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transistor saturation

Discussion in 'Electronic Basics' started by [email protected], Apr 15, 2008.

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  1. Guest

    So I'm reading up on transistor basics:

    http://www.kpsec.freeuk.com/trancirc.htm

    "A transistor that is full on (with R_CE = 0) is said to be
    'saturated'."

    Is saturated R_CE really ever 0, though? I mean, 0.000 ohms?

    My DMM routinely gives 0.1 ohm readings on bare wire, for instance.

    What are typical REAL values for saturated R_CE? Say, for a TIP31A or
    a 2N3055?

    The datasheets don't seem to give these saturated R_CE values,
    interestingly enough.

    Thanks,

    Michael
     
  2. Eeyore

    Eeyore Guest

    NO.

    For starters there's the intrinsic resistance of the semiconductor
    material. You can NEVER get below that. That's a big issue for low noise
    too.

    Graham
     
  3. Guest


    I had a feeling that's why they can get hot... I^2 R heat losses...

    Why are R_CE values not standard on datasheets though?

    Michael
     
  4. Guest


    <maximally_pedantic_mode_on>
    meant to say I^2 R power (not heat) losses.

    take care of your units, and your units will take care of you...
    <maximally_pedantic_mode_off>

    Michael
     
  5. Eeyore

    Eeyore Guest

    For sure.

    Because most bjts are used in the linear region I suppose ?

    Graham
     
  6. Jamie

    Jamie Guest

    Look harder on the data sheet.
    Saturation means it's at the lowest R level it can reach per that
    component.
    The spec's will tell you the voltage across the CE at that point.
    from there you can use ohms law to find the actual R and what ever
    else you need.



    http://webpages.charter.net/jamie_5"
     
  7. Eeyore

    Eeyore Guest

    NO. That will make NO DIFFERENCE.

    Graham
     
  8. Eeyore

    Eeyore Guest

    At what ratio of Ib/Ic ?

    You're talking out of your bottom as usual. There is no simple definition of
    saturation, it has to be specified by Vce and Ic and Ib/Ic. All THREE need
    to be specified.


    Graham
     
  9. Jamie

    Jamie Guest

    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    as usually, you're doing your norm. Looking at guys asses!
    Just don't get me going!.
    And I'll say it again for the other reader, "LOOK HARDER" the
    spec's are in the data sheet that will yield maximum saturation.
     
  10. Eeyore

    Eeyore Guest

    At WHAT RATIO of Ib/Ic ???????

    You DO NOT know what you're talking about as usual.

    Graham
     
  11. Jamie

    Jamie Guest

    Stop wasting my time ignoramus, you're way out of
    your league.

    http://webpages.charter.net/jamie_5"
     
  12. gearhead

    gearhead Guest

    A bipolar transistor doesn't have a fixed resistance any more than a
    diode does.
    Under ordinary operating conditions, as you gradually adjust the diode
    current, the voltage across it changes much less than does
    "resistance," defined as voltage divided by current. So with a diode
    you graph voltage versus current. Nobody talks about resistance.
    A bjt has diode junctions in it.
    If you're looking for a measurement that will stay fairly steady for a
    bjt, you are better off looking at the collector-emitter voltage
    drop. Even that will vary as you change the current. But get some
    numbers from a datasheet and do a little arithmetic and decide for
    yourself whether it makes more sense to talk about "resistance" or
    voltage across a bjt.
    If you want a semiconductor that has an operating region where it acts
    like a resistor, check out a mosfet. But be aware that for any given
    gate voltage, you have to keep the drain current below a certain
    maximum value or the device will stop acting like a resistor and start
    acting like a constant-current source.
    It's all very interesting, you could study this stuff for years.
    Now get started, ya bum!
     
  13. Bob Monsen

    Bob Monsen Guest


    In response to mdarrett, the way to ensure that a bjt takes a particular
    current is to use a resistor from emitter to rail. So, say you want about
    10mA. you do this:

    I |
    V
    |
    |
    |
    |
    |
    V = 0.7 + I*R |/
    .----------|
    | |>
    | |
    | |
    | |
    /+\ |
    ( ) .-.
    \-/ | | R
    | | |
    | '-'
    | |
    | |
    === ===
    GND GND
    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

    The thing you can depend on is that the NPN transistor has a voltage drop of
    about 0.7V from base to emitter under most load situations.

    So, you use that fact to make the current be a (relatively) linear function
    of base voltage.

    You never try to figure out what the actual current is going to be, give a
    base voltage or current, since it depends on too many factors, like
    manufacturing variance, temperature, etc. Also, the current is an
    exponential function of voltage, so it is very hard to get that voltage just
    right so the transistor passes the right current.

    Now, there are exceptions to every rule except one (this one).

    Regards,
    Bob Monsen
     
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