17V with -5 and +12

Discussion in 'Homework Help' started by Karan Khurana, Feb 7, 2005.

  1. Karan Khurana

    Thread Starter New Member

    Feb 7, 2005
    2
    0
    hi,
    i m trying to get 17V through a computer power supply for my project using the -5 and 12 connections in the power supply. i m using transistors as a switch which gets its on-off signal from a PIC(programmable processor chip) through a buffer. the problem is that the transistor can not take the -5V as its gorund for some reason. i m using the tp147/142 transistors. but they dont seem to work with the -ve voltage. is there anything else i can use as a switch that can handle the -ve voltage. voltages available from the power supply are: +/- 12V, +/-5V, 3V
    i need 15 or greater voltage input for my solenoid to get enough power to do whatever i need it to do.
    what can be done?
     
  2. pebe

    AAC Fanatic!

    Oct 11, 2004
    628
    3
    If your supplies are floating then for for the supply to your PIC chip, use -5V for the negative rail, and 0v as +5v for the positive rail. Then the PIC can drive a TIP 142 (npn) with emitter connected to -5V, with solenoid connected from collector to +12V. (Don't forget the diode across the solenoid). I assume that the centre taps of +/-5V and +/-12v are common - if not connect them.

    If on your supply, the commons are grounded and connecting the PIC in that way would present a problem, then connect the PIC to 0v and +5v for supply. Use a TIP 147 (pnp) with emitter to +5V. Solenoid between collector and -12V.

    TIP 142 will need a logic '1' to drive it but TIP 147 needs a logic '0'. Feed the bese through a resistor in either case.
     
  3. Karan Khurana

    Thread Starter New Member

    Feb 7, 2005
    2
    0

    i cant connect the pic to -5 because i need +5 for my other circuits.
     
  4. pebe

    AAC Fanatic!

    Oct 11, 2004
    628
    3
    Can you explain your other circuits, or why you cannot use the PIC in that way? How do they interface?
     
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