4002 Data

[johnallg]

Dane, lowering the resistor value from 10M will allow more bass note volume to get past the capacitor, ie, louder bass notes. Raising the value of the .01uF cap (like to .02uF) will also allow more lower bass volume, but there will still be a rolling off of bass but it will start at lower bass notes. Reducing the .01uF cap value, like to .0047uF, will effectively reduce bass note volume by raising the point the capacitor starts to affect the bass note's volume. In a nutshell, changing the resistor value will change how much influence the capacitor has on the lower bass notes. Changing the capacitor value will change when the rolloff of lower bass notes begins on the fretboard. Is that clearer?

[cjj]

I think it's interesting that with the 0.01 µF cap the frequencies lost about the Fc because of the (300k) resistor are probably more detrimental in terms of signal loss than the frequencies lost with the cap alone.

CJ, I know we've already wasted plenty of your time but you can also show us what 1 Meg would look like with those caps? And what a 0.047 µF cap would do? Please???

OK, since you said, "Please" i'll do it.

The following shows a 1M resistor in parallel to a series of caps all going to a 500k "volume pot" set for 500k. The first is a 0.0047uF (Vout0, lt blue, lowest trace), 0.01uF (Vout1, red), 0.047uF (Vout2, dk blue), and finally a 0.1uF (Vout3, green):

[cassius987]

Thanks CJ. Makes me want to do a big value series cap in place of a 0.0047 µF some time and see how useable it is.

Dane, lowering the resistor value ... will allow more bass note volume to get past the capacitor, ie, louder bass notes. Raising the value of the cap will also allow more lower bass volume, but there will still be a rolling off of bass but it will start at lower bass notes. ... In a nutshell, changing the resistor value will change how much influence the capacitor has on the lower bass notes.

And the other thing the resistor bypass seems to do is cut more mids than the cap's normal roll-off point would allow, if you look at the graphs. Less resistance seems to enhance this effect. It's a trade-off I guess. Less dramatic cut to the bass frequencies, but some mids drop out as well. That's why I would favor just using a really big cap on a single pickup (out of two) exclusively for the effect it has on comb filtering.

[cjj]

Speaking of comb filtering...
Anyone want to enhance their headache?


WARNING: Techno-Nerd™ ALERT!!!

There's a common misconception that an in-line cap produces a 90 degree phase shift. THIS IS NOT TRUE! Well, not in the way most people think. There IS a 90 degree phase shift between the CURRENT through a capacitor and the VOLTAGE across the cap. What happens to the voltage on the other side of the cap depends on what's on the other side and the frequency of the signal going into it.

For the arrangement where there's an in-line cap followed by a resistor to ground (the volume pot), we have what is called a high pass filter in that it passes frequencies higher than it's cutoff frequency and cuts stuff lower than that to varying degrees (it's not a proverbial "brick wall"). The cutoff frequency is given by:

Fc = 1/(2πRC)

So, for the 0.0047uF cap with a 500k resistor this is:

Fc = 1/(2*3.14159256*500e3*4.7e-9) The notation e3 means "times 10 raised to the 3rd power" or 1000, etc.
Fc = 1/1.476e-3 = 67.725Hz

Great, but what about that 90 degree phase shift? The phase shift is given by:

p = acrtan(1/(2πFRC)

And so, is dependent on frequency! At the cutoff frequency, the voltage phase shift will be exactly 45 degrees, below this, it slowly moves towards 90 degrees as the frequency approaches 0Hz (DC). Above Fc, it slowly approaches 0 degrees as the cap's impedance approaches 0. The cap's impedance is Xc = 1/ (2πFC), so you can see that as F gets really big, the denominator gets really big and the result will get very small. And vice versa.

From the above equations, we see that the voltage phase shift is also dependent on the cap's value. Here's a plot of the phase shift for 0.0047uF, 0.01uF, 0.047uF, and 0.1uF caps all going into a 500k resistor:

The dotted lines which start high on the left side show the phase and are read on the right hand scale. We can see that the 0.0047uF cap is only shifted about 73 degrees at 20Hz, the 0.1uF is a measly 9 degrees. All of them are down to less than 5 degrees by 2kHz.

So, by using a bigger in-line cap, you gain more low frequencies but you also lose the phase shift you were getting. So, does that mean you'll end up with all of the comb filter effects that you have with no cap? No, but the bigger the cap, the less effect it will have. In other words, it's gonna sound different with different size caps.

So, has your head split wide open yet? What happens if we add that 1M resistor back in?

Oh cool! That looks different! But wait! The maximum phase shift is now limited to 30 degrees! For the 0.0047uF cap this happens at about 60Hz, whereas in the first plot, without the resistor, we had about 49 degrees of shift at 60Hz for the 0.0047uF cap. This just makes it worse...

END of Techno-Nerd™ ALERT!!!
You can wake up now...

[bassduke49]

Must . . . resist . . . continually . . . opening . . . this . . . thread!

[ken_j]

CJ how does the input impediance of the amp effect all this?

[cassius987]

Interesting, so there's even a trade-off with cap size and phase shift. Wonder why I never thought of that being possible. I guess a 0.01 or 0.02 µF cap is a pretty good compromise overall (between retaining bass frequencies and getting away from heavy filtering).

Regarding input impedance I know I surely hate the sound of the 0.0047 µF cap on the bridge pickup plugged into a low impedance input. Talk about telephone sound.

[cjj]

OK, the amp's input impedance will essentially look like another resistor tied in parallel to the volume pot. So, in all of the above calculations, R would become the parallel combination of 500k and the amp input:

Rp = (500k * Rin)/(500k + Rin)

So, if the amp's input is 500k, it would effectively work like 250k.

The cutoff freq will respond according to the formula given above, Fc = 1/(2πRC)

Here's the plots for a 0.0047uF cap, the 500k "pot" and input impedances of infinite (lt blue, Vout0), 1Meg, 500k, and 250k (green, Vout3):

And sure enough, as R (the parallel combination) goes lower, the denominator gets smaller and Fc goes higher. The phase shift also moves higher in frequency to match.

[cassius987]

In that case I would say the 500k value used for most of the previous examples is probably not very applicable to a Ric or most other passive basses. Generally the final impedance load is 125-250k (from one or more pots valued 250-500k; e.g., P Bass is 250k, Jazz and 4001 is 125k, newer 4003 is 165k). The average amp input impedance seems to be 500-1000k nowadays, as well. I suspect the final load for most players is something like 90-100k, effectively shifting most of the curves higher in frequency. In my case, when I'm using a 4003, I get 143k.

Assuming the pot values are what they claim to be which of course is not true...

[ken_j]

Another factor is the capacitance of the cord.

[bassduke49]

Another factor is the capacitance of the cord.

Ohh, NOOOOOOOOOOOO!!!!

[jps]

Another factor is the capacitance of the cord.

Ohh, NOOOOOOOOOOOO!!!!

This info should be in your book, Paul.

[johnallg]

Another factor is the capacitance of the cord.

Ohh, NOOOOOOOOOOOO!!!!

This info should be in your book, Paul.

Yeah, you're getting all of this, right Paul??

[aceonbass]

John.....Thanks for your clarification. so lowering the Resister value will increase bass response, but at what point does it stop increasing I wonder? A conclusion I could draw from that is that having no resister would be even better. Of course a lot of this is subjective. One thing I know for sure is that the tone is a bit clearer than before in my 4003 with totally stock 4002 wiring, and that all three pickups together sound better. In fact, varying the tone controls on one pickup or the other in this configuration gets me a lot of very usable tones. I just wish I knew what caused the volume drop off. Maybe the volume drop off is really just a tone drop off that's so dramatic that it results in a drop in volume. On a different note, I pulled the ROS output wiring from the bass since I have no desire to ever run the bass in stereo, and added a standard mono Switchcraft output jack. I'm wondering if adding a lo-Z jack, like the 4002 has, is possible. In this case, I'd jump a signal off of the mono jack over to the lo-Z jack. I'm wondering if this would be too hot for a standard lo-Z or XLR input jack on a mixer or recording console? What would the 2nd ground be run from for the 3rd pin on the jack?

[johnallg]

John.....Thanks for your clarification. so lowering the Resister value will increase bass response, but at what point does it stop increasing I wonder? A conclusion I could draw from that is that having no resister would be even better. Of course a lot of this is subjective. One thing I know for sure is that the tone is a bit clearer than before in my 4003 with totally stock 4002 wiring, and that all three pickups together sound better. In fact, varying the tone controls on one pickup or the other in this configuration gets me a lot of very usable tones. I just wish I knew what caused the volume drop off. Maybe the volume drop off is really just a tone drop off that's so dramatic that it results in a drop in volume. On a different note, I pulled the ROS output wiring from the bass since I have no desire to ever run the bass in stereo, and added a standard mono Switchcraft output jack. I'm wondering if adding a lo-Z jack, like the 4002 has, is possible. In this case, I'd jump a signal off of the mono jack over to the lo-Z jack. I'm wondering if this would be too hot for a standard lo-Z or XLR input jack on a mixer or recording console? What would the 2nd ground be run from for the 3rd pin on the jack?

The lower the resistor goes, the closer to eliminating the capacitor you get. A zero ohm resistor (wire) gets the capacitor out of the circuit altogether, like when we bypass the 4003's .0047uF cap. Having no resistor would lose the most lower frequencies, it would be just the cap. Liken it to the .0047uF cap in-circuit on a 4003.

As to the lo-Z XLR out on a 4002 and trying to do it without a separate winding on the pickup, you would have to use a step-down transformer right off the pickup coil, custom wound to match the coil impedance and have an output impedance of 600 ohms. The transformer output winding would be the + and - of the XLR, with ground coming from the ground circuit.