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How to figure transistor base resistor?

Discussion in 'Electronic Basics' started by Glenn Ashmore, Feb 7, 2004.

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  1. This must be very basic but I just can't get a handle on it. Everything
    I can find on transistors is either simple stuff about the silicon or so
    complex I need to be a Greek major. I need a cookbook methodology for
    figuring out the proper values.

    Say you have a relay coil that draws 45mA at 12V and you want to hold it
    on with a 2N2222 transistor. The transistor is held on by the 12V
    supply and turned off by draining the base through another transistor.

    First, I keep hearing to "force the beta to 10" but the data sheet says
    the gain ranges from 35 to 100. Do I use 45/10, 45/30 or something
    else to find the current?

    What voltage do I need to saturate the base and how do I get it from the
    12V supply? What do I need to drain off to turn the transistor off?

    --
    Glenn Ashmore

    I'm building a 45' cutter in strip/composite. Watch my progress (or lack
    there of) at: http://www.rutuonline.com
    Shameless Commercial Division: http://www.spade-anchor-us.com
     
  2. John Larkin

    John Larkin Guest

    Presumably there is a resistor from +12 to the base to define the "on"
    base current. You can't just apply +12 to the base or it will fry.
    OK, technically if the min beta is 35, you'd need at least 45ma/35 =
    1.3 ma of base current to pull your collector load down to ground,
    which is all you really want to do to operate a relay. If you use more
    base current "just for luck", the actual ratio of load current to base
    current is called "forced beta" because you are force-feeding the poor
    thing. That's a good idea, since it guarantees a good low collector
    saturation voltage and provides lots of margin for error. So to
    "force" beta to 10, you'd need 4.5 ma of base current, or maybe a 2.2k
    resistor from +12 to the base.
    See above. Base voltage will be about +0.7 above emitter voltage when
    the transistor is on; but design to force *current* into the base, not
    voltage. Base current is an extreme (exponential) function of voltage,
    and varies a lot with temperature and between parts, so it's not
    practical to use base voltage calculations to get consistant results.
    So run a resistor from +12 to the base as above.

    Then base current Ib = (12-0.7) / Rb

    Actually, it's the collector you saturate, not the base. The
    transistor is "saturated" when the collector-emitter voltage is very
    low, below the base-emitter voltage at least.

    To turn it off, make the base current zero. You could do that by
    opening the resistor that's supplying base drive from +12, or you
    could short the base to the emitter (ground here, I guess) with a
    switch or with the collector of another transistor.


    John
     
  3. That is correct.

    That is kind of what I figured but where did you get the 2.2K? Figuring
    backwards: 2.2K*4.5mA = 9.9V so we get 2.1V into the base?
    Not sure I follow. Lets see: 11.3V/2.2K = 5.1mA A little off but if
    we had a 2.5K resistor it would be right on. OK so we are talking
    strictly current. That is where my brain got tangled up.
    That's what I was planning. So using the same process the controlling
    transistors need to drain that 4.5 to 5 mA and will need about 450 to
    500uA into its base.?
    --
    Glenn Ashmore

    I'm building a 45' cutter in strip/composite. Watch my progress (or lack
    there of) at: http://www.rutuonline.com
    Shameless Commercial Division: http://www.spade-anchor-us.com
     
  4. John Larkin

    John Larkin Guest

    The more-correct value is R = (12-0.7) / 0.0045 = 2511 ohms. I just
    sort of automatically pick the next lowest standard resistor value.

    No, using 2.2K will *not* increase the base voltage very much.
    Instead, base voltage will stay close to 0.7, and using a lower
    resistor will increase base *current*. To maybe 5 ma, in this case. No
    big deal.

    There are nice equations for Ib vs Vbe, but you don't need to be that
    precise for simple switching stuff like this. Just assume you'll have
    *about* +0.7 on the base when there's a useful amount of base current
    flowong.


    Yup. Since you're working in a boat, and they tend to get wet, as I
    understand it, try to avoid circuits that work with very small
    currents. If a transistor would turn on with a tiny amount of wet-pcb
    surface leakage current, you can add a base-emitter resistor to steal
    some of the base current and make it less sensitive. If you put, say,
    a 1K b-e resistor, it would swipe 0.7 ma of base current when the
    transistor was trying to turn on, so that would prevent small leakages
    from turning it on. Some people, myself included, never trust a
    transistor with a floating base to be "off", and always have a b-e
    resistor or equivalent; legacy of old leaky germanium transistors, I
    guess. Your relay driver stage is fine: a resistor pulls the base up
    hard, and a transistor yanks it down. The lower-level stage needs to
    be similarly robust.

    Salt water is *very* conductive stuff!

    John
     
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