Battery Charger (understanding it)

Discussion in 'The Projects Forum' started by AndrewCE, Jun 28, 2009.

  1. AndrewCE

    Thread Starter Member

    Jun 27, 2009
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    I found this schematic for a battery charger, and i'm wondering how exactly it works (i'm a noob to transistors and zener diodes). The site has a pretty good explaination, but I still have some questions.

    here is the page, the charger is at the bottom. http://www.owlnet.rice.edu/~elec201/Book/batteries.html

    First of all, is this the correct equivalent 9V circuit? [​IMG]
    would I have to change resistor values?

    Doesn't the LED require a current-limiting resistor? It seems like the LED would be destroyed. Or does one of the 16R act as that?

    Can someone explain each operating mode (what exactly it's doing when the battery is below 1.7v, between 1.7 and 9, and over 9)? The site explains it but i don't get a good idea of where current is going within the transistor(transistor noob).

    This would be a big help, so thanks in advance!
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    The LED is probably an end-of-charge indicator. Yep - I just look at the site, and that is what it is. They explain pretty well how things work. The 16 ohm resistor is the limiter, plus the fact that the zener soaks up 10 volts. The LED only sees 1.7 volts before it conducts.

    Current goes in the emitter and out the collector. The current in the base determines conduction. That is dependent on the voltage between the two resistors, which diminsihes as the battery charges.
     
    Last edited: Jun 28, 2009
  3. Audioguru

    New Member

    Dec 20, 2007
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    You are looking at an old 12 years old circuit for charging an old toxic Ni-Cad battery.
    A modern Ni-MH battery has much more capacity and is charged with a higher current then has a very low or no trickle charge current.
     
  4. AndrewCE

    Thread Starter Member

    Jun 27, 2009
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    o...kay, do you have another charger circuit to recommend me? preferrably one with no more than one transistor (noob, remember?)
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    The circuit you have will blow up a little 9V rechargeable battery.
    You should use a battery charger IC instead.
     
  6. AndrewCE

    Thread Starter Member

    Jun 27, 2009
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    well i doubt someone would post a battery charger circuit on the rice.edu website that would "blow up" a battery.

    not to mention the fact that i'm not only trying to charge a battery, i'm trying to LEARN about transistors/diodes etc. so a battery charger ic cop-out would be cheating myself.
     
  7. AndrewCE

    Thread Starter Member

    Jun 27, 2009
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    i've read some more info on this website, and so far here's what i'm thinking (and please someone correct me if i'm wrong, which i probably am):

    with no battery connected, the 9v wall wart won't have enough voltage to "breakdown" the zener, so it will be effectively an open circuit; no current flowing anywhere.

    with a low-charged battery connected, voltage from the wall wart will be directed to the collector. since there is a high resistance, voltage will be sent through the 392R to the base, opening up the transistor, allowing current through the battery (charging it).

    heres where i get superconfused. at some point the 2 9v sources add up to breakdown the zener and led?
     
  8. beenthere

    Retired Moderator

    Apr 20, 2004
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    The magic is in the wall wart - the output rating is only realistic under load. Lightly loaded, the output will be several volts high, and fudge the LED into lighting.
     
  9. Audioguru

    New Member

    Dec 20, 2007
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    They are charging a pretty big 12V battery at 520mA.
    Your circuit is blowing up a little 9V rechargeable battery with the same high current that they used.

    Your little battery should be charged at only 10mA for a Ni-Cad or at 17.5mA for a Ni-MH one. If your "9V" battery is actually 8.4V then your circuit will not fully charge it because it is 9.8V when fully charged but your charger will have an output voltage that is less. If the 10V zener diode is 5% low at 9.5V then your charger will have a max output of only 8.8V.

    Your circuit does not have two 9V sources. It has a 9V power supply and a "9V" battery. The zener diode is from 9.5V to 10.5V and the red LED adds 1.8V.
     
  10. AndrewCE

    Thread Starter Member

    Jun 27, 2009
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    interesting, what exactly is the phenomena that causes this? (i know the basics of adapters but i cant find a good schematic online)
    does it have a name? and how do i know how much of a load my adapter needs?
     
  11. AndrewCE

    Thread Starter Member

    Jun 27, 2009
    14
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    well i asked previously if i would need to change the resistance values. it seems that that would do the trick.

    not sure what you mean in the 2nd paragraph. are you saying that 9v batteries actually put out 9.8v?
     
  12. Audioguru

    New Member

    Dec 20, 2007
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    A "9V" rechargeable battery is either 6 cells of 1.25V= 7.5V or it is 7 cells of 1.25V= 8.75V.
    But each cell is 1.4V to 1.5V when fully charged. Then the voltage drops quickly to 7.5V or 8.75V.

    My "9V" Ni-MH battery is still charging and it measures 10.5V. Then it drops quickly to 8.75V then 8.4V.
     
  13. Audioguru

    New Member

    Dec 20, 2007
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    A cheap AC/DC adatpter is not regulated. Its internal circuit has resistance in series which reduces the voltage when it is loaded. So its voltage is made too high without a load so that when it drops with a load then the voltage is correct. Therefore without a load its voltage is too high.
     
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