Connect with us

a question (extreme newbie alert)

Discussion in 'Electronic Basics' started by Sean McIlroy, Feb 17, 2007.

Scroll to continue with content
  1. Sean McIlroy

    Sean McIlroy Guest

    hi all

    this will be my first attempt at formulating a sensible question about
    this stuff. you've been warned. here's the question...

    unless i've already misinterpreted something (http://
    webpages.ursinus.edu/lriley/ref/circuits/img96.gif) the following is
    an NPN transistor NOT gate:

    ================================================================

    -- V[B1] -- << R >> -- V[B1] -- << base >>

    -- V[C1] -- << R[C] >> -- V[C1] -- << collector >>

    << ground >> --------------------- << emitter >>

    ================================================================

    hence one has

    V[C2]

    = V[C1] - R[C] * I[C]

    = V[C1] - R[C] * I * gain

    = V[C1] - ( gain * R[C] / R ) * ( V[B1] - V[B2] )

    but the webpage (http://webpages.ursinus.edu/lriley/ref/circuits/
    node4.html) seems to be saying that there are m,b (m<0) with

    V[C2] = m * V[B1] + b

    perhaps V[B2] is fixed for reasons unknown to me?

    peace
    stm
     


  2. I can not parse what you intend by these strings of characters.


    Vbe, the forward biased base to emitter junction is not a
    fixed voltage, but varies over a small range for the normal
    range of input currents between something like .5 and .7
    volts. So the simplifying assumption is often that it is
    about .6 volts. If the input voltage (Vin) is 5 volts, the
    voltage across the base resistor,Rb, is roughly 5-.6=4.4
    volts. So the base current is roughly 4.4/Rb. The current
    gain of the transistor falls dramatically as the collector
    voltage gets pulled down lower than the base voltage, but
    the upper limit on the collector current is the current gain
    times the base current. So the collector current must be
    equal to or less than gain * (Vin-Vbe) /Rb.

    Does any of this help at all?
     
  3. kell

    kell Guest




    When using the transistor as an inverting buffer (NOT gate) you're
    supposed to drive it into saturation, so there's no point using the
    gain of the transistor to try to calculate collector current.
    Collector current is determined, in effect, by the collector resistor;
    voltage across the collector-emitter junction is minimal.
    You're using the transistor as a switch, it's only ever either ON or
    OFF.

    Not sure if that addresses your question, which is kind of hard to
    follow, but I did notice you used a gain calculation.
     
  4. Sean McIlroy

    Sean McIlroy Guest

    i think so. i'd like to check if you don't mind. i'll start by
    reproducing my diagram, this time with the voltages labelled
    correctly:

    ================================================================

    -- V[B1] -- << R >> -- V[B2] -- << base >>

    -- V[C1] -- << R[C] >> -- V[C2] -- << collector >>

    << ground >> --------------------- << emitter >>

    ================================================================

    here's what i intend by it:

    *) the base of a transistor T is at voltage level V[B2] and is
    connected to a resistor with resistance R. the other end of this
    resistor is at voltage level V[B1]

    *) the collector of T is at voltage level V[C2] and is connected to a
    resistor with resistance R[C]. the other end of this resistor is at
    voltage level V[C1]

    *) the emitter of T is at voltage level zero.

    hence the V[in] and V[out] of the webpage are respectively my V[B1]
    and V[C2]. this is the best i can do by way of a diagram. sorry.

    you seem to say that, to a decent approximation, V[be] = ( V[B2] - <<
    ground voltage >> ) = V[B2] is indeed fixed; specifically V[B2] = 0.6.
    consequently, if V[B1] = 5 then

    ( V[B1] - V[B2] ) = (5 - 0.6) = 4.4

    and

    I = ( 1 / R ) * ( V[B1] - V[B2] ) = 4.4 / R

    so that

    I[C] = gain * I = ( gain / R ) * ( V[B1] - V[B2] )

    implying

    ( V[C1] - V[C2] ) = R[C] * I[C] = ( gain * R[C] / R ) * ( V[B1] -
    V[B2] )

    and hence

    V[C2] = V[C1] - ( gain * R[C] / R ) * ( V[B1] - V[B2] )

    is this (with standard disclaimers about how a model is just a model)
    right so far? here are a couple other points you could clarify if you
    don't mind:

    i)
    range of input currents.

    by "input current" you mean "base current", right? can you give me an
    idea what is the "normal" range of input currents? (qualitative
    conditions, if they exist, would likely be more helpful to me at this
    point than quantitative ones.)

    ii)
    voltage gets pulled down lower than the base voltage.

    this constitutes a different biasing condition, right?

    peace
    stm
     


  5. Lots better!
    That's fine.


    Make that a less than or equal to. If the collector to
    emitter voltage approaches zero, the gain heads toward zero,
    so if this I(c) calculation shows a drop across R(C) greater
    than 5 volts, you just assume the collector to emitter
    voltage is less than .6 volts (the transistor is saturated on).


    All that assumes that the transistor does not get saturated
    on by this collector current through the collector resistor.
    But it wouldn't be a very good not gate if the transistor
    dropped a significant part of the supply and the collector
    resistor only dropped a part of it.
    The transistor gain curve versus collector current (at some
    fixed small voltage collector to emitter , like 5 volts, is
    often shown on the data sheet. There us usually some
    collector current that exhibits the highest point on the
    gain curve. For all collector currents with gain at least
    half of that peak gain current point, the base to emitter
    voltage will vary from about .5 to .7 volts. This is about
    the normal linear gain range of that particular transistor.
    Here is an example data sheet pulled at random to talk about:
    http://www.onsemi.com/pub/Collateral/2N3906-D.PDF

    Note figure 13, the current gain curve, with 1 volt
    collector to emitter (in this case, since this is a
    transistor designed for saturated switching, like your not
    gate). At 25C, the peak gain (normalized to a multiplier of
    1 on this graph) occurs at about 10 mA of collector current.
    It falls to .5 at about 50 mA. The low current end of the
    scale doesn't go low enough to find the low current
    equivalent gain point. But you can pretty well assume that
    if the transistor still has a little collector voltage, its
    base to emitter drop will be in the range I have said. Once
    the collector voltage falls below the base voltage, the base
    voltage rises from the high base current it takes to drive
    the collector hard into saturation. Figure 13 shows the
    actual base to emitter voltage at various saturated
    collector currents with the base current driven all the way
    to 1/10th of the collector current, effectively saturating
    the transistor till the current gain falls to 10. There is
    also a curve with the collector voltage forced to 1 volt,
    which drops the base voltage a little. But 1 volt is low
    enough that the current gain is already falling a little, so
    the base voltage is on the high end of what I have been
    estimating. Saturated switching needs extra base to emitter
    voltage, compared to linear amplification applications.
    Right. The gain is a process of diffusion of charges that
    are drifting along a very thin base layer, and if they
    wobble over into the reverse biased base to collector
    junction, the electric field of this reverse biased junction
    efficiently sweeps those charge carriers out to the
    collector terminal. Once that junction becomes forward
    biased, the internal electric field tends to head the charge
    carriers in the junction back toward the base, so they show
    up as increased base current, instead of increased collector
    current, hence, lower current gain.
    Yes. That bias condition is called "saturated on" or just
    saturated.
     
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.
Electronics Point Logo
Continue to site
Quote of the day

-