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Circuit Design Help

  1. May 11, 2009 #1
    I was wondering if anybody could help with this circuit I'm working on.

    It has a USB charger section as well as a main load section.

    A JFET ensures the charger isn't connected while the load is on.

    A Mosfet allows the tact switch to control the load (0.3ohm) running 12 amps.


    I think I have the load section of the circuit correct, but I'm not sure if the JFET is connected properly. The idea is that one tact switch turns on the load, and disconnects the charging section of the circuit. When the switch is open/off the load is off, and the charger is free to charge the battery
     

    Attached Files:

  2. jcsd
  3. May 11, 2009 #2

    vk6kro

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    It doesn't look like it would do anything.

    There is a capacitor in the line from the USB 5 volts. This will stop any DC.

    There is another capacitor in series with the DC to the JFET.

    That IC is a sophisticated battery charger chip, but it doesn't get to do anything.

    Sorry, but the rest of the circuit is just nonsense. The battery never gets charged.
    While it lasts, it will make a resistor get hot.
     
  4. May 11, 2009 #3
    Actually, those capacitors are not supposed to be in series. My mistake.

    The resistor getting hot is the load part of the circuit. That's useful.

    The question is....when the switch is closed and the resistor is getting hot, will the charger section of the circuit actually be cutoff? That's the reason the JFET is there. But I'm not sure it's setup correctly. Also, it may be better to use a depletion mode MOSFET.
     

    Attached Files:

  5. May 12, 2009 #4

    vk6kro

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    Looks a lot better without the series capacitors.

    No. If you join the gate and source of a JFET together, the JFET will conduct.
    So, with the switch open, the gate of the JFET is grounded and will be cut off. When you push the switch, the JFET will conduct. This seems to be the opposite of what you want.

    JFETs are low powered devices so you might like to look at a bipolar transistor for battery charging. You would have it in the same position as the JFET, but with a bias resistor coming from its collector to its base and a smaller resistor going from the base to the drain of M1


    Incidentally, you have shown the JFET as a P-Channel one. This would be unusual and you would have it reverse biased in this circuit. If you have it upside down (ie source at the top) it would be usual to say so in the diagram. I am assuming you didn't intend to do this.

    If you got a NChannel FET as your JFET, and connected its gate to the junction of the 0.3 ohm resistor and drain of the MOSFET, you might get the effect you are after. This is using the MOSFET M1 as an inverter.

    Doesn't the battery charger chip require any feedback to stop it overcharging the battery? If it doesn't, why not just have the USB line come in at the top of the JFET and leave out the IC?
     
  6. May 12, 2009 #5
    Thanks again for your feedback. I think the Nchannel FET is what I intended. Actually, I think another MOSFET should do. Connecting the gate between the load and the other mosfet should leverage the inverter behaviour that I'm looking for. When one mosfet is on, the other will be off, and vice versa.

    The charging IC gets feedback through the connection to the positive terminal of the battery. It monitors voltage, and then monitors current, as necessary for lithium polymer charging.
     
    Last edited: May 12, 2009
  7. May 12, 2009 #6

    vk6kro

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    You will need to limit the current in the JFET if you intend to keep using it. If you put it between a power source and a battery it could be destroyed in the first second. They can usually only cope with 20 mA or so.

    If the IC is detecting the condition of the battery by the voltage it is seeing at its output, there will be a problem with the voltage across the FET.
     
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