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Beginner's question: NPN common-emitter amplifier setup?

Discussion in 'Electronic Basics' started by [email protected], Feb 16, 2006.

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  1. Guest

    I have seen schematics with a resistor hooked up between the emitter
    and ground. In a book by Stan Gibilisco (I can't remember the name off
    hand) it stated that although the emitter was grounded from the signal
    that there was still a voltage due to the resistor. Is there a voltage
    because:

    When the collector draws current from the emitter, that current passes
    through the resistor creating a voltage drop? I don't think that
    polarities are correct.

    If this question doesn't even make sense to you guys, please let me
    know.

    Glenn
     
  2. Noway2

    Noway2 Guest

    A bipolar resistor can be thought of as a current controlled switch. A
    small current flowing into (out of) the base of an NPN (PNP) transistor
    causes a larger current to flow in the collector - emitter. While the
    collector current is not quite equal to the emitter current (due to the
    base current) for easy analysis they are assumed to be since the
    difference is usually very small. The ratio between base current and
    the collector current is called the transistor Beta (current gain) and
    is usually specified as a minimum value according to the datasheet.

    If on the resistor to which you are referring, one end connects to
    ground and the other end connects to the emitter, any current flowing
    through the emitter will flow through the resistor, bringing the
    emitter voltage above ground. In turn, the base voltage will be about
    ..6V - .7V above the emitter voltage, assumig the transistor is on. If
    enough current is flowing, such that the collector voltage drops to
    about or below the base voltage, the transistor will enter saturation
    which means that an increase in base current will produce little, if
    any increase in collector - emitter current. In this mode of
    operation, the CE voltage is typically about .2V to .3V. This assumes,
    that there is a collector resistor, which is generating a voltage drop
    from the supply to the collector terminal. A circuit of this
    configuration will exhibit an electrical "gain" equivalent to Rc / Re.
    The idea is to design the circuit to be dependant on the resistors
    rather than the inherent properties of the transistor which can vary
    significantly

    Sometimes a capacitor is placed in parallel with the emitter resistor.
    This is done so that at DC the emitter resistor is present to "bias"
    the transitor into a certain region of operation, while providing a
    high gain to AC signals. Note also, that the emitter has an inherent
    resistance associated with it that is proportional to the emitter
    current.
     
  3. Brian

    Brian Guest

    Take a look at http://www.fncwired.com/TransistorExample/

    Brian
     
  4. Some of it makes sense. You are referring to:

    : +V
    : |
    : |
    : |
    : |/c Q1
    : IN-----|
    : |>e
    : |
    : |
    : |
    : \
    : / R1
    : \
    : /
    : |
    : |
    : gnd

    In this case, when collector current flows out the emitter and through
    R1, a voltage is generated across it. Assuming that the base voltage
    is high enough that Q1 doesn't otherwise turn off, this voltage moves
    the emitter voltage closer to the collector voltage.

    But the usual 'view' of this is the other way around. One starts with
    the base voltage, drops it mentally by a presumed Vbe between the base
    and emitter (say, 0.7V, for example), assumes that this will be the
    voltage at the emitter, computes the current that would then flow
    through R1 based on that applied voltage, and then concludes that this
    is approximately the collector current -- less a small portion for the
    base current.

    Jon
     
  5. Graham Knott

    Graham Knott Guest

    Have a look at http://homepage.ntlworld.com/g.knott/elect225.htm
     
  6. Pooh Bear

    Pooh Bear Guest

    Yes of course. First principles.
    Uh ?

    An emitter R simply results in a degree of local feedback. It wil help
    stabilise DC operating conditions and will also reduce AC voltage gain (
    unless the emitter R is bypassed with a cap ).

    Graham
     
  7. That is exactly what happens. The emitter current is the sum of the
    collector current and the normally much smaller base current, so it is
    a way to produce a voltage related (mostly) to collector current that
    is applied to one of the transistor's input terminals.
    Makes perfect sense. Just remember that the input signal that
    controls the transistor is the voltage between base and emitter. So
    as you apply a signal to the base, the emitter resistor applies an
    output (collector) current generated voltage that opposes it, acting
    as negative feedback. This makes the output current much more
    accurately proportional to the base voltage variations. It also
    raises the base impedance and the collector impedance. And while it
    lowers the gain (as all negative feedback does) it extends the
    frequency range that flat gain occurs. And it helps reduce the
    effects of temperature (increased current gain and lower base to
    emitter voltage) on the bias situation.
     
  8. Guest

    Thanks for all the replies, everyone. I will definitely check out the
    websites that have been suggested.

    When you talk about current flowing through the emitter, is this
    current coming from ground? I think that current cannot flow from
    collector to emitter. If so, wouldn' t the voltage be higher at the
    bottom of the resistor (closer to ground)? That voltage got dropped
    across the resistor, leaving less voltage at the top.

    I'm confused, but a little less so thanks to everyone.

    Glenn

    PS: the books I've been reading don't seem to address this very
    clearly.
     
  9. Noway2

    Noway2 Guest

    Current will flow out of the base and into the emitter in a PNP
    transistor and yes, if the emitter were at ground, 0 volts, the base
    would be below this potential. PNP transistors have their place when
    the signals you want to work with are negative in amplitude such as an
    AC coupled amplifier, and when you wnat to switch something by bringing
    the voltage down rather than up, such as an active low signal.

    The emitter current will be a combination of collector and base
    currents. The collector-emitter current concept defies normal / basic
    thinking and is a function of semiconductor physics. The application
    of base current injects charge carriers into the C-E junction called
    the depletion region. If the right type of charger carrier is injected
    the depletion region gets narrower an if the other type is injected it
    gets wider. When the depletion region is narrower and voltage is
    apllied to C-E pair charge carriers are able to be swept across the
    depletion region - hence current flows. The amount of current that is
    able to flow depends on the width of the depletion region, hence more
    base drive more current can flow in the C-E. The ratio of base
    (current) drive to collector - emitter drive is the transistors Beta.
     
  10. Rich Grise

    Rich Grise Guest

    You have to decide early in your career if you're going to use electron
    flow, which hams, techs, and other people who live in the real world
    use, or "conventional" flow, which academecians, scientists, and other
    visionaries use. ;-)

    They're indistinguishable, except all of the signs are reversed.

    Conventional flow flows in the direction of the arrow at PN junctions:
    one in a diode, and the emitter/base junction in BJTs (bipolar junction
    transistors). Electrons flow in the opposite direction to the arrow.

    The conventional current in an electron beam goes upstream. ;-)

    With a negative-ground supply, the electrons come out the ground, flow
    up through your circuit, and get sucked into the positive terminal of
    the battery, where they get circulated by the force of the chemical
    reaction.

    All you have to do is change your point of view, and it's exactly the
    same as positive current flowing out of the positive terminal of the
    battery, down through the load, and to ground.

    Conventional current uses the right-hand rule, and electron flow uses
    the left-hand rule, but save that for when you start to get into
    electromagnetics. ;-)

    Hope This Helps!
    Rich
     
  11. Guest

    I have done my research and much of what you guys are saying makes
    sense!! Thanks!

    BTW: How do you set the voltage at the emitter so that it is 0.7 v or
    so less than the voltage at the base? I know that the base voltage is
    set my some sort of potential divider resistor setup. But how do you
    ensure that the emitter will be less than the base?

    Glenn
     
  12. ehsjr

    ehsjr Guest

    Getting a bipolar transistor to do what you want (within
    its capabilities) is all about controlling b-e current,
    and there are numerous was to do it, so there is no one
    single answer to your question "how do you ensure that
    the emitter is less than the base"?

    But there is an implication in your question that needs
    to be addressed. Your question implies that you may think
    the base must always be ~.7 volts higher than the emitter.
    You need to think of it a little differently.

    The amount of b-e current controls the amount of c-e
    current. It could be that you want *no* c-e current,
    in which case you would not want the emitter to be ~.7
    volts more negative than the base.

    What you want to do is control the b-e current and not
    have the base *always* ~.7 volts higher than the emitter.
    The reason .7 volts is an important number is that
    there will be no c-e current until the base voltage is
    raised to about .7 volts more positive than the emitter
    voltage in the NPN transistor. In a PNP, the base would
    need to be about .7 volts more negative than the emitter
    for there to be c-e current.

    Ed
     
  13. Well, this tells me about where you are at. I think I've been there,
    too. But it's been a long time and I may not remember how I was led
    out of it.

    For the moment, just imagine that you have placed two batteries in
    this way:

    : (2)
    : ,----------,
    : | |
    : | |
    : | |
    : (1) |/c Q1 | +
    : ,----| -----
    : | |>e --- B2
    : + | | -----
    : ----- | (3) ---
    : --- B1 | | 10V
    : ----- \ |
    : --- / R1 |
    : | 5V \ |
    : | / |
    : | | |
    : | | |
    : '------+----------'
    : (gnd)

    First off, before I go any further with the above circuit, note that I
    added a (gnd) at the bottom. It doesn't mean anything except to note
    where my "reference point" is at. There are three nodes in the
    diagram other than (gnd) and these are (1), (2), and (3). If I talk
    about the voltage at (1) as being +5V, you should keep in mind that I
    am talking about the voltage at (1) compared with (gnd). Voltages are
    always measured with respect to somewhere. And by common agreement,
    we learn to designate one particular node as being special and calling
    it (gnd). Which one makes the most sense to label that way will
    depend on things and it will take some experience before you will
    always pick the same place to label that way that a professional
    might. But it doesn't change the circuit if you put the (gnd) label
    somewhere a professional wouldn't -- it just means that folks
    (including you) may be a bit confused when you talk about some voltage
    here or there. But I could just as well have labeled the node above
    called (2) as (gnd) and relabeled (gnd) as (15), for example. The
    labels are entirely arbitrary and mean nothing as far as the circuit
    itself goes. It's just inside our heads and nowhere else.

    Okay. So trace out the above circuit in your mind until you agree
    with me that there are four places of common connections between
    components. We have 4 parts in the circuit; two batteries, one
    resistor, and one transistor. They are tied together with wire and
    since all places along a wire are considered to be at the same
    potential, we can mentally consider everywhere that only wire exists
    as being the same "point." I just labeled those points as (1), (2),
    (3), and (gnd). I hope you agree with that much.

    Now, what is the potential at (1)? It's +5 (with respect to (gnd), of
    course.) That's because there is a battery there and its job is to
    make sure that one end of it is at a fixed potential away from the
    other end. In this case, the plus (+) side is tied to the base of Q1,
    so the base of Q1 _must_ be at +5V because the battery will make sure
    of this fact. That's its job.

    Hopefully, we may agree that (1) will be at +5V (with respect to gnd.)
    This means Q1's base is at +5. Now think about the effective diode
    between Q1's base and Q1's emitter.. Ignoring the rest of the
    circuit, that looks about like this:

    : (1)
    : ,------,
    : | |
    : | ---
    : | \ /
    : | V
    : | ---
    : | |
    : + | |
    : ----- | (3)
    : --- B1 |
    : ----- \
    : --- / R1
    : | 5V \
    : | /
    : | |
    : | |
    : '------'
    : gnd

    If you ignore R1 for a moment and think of it as just a wire, I think
    you can see that the diode will be forward biased. This means it will
    allow the flow of current, readily. Without R1 (with it shorted out
    by a wire), the voltage across this Q1's base-to-emitter diode would
    be the full 5V and a lot of current would flow. Way too much, and it
    is almost certain that Q1 would self-destruct. Now, if you re-imagine
    R1 included, what happens? Well, as more and more current is
    considered to be flowing through the diode _and_ R1, R1 starts
    dropping more and more voltage according to I*R1. Eventually, of
    course, all of the 5V would be used up by R1. In other words, if you
    imagine that "I" is large enough, the product of I*R1 would be exactly
    5V. When that happens, there would be no voltage left to exist to
    forward bias the diode. And no current (not much) would be able to
    flow. So that would mean that that much "I" is too much to be
    reasonable. So we are sure that "I" must be less (<) than this. In
    other words, I*R1 < 5 or written another way, I < (5/R1).

    But how much less?? Well, it turns out that the value of "I" will
    automatically find itself about right when I = (5 - .7) / R1. In
    other words, the diode itself "wants" to see about .7 volts across it
    in order to work in the forward conduction region. It might be a
    little bit less than .7 volts or a little more than .7 volts. But not
    a lot different. This is because even a small increase in voltage
    across it (say, about 0.06V) means a multiplication by 10 of the
    current through it. So you can see that being off even by a tenth of
    a volt one way or another might mean changes in current spanning a
    factor of 100 or 1000 times. So diodes will tend to find their
    forward drop to be pretty close to 0.7 volts most of the time. I see,
    in transistors, between 0.6V and perhaps 0.9V depending on their use.
    But quite often very close to the .7V.

    So if it is true (and I tell you it is) that the diode, forward biased
    as it is, will drop about .7V then we can calculate from that fact
    that the voltage at node (3) should be .7V less than +5V or about
    +4.3V. And it's true, it will be about that.

    Now, before I continue, let me redraw the circuit and hope that you
    still agree with me that it is the same circuit:

    : (2)
    : ,----------,
    : | |
    : | |
    : | |
    : (1) |/c Q1 | +
    : ,----| -----
    : | |>e --- B2
    : + | | -----
    : ----- | (3) ---
    : --- B1 | | 10V
    : ----- \ |
    : --- / R1 |
    : | 5V \ gnd
    : | /
    : | |
    : gnd |
    : gnd

    I just broke the displayed wires, but conceptually this is the same
    circuit. In practice, of course, you'd have to wire those ends
    together. But for understanding the circuit, you don't need all that
    wire. And in fact, it will tend to confuse you a little because
    you'll start wondering how various components attached to it might
    affect things. Breaking the wire and just labeling each end the same
    is my way of forcing you to stop thinking that way and to watch for
    the important stuff and ignore stuff that doesn't matter.

    Now you were asking how we can know (or guarantee, somehow) that no
    matter what happens in the circuit that node (3) will still be about
    +4.3V? Well, it turns out that if you disconnect B2 entirely from the
    circuit that it will be true and that if you hook up B2 it will still
    be true. In fact, if you change the voltage for B2 from 10V to 20V,
    it will still be true. This will remain just magic to you until you
    really think more closely about things. But it does work that way,
    close enough for now, anyway.

    The only thing that matters here is that the base-emitter diode of Q1
    is forward biased by the base-emitter circuit, which includes B2 and
    R1, and that this is the only important detail in knowing that the
    voltage at the emitter will be about 4.3V or 0.7V less than the base.

    If you don't connect up the Q1 collector and leave B2 out of the
    circuit, then all of the current flowing through R1 in order to
    produce a voltage of 4.3V across it (I*R1 = 4.3V) will come from B1
    via Q1's base. In other words, the base will be supplying all of the
    current equal to (4.3V/R1). But if you now connect up B2 to Q1's
    collector and include it as part of the circuit, then most of that
    current (4.3V/R1) will actually come from the collector and B2 instead
    of from B1 via the base. In other words, the base current will
    suddenly and sharply decline to a tiny percent of its original value.
    This is because Q1 will start conducting collector current to replace
    the base current. R1's current will still be the same, 4.3/R1. But
    most of it, about 99% of it, will come from the collector instead of
    the base, when B2 is added and hooked up to the collector of Q1.

    Does all that make sense, still?

    Jon
     
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