A Passive Equalizer Circuit

Discussion in 'DIY Pro Audio Forum' started by NewYorkDave, Feb 3, 2004.

  1. NewYorkDave

    NewYorkDave Guest

    Not content to build a Pultec copy, I've been thinking of my own circuit for a passive EQ. My criteria are low loss and lack of undue loading on the device preceding the EQ. I also wanted something that could be built small, with an uncluttered control layout--suitable, perhaps, for building into a console. Here's the general outline of what I've come up with:

    [​IMG]

    As you can see, this is more of a theoretical construct than a DIY project, unless you like to do the math yourself.

    The insertion loss of a passive equalizer is equal to or greater than the maximum amount of available "boost"--in other words, you're not really boosting, you're just decreasing the insertion loss in a particular frequency band. I reasoned that since I rarely use anything near 20dB of boost when equalizing, it would be wasteful to throw away that much signal in a passive EQ implementation.

    This EQ is unusual, I believe, in that it uses the input impedance of the device connected after it as part of the voltage divider. Or, if running into a very high impedance (such as the grid of the tube), RL could be a resistor included in the EQ itself. The input impedance of the EQ never falls below the value of RL as long as the resonant frequencies of Z1 and Z2 are spaced far enough apart. In this condition, the input impedance ranges from RL (at maximum boost) to 4RL (at "flat").

    The unusual "two part" construction of the series and shunt resistance arms is because with a conventional voltage divider (one continuous string of resistors), RL would have to be 10X or greater the total value of the shunt arm. This circuit allows RL to be of a much lower value, about one-third that of the series arm. The resistances in the shunt arm are selected to give the appropriate attenuation when placed in parallel with RL. The formula for attenuation at frequencies where the reactance of Z is very low compared to RL is:

    20 * log (Rshunt / Rshunt + Rseries)

    (Rshunt = shunt arm in parallel with RL).

    I'm too much of a lazy **** to calculate actual resistor values now. You get the idea, right? :D

    The basic operation of the EQ is thus... When in boost, the reactive element (Z1 or Z2) shunts some or all of the series arm and reduces the insertion loss in that frequency band. In "flat" the reactance is out of the equation, and the circuit acts as a simple 10dB voltage divider. When in cut, the reactance acts as a frequency-dependent shunt across the lower arm of the voltage divider, increasing the insertion loss beyond 10dB in the appropriate band of frequencies.

    Since I never seem to find the time to actually build this thing, I thought I'd put the basic circuit concept out there for review and comments. Thanks.
  2. bluebird

    bluebird Guest

    Hey very cool. so then I assume you would have a selector for the filters also? or would this be more of a graphic type of eq?
  3. bluebird

    bluebird Guest

    Now that I think about it... I could incorporate this metod into the pultec style eq I'm making.

    with this I could combine the mid cut and high boost sections together...and GAIN a hi cut. Id need a 2 pole 11 position for slecting 11 different filters. and then a 1 pole 12 pos. for gain cut and boost...

    DAMN I have already drilled 42 holes in the front of my chassis...

    I'm very excited about this idea. I'm almost done setting up the six tube gain makeup stages, mounted all output TX's. power supply mounted.

    Dave you may be able to see if this work sooner that you thought [​IMG]
    Last edited by a moderator: Mar 31, 2014
  4. gyraf

    gyraf Guest

    This looks right - it's more or less what I do in our G14 parallel-passive eq.

    Use a relatively low R(Load), or you run into VERY large inductor values. But not so low, that you can't drive it from the previous stage - remember, that at max. boost, the input will be looking directly into R(Load) at resonance frequency.

    To be able to do the shelving-switch-stunt, you'll need both an individually optimized inductor AND a series capacitor for every single frequency. Possible, but expensive.

    Jakob E.
  5. NewYorkDave

    NewYorkDave Guest

    Guys, thanks for the comments. There are two things I should mention that might not be obvious from the drawing. First, there is only ONE set of resistor strings, not one set per switch. Both switches connect across the same set of resistors. That's why it's important for the resonant frequencies to be spaced fairly far apart. (This is usually the case in a practical treble/bass equalizer, anyway).

    Also, the impedance of Z1 or Z2 not only needs to be RL/10 or less at the nominal cut/boost frequency, but it should be rather HIGH at mid-frequencies.
  6. cjenrick

    cjenrick Guest

    Get a rough schematic and put it on a simulator
  7. NewYorkDave

    NewYorkDave Guest

    I don't have one. I "simulate" my circuits the old fashioned way--first with paper and calculator, then in real life. I'd be willing to play around with simulator software, but I'm not willing to spend money on something that I would never really trust, anyway.
  8. NewYorkDave

    NewYorkDave Guest

    All right... I downloaded the student version of Circuitmaker. I need to spend more time to learn the program--time which I really don't have--but I did manage to run some successful simulations. The results look good. It'll just be a matter of calculating the particular component values.

    By the way--for anyone else here who uses Circuitmaker--does you know how to get the Bode plot to read out in dB instead of directly in voltage? Also, is there a more convenient way to read the circuit's input impedance besides inserting a resistor in series with the signal generator and measuring the voltage after it?

    [ February 04, 2004, 05:30 PM: Message edited by: NewYorkDave ]
  9. PRR

    PRR

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    Beyond a few wide-space low-Q networks, your load impedance has to be VERY low (like a summing junction) or you will get unexpected interaction and dips between peaks. Even the simple R-C bass-treble control has some extreme kinks from bass/treble knob interaction.

    > how to get the Bode plot to read out in dB instead of directly in voltage?

    In PSpice, you can either use Marker, Advanced, dB; or on Probe you can change "V(out)" to "Vdb(out)".

    > read the circuit's input impedance besides inserting a resistor in series with the signal generator and measuring the voltage after it?

    Put a voltage and a current marker on the input wire, have Probe plot "V(in)/I(in)" (or "V(in)/1000*I(in)" to get results in KΩ). In some implementations, you can't put a current marker on a wire, only a pin; insert a zero-volt batter or a 0.0001Ω resistor and mark the current on its pin. (Hmmm... may be possible to get the current out of the signal generator.)

    Don't know how this translates to CircuitMaker.
  10. bluebird

    bluebird Guest

    How does this differ from the pultec mid eq circuit? I guess for the cut it would be different.

    http://www.gyraf.dk/gy_pd/pultec/meq5.gif

    Where you have the 1K Load for the bass boost and the 5.1K load for the hi boost.
  11. NewYorkDave

    NewYorkDave Guest

    I did manage to figure out how to get a dB scale. It was just a matter of clicking on an icon and opening up an options dialog in simulation mode.

    I simulated the circuit as a simple high-low shelving EQ (widely-spaced low-Q networks, like PRR says) and it "works" exactly as my calculator, pencil and paper predicted :D . This made me feel good because calculating the values made for a damned tedious couple of hours. (I'm slow at this stuff).

    I added series resistance to the low-frequency inductor, but otherwise didn't try to account for parasitic effects. If I were being really hardcore about it, I could add a tiny inductance and capacitance to the high-frequency shelving cap and also a tiny cap across the inductor windings--but I don't think these parasitics are going to be worth worrying about in the audio band (he said, quite confidently ;) ).
  12. bluebird

    bluebird Guest

  13. NewYorkDave

    NewYorkDave Guest

    The tedious part was calculating the values of the resistors in the voltage divider. The upper arm was easy, but the lower arm was a pain because the parallel resistance of the load must be part of the equation. Like I said, I'm not a math wiz.
  14. NewYorkDave

    NewYorkDave Guest

    For Circuitmaker users (and possibly other SPICE users as well--I don't know how universal the file format is), here's a "rough draft" of the circuit. It's a .ckt file, but I saved it as .txt so that it would display in a web browser. Simply save the file and change the extension to .ckt if necessary:

    Equalizer

    The resistor values were calculated, but the inductors and caps were just chosen arbitrarily to give usable-looking curves on the simulator. For convenience in calculating, I normalized the values to 1K--but it could be adjusted easily to another impedance (600 ohms, for example) by simply scaling the components appropriately.

    The resistor values represent the exact numbers from the calculations and, of course, one would have to use the nearest standard resistor value in real life. The dB figures are nominal, anyway, especially in the "boost" positions, since it's limited by the generator output impedance, the DC resistance of the inductors, and the Q-broadening resistance added to the midrange circuit.

    The circuit is drawn as a high-low shelving with peaking midrange EQ. It's drawn with all three bands in maximum boost. To change the settings, simply snip one of the wires and reconnect it to the appropriate point on the voltage divider. One end of the resonant circuit must always be connected to the output (the high side of R7). Be sure to read the output at this point as well.
  15. bluebird

    bluebird Guest

    I'm not sure if I did this right but on your simulation with the 1K signal I got an amplitude change from 750mv boost down to 200mv cut.
    when I applied a 5K signal the top of the boost was only at 500mv and the cut only went down to 250mv.

    I'm assuming you had the filters set for a 1K resonance?

    so from 1K to 5K there only seems to be a 250mv difference in boost. and less on the cut side.

    will this be substantial in a real world situation?

    When testing the filters using the little circuit CJ mentioned, by putting a 1K resistor in series with the cap/inductor filter, and placing the scope across the 1K resistor I was able to see, by sweeping the signal generator where the resonance point of the filter was.
    But it seemed very subtle like on your simulation. very wide Q. looking at it with my eyes I would not think this would sound like much...

    still learning...
  16. NewYorkDave

    NewYorkDave Guest

    No, that's totally wrong. Are you using Circuitmaker? I created the file at home and mailed it to myself at work, and it ran the same here as it did at home.

    The simulation should be set up for a frequency sweep from 10Hz to (if I recall correctly) 50kHz. It's not a single-frequency simulation. Also, the difference between maximum cut and boost is 22dB, which is pretty substantial.

    I created some screen shots of curves, but I can't post them yet because I can't access my file hosting account for some reason.

    By the way, I got a more useful-looking set of midrange curves using .47uF, 100mH and 100 ohms in series for the midrange circuit. I'll update the ckt. file with that data as soon as I can access my account again.
  17. cjenrick

    cjenrick Guest

    sometimes you have to use a different series resistor value when sweeping filters, depending on the reactance of the cap or inductor at resonance. a 10k might be better,
    experiment.

    if you build up voltage to the the center frequency from the bottom end, and then as you sweep past the center towards the higher freqs, and you see that the voltage does not go back down, that means you have too much capacitance in the indutor, talkin bout real world, not simulator of course...
  18. bluebird

    bluebird Guest

    Oh my bad, I WAS doing it totally wrong. I wasn't aware of the sweep thing. Still learning this stuff, go easy on me.

    I'll try again...
  19. NewYorkDave

    NewYorkDave Guest

    Don't feel bad. I'm new to this program (and simulators) myself, and I'm still stumbling my way through this.

    The updated .ckt file is available here:
    Right-click and "save"

    Here are some example curves:

    All controls on full boost
    [​IMG]

    Mid flat, bass and treble full boost
    [​IMG]

    Bass and treble flat, mid full boost
    [​IMG]

    B and T flat, mid full cut
    [​IMG]

    B and M flat, treble full cut
    [​IMG]

    Mid flat, B and T full cut
    [​IMG]

    B and T full boost, mid full cut
    [​IMG]

    T and mid flat, B full boost
    [​IMG]

    Of course, there are many other possible combinations. I used full boosts and cuts here just to give you an idea of the range.

    You'll notice that the boosts never quite achieve zero insertion loss. That's because of the output impedance of the signal generator (which I set to 100 ohms in the model). If you drive it with a lower impedance source--say, the output of an op-amp follower--you get a lot closer to zero insertion loss at full boost. It will still never get to 100% at the low end because of the DC resistance of the inductor. (I obtained the value of 75 ohms for the 1H inductor from the Wilco online catalog).
  20. Kev

    Kev Guest

    Are these curves from CM2000 ??
    ... or is this another Spice simulator ?

    Good stuff Dave ... I want to see more of this sort of thing in the future. [​IMG]
    Last edited by a moderator: Mar 31, 2014
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