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Common Emitter Amplifier with 2N3904

Discussion in 'Electronic Basics' started by Lily Bepant, Jan 10, 2004.

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  1. Lily Bepant

    Lily Bepant Guest

    I have been up to a question like this:
    "Design a common emitter amplifier with an output impedance of 4.7Kohm
    and a gain of -10 using 2N3904 powered by 12V DC power supply."

    I got the pencil and paper started to design the thing. (I'm not an
    electronics major)

    I have jumped through many equations to find that: (I'm novice, so
    don't assume that these are correct, please correct wherever possible)


    Ic(max).. I looked up 2N3904 manual and found that Ic(max) is 200 mA
    for it. With 12V Vbb, it means Rc+Re= 60 Ohms. We want a gain of 10,
    thus Rc/Re should be 10, which yields to the result that Re = 60/11 =
    5.4 ohms Rc = 55 ohms (rounded these).

    These values seem to be weird to me. What am I doing wrong?

    Also, what would I take B (beta) of this thing? This is a design
    question, therefore I understand that I should design something
    independent of the B (beta) value. But how, when

    R1+R2 = [(Vbb)]/[(10Ic)/B]

    or not?
    I'm interested in electronics but don't have enough knowledge on it.
    Thanks for your guidance.

    Best Regards,
  2. beta varies with the wind. Is there any reason you can't use an opamp for
    your amplifier? Mucho easier.
  3. A maximum rating is something you have to be sure your design does not
    exceed. Staying well below the maximum current rating for the
    transistor improves almost every aspect of operation, including power

    Some very rough rules of thumb:

    1. The collector impedance of a common emitter amplifier is generally
    high for low frequency operation, so you can approximate the output
    impedance of the amplifier by assuming that it is made up mostly of
    the collector load resistance.

    2. The voltage gain cannot be higher than the collector resistor
    divided by the emitter resistor. It will actually be a bit less than

    By 1, you should start with a collector resistance of about 4.7 k

    By 2, you should start with an emitter resistor of 470 or a bit less
    (390, perhaps).

    Note that these two choices will give an output impedance of about 4.7
    k ohms, but a voltage gain of about 10 only when the amplifier runs
    with no external load. If that is how you read the specification,
    then you can continue with the rest of the design. If the spec
    actually means that the voltage gain is to be 10 when the amplifier
    has a 4.7 kohm external load connected then you have to design for an
    unloaded gain of 20.
  4. I read in that Lily Bepant <>
    No, this is not correct. Vbb would be the voltage of an independent base
    bias supply. And you do not design every stage to run at Ic(max)!
    Using that wrong equation. Begin with Vcc - the *collector* supply
    voltage = 12 V. Now, the terms of the question do not include anything
    that allows you to calculate a required collector current. So you have
    to choose one. With a 4.7 kohm collector load, and 11 V available once
    you have dropped 1 V across the emitter resistor (assumed for the moment
    it is decoupled by a parallel capacitor at signal frequencies) to get
    reasonably stable d.c. operating conditions, you would get as much
    output voltage swing as possible by dropping 5.5 V across the collector
    resistor, which gives you a collector current of 5.5/4.7 = just under
    1.2 mA. A standard value 820 ohm emitter resistor gives you 1 V on the
    emitter then.
    Well, again, it's not Vbb but Vcc (unless you REALLY have a separate
    base bias supply). What you do is to choose R1 and R2 using an average
    value for beta and then see what happens to Ic with maximum and minimum
    beta devices. In my book, beta min is 100 and beta max is 300 at Ic = 10
    mA. I don't have a beta against Ic curve, so I'm going to assume that
    the beta doesn't vary with Ic sufficiently to create a problem. The
    average beta is 200.

    Here, we have 12 V/(12 mA x 200)= 200 kohm. Very neat! Now there is 1 V
    on the emitter and thus 1.7 V on the base. We can assume Vbe = 0.7 V
    because it doesn't vary much between devices and Ic varies strongly with
    it, making our d.c. stabilising circuit (emitter resistor and base
    potential divider) a high-gain control loop.

    So R1 + R2 = 200 kohm and R1 has 11 V across it while R2 has 1 V across
    it. Strictly, it also has 0.5% less current in it, because that goes
    into the base, but we won't be using high-precision resistors, so we can
    neglect that. Thus R1/R2 = 11, and that's 2 equations with 2 unknowns,
    R1 and R2. We get R2 = 200kohm/12 = 16.7 kohm and R1 = 183 kohm. We
    would choose the standard values 180 kohm and 16 kohm.

    I'll leave it to you to calculate what collector current you actually
    get with 180 kohm, 16 kohm and 820 ohms, with beta values of 100, 200
    and 300. For extra credit, what is Ic with a beta of 200 and the
    transistor immersed in boiling water?

    However, this stage has a gain of much higher than 10. The gain is
    RcIe/26, with Ie in milliamps, giving a gain of 217. To get a gain of
    just 10, we split the 820 ohm emitter resistor into Rc/10 = 470 ohms and
    820-470 = 350 ohms (use standard value 360 ohms), and decouple (connect
    a capacitor in parallel) the 360 ohms only. We could have just used the
    470 ohms as the emitter resistor but that would have given only about
    0.5 V on the emitter and if you do the calculations you will find that
    Ic varies a lot more with beta than it does with Ve = 1 V.
  5. Ian Bell

    Ian Bell Guest

    Quite possible but it is a pretty vague specification. Is it and ac
    amplifier or a dc one? (I suspect it is ac). If it is ac over what
    bandwidth should it have this gain and what is the impedance of the source
    signal. if you have this info a basic design is quite straightforward.
    Don't know where this came from. it's not a standard one AFAIK and Ic(max)
    is not overly important for this design.
    First the IC(max) is a red herring.

    Gain is roughly RC/Re. Output impedance is roughly Rc so make it the 4.7K
    specified. So Re is 470 ohms.
    The gain being ruoghly Rc/Re comes from assuming beta is large and
    conversely the negative feedback due to Re ensures the gain is relatively
    independent of beta.
    All you need to do now is to work out the bias circuit values to ensure dc
    stability and give a sufficiently high input impedance.


  6. It's not mandatory to make Ic the maximum 200mA allowed. The wanted
    output impedance is 4.7K, so the collector resistor should be 4.7Kohm.
    The emitter resistor 4700/10 -> 470 ohms.

    To give the amplifier maximum swing, the output needs to be at 6V
    with no input signal. Ic = 6V/4700 -> 1.2mA
    Voltage at the emitter is 1.2mA * 470ohm -> 0.6V
    Vbe should be ~0.7V, so the voltage at the base needs to be 1.3 volt.
    For this, you can use a voltage divider of 47K and 5K6.

    Hope this helps.

    Thanks, Frank.
    (remove 'x' and 'invalid' when replying by email)

  7. Yeah, but no one seems to have mentioned that if the emitter resistor
    is bypassed (as it may well be) then the the Vgain will be RC/RE+re
    where re is the internal resistance of the emitter diode and typically
    quite small - but worth remembering nonetheless. You're also assuming
    the output is unloaded, which is a bigger assumption. If it's loaded,
    the Vgain will be RC||RL/RE+re.
    A Ve of 0.6 seems very low. Should be at least a volt, IMHO.
  8. I'm only considering/answering the topology the OP presents, which
    uses just a emitter resistor and nothing else. Furthermore, the OP
    wanted an output impedance of 4K7 with a gain of 10. For design purposes
    that means the load is assumed as not present.
    Yep, zip, nada, nothing ;)
  9. Lily Bepant

    Lily Bepant Guest

    Hello again,
    I didn't understand the maximum output voltage swing concept. I have
    (tried to) built DC and AC circuits, tried some equations and loops
    but couldn't come up with the values you guys have found. (I tried to
    find the balance by calculating the intersection point of AC and DC
    load lines to find the quiescent operating point. Won't the maximum
    output voltage swing will be achieved where AC and DC lines
    I think I'm missing something again here.

    Thanks for the answers!
  10. Fred Bloggs

    Fred Bloggs Guest

    This is a troll...
  11. Bill Sloman

    Bill Sloman Guest

    The output impedance of a 2N3904 transistor in a common emitter
    configuration is of the order of 100k, so if you use a 4k7 load
    resistor, the output impedance of the amplifier will be close to 4k7.

    You want to use up some of that 12V biassing the transistor to get a
    reasonably predictable collector current, so 1mA through 4k7, giving a
    nominal (average) output voltage of 7.3V is probably okay.

    At 1mA, the incremental emitter resistance of transistor is about 30R.

    To get a gain of -10, you need an extra 440R between the emitter adn
    0V. 430R is probably close enough.

    1mA through 430R is 0.43V.

    The Fairchild data sheet for the 2N3904

    gives the base-emitter "on" voltage as about 0.67V at 25 degrees
    Celcius, so you need to bias the base at a nominal 1.1V to have 1mA of
    collector current.

    You can do this with a voltage divider. At 1mA and 25C the typical
    current gain is about 230, minimum 70, so the base current could be as
    high as 14uA, though 4.3uA would be typical.

    So design the divider for an impedance of around 10k, and an unloaded
    voltage of about 1.15V - 10k and 100k would be close enough.

    If you want closer control over the bias voltage and the collector
    current, you can bias the base further above the 0V rail, and use a
    bigger resistor from the emitter to the 0V rail to define the nomimal
    current. Thios obviously gives you less gain, so you by-pass this
    resistor with a second, smaller resistor in series with a capacitor to
    the 0V rail.

    The capacitor blocks any DC current, but - for sufficiently high
    signal frequencies - serves as an AC short circuit, so that you high
    frequency gain is defiend by the parallel sum of the two resistors.
  12. This was class A. wasn't it? You want to accommodate maximum output
    voltage swing by 'setting' the collector voltage at midway between Vcc
    and Vb. It's not critical with smaller signals but becomes
    increasingly important as signals get larger (to avoid clipping).
  13. Lily Bepant

    Lily Bepant Guest

    What makes you think that I am trolling? I'm trying to gather some
    information about the subject and still couldn't succeed.
  14. I'm afraid there are just some folks here who take up perfectly good
    space for no apparent reason, Lily. "Mr. Bloggs" is one of them.
    Fortunately, most people here are very helpful and will do their best
    to assist you in any way possible. Feel free to ask away as much as
    you need to in order to gain a better insight into your problem with
    this amp. It's what Usenet *should* be about.
  15. I read in that Lily Bepant <>
    The way you specified your question, '4.7 kohm load', doesn't give
    enough information to justify an assumption that this load is AC-
    coupled. That's why I, and others, assumed it was the collector
    resistor. If it's AC-coupled, then the data you provided is
    insufficient to produce a solution without yet more assumptions.
  16. Fred Bloggs

    Fred Bloggs Guest

    I have a solid record of providing REAL help to REAL people. The same
    cannot be said of you- you are worthless trash who thinks SED is your
    personal chat room. You have absolutely NOTHING to offer-and the other
    contributors have already told you to take your elementary drivel to the
    basics newsgroup. You never built anything in your life- you are a
    dilettante and talker- too dumb to even run simulations and learn
    electronics at the community college first semester level. You are a
    total waste of time and it is people like you who are responsible for
    USENET being abandoned by all but the trite-minded pedants. Like I said-
    you are supeficialsuperficiala retiree idling away the last hours of
    your waste of life- be sure and donate your organs or something useful.
  17. Fred Bloggs

    Fred Bloggs Guest

    Isn't this typical- the damn jerk-off struggles to get a battery snap on
    right- and now proceeds with technical critiques of working engineers.
    He's a pathetic UK retiree maggot rotting in solitude in a slum
    somewhere- you can see from his prose that he was probably a
    slick-mouthed sales trash of some kind- and obviously a total zero when
    it comes to anything constructive
  18. Fred Bloggs

    Fred Bloggs Guest

    Lemme see- you know about the various load lines, gains, and I/O
    impedances, and reading data sheets- but you pretend you can't do a DC
    bias? This makes no sense- you are a troll. Trolls are generally of low
    intelligence, love to pseudo-intellectualize over definitions, and
    usually deal with an elementary circuit that just will not work for some
  19. John Larkin

    John Larkin Guest

    Probably the reason nobody mentioned it is that it's not true.

  20. I was under the impression he was a 17 yr. old kid. Now, about battery
    snaps and the like, don't you hate those battery compartiments where
    all the AA(A) batteries point in the same direction v.s. the traditional
    zigzag arrangement? Got me fooled again, today ;)
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