Increase in Current ??? (power)

Discussion in 'General Electronics Chat' started by precious_ain11, Aug 10, 2011.

  1. precious_ain11

    Thread Starter New Member

    Aug 10, 2011
    guyss.... i have a 50V DC (200mA) source of i want to make an inverter which will convert it to 220V AC and power nearly 500 watt.
    can any1 help me out ???
    because the current is very low. so how can i get the output power of 500watt.
    plzz guys help me its really important project for me.
  2. JDT

    Well-Known Member

    Feb 12, 2009
    50V at 200mA is only 10W.

    Can't be done.
  3. tom66

    Senior Member

    May 9, 2009
    Unless you invent a perpetual motion machine.

    Why do so many people ask for 220-240V inverter circuits? If you have a 50V @ 200mA source, where is that coming from? What battery series produces 50V but at only 200mA?
  4. SgtWookie


    Jul 17, 2007
    I think the OP wants to power household items (lights, fans) using electricity from the telephone company.

    It will not work, and your telephone company will disconnect your telephone service.
  5. Crispin


    Jul 4, 2011
    Available now at Farnell
  6. tom66

    Senior Member

    May 9, 2009
    Ah yes, 50V.... should have been obvious. You can draw a few mA from the phone line before you get significant voltage drop. And your phone company will fine you if you happen to overload the network.

    To give an idea of how little power you can draw from the phone lines, consider this. Our phone base station provides 9V(dc) @ 300mA. The phone can operate from the phone line, without a power supply. However, the LCD is barely visible (especially compared to when plugged into the 9V power supply), and that only requires a few mA to operate.

    Also, you are billed for electricity usage on your phone. You might not realise it but look at the "line rental" or "fixed monthly fee"; that includes the cost of the electricity that might be used by any ringers and other devices.
  7. iONic

    AAC Fanatic!

    Nov 16, 2007
    You have a 1lb rock sitting on a ledge 10 ft high. It has A potential of "X" (10W) The only way to get "Y" (500W) is to increase the size of the rock to 50lbs or raise the height of the ledge to 500ft. Either way you look at it you have to add energy to the system. You can't just have a 1.5V AA battery in your left hand, say "hocus pocus" and now have a 40lb 12V 50AH car battery in the left hand.