Connect with us

In need of some help!

Discussion in 'Electronic Basics' started by Alistair, Oct 17, 2003.

Scroll to continue with content
  1. Alistair

    Alistair Guest

    Hello again,

    I still have a problem with this transistor setup. 4017 (5 Stage Decade
    Counter) output to an NPN (BFX34) to drive 2 21 watt 12 volt lamps. I have
    tried a suggestion of using a PNP FET (BUZ11A) whit a resisitor of 20 ohms
    onto the base. The transistor gets hot and so does the resistor. I have also
    tried a TIP42A, and found that with it mounted to the metal structure, it
    shorts out to earth - which the lamps are connected to and eventually the
    vehicle. So what I really need is a transistor that can be driven either
    straight from the 4017 or via the NPN transistor from the 4017, capable of
    sinking 5 amps without getting hot,so I dont have to mount it to the steel
    frame.Please dont suggest using relays as they are totally out of the
    question.

    Many thanks

    Alistair
     
  2. John Fields

    John Fields Guest

    A BUZ11 is an N channel MOSFET, (not a PNP FET???) and in order to use
    it as a high side driver you need to connect the drain to +12V, the
    source to your lamp load, and drive the gate to substantially higher
    than 12V in order to get it to work without dissipating a lot of power.

    If you use a MOSFET in a TO-220 package the tab will almost certainly be
    connected to the drain, (check the data sheet) so if you're using a P
    channel MOSFET with the tab connected to chassis ground and the source
    connected to +12V what you'll be doing when you turn the FET on is
    connecting +12V to ground, NOT a good thing. You might want to try
    this:

    +12 +12
    | |
    | |
    [1K] |
    | S
    +--------+-----G IRF4905
    | |K D
    C [Z15V] |
    4017OUT>--[10K]--B 2N4401 | [LAMPS]
    E | |
    | | |
    GND>--------------+--------+------+

    The 10k base resistor is assuming you're running the 4017 on 12V...

    An IRF4905 has a channel resistance of 0.02 ohms, so fully enhanced the
    MOSFET will dissipate about 1/2W with a 5A load, so you should need no
    heat sink. The FET is rated for 260A max, (check the data sheet for
    your 5A VS temp) so you should be OK even with a lamp inrush current of
    50A.
     
  3. the Wiz

    the Wiz Guest

    That's why they sell insulated mounting kits with mica or teflon washers to
    allow heat transfer to the metal and prevent an electrical connection.
    More about me: http://www.jecarter.com/
    VB3/VB6/NSBasic Palm/C/PowerBasic source code: http://www.jecarter.com/programs.html
    Drivers for Pablo graphics tablet and JamCam cameras: http://home.earthlink.net/~mwbt/
    johnecarter [email protected] mindspring dot.dot com. Fix the obvious to reply by email.
     
  4. CFoley1064

    CFoley1064 Guest

    I have also
    Hi, Alistair. You're almost there. You need a heat sink. Use the NPN
    darlingtons you were directed to in another post, and purchase two RS Premium
    Heat Sink $1.59 USD Catalog #: 276-1368. Since a darlington being used as a
    switch has about 1V from collector to emitter, you will have to dissipate about
    1V * 5A = 5 Watts of power. The heat sinks will do the job -- just make sure
    they're not touching the chassis or anything that's grounded.

    Good luck.
    Chris
     
  5. Alistair

    Alistair Guest

    Well thank you all for the overwhelming response to my tiresome problem!!!
    :eek:) I will let you all know the outcome of the steps Ive taken.

    Alistair
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Similar Threads
Loading...
Electronics Point Logo
Continue to site
Quote of the day

-