Boost Converter shutdown without a switch (automatic)

Discussion in 'The Projects Forum' started by Whiteheat, Aug 7, 2013.

  1. Whiteheat

    Thread Starter New Member

    Jan 21, 2013
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    I have a solar USB charger project I am working on with my 14 year old son to use when we are camping. I could buy one but what fun is that.
    The boost converter has an ENABLE CONTROL (active high (>.96 V)). I currently have that tied into the Vcc (Power in) per the data sheet. I notice after I unplug the phone from the charger, the circuit is still pulling about 4 mA and drains the battery if not used at night. Also there are no LEDs in the circuit that may be drawing any power.
    Is there any easy way I can make the Enable LOW (< .6 V) when the system is drawing say less than 10mA? This is such a small amount of power but when charging from solar, I need all I can get. I don't want to use a switch if at all possible and the simpler the better.:rolleyes:
    You help would be appreciated!
     
  2. richard.cs

    Member

    Mar 3, 2012
    162
    31
    If you really want to avoid a switch entirely you'll need a boost convertor that draws less idle current. What you could do though is make a circuit where you have to press a push button to start the convertor and then the convertor automatically shuts down when the current drops below a threshold.
     
  3. Whiteheat

    Thread Starter New Member

    Jan 21, 2013
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    0
    The boost converter I am using (Feeling Technology FP6291 - data sheet attached - the boost part of the circuit is the same as page 10 of the datasheet ) - (I got this chip off a cheap ebay boost converter that the circuit board cracked)
    It says the shutdown current is .1uA - the problem is how do I drive the shutdown pin low automatically when say the current draw is under 10-40mA and then start back up when current draw is more or is this not possible?
     
  4. richard.cs

    Member

    Mar 3, 2012
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    The second part is the problem. You can't measure current draw if there is none with the boost converter off. With the convertor off the load sees your input voltage less about 0.3V - if your load is resistive then some current will be drawn but most electronic loads you might be charging are unlikely to draw any current. If there's no current you can't detect it.
     
  5. Whiteheat

    Thread Starter New Member

    Jan 21, 2013
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    UGH :(.
    The more I got to thinking about it, I was thinking this would be the problem (how do you sense something that isn't there...).
    Is there any way you can think of to lessen the current draw when idle? I was going to increase the resistor size (from K to M) of the feedback resistors but read somewhere that that may cause noise and instability problems with the feedback pin causing noise at the output as well as output voltage instability.
    I would be OK if I could get the idle current draw at like 1-2mA if possible.
    Any suggestions you can give on this would be appreciated.
     
  6. richard.cs

    Member

    Mar 3, 2012
    162
    31
    What are your feedback resistors now? Do you think they're responsible for the current draw?
     
  7. Whiteheat

    Thread Starter New Member

    Jan 21, 2013
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    0
    They are 88K and 12K, which should equate to about 50 uA or so going through the 12K resistor to ground, so I don't think that is it. I am going to have to start poking around the circuit to find the 'vampire' that's sucking the little bit of current - if I can get it to 1-2 mA, I will be happy.
     
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