Designing a Common Emitter Amplifier

Discussion in 'Homework Help' started by mygumballs, Apr 28, 2008.

  1. mygumballs

    Thread Starter New Member

    Apr 28, 2008
    hi everyone, this is my first post/thread and i hope you all don't mind me jumping right in to asking a homework help questions.

    basically i'm performing a lab experiment where we are to design a common emitter amplifier using a 2n2222 bjt as in this diagram i hastily drew up linked below:

    the specifications are minimal: a Vcc of 10V, a collector current of 1 mA, and a gain of -14. other than that, we must design the amplifier by choosing the resistor values to obtain this. my question, essentially, is how do i do this?

    now i generally have a good idea of what i need to do, but my professor is so demanding and so unhelpful, that i know there will be bad consequences if i don't get it completely correct. so here's what i know:

    i have on that figure a beta value of 400. i'm not sure if this is correct, but i recall this is the value we determined when the class viewed the characterstics of a 2n2222 on the cure tracer. regardless, if possible, it would be nice to make my calculations beta independant. though i seriously doubt this is possible.

    using the ac analysis technique, i found the following equation for hte gains:
    Av = (-Rc*Beta)/(Beta/gm + (1+Beta)*Reac)
    which will give me the relation:
    Rc = 364 + 14*Reac

    at this point, i don't know how to progress to find these two resistor values without just pulling on of their values out of the air arbitrarily. if i do have to choose values for one of these resistors, is there any criteria i should use?

    after reading some threads here about similar problems, i've read about choosing a q point in which either Vce or Vc should be half of what their saturation equivalents would be. but because of the ambiguity between those two, i'm not sure what rule to use.

    and furthermore, after possibly finding the values of Rc and Re, my professor insists that there is a sure way to find exact values for R1 and R2, a process that DEFINATELY eludes me.

    i really need help with this. i've been finding it extremely difficult to find someone who can help me with transistor analysis and if someone here can show me a step by step process of how to design this amplifier, i know i can use it to finally understand how these things work in general. i will be thuroughly thankful if you can help me and i'll definately be a regular subscriber to this forum. thanks!
  2. Caveman

    Active Member

    Apr 15, 2008
    First of all read the datasheet. hfe is the same as beta. You will see that they only define a minimum. You will notice that they only define minimums, and these are each different for specific situations.

    The simplest analysis goes like this:
    1. Vin increases by dVin.
    2. Because the Vbe is about 0.7 and doesn't want to move, the emitter voltage rises by the same amount, dVin.
    3. Therefore the emitter current increases by dVin/Re.
    4. Since the collector current is about the same as the emitter current, the collector current also increases by dIc = dVin/Re.
    5. Therefore the collector voltage (vout) decreases by dVout = dIc*Rc = dVin*Rc/Re.
    6. So, the gain is dVout/dVin = -Rc/Re.

    However, a lot of assumptions were made here. First of all we assumed that the transistor is biased in the linear range of operation. This is done by setting the bias voltage divider at the base to the right value. Second, we assumed that the Vbe is 0.7V and doesn't change. This is a pretty good assumption if Re isn't too low.

    Does that get you there? This should all be in your textbook.
  3. mygumballs

    Thread Starter New Member

    Apr 28, 2008
    that gets me part of the way there. one of the problems is there is that capacitor over the second emitter resistor, in turn, the gain formula is actually:
    gain = -Rc/Reac
    which gives me the relation between the two resistors, but not their individual values. i understand that this is a design problem, so i will probably have to choose some resistor values myself, but i want to confirm that i would have to in fact choose either Rc or Reac myself in this situation.

    Also, how do i next find Redc? i presume i need to choose a q point for that aspect, but i am unclear as to how to go about doing this.

    And furthermore, you mentioned that i would need to set "the bias voltage divider at the base to the right value." part of my ultimate question is how do i do that? i presume you mean choosing the correct R1 and R2 values, but what criteria do i use to do this?

    i really appreciate this help. i assure you, i've tried going through my text, but i just can't seem to find a straight asnwer to my problem. i'll be in your debt if you can help me with the rest of this design.

    btw, we had already performed the experiment in the laboratory using the following resistor values: R1 = 30K, R2 = 10K, Rc = 5K, Reac = 330 ohms, and Redc = 1.465K
    and our experimental gain, along with the PSPICE simulation, gave us about -14. so there are atleast these values that i know work, but i need to show how i can determine these values with analysis alone. regardless, even if my theoretical analysis doesn't result in these same resistor values, i would atleast like to know how to perform the design procedure anyways.

    thanks again!
  4. Caveman

    Active Member

    Apr 15, 2008
    How about I go about it this way. I'll take your values and analyze what I expect. Then you can turn the math around to get back to the values. However, I will need your capacitor values as well.

    First of all let's do a DC bias analysis. We've gotta make sure that it is correct here first.
    1. R1 = 30k, R2 = 10k. So if we assume that the base current is small, then we can just divide it down to find the voltage, so Vbase = 10V*10/(30+10) = 2.5V.
    2. The emitter voltage is about 0.7V below that, so 1.8V.
    3. The emitter total DC resistance is (1.465k + 0.33k) = 1.795k. So the emitter DC current is 1mA.
    4. The collector current is about the same as the emitter current, so 1mA. That's what you wanted.
    5. The voltage on the collector is 10V - 1mA*Rc = 5V.
    6. The DC gain is about Rc/Re = 2.79.

    Now we did assume that the base current was not significant. If beta is about 75 on the low end, then Ib = Ic/beta, so Ib = 1mA/75 = 13uA. The biasing current through the R1 + R2 = 40k resistance is 250uA, so our assumption is pretty good.

    Now for the AC analysis, we need to know what the lowest frequency that you want. But a simple analysis is as follows. Let's just say you want 1kHz as your low frequency cutoff.

    First you need to calculate the input and output capacitors.
    For the input capacitance, the frequency of cutoff is fc = 1/(2*PI*(R1+R2)*C). So for 1kHz = 1/(2*PI*(40k)*C) gives C = 4nF.

    For the output capacitor, the formula is the same, but R is Rc, so this gives 32nF. Note that if you had a load resistor, you would need to use the parallel of Rc and RL as the R in this equation.

    Now, if you assume that CE is sized such that Redc is effectively shorted at 1kHz, then the gain is just Rc/Reac = 5k/330 = 15.15. To make this happen, we want the impedance of CE in parallel with Redc to be much smaller than 330 at this frequency, let's say about 10% or 30 ohms. Since Redc is much larger than this, basically we want CE's impedance to be about 30 ohms. So 1/(2*PI*f*C) => C = 5.3uF.

    Does all of this make sense?
  5. Audioguru

    New Member

    Dec 20, 2007
    The transistor has an internal emitter resistance of 0.026/1ma= 26 ohms which is in series with the unbypassed 330 ohm resistor, so the AC gain is 5k/(330 + 26)= 14.0.
  6. mygumballs

    Thread Starter New Member

    Apr 28, 2008
    thanks everyone, but i think i figured it out now. i talked to a different professor and found out some basic assumptions in design that really help out.

    in case you wanted to know them, one is that we can simply set Vb to 1/4Vcc in order to make sure the junction between Vb and Vc is in reverse bias. another assumption he mentioned was that R2 must be much less than Beta*Re. with these assumptions and the advice you've all given me, i've found a way to properly determine those resistor values i mentioned above.

    again, thanks for all your help.