Connect with us

common emitter configuration- voltage divider biasing.

Discussion in 'Electronic Basics' started by Jenny, Jul 26, 2004.

Scroll to continue with content
  1. Ian Bell

    Ian Bell Guest

    Very true, but this was not the OP I thought it was worth a quick mention.
    The bottom line really is that a CE configuration is an excellent vehicle
    for a tutorial because its limitations soon become clear, which leads
    naturally to other transistor configurations.

    I am very pleased people are prepared to question basic assumptions in oder
    to better there understanding. What really irks me is that you still see
    really poor CE designs in the popular hobby electronics magazines.

  2. Joe

    Joe Guest

    Hi Jon,

    Yes, I read it, and some of it is below, in quotes:

    "What I didn't like about the web site is that the C they used simply
    their emitter R, so that the transistor's emitter then went straight to

    It's not a typical arrangement for audio. I'm not even sure what it would
    be a
    good approach for."

    I think I understand your statements. I am just wondering, the equation they
    use to get the value of the emitter cap is:


    where F is the minimum frequency you want to pass, and Re is the emitter
    resistor. Also, they say that this would be the minimum value cap you
    would want to use, so I usually use the next higher standard value.

    Is this equation incorrect? If so, do you know what it should be?

    I am a hobbyist, not an engineer, but I like to find out the best ways to
    design different things, and I can imagine that with some situations, maybe
    it is a little 'black art', or engineering intuition that goes into a
    design, but I do not possess any of that so I have to rely on equations.

    Best Regards,
  3. The whole idea on that page is wrong, I think. The very topology is wrong, to
    my mind. Since I don't understand what their application might be, I cannot
    possibly correct their equation. I'd need to know what in the heck they were
    thinking, before I could even consider it.

    I can, however, tell you how to calculate it for the topologies I had in mind
    and pointed out in another post... because I think I understand them to a
    I'm just a hobbyist, too. Zero professional design experience. Perhaps like
    you, math and physics are my joys.

  4. Joe

    Joe Guest

    Hi Jon,

    This is becoming an interesting discussion. From what I can understand, the
    emitter cap is there to shunt (if that's the right word) certain frequencies
    to ground so they don't get amplified. But when you figure out the cap, you
    have to use the lowest frequency you wish to eliminate (shunt?), and then
    all frequencies higher than that also get eliminated. So is it sort of a
    low pass filter? One other thing I noticed about using their equations is
    that the emitter resistor is almost always 10% of the collector resistor. I
    started using their method because I didn't know how to calculate the bias
    voltage divider for some transistors I couldn't find the curves for. Then I
    started to incorporate all their equations into my spreadsheet so that now I
    can build one from input to output. They made it easy, all you need to enter
    is the hfe, Vcc, desired collector current and frequencies to pass.

  5. No. Quite the oposite.
    No, the lowest frequency you want to *include*.
    The circuit forms a HP filter.

    The object is to get the input signal directly across the base-emitter
    junction. This maximises the signal to the transistor, hence the current
    that will flow through it. The emitter impedance attenuates the signal
    to the base emitter such that Vbe = Ze wants to be as
    low as possible.
    As I have noted, this can all be done automatically in SuperSpice,

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

    Ian Bell Guest

    Joe wrote:

    As the capacitor is across the emitter resistor which provides negative
    feedback, the frequencies that the emitter capacitor shunts are amplified
    *more* not less.

  7. Like I said before, I don't really understand the use of the topology shown at
    that site. Given what they drew there, of course, as the frequency increases,
    the shunting gets more pronounced.

    But the problem as I see it in that example is that if the capacitor is sized to
    effectively bypass the emitter R at some low frequency and for those frequencies
    above that point (until other factors dominate and change what I'm about to
    say), then the emitter is essentially grounded above that frequency point. So,
    let's say you size it to bypass the emitter R at 100Hz and above, then for those
    frequencies the Z of the capacitor will be diminishingly tiny and the emitter
    will be, in effect, at ground.

    This isn't a good thing, usually. What it means is that the gain is then based
    on R(C)/(r(e)+Z(c-bypass)) and, since r(e) is highly dependent on I(C), the gain
    will be varying all over the place. Similarly, if the Z of the capacitor hasn't
    yet diminished to the point where it is very much smaller than r(e) itself, then
    the frequency will also be varying the gain all over the place, too.

    It's not good.
    You would use the lower frequency that you want to shunt, yes. But that is the
    reverse way to look at it. You are actually setting the lower end of the
    __desirable__ frequency band. Frequencies less than this will get smaller gains
    and will therefore diminish in their relative presence in the output because
    they are NOT bypassed by the C.

    Here's a better topology (not the only one, though):

    V+ V+
    | |
    | |
    | |
    \ \
    / Rb1 / Rc
    \ \
    | |
    | +--------> OUT
    | |
    | |
    | b|/c Q1
    +-------| NPN
    | |\e
    IN >-----+ v
    | |
    | +-------,
    | | |
    | | \
    | | / Rac
    | \ \
    \ / Re |
    / Rb2 \ |
    \ | |
    | | --- Cac
    | | ---
    | | |
    | | |
    --- --- ---
    gnd gnd gnd

    Without getting into more complex models, gain is:

    Rc / ( r(e) + ( Re || (Rac + Z(Cac)) ) )

    Here, you can see that Cac acts similarly, but that when its Z is tiny, then the
    resulting gain is set by Rc/(r(e)+(Re||Rac)). (|| means the parallel equivalent

    In other words, you design the DC operating point by ignoring Rac and Cac, at
    first. Once the DC operating point is established in that way, then you design
    Rac and Cac so that when the frequency rises above some given point, Cac will be
    an effective bypass, tying Rac to ground. But NOT the emitter itself!

    That way, you can design this to have a predictable gain that doesn't vary much
    over I(C) or over frequency, once that low frequency is reached (and beyond)
    because you can arrange things so that (Re||Rac) is still large compared with
    r(e) "little r-e" and also that Rac is large compared to the Z of Cac at and
    above some design frequency. Since (Re||Rac) also isn't dependent on the
    frequency at all because the Z of Cac is also diminishingly tiny by comparison,
    the gain can be made relatively stable over a wide range of frequencies.
    No, the other way around. Since frequencies lower than this set a high Z, the
    gain of the amplifier at those still lower frequencies declines until the Z is
    very much greater than the emitter resistor and that emitter resistor then sets
    the low gain limit. Since higher frequencies have a higher gain (and in that
    web site's design, a continually rising gain that doesn't make sense, usually),
    they appear more dramatically in the output.
    Well, in general, you usually want some gain! If the emitter resistor were
    larger, you'd be attenuating the signals instead of increasing them. Sometimes,
    that's what you want. But not here, I think.
    I went through some of the calculation steps in detail, earlier. Did you read
    them? Did they make any sense? Also, most small transistors (BJTs) can work
    just fine on a quiescent I(C) of anywhere from 0.4mA to 20mA, and probably even
    wider than that. If you want to just play and aren't designing anything in
    particular, I'd pick a quiescent I(C) of 1mA and build up a circuit on that

    Like others have pointed out, a real design will need to account for what is
    driving the circuit and what the circuit is supposed to drive, itself. For
    example, if you are driving your 'test' amplifier circuit from an oscillator or
    even a square wave generator, then how are you going to hook that up? Do you
    imagine just tying one end to ground and the other end to the supposed input at
    the BJT base? Through a resistor of some kind? Through a capacitor of some
    value? How? And some inputs, like say a microphone or some other kind of
    transducer, may have very particular requirements in order to operate well.

    Usually, you have some kind of circuit designed to accommodate your transducer
    and allow it to operate well. So that circuit is designed to meet the need of
    the transducer and reflect the signal accurately into the next stage, which
    often IS designed for some gain. But the first stage/circuit is more designed
    for the transducer. Then you might have several stages of voltage amplification
    after that. Finally, followed by yet another circuit designed for the type of
    output transducer -- for example, such as a speaker. That last stage will again
    be tailored for the output transducer and not for voltage gain, itself.

    In the above case example I gave, where the amplifier is designed for voltage
    gain at some AC frequency and above, it might look more like:

    V+ V+
    | |
    | |
    | |
    \ \
    / Rb1 / Rc
    \ \
    | |
    | +--------> OUT
    | | (perhaps to another capacitor input)
    | |
    | b|/c Q1
    +-------| NPN
    || | |\e
    IN >----||---+ v
    || | |
    | +-------,
    Cin | | |
    | | \
    | | / Rac
    | \ \
    \ / Re |
    / Rb2 \ |
    \ | |
    | | --- Cac
    | | ---
    | | |
    | | |
    --- --- ---
    gnd gnd gnd

    Cin would be sized, like Cac, to pass frequencies above some point very well.
    Using a capacitor there allows the DC biasing network to do it's job and
    correctly bias Q1 and to maintain that bias as higher frequency signals are
    passed on by Cin. Any DC setpoint from the input side of Cin won't pull down or
    pull up the DC biasing -- it will simply charge Cin and leave it charged.
    I've done similar things with LTSpice.

  8. (snip)

    That depends on what you need. Its big advantage is maximum (if
    variable) gain.
  9. Joe

    Joe Guest

    Hi Jon,

    Ok, so, judging from the other responses, I have been looking at things sort
    of bass ackwards :)

    I was thinking that the emitter cap would in most cases be shunting out
    unwanted frequency. Say for audio, you want from 20hz to 20Khz but get rid
    of anything above 20Khz, so you calculate your Ce for 20Khz. Now I can
    understand a little better that the Ce is actually feeding back those
    frequencies to get more gain. OK, again I was looking at it wrong. In this
    case, you would choose the emitter resistor to pass all frequencies above
    20hz to get your audio amplification.
    OK, got it.
    So do I read the above correctly? Cac is in series with Re? I am not too
    good at reading ascii schematics.

    OK, understood, the problem I was having was with choosing the DC operating
    point. I use mostly small signal transistors , but usually just as a switch.
    I wanted a general set of equations that I could use for any common emitter
    Probly the best thing for me to do is LTspice some of these circuits and
    look at the output waveforms, and then possibly breadboard them and check
    them on the scope. I have not built many common emitter amps so maybe I need
    to do that.
    My point was just why 10%? If it were 20 or 30 or 50%, there would still be
    some gain. I am probly trying to generalize too much to make designs easier.
    Yes, they made sense to me, but, as you can see, I was not thinking quite
    right. :-(

    Also, most small transistors (BJTs) can work
    That is probly a good idea, as I mentioned above, there's no substitute for
    They show how to compute the value of the input cap, and output cap, but
    don't really explain if a resistor is needed.

    How? And some inputs, like say a microphone or some other kind of
    I think I am getting lost now. I cannot read the above schematic, but I have
    never seen a DC setpoint on the input side of an input cap. The cap would
    block it wouldn't it? Then there would be no biasing on the transistor, it
    may just act like a switch depending on what frequency is input to it.
    I think I will be using LTSPICE more often (it is quicker than
    breadboarding, but I figured it wouldn't always reflect the real world) and
    trying different configurations using the equations and then just picking a
    bias current as you stated earlier and see what the differences are.

    Thanks for your patience explaining this to me. Thanks to everyone else who
    replied. I am getting clearer on this now.

  10. Joe,

    Are you using a FIXED SPACED font when you read the ASCII schematics??? This is
    seriously a BAD THING if you are using a proportional spaced font. Try one of
    the Courier, Pica, or Prestige Elite fonts, for example.

  11. some frequency and above. Of course, I'm still struggling to understand
    an application where this is "good." -- both setting up a DC operating point
    like this AND arranging for an 'uncontrolled' AC bypass like this? Know of a
    serious one? I'm curious.

  12. Rich Grise

    Rich Grise Guest

    I slapped together a little mic preamp by the seat of my pants doing
    exactly that. I didn't care too much about gain, just so it was "enough."
    (maybe 3-5, actually.) I used a 10K pot for output, 1K emitter R, about
    220K and 820K for base bias, and I think a 10 uF right across the 1K Re.

    It was "good enough" for voice, anyway.

  13. Well, why wouldn't you add a resistor in that leg so that you got a nice, flat
    gain over a range of frequencies, rather than a highly distorting gain depending
    on the signal level's voltage AND also it's frequency? I mean... a resistor is
    all it takes, for gosh's sake!

    And we aren't talking about 'good enough' for some uses. That web site is a
    little more than about hacking something together that might vaguely work... in
    some kind-of way. They are trying to explain things and provide design
    explanations. So again, why didn't they include the resistor?

    Remember, I'm asking about good design. In what application is that circuit
    "good," as a matter of design?

    That site still just doesn't make sense to me.

  14. Just to throw in my 2 cents, as I understand it, the emitter resistor
    bypass cap is there to increase the gain as the frequency increases.
    The gain of the amp is equal to the collector resistor divided by the
    effective emitter resistance. If you put in a cap parallel to the
    emitter resistance, what happens is that at higher frequencies, that
    effective emitter resistance decreases (the cap 'shunts' or shorts the
    higher frequency currents to ground.) Thus, doing this increases the
    gain of the simple common emitter amp.

    Somebody asked why you might want to do this. I think the answer is
    that these days, if controlled gain is required, its generally
    accomplished by using negative feedback. Negative feedback is more
    effective if the open loop gain is higher, so getting the highest gain
    you can is a good thing, even if you aren't able to control it very

    All of this is described quite well in "The Art of Electronics", 2nd
    edition, chapter 2. If you really want to understand how all this
    stuff works, or at least get a basic understanding of how to work the
    formulas and bias an amp, go check out that book from the library and
    read chapters 1 and 2. The feedback discussions are in chapter 4, I
    think, and are also well worth the work.

    Bob Monsen
  15. Its done *all* the time when you use overall negative feedback. In this
    case, you don't care what the gain is, only that it is as large as
    possible. Indeed, all op amps are pretty much designed to have
    'uncontrolled' dc and low frequency gain. Its only the compensation
    capacitor that reign in the gain.

    Its also done when you have very low signals and desire the utmost in
    low noise. Any resistance in the emitter circuit will add noise.

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

    Ian Bell Guest

    I think you should consider this topology simply as an example to enable
    people to understand the basic trade-offs in transistor circuit design. it
    is *not* a topology that would normally be used in a proper design.
    However, once you understand the limitations of a simple CE circuit you can
    start to grasp the more common multi-transistor configurations.

  17. Ian Bell

    Ian Bell Guest


  18. Rich Grise

    Rich Grise Guest

    Well, like I said, it was good enough for the app. I just didn't bother,
    I guess. I hope I'm forgiven. :)
    Others have mentioned a gain stage in an amp with overall negative
    feedback, which makes sense to me, but I can't take any credit for
    noticing it. :)

  19. Thanks, Kevin. In other words, when you wrap the stage(s) with a feedback loop.
    Ah. Of the (4kTBR)^.5 variety.

    I can think of cases, rather easily, for your first paragraph. So I'm with you,
    there. But in the second case, can you suggest a specific application that I
    can consider more closely (one where the overall negative feedback is NOT used,
    but where noise needs to be minimized in this way?)

  20. Joe

    Joe Guest

    Hi Jon,

    My default font says it is 'courier new' whatever that means.

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