Calculate Frequency/Q of Transistor High Pass

Discussion in 'General Electronics Chat' started by tactile, Feb 19, 2010.

  1. tactile

    Thread Starter New Member

    Feb 19, 2010
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    Hi All,

    I'm working on a software model of the Roland TR-808 Drum Machine, based on the schematics. I'm having difficulty working out the frequency and gain of this filter.

    [​IMG]

    Any explanation or information would really be appreciated.

    Thanks in advance,
    Jonathan.
     
  2. Ron H

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    Apr 14, 2005
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    I won't attempt to analyze it. Others here might (and I'm sure they can).
    Here is the simulated response.
    I had to add a power supply, and a bias divider on the base.
     
  3. Ron H

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    I suspect you have swapped the 2.7k and the 27k. Here is the result of swapping them.
     
  4. The Electrician

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    Oct 9, 2007
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    Here you go.
     
  5. eblc1388

    Senior Member

    Nov 28, 2008
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    Hi Ron,

    In your above BODE plot, the phase angle is shown positive (e.g. +80° at 300Hz) by LTSpice.

    This got me puzzled if this is equivalent to saying the angle is -280°. Sometimes when I select the frequency response in LTSpice, I would get negative degrees, some even larger than -360° and sometimes positive.

    How do we interpret this value as related to the input signal?
     
  6. tactile

    Thread Starter New Member

    Feb 19, 2010
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    Thanks for taking the time to help.

    The values of the resistors are as posted R3 is 2.7K and R1 is 27K. The filter is part of a cymbal voice, and the high Q is desirable (makes a metallic ping sound).

    Ron, I notice that you've changed the value of R1 (R2 in your circuit) from 68K to 160K. Was that by design?

    Electrician, thanks for the formula for Q.

    I'm really baffled that both of you get ~800Hz as the frequency. I was expecting around 8000Hz! I've attached the original schematic - the filter in question is Q26. I obviously didn't give enough information in my first post, and to be honest, I don't know what's relevant.

    Thanks,
    Jonathan.
     
  7. Ron H

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    I'm not sure what you're asking. In this case, we have a high pass filter. Think of a simple single pole RC high pass. The output voltage will lead the input by nearly 90° at frequencies well below the -3dB frequency, and will be asymptotic to 0° well above the -3dB frequency.

    BTW, you misread the Bode plot. The phase shift at 300Hz is around +165 to +170 degrees, depending on which circuit you were talking about.
     
  8. Ron H

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    It's R1 in my circuit, R2 in yours.
    As I said in my first post, I had to add a bias divider (R1 and R4), because I used a single power supply. The resistance of R1 in parallel with R4 is ≈68k. If I had used two supplies, as in the PDF you posted, I would have simply used 68k.

    The capacitors in the PDF are 1.5nF. In the circuit you first posted, they are 15nF. This accounts for the factor of 10 difference in the corner frequency.
     
  9. eblc1388

    Senior Member

    Nov 28, 2008
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    °
    It's the second Bode plot I'm referring to. I checked again the phase angle is really 80° at 300Hz.
     
    Last edited: Feb 20, 2010
  10. Audioguru

    New Member

    Dec 20, 2007
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    The filter is fed from a fairly high impedance so its response is completely different from being fed from a very low impedance.
    Also the filter is fed into another highpass filter (470pF and 39k ohms in series).
     
  11. The Electrician

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    Oct 9, 2007
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    As Ron H pointed out, the capacitors in the PDF are 1/10 the value you posted here. Taking that into account gives the higher frequency corner you were expecting.

    But, there is more to the circuit that is relevant. On the PDF schematic, R144 and C64 are part of the filter, and you can't separate the response of those two components from the rest of the Q26 circuit because the output impedance of the R144/C64 pair is not zero. That means that since the input of the Q26 circuit you posted in post #1 is not driven by a low impedance (it's driven by the R144/C64 pair), the response is not just the sum (in dB) of the two parts taken separately; you have to calculate the whole thing at once.

    Doing so, we find that the R144/C64 pair adds a real pole to the gain function. It also shifts the frequency of the complex pole pair, and changes the Q of the complex pole pair.

    The denominator of the gain function is now a cubic instead of a quadratic, and getting a symbolic expression for the Q is impractical. However, the Q can be calculated numerically. The pole Q of the circuit you originally posted is 2.51, and with the R144/C64 pair added, the pole Q becomes 1.76.

    In the attachment are shown several responses. The red curve is the response of the isolated Q26 circuit you originally posted, with the cap values corrected.

    The blue curve is the response of the full Q26 circuit, including R144 and C64.

    There is another frequency sensitive circuit following the Q26 circuit; the VR7 circuit, apparently a level control, has a gain function that also varies with frequency. It also has a variable resistor labeled vr7. That one is easy to calculate; it's just an inverting opamp circuit. The green curve shows the response of the VR7 circuit with vr7 having a value of 50k.

    The black curve is the composite of the full Q26 circuit and the VR7 circuit.

    There's another complication. It appears that the Q27 circuit may cause changes in the response of the Q26 circuit as a result of other control signals.
     
  12. Ron H

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    I forgot that the frequency scales were different on the two plots, so I was reading 300Hz on the first one, and 30Hz on the second one. My bad.
     
  13. tactile

    Thread Starter New Member

    Feb 19, 2010
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    Thanks for the responses,

    >> As Ron H pointed out, the capacitors in the PDF are 1/10 the value you
    >> posted here.

    Sorry about that. :X

    >> the response is not just the sum (in dB) of the two parts taken
    >> separately;

    Does this mean that the formula to calculate the frequency is hideously complicated?

    >> The pole Q of the circuit you originally posted is 2.51, and with the
    >> R144/C64 pair added, the pole Q becomes 1.76.

    I got 2.51 plugging the values into the formula you gave earlier, and I understand that the value changes because of the R144/C64 pair, but how did you arrive at 1.76?

    Regards,
    Jonathan.
     
  14. The Electrician

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    Oct 9, 2007
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    It's not too hideous. :)

    To get the pole Q, you have to find the roots of the denominator of the gain function. The Q is a simple function of the complex pole.

    I let R0 and C0 stand for R144 and C64.

    See the attachment for the gain function and pole Q.
     
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