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Transistor biasing

Discussion in 'Electronic Basics' started by BradBrigade, Aug 30, 2005.

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  1. BradBrigade

    BradBrigade Guest

    I know this question has been asked a million times, but I've read all
    the answers and still don't understand. I've been studying this off
    and on for years and can't seem to get it. I just give up for a few
    months and come back.

    I keep reading that in order to properly bias a transistor, you need to
    first determine Ic then figure out Ib=Ic/Hfe. The problem is Hfe is
    all over the place and all the datasheets I look at give the transistor
    Hfe between 25 and 300. That doesn't help me. Maybe if it was between
    50 and 60 I could average, but not 25 and 300. Plus, that data is for
    a 10V supply, which I might not want, and give no data for other supply
    voltages. I want Ic to be 100mA, but with no idea of what Hfe is, how
    do I go from there and be confident that my design will work and not
    explode in my face when I hook it up?

    Ok, so Hfe varies with Ic. So the solution is to put a resistor before
    the emitter and have negative feedback stabilize the gain. But how do
    I determine the value of that resistor? And for that matter of the
    potential divider on the base or the collector resistor. I search and
    search and all I get are specific examples of biasing with no clue how
    the values were determined. Do I really have to trial and error this
    stuff? Is it even possible to plan a circuit on paper and not have it
    behave wildly different when it is built?

    One more thing. The negative feedback is supposed to keep the gain of
    the amplifier constant regardless of the transistor inserted. But does
    that mean regardless of the differences between individual transistors
    of one type (like 2N222) or can I plug in ANY transistor of any type
    and the gain should stay about the same?

    Any help is appreciated. Maybe it'll jar my brain into finally
  2. Ralph Mowery

    Ralph Mowery Guest

    Look here for an explination.

    Forget about Hfe of the transistor. It mostly maters only in that you have
    enough. The ratio of the collector and emitter resistors determin the DC
    gain of the circuit. The AC gain can be improved by a capacitor across the
    emiter resistor. Also do not try to calculate the gain to the smallest
    fraction or many decimal places as most formulars are only close
  3. john jardine

    john jardine Guest

    The general idea is to design for worst case conditions. If your minimum Hfe
    looks likely to be a ball park 25, then design to this value (or preferably
    In practice the Hfe value is only of interest when setting the base bias
    resistors and TBH it's usual to design most amplifiers without even looking
    at a data sheet; just assume all small transistors have a Hfe of 100 and all
    power transistors a Hfe of 10. Your Hfe estimates may be way off the mark
    but the circuit will still work, (that's why the biasing resistors are
    there). Only noticable change would be a small shift in the static collector
    The emitter resistor is simply set to a value that that will give as high a
    emitter voltage as you can get away with. The higher the voltage the more
    stable the biasing. A couple of volts or more is OK (say 22ohms at 100ma),
    0.5V is pushing it a bit.
    The potential divider at the base is set to pass around 4x to 10x times the
    expected base current (your choice). The expected base current depends of
    course on that Hfe but is now of little consequency as the base current
    loading (ie due to Hfe differences) can vary over a massive range without
    upsetting the base voltage (hence emitter voltage, hence collector voltage)
    too much.
    And yes, plug in pretty much any transistor and it'll work the same. Which
    it must, seeing as you've just cornered the gain differences, so there's
    pretty much nothing left but basic 'transistor action'. Which is the same
    Easy eh!.
  4. padmow

    padmow Guest

    I think, we need "beta" as high as possible so we can use only
    resistors to determine the GAIN of the amplifier.

    I'm confused with your explanation here, could you contrasted to using
    single base bias resitor only, without using Emitter resistor as
    negative feedback element.
    Using potential divider bias, hence Thevenin source and Therenin
    resistor, I can't understand how that would stabilize Ic.

    thank you to clear up my confusion
  5. Jasen Betts

    Jasen Betts Guest

    it's easier in a spreadsheet, kirchoff's laws can be put in and you can see
    the effects of different choices...

    as long as circuit gain is significantly less than the Hfe, and the
    replacement transistor can handle the conditions, (current voltage wetc)

    Trying to use a transistor with a Hfe of 25 in a circuit with a gain
    of 20 probabl;y wouldn't work too well...

  6. Ralph Mowery

    Ralph Mowery Guest

    I'm confused with your explanation here, could you contrasted to using
    Unless you want to measuer each transisitor you use, you can not use the
    single base bias resistor. Even then as the transisitor heats up the
    circuit will often bceome unstable (thermal runaway) due to the changing of
    the charistrics of the transistor. There are some circuits that try this
    with a diode mounted on the same heatsink as the transistor to compensate
    for the heating of the transistor.

    The formular in the text book is for a "perfect" transisitor where the
    transistors parameters are known and do not change due to heat and other

    There are 3 common circuits that can be used to bias a NPN transistor. The
    single resistor from V+ to the base, The one that adds an emitter resistor,
    the one that adds a resistor from the base to the negative voltage. Each
    one is more stable than the previous. The single resistor to the base is
    way too unstable for mass production.
  7. The point of view that makes base biasing sensible is to understand
    that junction transistors are turned on by voltage. That is, the base
    to emitter junction must be forward biased by some particular voltage
    to turn on some particular collector current. But while this bias
    voltage is doing its job, the base emitter junction is also leaking
    diode current. So there is an incidental ratio of collector current
    to base current (that varies with conditions and with different
    devices) and that ratio is called beta.

    The resistor divider method is a way to produce the appropriate base
    voltage, while also making a voltage source that has a low enough
    effective (Thevenin) series resistance to provide the needed base
    current without much distorting that voltage. If the divider passes 4
    to 10 times as much current as the worst case (lowest beta) base
    requires, then its voltage is sufficiently stiff to hold the bias
    voltage acceptably stable for devices of higher beta, as well.

    The emitter resistor is a negative feedback mechanism that provides an
    other input voltage (remember that the input is the voltage between
    the base and emitter, so both those terminals are inputs) related to
    collector current. The difference between the voltage produced by the
    emitter resistor and the base divider voltage is the actual input bias
    voltage for the transistor.
  8. john jardine

    john jardine Guest

    Last month I bought 200 off (cheap) BC557C transistors and in an idle moment
    measured the Hfe of one of them. It came out at a rather pleasing 450.
    Curious, I started to measure the others in the packet. Surprised to find
    all of them had values of 450 +/-5% !!. (still though design 'em in as if
    they have an Hfe of 100).
    The single resistor bias method is theoretical, only to be found in
    textbooks and is worthless in practice. (sorry :)

    John P' knows the 'current/voltage' dichotomy and his provided explanation
    would also be mine.
  9. padmow

    padmow Guest

    :The resistor divider method is a way to produce the appropriate base
    :voltage, while also making a voltage source that has a low enough
    :effective (Thevenin) series resistance to provide the needed base
    :current without much distorting that voltage. If the divider passes 4

    :to 10 times as much current as the worst case (lowest beta) base
    :requires, then its voltage is sufficiently stiff to hold the bias
    :voltage acceptably stable for devices of higher beta, as well.

    for same type of transistor, would the one with higher beta draw more
    Ib compared to one with lower beta.

    :The emitter resistor is a negative feedback mechanism that provides an

    :eek:ther input voltage (remember that the input is the voltage between
    :the base and emitter, so both those terminals are inputs) related to
    :collector current. The difference between the voltage produced by the

    :emitter resistor and the base divider voltage is the actual input bias

    :voltage for the transistor.

    wow....thats a clever idea to regards those terminals as inputs. I
    could never have thought that.

    Based on your explanation I tried to simplify so my lame brain can
    digest it. I think, the variation on voltage across, BE junction, due
    to R-emitter, provide the negative feedback.

  10. bg

    bg Guest

    BradBrigade wrote in message

    The gain of an amplifier using feedback is given by --- A' = A / 1+BA
    A' = Gain with feedback
    A = Gain without feedback (open loop gain)
    B = fraction of output fed back to the input (feedback factor)

    Here are a few examples -
    If A = 50 and B=1% Then A' = 50/ 1+.05 = 47.6
    If A = 100 and B =1% Then A' = 100 / 1+1 = 50
    If A = 200 and B = 1% Then A' = 200/ 1+ 2 = 66.7

    Note that with 1% of feedback, the gain with feedback varied by about 40%
    while the open loop gain varied by 400%. If this had been a transistor with
    beta varying from 50 to 200, by using feedback we could design a stage with
    a current gain that would fall somewhere between 47 and 66.

    Now look what happens if we increase the feedback to 50%
    If A = 50 and B=50% Then A' = 50/ 1+ 25 = 1.92
    If A = 100 and B =50% Then A' = 100 / 1+ 50 = 1.96
    If A = 200 and B = 50% Then A' = 200/ 1+100 = 1.98

    For all practicle purposes, with 50% feedback, the stage gain is 2 while
    the open loop gain varies by 400%. Again, if this was a transistor
    amplifier, any transistor in the lot would make an accurate current gain of

    The downside to negative feedback is that we waste alot of gain but the
    upside is that we can make the gain very predictable. The reduction in gain
    is a direct indication of the effectiveness of the feedback. .

    Current gain -
    It is typical to use enough feedback so that the stage current gain drops to
    1/4 or less of the gain without feedback. All we need to know is the minimum
    beta to expect. For example, if you were using a transistor type, where
    beta varied from 100 to 200, obviously you would never design a stage
    having a gain of 150 unless you hand picked the transistors, but you could
    design a stage with a current gain of 20 and expect all of the transistors
    to work.

    In a common emitter transistor amplifier -
    Rl is the collector load resistor
    Re is the emitter resistor
    Ra is the resistor from Vcc to the base
    Rb is the resistor from base to ground

    The ratio Rb/Re is the stage current gain. This ratio is kind of equivalent
    to B. That is, as Re gets larger, or Rb gets smaller, we are using more
    negative feedback and the stage current gain is reduced. If we had a
    transistor with a minimum beta of 100 , it would be typical to set Rb/Re to
    about 25. A higher ratio is acceptable if you can tolerate the gain
    variations, or a lesser ratio, if really tight control of the gain is
    needed. It would also be possible to use a feedback resistor from the
    collector to the base or global feedback form another stage to stabilize
    this stage but that is a different story. For now we will look at biasing
    when the feedback is due to Re.

    A typical transistor will have a maximum beta at a certain collector
    current, or a fairly constant beta over a range of collector current. This
    range of collector current is where the transistor was designed to operate.
    When Vbe is at .5 volts, the transistor collector current is at its lowest
    range. When Vbe is at .6 volts the collector current is just about right,
    and when Vbe is at .7 volts, the transistor is conducting heavily or near
    saturation. We don't know the exact value of IC for a given Vbe but we can
    get an idea of where IC is, over it's usefull range. I always shoot for a
    Vbe of .6 volts and then tweak as neccessary. If we have an emitter resistor
    of 1K and we want to operate at a collector current of 1ma, then the voltage
    across the emitter resistor is Ie x Re = 1 volt Note that Ie is Ib+Ic, but
    if Ib is a small part of Ie which usually is only about 1% of Ie, then for
    all practicle purposes Ie = Ic. The base voltage is .6 volts higher than the
    voltage across the emitter resistor or 1.6 volts. This base voltage appears
    across Rb. Ra then has to drop Vcc - Erb at a current of Irb. This gets you
    close to the value of Ra. Do not expect it to be exact. the reason being,
    is that Vbe increases 60mv for a tenfold increase in Ib. It is an
    exponential relationship and temperature dependent. Therefore the value of
    Vbe is somewhere between .5 to .7 volts over the usefull range of Ic. If you
    want an equation for Ra, try this approximation -

    Ra = E/I = (Vcc - Erb) / Irb
    And Erb = Emitter voltage + .6 volts
    Emitter voltage = Ic * Er

    Up to this point, you should have a grasp on the ratio Rb/Re to set current
    gain, and how to get a ballpark value for Ra which supplies the forward bias
    base current, but we have said nothing about the value of Vcc or the
    voltage gain. Providing we use enough feedback to reduce the stage current
    gain, the voltage gain will be Rl/Re. Our AC input voltage is divided up
    across the base emitter junction and Re, and the output voltage is across
    Rl. In terms of the AC signal voltage , voltage gain = E out/Ein = Collector
    Ac voltage / Base to Ground AC voltage. If the signal developed across the
    base emitter junction is small compared to the siganl developed across Re
    (which it will be with enough feedback from Re), then the Ac voltage gain is

    Ideally we can vary Vcc without significantly changing Ic provided that Ra
    is tied to a supply other than Vcc, because if we vary Vcc, then we vary Ib
    which will vary Ic. The value of Ra is probably best calculated after a
    value is chosen for Vcc. So far we can set current gain with the ratio Rb/Re
    and voltage gain with the ratio Rl/Re. To select Vcc we need to consider the
    Q point collector voltage and the transistor power dissapation. If we had Ic
    with no input signal = 1ma and Rl = 10K and Re = 1K then we drop 10 volts
    across Rl and 1 volt across Re. Lets assume Vcc = 24 volts. When the
    transistor is not conducting (open circuit), Vc = 24 volts. When the
    transistor is fully conducting (short circuit) then Vc = 2.2 volts.
    Visualize this as a simple voltage divider. When the transistor is an open
    switch, the collector voltage is Vcc and when the transistor is a closed
    switch we have a voltage divider of Rl in series with Re. The transistor
    will not be a perfect switch with 0 ohms but this is close enough for an
    estimate. Therefore our output signal can vary from 2.2 volts to 24 volts.
    We can have a peak to peak output signal of 24-2 or 22 volts. The largest
    voltage swing we can have at the output would occur when the collector Q
    point voltage sits in the middle of this swing or 2 +11 volts = 13 volts.
    This is class A operation. In a typical amplifier, we would not want to
    swing the out put signal this close to its limits because it would distort.
    If you
    needed a 22 volt peak to peak signal you would have to raise Vcc to provide
    headroom. But this example might be perfectly ok for say a 20 volt or less
    peak to peak swing. As the value of Vcc goes up, the transistor power
    dissapation also goes up which causes the transistor to heat up or smoke or

    Hopefully this gives you a better understanding of the DC bias. Set the
    current gain Rb/Re, so it is small compared to beta. Set the voltage gain
    Rl/Re to any desired value, but remember that a high voltage gain means Rl
    has to be large compared to Re and this means that Vcc also has to be
    higher. The collector Q point voltage puts a limit on just how far the
    output signal can swing. The largest swing for class A operation is when the
    collector Q point voltage is midway between Vcc and Ve when the transistor
    is shorted. Typically, voltage gains max out at 20 to 30 in practicle
    amplifiers. Also when the voltage gain is high, the bandwidth is reduced and
    the output Z is raised, but these are compramises you don't need to deal
    with just yet.

    For low frequency or DC operation, the input resistance is Ra and Rb in
    parallel and the output resistance is Rl.
  11. padmow

    padmow Guest

    Kevin Aylward
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.


    I would strongly suggest you write a book about that. It sound more

    It doesn't have to be thick but its enough to emphasize that transistor
    is NOT a current controlled device.

    I believe all of us here will support you.

    best wishes
  12. I have played with the idea of doing a book on and off over the years. I
    might have a go after I finish the SS update I am working on.

    I think that there is a place for something in-between Wins Art of
    Electronics and the high brow, pretentios academic texts.

    Kevin Aylward
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
  13. BradBrigade

    BradBrigade Guest

    Thank you to everyone that replied. I think my biggest problem was
    that I didn't realize that the transistor Hfe was not the same as the
    amplifier gain. That cleared a lot of confusion. Also, despite
    reading it over and over, it just never really sunk in that a diode
    drops about .7V regardless of the current through it. I was really
    mixed up with that when trying to set the base current. I've been able
    to design a few amplifiers on paper now and have them perform as I had
    calculated (or very nearly so). I feel much better now.

    Thanks again!
  14. If there is no emitter resistor, both bases look like a diode to the
    emitter voltage, so, the base current would be independent of beta.
    What would change would be the collector current. If there is an
    emitter resistor, also, then as the collector current tends to rise,
    so does the emitter voltage, so the higher beta unit would draw less
    base current and the collector currents would be more similar (though
    the high beta unit would still have a somewhat higher collector
    current than the lower beta unit).

    Except that there are two kinds of negative feedback possible. One is
    based on the collector current (That is mostly what the emitter
    resistor reacts to. Of course, the smaller base current also passes
    through it.) The other kind is based on the collector voltage. You
    can add the second kind by deriving the base voltage divider from the
    collector voltage, instead of from the supply voltage. That way, if
    the collector current increases, it lowers the voltage applied to the
    base divider and reduces the collector current increase.

    You can use both types of negative feedback in the same circuit or one
    or the other. Current negative feedback raises both the input and
    output impedance, while voltage feedback lowers both. So you can
    combine them in various combinations to lower the gain (and stabilize
    it for beta variations) and also get the input or output impedance
    where you want it.
  15. Jasen Betts

    Jasen Betts Guest

    transistors from a single batch are likely to be well matched
    my (non authporitive) reference lists BC557 has Hfe 110-330, I think
    A,B,and C variants of a transistor model usually offer higher Hfe,
    I use it here, well a single bias resistor from the collector to the base,
    it works great to amplify a surplus 600 Ohm dynamic microphone connected
    to a soundcard input that (I think) expects an electret microphone.

    admittedly I did experiment with different base resistors until I found one
    that worked well.

    +-[4M7]--- C -------------->
    22uF | to soundcard
    || | Q1
    --->>---||---+---- B BC547B +--------->
    mic ||+ |
    600R |
    --->>------------------ E ----+

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