/8W or higher resistor.

    If you want 150mA, then:
    R = 1.25/0.150 = 8.333... Ohms.
    Power: 1.25 x 0.15 x 2 = .375 Watts. Use a 1/2 Watt or higher.

    You might have problems finding an 8.333... Ohm resistor, as it isn't a standard value. You could use a 10 Ohm resistor instead.
    Your output current would then be:
    I = 1.25/R
    I = 1.25/10 = 125mA.
     
  6. someonesdad

    Senior Member

    Jul 7, 2009
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    You can go to Harbor Freight and get a decent (but not necessarily great) DMM for under $10 (our local store used to sell them for $4; don't know what they're at now).
     
  7. Hobbles

    Thread Starter New Member

    Jan 11, 2009
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    Man thats perfect thanks alot I actually happen to have a couple of LM317s around so cool sounds perfect just need to find a 10 ohm resistor now. So the circuit should look like this pretty much right I also calculated this should charge it in 1h and 48 minutes at 125mah for a 150mah battery and will probably need something to watch the tempature and should stop the timer and shut off the rest of the circuitif it gets over a certain temp. just in case.

    [​IMG]
     
    Last edited: Nov 26, 2009
  8. SgtWookie

    Expert

    Jul 17, 2007
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    I didn't realize that you were charging a 9v "transistor" battery.

    You should probably charge those at a low current. I suggest that you don't try to charge it any faster than a 4 hour rate, but longer is likely better. The faster you try to charge it, the more heat will be generated.

    If your battery is rated 150mAH, then a 4-hour charge rate would be 37.5mA.
    R=1.25/0.0375 = 33.333... Ohms.
    33 Ohms is a standard value of resistance. So, let's figure out the current flow through it:
    I = 1.25/33 Ohms = 37.9mA - close enough. Let's see what Watt rating you'll need.
    P = EI, so 1.25v x 0.0379mA = 47.375mW, doubled it's 95mW (rounded off). You can use a 1/10 Watt or higher rated resistor.


    Tell us more about the battery - what kind is it? NiMH? NiCD?
     
    Last edited: Nov 26, 2009
  9. Wendy

    Moderator

    Mar 24, 2008
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    I'd be interested in this. B.G. Micro has a sale on them.
     
  10. Hobbles

    Thread Starter New Member

    Jan 11, 2009
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    Its a 9v NiHM well 8.4v 150mah heres the battery, I will probably go with 4 hours charge time then. If im charging it that long do you think I still need to watch the temperature? Im gonna have a small microcontroller to shut it off after 4 hours of charging which could also control the temp. if needed.
     
    Last edited: Nov 26, 2009
  11. SgtWookie

    Expert

    Jul 17, 2007
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    Much more information on this page:
    http://www.powerstream.com/NiMH.htm

    Looks like a C/10 rate would be much better.

    Since you're using a uC to control this charger, you have a lot more options. You could have the uC discharge the battery using a light bulb until it read 1v/cell, which would be 6v - and then charge it at a fixed current for a set time.

    As the page says, NiMH batteries have a 66% charge efficiency; that efficiency will go down if you try to charge them fast.
     
  12. Audioguru

    New Member

    Dec 20, 2007
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    A real Ni-MH battery charger measures the battery voltage during charging to avoid over-charging a battery that is not discharged completely and to shutoff the charger when the battery is fully charged.

    I have a simple battery charger that has only a timer. It overcharges a battery that already has some charge.

    A Ni-MH or Ni-Cad battery that becomes overcharged at the high current you are using gets hot enough to melt or blow its vent.
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    What is the voltage that your microcontroller runs on?
     
  14. SgtWookie

    Expert

    Jul 17, 2007
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    OK, finally tracked down a datasheet for that battery:
    http://www.akkudaten.de/datenblatt/15f8h.pdf

    Recommended charge rate is 15mA for 14 hours.

    If charged at this rate and overcharging occurs, the battery won't vent/leak.

    So, need to re-visit the LM317 resistor value.
    R = 1.25/15mA = 1.25/0.015 = 83.333... Ohms.
    82 Ohms is the closest standard value of resistance.
    I = 1.25/82 Ohms = 15.244mA - I'd call that close enough.
    You could use a 1/10Watt or higher rated resistor.

    The LM317 requires a minimum 10mA current for proper regulation; you'll be drawing enough.
     
  15. Hobbles

    Thread Starter New Member

    Jan 11, 2009
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    Thanks alot for all the help!
     
  16. Audioguru

    New Member

    Dec 20, 2007
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    You cannot use your 9V power supply to activate the LM317 current regulator plus charge the 8.4V battery.
    The LM317 needs a minimum of 2.75V and the 8.4V battery is 9.8V to 10.5V when fully charged so you need a minimum of 13.25V.
     
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