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Transistors (BJTs) confusion

  1. Jul 16, 2007 #1
    Hi all,

    I hope you guys can help me with my confusions here:

    1) I don't understand the usage of transistors as switches. Why would you use transistors as switches, and how does that work??

    2) Why would you choose a PNP over an NPN transistor. I've been reading this book and I don't understand what this guy says.


    any help is greatly appreciated. thanks!
  2. jcsd
  3. Jul 17, 2007 #2


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    1) BJTs are used as switches because they can perform
    a switching function. In a switch one generally has
    a control input, and then a switched circuit path connection.
    It is often desirable for a switch to take a relatively small
    amount of (voltage, current, or power) on their
    switch control inputs in order to control a relatively large
    amount of the same quantity on their switched circuit
    path. Therefore it is useful for a switch control circuit to
    have high "sensitivity" or high "amplification" in order to
    control a large signal by using a small signal.
    Active devices have such "amplification" capability, so
    such devices as BJTs, FETs, IGBTs, SCRs, TRIACs,
    and various kinds of electron tubes are consequently useful
    for switching functions because they permit a small control
    signal to have a large effect on the controlled / switched

    2) Well the point that's being made is a little esoteric and
    really depends on a set of operating conditions and
    concepts that are pretty unique to the circuits described.
    Actually I find the way the information is presented to be
    a bit inaccurate and confusing in a couple of senses.
    I suppose the main point is that it's better for some
    concerns to have the switched load on a BJT's collector
    rather than in the emitter circuit because the voltage drop
    across the load doesn't effect the base/emitter biasing
    and similarly the base-emitter voltage doesn't couple
    strongly to the collector-emitter voltage. Typically in
    a saturated transistor the VBE voltage may well be in the
    0.5 to 0.8V range, and the VCE voltage can be quite low
    like 0.25, 0.35V range. If you applied only 0.35V to
    the base relativeyou might not have enough base current
    to saturate the transistor, so you'd end up with more
    resistance (lower conductance) in the collector/emitter
    circuit, and the collector/emitter voltage drop would
    consequently be higher, so you'd have less 'head room'
    for your signal and less current on your load. It's really
    mainly an issue at quite low collector-emitter voltages.
    If you're just trying to understand BJTs, work on
    understanding their principles of beta hFE, alpha, IB, IC,
    VBE, VCE, and then once you grasp those start analyzing
    various circuits that use BJTs as switches, amplifiers,
    etc. and then those analyses will make more sense for
    you. Perhaps you should get a free circuit simulator
    program to help you analyze the way a simple circuit like
    the one presented will function across a variety of
    operating load resistances, supply voltages, etc. and then
    things will be more clear to you qualitatively and
  4. Jul 17, 2007 #3


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    Oh and don't read too much into PNP vs. NPN from that
    author's comments... PNP is really the same as NPN in
    the sense of both topologies offering parts with comparable
    electrical performance characteristics. The only difference
    is the polarity of the required biasing and current flows.

    In some cases it's preferable to switch the 'low side' of
    a circuit by switching between GROUND and a LOAD
    connected to some higher voltage.

    In other cases it's preferable to switch the 'high side' of
    a circuit by switching between the
    HIGH VOLTAGE SUPPLY and the load whose other terminal
    is grounded.

    If your switch control logic is all running off of low voltages,
    and you're switching a load connected to some much
    higher positive voltage, then it's easier to make a low side
    switch with a NPN transistor since its control lead (base)
    need only be able to operate in the (0 to +2VDC) range
    even though the collector voltage in the off state may
    be positive dozens or hundreds of volts.

    If you were to try to use a NPN in the same circuit as
    a high side switch, the control voltage would have to
    be able to go up to very near the maximum positive
    supply voltage which could be positive dozens or hundreds
    of volts, which is too high for common logic chips to
    easily work at directly.

    Similarly a PNP works well as a low-side switch for a
    load whose other terminal is a negative supply voltage
    since then the emitter can be grounded, the base voltage
    can operate in the -0 to -2V range, and the collector can
    switch negative dozens or negative hundreds of volts

    So you see in many cases the only choice criteria
    between NPN and PNP is the polarity of the available
    circuit voltages for use as control and supply voltages.
  5. Jul 17, 2007 #4
    WOW! thanks for the very long explanation. i appreciate you taking your time. that'll take some time for me to digest. i think you're right, i dont have a good grasp on the concept of BJTs just yet. many things u said in #2 still sound like gibberish to me, like this one:

    why does the voltage drop across the load not affect the base/emitter biasing, and why does the base-emitter voltage not couple strongly to the collector-emitter voltage. im not even sure what biasing and coupling mean... :cry: god i suck...
  6. Jul 17, 2007 #5


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    You're welcome. Don't worry, it'll all make a lot more
    sense after you study a little and start to intuitively
    understand the way the circuit voltages and currents

    In a BJT you have:

    IB = Base current, which flows through the base, out
    the emitter, so the IB flows both in BASE and EMITTER
    since that's the path it must take.

    IC = Collector current, which also flows in the EMITTER
    since the path of the collector current is out through the

    IE = Emitter current
    This is the sum of the base current and collector current,
    and is the actual current flowing out the emitter.

    VBE = Voltage between base and emitter
    VCE = Voltage between collector and emitter
    hFE = beta = current gain = IC/IB

    The base-to-emitter junction is a diode, and so there
    will not be much base current until VBE is starting to
    get to around 0.5V or so in the correct polarity.
    Between 0.5 and 0.8 VBE (of the appropriate polarity)
    there's a very rapid (square law or exponential) increase
    in base current with respect to added VBE. So that's
    why that other article said it takes ~ 0.7VBE to 'turn on'
    and 'saturate' a single BJT since at that level of applied
    voltage there's enough base to emitter current flowing
    to cause the transistor to conduct very heavily and
    essentially 'saturate' where additional VBE or base current
    cannot cause appreciably more collector current to flow,
    and in this saturated state with high IC flowing,
    VCE is at the minimum value it may typically attain.

    The BJT is said to be a current operated device since
    currents in one path are related to currents in another
    path by relatively linear relationships. The voltages are
    secondary and obey non-linear relationships and are
    basically there to get the currents to flow.

    When IB = 0 then IC = 0, and IE=IB+IC=0
    and the transistor is 'off', and any load resistance in
    the collector circuit sees no current.

    A typical BJT might have hFE = Beta = IC/IB = 200,
    though certainly some may be 20 and others could be
    300, but let's say it's 200 for the example.

    When you increase IB to perhaps 10mA, then
    IC=IB*Beta = 10mA * 200 = 2000 mA so a small current
    IB has switched on a large IC current 200 (Beta) times
    its size!

    If you put a resistor in series with the emitter the
    current flow there will be IE = IC + IB, and of course
    that current causes a voltage rise across the emitter
    resistor of IE * R.. So if R = 100 ohms and IB=10mA
    and IC=(Beta 200) * IB = 2000 mA, the voltage across
    that resistor must be V=IR = 2*100 = 200 volts.
    But, whoops, the BASE current must flow OUT through
    the EMITTER by the BASE being about 0.7V closer to
    the collector voltage than the emitter is. But now with
    such a current flowing through an resistance in series
    with the emitter the VE = 200V, so
    VB must equal VE = VE + 0.7 (assuming NPN where
    the VBE = +0.7 since in PNP VBE = -0.7 due to the
    opposing polarity).
    So if you had such an emitter series resistor you'd have
    to supply a voltage of nearly 200.7V to the base so
    that 2A could flow in your 100 ohm emitter resistor
    leaving VE = 200V, and that's a bit inconvenient to have
    to supply such a large base voltage in many cases.

    If your resistor 100 ohm load was in series with the
    collector, and your NPN BJT had the emitter at 0V,
    grounded, you would apply VBE = around +0.7V to get
    IB=0.01A = 10mA, so VE=0V, VB=+0.7V,
    IC=IB*(Beta 200) = 2A, and VC = 0.4V,
    IC=2010mA = (IB*Beta+1), and your
    collector series resistor has still the 2A running through
    it, and the top of the collector resistor must go to
    a supply voltage of VCE+IC*R = +200.4V.
    This is more convenient since now the VB only has
    to be 0V = off, 0.7V = switched on, and VE never
    changes from 0V, and VCE in the on state saturates
    at only +0.4V, so the transistor doesn't consume much
    power though it's switching on/off a 200V load at 2Amps.

    The main difference between the cases is that the
    base and emitter voltages don't rise due to the voltage
    drop across the load resistance if you just leave the load
    in the collector circuit and hence is why I said that in
    such a case the base and emitter biasing voltages are
    not affected by a load in series with the collector whereas
    they're very much effected by a large resistive load
    were it in series with the emitter since IB must flow
    through IE along with IC flowing in IE (IE=IC+IB).

    Biasing? Well that's just a term for whatever voltage
    or currents you apply to various points on a circuit
    under a given operating condition.

    Switched off VB=0V, IB=0mA, so 0V is your off-state
    VBE bias, and IB=0mA is your off-state base current

    To switch on, bias (cause to exist from some external
    circuit connection) VBE=+0.7V for NPN, to cause a
    flow of IB through base and out the emitter =10mA.

    Check this link, it can be handy to understand some
    of the operations:
  7. Jul 17, 2007 #6
    wow ok this is much clearer now. some bulbs light up in my head.

    so, in this example's PNP circuit, VE = VB + 0.7v. Since there is a Resistor in series with the emitter, assuming
    R = 100 ohms,
    IB = 10 mA,
    IC = 10 mA * 200 = 2000 mA,
    IE = 2010 mA,

    VE = 100 ohms * ~2A = 200 V, and thus
    VB = 199.3 V,
    and so it is inconvenient to have such a large base voltage.

    on the other hand, in the example's NPN circuit, VB = VE + 0.7v
    since the emitter is grounded,
    VE = 0v
    VB = 0.7v, and obviously this is much more convenient than some 200 Volts.
    I'm not sure where you got VC = 0.4v tho. Is that an arbitrary number you chose? and also IC = (IB * Beta + 1) ?

    thanks very much tho. i think this is much clearer now.
  8. Jul 17, 2007 #7


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    Yes, that's all about right.

    Maybe I made a typographic error.
    IC = IB * Beta
    IE = IC + IB = IB * (Beta + 1) just since
    IE = (IB+Beta) [which is IC] + IB.

    VCE_sat ~= 0.4V is just a typically low number
    of the VCE differential voltage when a transistor is
    fully saturated in a heavily conducting state.
    It's not uncommon to see VCE_sat numbers in the
    range of 0.2 to 0.5V.

    VC on the other hand will of course just be whatever
    it's going to be based on the circuit, but the differential
    voltage between the collector and emitter (VCE) will
    be small when saturated on. When the transistor
    is off (IB=0) then IC=0 (or close to it) and the VCE
    will presumably have a very large difference as if
    the transistor weren't even in the circuit since the
    collector is like an open non-conducting circuit in the
    off state.

    Another useful parameter of BJTs to know is
    VBR_CEO which is the breakdown voltage of the
    collector-to-emitter when the base is disconnected
    (open circuit).. if you applied a voltage higher than that
    the transistor's non-conduction would break down
    and current would start flowing even though there's
    no base current since you've exceeded the maximum
    allowed voltage the particular transistor can keep at bay.

    Here are a couple of links to actual data sheet
    specifications for very common real transistors.
    It may be instructive to look at the parameters and
    a few of the graphs to get an idea of how the
    voltages and currents really perform.

    Small signal, fairly high gain general purpose switching

    Small signal, fairly high gain general purpose switching

    A medium power NPN:

    A medium power PNP:

    Hopefully those are pretty detailed with graphs
    etc. though sometimes they're a bit skimpy on details
    depending on the manufacturer.
  9. Jul 17, 2007 #8


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    oops another typo:
    IE = (IB+Beta) [which is IC] + IB.

    Should have read:
    IE = (IB*Beta) [which is IC] + IB.
  10. Jul 17, 2007 #9


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    Ah too bad; I just looked at the fairchild data sheet
    and it's really lacking a lot of data that it could present.
    Motorola et. al. had much nicer data sheets for the same
    parts once upon a time.

    Anyway, I found some other data sheets for other parts
    that are a lot more useful for learning how the
    voltages and currents and parameters relate:

    And others e.g.:
  11. Jul 17, 2007 #10
    Last edited: Jul 18, 2007
  12. Jul 17, 2007 #11


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    Where's the diagram / schematic of the inverter? I don't see any image, link, drawing, or so on.
  13. Jul 18, 2007 #12
    link added! sorry...
  14. Jul 18, 2007 #13


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    a) "When Vin = 5V, how is V3 (or V base) = 0.83V ? How do we know what the current and the voltage drop across that resistor is?"
    Well there is a resistor in series with the base (RB=10K),
    and the base to emitter voltage is given as 0.83V in
    this case, and we know that the applied voltage to
    the series circuit of base_resistor + base_emitter_diode
    is 5V. So since VBE=0.83 then 5-0.83=4.17V must
    appear across the 10K resistor in series with the base.
    So 4.17V/10K = 417uA of current in both the base
    and the base resistor with 4.17 across the resistor and
    0.83 across VBE to make the total of 5V which is
    supplied to be the base-resistor-emitter circuit.
    The reason VBE=0.83V at IB=417uA is just because
    "that's what it is" for that particular model of a transistor
    device. Different transistor devices (or models of them)
    could have somewhat different values of VBE vs IB vs
    temperature. Generally they'll all be in the range of
    0.4 to 0.9V for reasonable values of base current, though,
    for single BJTs. Darlington pairs or unusual semiconductors
    (i.e. not Silicon) will have different properties.

    b) "how does the output V4 become 0.019 Volts?"
    RC = 1000 ohms in series with the collector
    because that's what the page says.
    VCE_sat=0.019V when IB=417uA, and the page
    says that the conduction of the transistor is in the
    saturated region. In saturated conduction VCE is
    always a 'small number' i.e. pretty much the smallest
    possible number for that particular transistor with
    that amount of IC flowing.

    Since we know that V_supply=5V to the circuit
    containing 1000 ohm resistors in series with VCE=0.019V
    we realize that 5V (V_supply) = VCE (0.019) + V_resistor
    so V_resistor = 5-0.019=4.981V across a 1000 ohm
    resistor, so I_resistor = V_resistor/R = 4.981/1000=
    4.981mA must be flowing in that collector circuit resistor,
    and since it's in series with the collector, 4.981mA
    is also IC.

    So IE=IC+IB,
    and IC=IB*Beta = 417uA * 11.9, so in this transistor
    model under those conditions of modeled temperature
    and base current, collector current, we see that the
    beta is around 12 in saturation which is not unreasonable
    since the very definition of saturation is that the collector
    circuit is already passing essentially the maximum current
    that it could and that no further increase in base current
    is capable of producing any significant further increase
    in collector current. VCE is already approximately
    as small as it can get in saturation mode.

    One could keep adding more IB and not get much any
    more IC, so the apparent gain goes down during
    saturated conduction since you're already applying
    'more than enough' base current, and any additional
    base current doesn't really increase collector current much.

    Let's look at a case where the BJT is said to be
    in forward active mode near the page bottom:
    IB=I_RB since RB and IB are in series.
    IC=I_RC=3.934mA since RC and IC are in series.
    This seems suspicious since Beta is about 100x higher
    than it ought to be. Maybe there's a typographical error somewhere.

    FYI in the listed PSPICE model for a BJT 'BF' is the forward
    gain for the transistor model being defined, and in
    the web page you linked, BF=80, so that should
    approximately equal beta during the
    forward region of relatively linear relationship between
    IC and IB i.e. above cut-off and below saturation.
    Apparent Beta=12 or really any value LESS than the
    maximum forward region Beta is believable in saturation,
    but having Beta *extremely* higher than the nominal
    value in the forward region isn't usual unless you're
    dealing with a transistor operated WAY outside of its
    normally specified operating environment e.g. a
    RF transistor designed to work at frequency 3GHz but
    being used only at 1MHz, so of course the Beta might
    be quite different over those extremely different frequencies.

    Let's look at another listed case where the BJT is in forward active mode:
    IB=I_RB=10nA since RB and IB are in series.
    IC=I_RC=99uA since RC and IC are in series.
    That doesn't make much sense either since the
    expected forward linear region Beta=PSPICE BF=80,
    and this is more than 100x too large and yet it's listed
    as being an example of forward active mode conduction.

    I can understand why you'd be confused by the way these
    listed voltages and parameters relate. They're not
    what I would expect from a BJT model that ostensibly
    has a Beta of 80; I would think IB would be much higher
    than it seems to be at the given VBE values
    in the various modeled scenarios I've done the calculations for here,
    and I'd think that the apparent BETA (obtained by IC/IB calculations)
    would be much closer to the expected 80 in the forward region, and less
    than that value in the saturation region.

    Though the particular numbers of
    VBE vs IB and apparent BETA aren't what I'd consider
    very typical of many real transistors, the overall topology of the
    circuit would indeed perform a logic inversion type of function for
    reasons that are intuitively straightforward even if the
    particular parametric examples are in part bogus (IMHO).

    If you apply 0 voltage to the base, IB=0, and
    the transistor doesn't conduct and pull the voltage output node
    at the collector to ground.
    With no conduction in the collector, or base the transistor
    may as well not be in the circuit, and the output voltage will be
    basically the same as the collector supply voltage since
    there's no real pull-down load absent collector current.

    If you apply a logic high voltage e.g. 5V to the base
    circuit, you'll get several milliamps of base current
    through the series base resistor, which could produce a
    heavy collector current by multiplication by Beta,
    and the transistor will be saturated in conduction with
    a very small VCE_sat voltage well under 1V appearing at
    the voltage output collector terminal.

    Thus a 0V on the base supply -> 5V at the collector,
    and 5V at the base supply -> nearly 0V (VCE_sat) at the collector,
    and an inversion of logic levels from base to collector
    has taken place.

    If you look at the data sheet graphs for the real ZETEX
    transistors I linked to, you can see that at very small
    collector currents below 10mA, VCE_sat does indeed have
    very small values, so the listed numbers in the examples
    on your web page aren't horribly out of line with what
    one might expect at small collector currents.

    The model's VBE_on vs IB seems rather unusually large,
    though, as does the apparent Beta. That's almost
    the kind of BETA one might expect for a cascaded PAIR
    of BJT's in a darlington configuration, and in that case,
    the 'VBE' would be close to double what one would expect
    for a single normal BJT too, but that doesn't explain
    why BF=80 is listed in the PSPICE model on the page...

    So who knows what the model really is for that transistor
    and why the modeled parameters seem a bit unexpected in
    some cases relative to the model parameters shown.

    It's also strange that they didn't do a MUCH better job
    indicating the CURRENTS in their graphed data points
    since this is about understanding BJTs, which are really
    current operated devices even though it's true that
    a logic inverter is usually designed to be
    defined by the voltages of its inputs and outputs.
  15. Jul 20, 2007 #14
    Xez, thanks for your reply. I havent had the time to digest everything you wrote, yet, but i just wanted to say thank you :)
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