Output Impedance of Emitter Follower

Discussion in 'Analog & Mixed-Signal Design' started by bradwhitlock, Jun 17, 2017.

  1. bradwhitlock

    Thread Starter New Member

    Feb 28, 2005
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    When I calculate the output impedance of the following circuit, I get about 8 Ohms at 1kHz. However, when I simulate the design, in order to attenuate the output by 1/2, I have to use a 16 Ohm load. My understanding is that if I match the output resistance to the load, I should get 1/2 the unloaded voltage swing.

    The transistor model states Beta of 300. My thinking is that Rsource is dominated by the parallel combination of R19 and R20 which is in series with C1, R18 has minimal effect. I get 2.5k Ohms. 2.5k / Beta = 8 Ohms. So, I am expecting that a load of 8 Ohms will yield a voltage amplitude of 50mV, but I actually need 16 Ohms to achieve 50mV swing.

    What am I missing here? Why does the simulation not agree with my calculations?

    Unloaded Circuit:
    Screenshot from 2017-06-16 22-21-35.png

    Circuit loaded to get 1/2 the voltage swing:
    Screenshot from 2017-06-16 22-07-54.png
     
  2. DickCappels

    Moderator

    Aug 21, 2008
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  3. bradwhitlock

    Thread Starter New Member

    Feb 28, 2005
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    Thank you for the quick reply, I'll take a look at the thread you referenced
     
  4. Bordodynov

    Well-Known Member

    May 20, 2015
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    See
    Calc.png
     
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  5. Bordodynov

    Well-Known Member

    May 20, 2015
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    The difference in the calculations is due to the fact that in the theoretical calculation I did not take into account the complex (imaginary) character of the resistance of the capacitor. I am sure that taking this into account will give the same result. But you have to mess with complex numbers.
    Rs=900-j1600
     
    Last edited: Jun 18, 2017
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  6. DickCappels

    Moderator

    Aug 21, 2008
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    So...the impedance is about twice the real part, about what @bradwhitlock observed. Very good.
     
  7. bradwhitlock

    Thread Starter New Member

    Feb 28, 2005
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    Thank you Bordodynov

    My take-away is that I failed to consider re which you found to be 11 Ohms. Also looks like you found Beta to be 330, which also affects the calculation a bit.

    I'd like to switch gears and ask another question. I'd like to create a low pass filter on the output of the circuit. I want a -3dB point of 200kHz. So, I calculated Rs at 200kHz, and I find that to be about 3 Ohms and using re of 11 Ohms as you found I get output impedance of 14 Ohms.

    So I calculate a value for a capacitor to shunt the output to ground with a value of 57n. That should give 14 Ohms at 200kHz.

    When I run the simulation, I get -3dB at 100kHz instead of the expected 200kHz. So, again, I have a factor of 2 error. I verified that a cap with value of 28n gives me -3dB of 200kHz, which would lead me to believe that my output impedance at 200kHz is actually about 28 Ohms.

    Am I to believe that internal capacitance in the BJT is on the order of 28nF? That seems high to me, but that would account for the extra 14 Ohms which would explain the factor of 2 error that I see in the simulation.

    If not, what am I doing wrong here?

    Screenshot from 2017-06-17 01-41-22.png
     
  8. Bordodynov

    Well-Known Member

    May 20, 2015
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    See ie ==> Re=29Ohm
    Test_beta2.png
     
  9. MrAl

    AAC Fanatic!

    Jun 17, 2014
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    Hello there,

    Unless you plan to run a purely resistive load you've made a mistake grande.

    First, you should be testing with a large capacitor in series with the resistive test load. That prevents the DC bias point from changing once the load is connected.

    But even more important, you cant expect to get an 8 ohm output impedance looking into a 6k resistor combined with an NPN emitter. That is because the emitter can not sink any current it can only source current, so if your load is resistive it might work, but when you go to a real load like a speaker with a series capacitor (as is also required) the amplifier will not be able to drive the somewhat reactive load properly, and that is because it needs to be able to sink as much current as it can source. Without that ability the output impedance will be asymmetrical and that means it's not really a true impedance. This means if your application requires +2ma peak then the circuit also needs to be able to sink 2ma. Obviously with a 6k emitter resistor and approximately 6vdc output voltage bias point it can not sink 2ma, and that's not very much current for an 8 ohm load. If the output can not sink enough current the output will be distorted. What this means is the emitter resistor needs to be lowered, and the amount will vary depending on the output voltage peaks.

    So to start, add a series output capacitor of a large value in series with the load resistor. It has to be a high enough value so at the test frequency it does not drop much voltage. At 1KHz probably 2000uf is good with maybe down to a 4 ohm load.

    Remember to look for distortion on the output.
     
  10. bradwhitlock

    Thread Starter New Member

    Feb 28, 2005
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    Thank you MrAl, your comments have helped me to understand issues I have seen in the past.

    Thank you again Bordodynov, I took more time to understand those .meas statements as I have never used them before. I see that I can run the simulation and then see the output from the .meas statements in the Spice error log (View>>Spice Error Log). This is an epiphany!

    Also, hooking up the current source, with AC amplitude of 1 and then measuring the voltage at the emitter actually yields the impedance. Another epiphany!

    With these new tools, I can validate my hand calculations and figure out what I am doing wrong.

    Thank you again!

    Best Regards,
    Brad
     
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