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Current amplification without external sources

  1. Jan 22, 2013 #1
    i want to amplify my current from 300mA to 600mA. i m making a solar charger and the specifications say 6v 400mA. however i m unable to receive 400mA.. is there a way to amplify current to 600mA from 300mA.. i can lower my volt to 5V from 6V. i dont want to use any external dc or ac supply
     
  2. jcsd
  3. Jan 22, 2013 #2
    No there isn't.

    If you want to create 600 mA you must find a combined 600 mA from your circuit. If the only thing you can get out your solar charger is 300 mA, then all you can get is 300 mA.

    Think conservation of charge. If you have 300 mA running into your "black box" but 600 mA running out, then your "black box" is somehow magically creating charge. So you would need external circuitry to create a current amplification.
     
  4. Jan 22, 2013 #3
    i know that conservation of charge principle.. i m asking for a circuit which can convert me voltage to current somehow... decrease my volts from 6 to 5 and increase current... i increased my current from 30mA to 300mA somehow using other circuitry.. but i want further increase
     
  5. Jan 22, 2013 #4

    sophiecentaur

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    You won't get more Volts times Amps (Watts) out, whatever you do. The only suggestion is to use an inverter to produce 3V @ 600mA, if you already can supply 300mA at 6V. This is for an ideal inverter, with no losses. In reality, with such low voltages involved, I doubt that you'd get more than 60% of what you put in.

    This is a case of "If I were you, I wouldn't start from here", I'm afraid. You either need a different set of PV cells or design a different device to use the volts and current that the ones you have can supply.

    How have you managed to increase your current from 30mA to 300mA? Was it by going from series to parallel with your PV array? That implies you must have started with a 60V array (??). If this is the case, you may be able to design a more efficient switched mode supply, based on the 60V supply that will give the current output you want - but you still won't get more Watts out than you put in.
     
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