Passive VT resistance question (light math maybe?)

Discussion in 'Pickups & Electronics [BG]' started by cdlynch, Feb 5, 2018.

  1. In the most simple passive VT circuit, where both 250K pots are dimed, how would you simulate that system with no pots and one RC filter? There's still some HF bleed to ground with the controls all the way up, but is it just HF bleed from the tone pot, or is there a bit more from the the volume pot as well? I think there must be some HF bleed from the volume, otherwise people wouldn't use no-load pots for volume.

    I did this for a custom project with no knobs, and I went with 250K for my resistor in this schematic, but it doesn't account for the addition of a volume pot. If the volume was another tone contol, I'd assume I'd want 125K for my single value, but since the volume doesn't have a capacitor path to ground, I assume that changes the calculation, and I'm not sure how to measure it. Any ideas?

  2. Crater


    Oct 12, 2011
    Dallas, TX area
    Hot wire from the pickup would go to a 250 k ohm resistor in series with a 0.047 capacitor in series with that, and the far end of the cap tied to ground. Then another 250 k ohm resistor between 'hot' and 'ground'. I don't think using a single 125 k Ohm resistor will sound the same.


    R1 & R2 = 250 k Ohm
    C1 = 0.047 or 0.05 microfarads
    cdlynch likes this.
  3. Thanks @Crater, that diagram makes it much easier to visualize. So I'm trying to figure out how to make what you've drawn sound the same as an identical circuit, but without R2, and without that path to ground, so just hot to R1, to cap, to ground. If the cap wasn't there, the effective resistance would be half, and I don't know how the cap changes that, but I still assume there'd be a reduction of resistance because of the parallel resistors. Is that wrong?
  4. Crater


    Oct 12, 2011
    Dallas, TX area
    Changing R1 to a smaller value and omitting R2 will be like having no volume control, but with the tone control on 5.

    You asked for an exact "no knobs" replication of single-pickup bass electronics, that's what I provided.

    Why the obsession with eliminating a single fixed resistor? They cost pennies and are small enough to fit in a control cavity. Using a single resistor and cap won't give the same EQ curve. It might be close enough for your use, but there's no simple math to calculate this because the source (pickup's) impedance and the load (amp's) impedance affect all this. Not every circuit can be simplified down to two components, because believe me, if they COULD, they WOULD and would pocket the savings from reduced parts count and labor to install them.

    Get to figurin', then. I'm out.
    cdlynch likes this.
  5. I see now that the solution isn't to find a lower resistance, but to add a second 250K resistor to ground. That will work well in my current switch layout. I started with the assumption that a single resistor solution would be easier, because I am very new to electronics, otherwise I wouldn't have asked. I didn't mean to annoy you. Thanks for taking the time to reply.
  6. The resistors are no longer strictly in parallel in the diagram posted, since there's a capacitor in series with one of them. You now have an impedance rather than pure resistance in your path to ground, meaning that total voltage drop depends on frequency, whereas resistors have the same effect at any frequency.

    Edit : to clarify, the two paths to ground are in parallel still. Don't want to cause confusion.
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  7. Thanks, impedance is still a tricky concept for me, but being able to relate it to this project will help me when I look it up again Right Now. ;)

  8. Here's a link to a really good article on capacitors and how they work in terms of reactance (which is resistance for a capacitor, really - it is measure in ohms as well)

    Capacitive Reactance - The Reactance of Capacitors

    The quick breakdown of the posted circuit : As you go up linearly in frequency, the reactance of the cap goes down linearly. This means the higher the frequency, the lower the resistance to the ground path in the left branch of the diagram.

    At DC (frequency = 0) the cap essentially is an open circuit, and at very high (infinite if you will) frequency the cap starts to act like a wire.

    I would, however, argue that this circuit will have absolutely no effect on your signal! You haven't actually formed a filter here, which means that the signal at input will be identical in voltage to the circuit's output.

    I put it through LTSpice to confirm my hypothesis and yes, the voltage frequency response plot of that circuit is flat.
    cdlynch likes this.
  9. Interesting and confusing... Is it not an accurate model of a simple VT circuit? Shouldn't it bleed some HF to ground even at full resistance? My tone circuit as it stands exists without a volume control, and can be set for no load, max load (dimed), and min load (full LPF). If I add another 250K resistor to ground, would that just bleed off some volume and not change the HF content? Without the extra resistor it works as expected.
  10. JKos

    JKos Supporting Member

    Oct 26, 2010
    Surprise, AZ
    With a real source (a pickup), it certainly is. When you modeled it, did you include a model of a pickup as the source? A pickup has resistance and inductance. If you simply used a perfect source, then, no, it wouldn't behave like a filter.

    Check this out. CircuitLab - Editing "Guitar Pickup"
    pickup model.PNG

    - John
    Last edited: Feb 5, 2018
    cdlynch likes this.
  11. Clever. I only modeled it with a simple AC sweep, ignoring inductance entirely. Maybe I'll try it once I'm home. However, even with an accurate pickup model, all the HF bleed even be noticeable? I personally doubt it. Again, will check once home!
  12. Okay, here's a response plot using the pickup model suggested by @JKos
    Blue line is the pickup's direct output in response to a sweep;
    Red line is the output using one 250k resistor in series with a .022uF cap to ground;
    Green line is output with the 250k/.022uF in parallel with a 250k, again to ground.

    There's around a 3dB difference between the pickup and the proposed circuit (green) right around 2.5kHz. To our ears, a small but easily noticeable high-end roll off. The difference at 10kHz is roughly 8dB, or nearly half the perceived volume.

    @cdlynch looks like we're both learning today :D

    Edit: Focus on the solid lines - the dotted lines indicate phase shift.

    Attached Files:

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  13. It's absolutely audible. Try a no-load tone pot and you will hear a little extra brightness for sure. There is an obvious little jump in brightness when the pot reaches full clockwise. Or just tack in a switch inline with the cap. You can switch the load of the tone pot and cap in and out.

    If you are simulating this, make sure you also add the load of the input z, plus the capacitance of a typical 4m guitar lead.

    Of course, the problem with spice and these things is that it only really gives you a potential frequency response. It's an AC sweep creating your graph, not a plucked bass string. If you don't have a horn in your bass cab, or you play flatwounds for example, this stuff is purely academic.

    And speaking of academic,
    You'll struggle to find 250k resistors, especially in small quantities, but 240k or 270k will be fine.
    Last edited: Feb 6, 2018
    cdlynch likes this.
  14. Awesome, so adding the extra resistor to my tone control will better simulate a normal circuit. If you're still in a simulating mood, it might be interesting to compare the green line with the red, but with a lower value single resistor.

    Yep, I went with 200K and 50K in series.

    Here's a before and after; the left is the circuit in my bass now:
    There's a difference between no load and LPF min, but I thought it should be a bigger difference. R2 simulating a volume control goes the extra mile, so I just have to pop R2 in there. The simulation shows adding that only drops the volume by under half a dB, not bad.

    Thanks for your help and insights!
  15. Of course it's audible. That's what we both said.

    As far as "academic", the simulation legitimately illustrates the difference between having the circuit in place on the output and not. The difference between the two still applies to reality.

    Good thinking on the cable and amp input. Unfortunately I can't really sim the amp input given that I don't know what amp being used, not to mention lots of amp manufactures won't list such a spec in a way that could be used.
    cdlynch likes this.
  16. And here's a graph with a 120k ground resistor compared to the other 3. Colours have changed, the names of each line will tell you what's what.

    Attached Files:

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  17. Ground resistor - does that mean there's a 250K + cap on one leg, and a 120K leg to ground? Or is it 120K + cap and no second leg? The latter is what I'm curious about. I expected the shape to be different without a parallel path to ground based on what Crater was saying.
  18. That was the first of the two you mentioned here.
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  19. Cool, thanks. That makes sense that it'd roll off more with the same curve, and almost 1 dB lower volume.
  20. Ooooo, maybe I should splurge on these fancy fellows!