CE - Full Emitter Resistor Bypass - Computing Capacitor Values

Discussion in 'Analog & Mixed-Signal Design' started by elec_eng_55, Aug 23, 2018.

  1. elec_eng_55

    Thread Starter Member

    May 13, 2018
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    I have some more questions in my ongoing saga re CE amplifiers.

    (1) My Heathkit course manual calculates C1 And C2 using the following:

    C1 = 3.18 / (f1 * Rin) = 3.18 / (100 Hz * 3.94K) = 8.07 uF
    C2 = 3.18 / (f1 * RL) = 3.18 / (100 Hz * 5.6K) = 5.68 uF

    Why wouldn't they use 0.159 rather than 3.18 for C1 and C2?

    C3 = 0.159 / (f1 * RNF) = 0.159 / (100 Hz * 5 Ohms) = 318.47 uF

    (2) I came up with a formula for computing the value of RNF:

    RE = 500 ohms
    RL = 5.6K
    RC = 2.5K
    re' = 37mV / 3mA = 12.3 ohms
    Av = 100

    RNF = (RE * ((RC * RL) - (Av * RC * re') - (Av * RL * re'))) / ((Av * RC * re') + (Av * RC * RE) + (Av * RL * re') + (Av * RL * RE) - (RC * RL)) = 5 ohms

    I hope that is correct.

    (3) How do I compute the ac current flowing in RNF as I want to calculate the wattage
    for this resistor.

    I think that I am now in the home stretch!

    David
     
    Last edited: Aug 24, 2018
  2. Jony130

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    Feb 17, 2009
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    Safety margin due to the fact that the electrolytic capacitor has a very large tolerance +/- 20% or more and lose much of his initial capacitance when aging (30% loss in capacitance or more). So this is why they decide to use to this safety factor 3.18/0.159 = 20.

    So, using this equation C = 3.18/(f*R) you get 20 times larger capacitor then you need.

    Also, can you explain to me why didn't to use the standard capacitor values (1; 2.2; 3.3; 4.7; 6.8 )?


    What can I say OMG? What a equation?

    I will do it this way (in steps)

    Desired voltage gain is 100V/V

    Rc'' = 2.5kΩ||5.6kΩ = 1.72kΩ

    Re'' = 1.72kΩ/100V/V = 17.28Ω

    RNF = 17.28 - 26mV/3mA = 8.618Ω

    And becouse RE >> RNF
    I pick RNF = 8.2Ω as a standard E24 value. And the gain in simulation is 98.57V/V, not so bad.

    And why your thermal voltage is 37mV?

    This current will be very small so you do not have to worry about the wattage

    V_RNF_peak ≈ Vin_peak * RNF/(RNF + re) = 1mVp * 8.2Ω/(8.2Ω + 8.67Ω) = 0.48mVp

    But in terms of a power, we are interested in RMS value not in peak
     
  3. elec_eng_55

    Thread Starter Member

    May 13, 2018
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    Hi Jony:

    Thanks for your reply.

    I was just showing the theoretical values first and then I will convert to the standard values once
    I have a good design. At least that was the thought behind it.

    The 37mV came from my old Heathkit course manual but I will rework using 26 mV.

    That crazy formula comes from reworking Av = (Rc//RL) / (re' + RE//RNF) and I must admit that
    I couldn't isolate RNF by hand so I used a website to do the manipulation for me.

    David
     
  4. elec_eng_55

    Thread Starter Member

    May 13, 2018
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    I guess I have one last question (just call me Columbo). I am trying to understand how
    one arrives at the maximum unclipped, lowest THD output voltage that such an amplifier
    can produce. My course material says that this works.

    The lesser of

    vO_Plus = ICQ * rL = 3mA * 1.73K = 5.19 V peak

    or

    vO_Minus = VCEQ = 6 V peak

    whichever is less.

    Therefore the maximum unclipped low THD output should be 5.19 V peak.

    When running a simulation, a Vin of 4mV yields a Vout of 378 mV and the THD
    has just exceeded 1%. There is, however, no apparent clipping.

    Is there a mathematical approach to find the "real" maximum ac output voltage?

    It seems that having a VCEQ as high as 6 volts is not necessary!

    That's me just thinking out loud.

    Thoughts?

    Sincerely,

    David
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Because you still forget that the BJT is a very nonlinear device. And that the voltage gain is not constant but will change with the input signal exponentially.

    Av = Vout/Vin = - (Rc * Is)/Vt * exp^(Vin/Vt) (with CE capacitor across RE ressitor)

    And this is why single stage BJT amplifier produces alot of distortion. We get about 1%THD per 1mV at the base (10% for Vin =10mV) .
    https://www.researchgate.net/public...C_DISTORTION_IN_LOW_FREQUENCY_POWER_AMPLIFIER

    http://www.kevinaylward.co.uk/ee/bipolardesign2/bipolardesign2.xht

    You can improve the distortion by adding the emitter resistor (without CE capacitor). But to get "low THD" you need to pick RE >> re.
    But it is impossible to get high voltage gain and low THD at low load resistance in such a simple circuit.
    As you can see you're expecting too much from this simple circuit. And this is why no one uses this circuit as "low THD" amplifier.
     
    ericgibbs likes this.
  6. elec_eng_55

    Thread Starter Member

    May 13, 2018
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    I have run a simulation with the schematic changed to reflect standard component values (attached). I also
    attached the resulting AC Analysis.

    My hand-calculated analysis (based on Heathkit manual) is as follows.

    Vin = 1mV @ 1 Khz
    Iin = Is = Vin / Rin = 1mV / 4.22K = 237nA

    ie = Vin / (RE + re') = 1mV / (470 ohms + (26mV / 3mA)) = 2.09uA
    ib = ie / hfe = 2.09nA / 100 = 20.9nA
    ic = ie - ib = 2.09uA - 20.9nA = 2.07uA

    Pload = (VCEQ * 0.707) * (ICQ * 0.707) = (6V * 0.707) * (3mA * 0.707) = 9.00mW

    Pin = Pload / Ap = 9.00mW / 100 = 90uW

    Av = (RC * Is) / (Vt ^(Vin/Vt) = (2.5K * 237nA) / (0.7^(1mA/0.7V))
    = 0.0006325 / 0.7^0.00142857 = 0.0006325 / 0.999 = 0.0006

    I am confused with the difference between my results and the LTSpice ac analysis.
     
    Last edited: Aug 25, 2018
  7. ericgibbs

    Moderator

    Jan 29, 2010
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    hi 55,
    Please post, a simulation with the schematic changed to reflect standard component values (attached).
    E
     
  8. elec_eng_55

    Thread Starter Member

    May 13, 2018
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    Sorry .... I forgot to do it.
     
  9. ericgibbs

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    Jan 29, 2010
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    hi,
    As you say thats well out.
    Could you post a clip of the Heathkit manual that shows the equations.?
    E
     
  10. elec_eng_55

    Thread Starter Member

    May 13, 2018
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    They are strewn throughout the text I would have to scan a lot of pages and my
    scan software is not working. Sorry E.
     
    Last edited: Aug 25, 2018
  11. Jony130

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    What you are trying to calculated?
     
  12. elec_eng_55

    Thread Starter Member

    May 13, 2018
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    0

    I have run a simulation with the schematic changed to reflect standard component values (attached). I also
    attached the resulting AC Analysis.

    My hand-calculated analysis (based on Heathkit manual) is as follows.

    Vin = 1mV @ 1 Khz
    Iin = Is = Vin / Rin = 1mV / 4.22K = 237nA

    ie = Vin / (RE + re') = 1mV / (470 ohms + (26mV / 3mA)) = 2.09uA
    ib = ie / hfe = 2.09nA / 100 = 20.9nA
    ic = ie - ib = 2.09uA - 20.9nA = 2.07uA

    Pload = (VCEQ * 0.707) * (ICQ * 0.707) = (6V * 0.707) * (3mA * 0.707) = 9.00mW

    Pin = Pload / Ap = 9.00mW / 100 = 90uW

    Av = (RC * Is) / (Vt ^(Vin/Vt) = (2.5K * 237nA) / (0.7^(1mA/0.7V))
    = 0.0006325 / 0.7^0.00142857 = 0.0006325 / 0.999 = 0.0006

    I am confused with the difference between my results and the LTSpice ac analysis.
     
  13. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Ok I see

    LTspice shows that β = 310; Ic = 3.17mA; hence re ≈ 8.2Ω

    Rin = R1||R2||[ (β+1)*re+(RNF||RE)] ≈ 2.23kΩ

    Vin = 1mV/Rin = 448nA

    Ie = 1mV/ (re + (RE||RNF)) = 1mV/(8Ω + 8.2Ω) = 61.72uA
    Ib = Ie/(β+1) = 200nA

    Ic = Ie - Ib = 61.5μA

    IL = Ic * Rc/(Rc + RL) = 61.72μA * (2.7/(2.7 + 5.6)) = 20μA ; So Vout = 20μA*5.6kΩ = 112mV

    So, the Av is around 112V/V

    Not bad, LTspice shows 107.7

    What you are trying to calculate here is beyond me.


    I think that this is a time for you to throw away your "Heathkit manual" and start to learn form another source. Because what I see is that you are trying to fit the "equation from Heathkit manual" to every circuit you met. And this approach will not work in the long-term.
    http://www.ittc.ku.edu/~jstiles/412/handouts/5.6 Small Signal Operation and Models/section 5_6 Small Signal Operation and Models lecture.pdf
     
    Last edited: Aug 25, 2018
  14. elec_eng_55

    Thread Starter Member

    May 13, 2018
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    Thanks again. I will study this info and you are certainly right about the Heathkit manual.

    Jony:

    Where does β = 310 come from? Can it be calculated without referring to the LTSpice simulation? Or is β just from a 2N3904 data sheet?

    I am actually writing a program in VB to design circuits, the CE being the first. In fact, these are the last calculations to add to the program. I plan to use my program to design many different circuits, then run a sim, if the two are in agreement I would then run another sim using standard component values and then actually bread board the circuit for bench testing.

    David
     
    Last edited: Aug 26, 2018
  15. elec_eng_55

    Thread Starter Member

    May 13, 2018
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    Hi Jony:

    RNF = (RC//RL / Av) *re' = 9.5Ω ≈ 10Ω

    CNF = 0.159 / (f1 * RNF) = 0.159 / (100Hz * 10Ω) = 150uF

    THD = Vin * (re' / (RNF + re'))^2 = 1mV * (8.7Ω / (10Ω + 8.7Ω()^2 = 0.02% ⇒ LTSpice says THD is 0.65%

    David
     
  16. Jony130

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    Feb 17, 2009
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    THD for 1mV at the input will be around 1 * (8.7Ω / (10Ω + 8.7Ω))^2 = 0.216%
    But for 10mV it will become 10 * (8.7Ω / (10Ω + 8.7Ω))^2 = 2.16%
    Try it in LTspice. And do not expect to much from this super simplified formula (2nd harmonic only). No one calculates the THD in more complicated amplifiers.
     
  17. elec_eng_55

    Thread Starter Member

    May 13, 2018
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    In LTSpice it says 0.65%.

    The formula for calculating CNF doesn't even come close to a good value. I don't understand.
     
  18. Jony130

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    Strange I see this results

    2.png


    Which formula ??

    This one RNF = (RC//RL / Av) *re ?

    Try this one

    RNF = (RC//RL / Av) - re
     
  19. elec_eng_55

    Thread Starter Member

    May 13, 2018
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    RNF = (RC//RL / Av) - re is the one that I used and got 9.5Ω ≈ 10Ω

    CNF = 0.159 / (f1 * RNF) = 150 uF

    The THD is then 0.65% in LTSpice.
     
  20. ericgibbs

    Moderator

    Jan 29, 2010
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    hi,
    My LTS result is close to Jony's
    E
     
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