Could someone Check My Working out, for a Common Emitter Amplifier

Discussion in 'Homework Help' started by Denny1234, Apr 25, 2008.

  1. Denny1234

    Thread Starter Member

    Feb 17, 2008
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    Hi I am trying to design a CE Amplifier with a spec as so, so far I have got the working shown below, but I have two problems, it dosent get me the lower cut off freq I desire (40hz) and two of my capacitors CL and Ce have no effect on my lower cut off no matter what vlaues I put into them which seems wrong. Could someone run through my theory so far and point out where I may be going wrong?

    Circuit I refer to (their values are different)
    http://www.ecircuitcenter.com/Circuits/trce/trce.htm

    My spec

    Voltage gain: at least 20 dB at mid-band into 600 Ω
    Frequency response:
    Lower cut-off frequency: 40 Hz
    Upper cut-off frequency: 1 MHz
    Input impedance: 600 Ω ± 10%
    Output impedance: 600 Ω ± 10%
    DC supply voltage: 12 V
    DC supply current: not more than 15 mA
    Output power: not less than 2.0 mW into 600 Ω at mid-band.

    Note : thanks for Caveman for his help so far in explaining to me how the circuit works

    My Calcs are below
     
  2. Denny1234

    Thread Starter Member

    Feb 17, 2008
    27
    0
    What I have done

    Working out the value of C1 for my Input filter ….

    I determined my value for Input Impedance of transistor like so...

    re (emitter resistance) = 1/40Ie
    which is 1/ 40*10mA = 25 ohms

    My Data sheet says
    hfe (beta) for IC = 10 mA ; VCE = 10 V; (which are my conditions) hfe = 75

    So 75*25 =1875 ohms for my input impedance of my transistor
    R1 = 1950 ohms
    R2 = 870 ohms
    Rs = 600 ohms

    So for a 40hz low frequency cut off, I just did 1 / 2pi*f*R

    Total R being 600 + (870//1950) + 1875 = 3075

    So C1 = 1/2pi*40*3075 = 1.29uF

    Does that seem okay, will that alone determine my lower cut off frequency in theory??, because I actaully get 20hz cut off with that capacitor which is half of what I want.

    My other two high pass filters I think are CL and Ce.
    So to get a low freq cut off of 40 hz I just did

    For CL = 1/ 2*pi*40*R
    R being RC + RL = 1200
    So CL = 1/ 2*PI*40*1200 = 3.3uF

    For Ce = 1 / 2pi*40*Re2
    = 1 / 2*pi*40*270
    = 14uF
     
  3. Audioguru

    New Member

    Dec 20, 2007
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    Now that the required gain is a lot more than the original circuit, everything must be re-designed and re-calculated. The output load before was 100k and now it is 600 ohms.
     
  4. Denny1234

    Thread Starter Member

    Feb 17, 2008
    27
    0
    Hi, but i never used that circuit (with the link) at all, all i did was to use it as a reference for asking which components refer to what. My calcs are based on the spec, which has a 600 ohm load resiatcne and 600 ohm source resistance unlike the links components and a 22n2222 amplfiier.
     
  5. Denny1234

    Thread Starter Member

    Feb 17, 2008
    27
    0
    Also can i ask you something about something you mentioned before, and that i only just picked up on!

    "The high frequency cutoff frequency is when the reactance of the transistor's output capacitance plus the stray capacitance is equal the the value of collector resistor that is in parallel with the load impedance. 16pF and 10k ohms is 1MHz. 16pF and 1k ohms is 10MHz."

    Having done that with my 600 ohm load resistor i got 265pF as my cap and that has given me a cut off of 4mHz on my plot. Its pretty good as im getting their, but just to ask, is this the kind of high frequency that the Miller effect of capacitive feedback from the collector to the base would chnage the calc?
     
  6. Caveman

    Active Member

    Apr 15, 2008
    471
    0
    You've got an early math error, and then some formula errors.
    The formula for re is
    re = 1/(40 * Ie), where Ie is in amps and re is in ohms. So,
    re = 1/(40 * 0.010) = 2.5 ohms.

    Now re reflects to the base of the transistor as re' = beta * re = 187.5 ohms.

    But that is just your intrinsic emitter resistance. Your emitter resistors also reflect back. So your actual base reflected transistor re' is
    re' = beta(re + RE1 + RE2). You didn't really specify RE1 and RE2. I figure RE2 = 270 ohms and RE1 = 60 ohms. So re' = 25 kohms. Compared to the R1 and R2 values, re' is so large that it is swamped out. You can pretty much ignore it. This is good because it depends on the transistor's beta which is not well controlled.

    So, your effective input impedance is R1 || R2 || re' = 601 ohms.

    I'm just curious, did you pick these values (or get them from someone else), because this is really close to matching but your post doesn't calculate right?

    Now the other problem that I see is your filter calculation which should be 1/(2*PI*R*C), where R is the effective input impedance plus the source impedance. You don't add re', it is parallel. So the frequency you actually calculated is:
    f = 1/(2*PI*(600 + 600)*1.29uF) = 103Hz.

    For 40Hz, you need:
    C = 1/(2*PI*(600 + 600)*40Hz) = 3.3uF.

    You've got CE and CL right.
     
  7. Denny1234

    Thread Starter Member

    Feb 17, 2008
    27
    0
    Thanks for that, ive rejigged my calcs and with those values and typical of my circuit my cut off freq is further off now, around 8hz. Im glad ive got the theory correct though on my lower cut off, just for some reason my multisim isnt working. Do you have multisim?
     
  8. Caveman

    Active Member

    Apr 15, 2008
    471
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    Now, I have a bit of a pulpit to get on here about something. One of the most misunderstood simple concepts in electronics is the whole matching of source and load resistances.

    All of the confusion starts with the Maximum Power Theorem, that states that the maximum amount of power that can be drawn from a voltage source, Vs, with source resistance, Rs, is 1/2*(Vs^2)/R. And that this is accomplished when the load resistance Rl equals the source resistance Rs.

    Everybody takes that statement and just matches everything they can get their hands on.

    Look at this circuit: http://www.ecircuitcenter.com/Circuits/trce/trce.htm.
    Assuming you have a source resistance of 600 ohms and a constant load resistance of 600 ohms. What should you design your effective input resistance and output resistance to be? 600 ohms on both, right? WRONG!

    But if the Maximum Power Theorem is true, how can this be? The reasoning is different for the input resistance and the output resistance.

    For the input resistance, you must remember that you are not amplifying power with this circuit. You are amplifying volts. And so you want the most volts possible out of this circuit. This is done when you load the source as little as possible. So the goal is to maximize the effective input resistance, not match it. If you put a 1mV input into the matched 600 ohms, you would get 0.5mV out. If you put it into a 2400 ohms input, you would get 0.8mV out. That is 4dB more signal, which is 4dB less that the transistor must supply.

    For the output resistance, you may be supplying power, so that logic doesn't work. But, the theorem doesn't tell how to pick Rs, only how to pick Rl with a given Rs. You want Rs to be as low as possible because that means more power is going to the load and less is burned up in a source resistor that needn't be there.

    There are good reasons to match loads, and they are valid.
    One example is an audio speaker. A speaker's loudness is defined by the amount of power that is driven through it. Most are designed for 8 ohms. That is because it is fairly easy to define consistent 8 ohm output drivers. The extra source resistance helps to hide the inconsistencies in transistor output resistances to make the system functionality more consistent and reliable as well as provide some current limiting. But if you want a louder amplifier, drive it with a 4 ohms driver. You will get 30% more power to the speaker.

    Another example where matching is actually important is in high frequency and transmission line applications. This is due to the need to reduce reflections, not the maximum power transfer theorem. This is the primary reason that function generators have 50 ohm outputs (the other is current limiting).

    Read the wikipedia entry for Maximum Power Transfer for more info.
     
  9. Audioguru

    New Member

    Dec 20, 2007
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    896
    No.
    Old vacuum tube amplifiers matched the speaker's impedance.

    Solid state amplifiers for the last 50 years have an extremely low output impedance because they do not have an output transformer and have plenty of negative feedback. A good amplifier has a "damping factor" of 200 or more. Then the output inpedance of the amplifier is 8/200= 0.04 ohms or less. These amplifiers damp speaker resonances very well.

    The output voltage of the modern amplifier does not change if there is no load or if the load is 4 ohms or even 2 ohms for some amplifiers.
     
  10. Caveman

    Active Member

    Apr 15, 2008
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    I have apparently been misinformed on the audio stuff. However, the statement about load and source matching still stands.
     
  11. Audioguru

    New Member

    Dec 20, 2007
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    High frequency signals in coaxial cables have the source and load impedances matched to avoid reflections in the cable.
    Audio circuits always have an extremely low output impedance and the input of the stage it drives is much higher impedance for the best voltage transfer.
     
  12. Caveman

    Active Member

    Apr 15, 2008
    471
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    I agree and that was my point. Everyone gets it when they are doing opamp circuits, but for some reason, whenever the circuit has discrete transistors, a lot of people will try to match the stages. They throw away 6dB every stage while they keep tacking on more to get the total gain up. All because of a misunderstanding of a very simple theorem.
     
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