Why 1 meg pots?

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timtam
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Re: Why 1 meg pots?

Post by timtam » Sat Jul 06, 2019 6:34 pm

oid wrote:
Sat Jul 06, 2019 1:39 pm
The Jaguar's extra resistor is part of the strangle switch, it is the R in the RC circuit along with one side of the tone pot, they are parallel, if memory serves correctly it is completely out of circuit when the strangle is engaged and has no effect.
We mused on the 56k resistor here ...
http://www.offsetguitars.com/forums/vie ... 7&t=109553
Note that the schematic I found and linked there has the tone and volume pots reversed in the rhythm circuit. It's the tone pot that is 50k.

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gusgorman
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Re: Why 1 meg pots?

Post by gusgorman » Sat Jul 06, 2019 11:08 pm

oid wrote:
Sat Jul 06, 2019 12:34 pm
The 1Meg pots in the lead circuit are there for exactly as previously stated, brightness, the higher value pots loads down the pickups less so more high frequency gets through. The simple explanation is that resistance (impedance) is not constant over frequency, that 1Meg is 1Meg at DC, frequency of zero, when we feed it an AC signal like the output of a guitar pickup we get a different resistance for our E2 6th string than our E4 1st, so we get different amplitudes for different frequencies. Despite popular belief, turning your volume down does not make it mimic a lower value pot, if you measure the resistance between the outer two lugs of a pot you will see that the resistance is always the same regardless of the knobs rotation and this is the value the pickup sees, a parallel resistance of constant value to ground. There is some effect caused by the fact that the output of the guitar sees two varying resistance, one in series with the pickup and one to ground, but this again has no relation to a lower value pot, it does have more effect on the sound than the other end, but it still is not mimicking the lower value pot, the overall resistance remains constant. That is purely academic anyways, all that matters is that we loose some high end as we turn down the volume and larger value volume pot will give us more highs assuming everything else stays the same.

The 50k pot in the rhythm circuit is because it is not your standard tone control. Most guitar tone controls sweep the cutoff frequency of the filter, so with the tone rolled back all the way the filters cutoff may be 100hz, all the way up the cutoff is essentially infinity and we can sweep that frequency anywhere between those two extremes, it is a variable frequency lowpass filter, the amount of cutoff is constant but the frequency at which the cutoff happens is variable. In the rhythm circuit that filter is fixed in frequency at about 160hz (E3) for the JazzMaster and 320hz (E4) for the Jaguar, as we roll back the tone we vary the amount we cut the frequencies above that cutoff point, it is a fixed frequency lowpass filter, the amount of cutoff is variable but it always happens at the same frequency. So if you change the value of the 50k pot you change the cutoff frequency of the filter, we can easily calculate new values with:

F=1/(2piRC)
F in Hertz
R in megaOhms
C in microFarads
Thankyou for that! Very interesting. I never knew that about the tone control on the rhythm circuit.

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jorri
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Re: Why 1 meg pots?

Post by jorri » Sun Jul 07, 2019 5:58 am

oid wrote:
Sat Jul 06, 2019 1:39 pm
My plot is to show the difference between the filter responses, not the frequency response of the entire system. The plots you posted look to be the plotting the effects the circuit has on pickups own response, they do not in anyway reflect the actual output of the guitar, the guitars tone circuits can and will completely remove those resonant peaks, they are only there as shown when the tone and volume are at max and other than the no-load plots do not say what the load is they are driving which also affect that resonant peak.

There is a resonant peak on the rhythm circuit as well but it only shows itself when the tone control is fully open. The volume drop of the rhythm circuit is only above the cutoff of the filter, this has an effect on the volume of the notes below the cutoff since those notes have harmonics which are above the cutoff point. The volume drop is simply because you are filtering out signal, there is no drop if the tone is full open.

The Jaguar's extra resistor is part of the strangle switch, it is the R in the RC circuit along with one side of the tone pot, they are parallel, if memory serves correctly it is completely out of circuit when the strangle is engaged and has no effect.

You seem to be confusing the pickups response with overall response and the filters response, they are all separate things.
disagree on both those counts.

the resistor is connected in parallel to the half of the tone pot that's in the signal chain, in order to change the taper and resistance of such section. there is nothing connected to the strangle. Its always in there! I've installed strangle switches with no resistor and they are pretty much the same! What it may do is ensure that there's more consistency with the strangle switch, but when the tone is turned down it affects that too.

they simply aren't separate things: that's the point- its an interconnected circuit, so why track half the circuit only- if i only track the resistance of a tone pot, i wouldn't get a tone pot at all, and i could do that lol. Its not a simple RC filter, but the pickups change such filter's response, acting as resistance and inductance load. but yes, its the overall response on those graphs i posted.

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jorri
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Re: Why 1 meg pots?

Post by jorri » Sun Jul 07, 2019 6:06 am

*
timtam wrote:
Sat Jul 06, 2019 6:34 pm
oid wrote:
Sat Jul 06, 2019 1:39 pm
The Jaguar's extra resistor is part of the strangle switch, it is the R in the RC circuit along with one side of the tone pot, they are parallel, if memory serves correctly it is completely out of circuit when the strangle is engaged and has no effect.
We mused on the 56k resistor here ...
http://www.offsetguitars.com/forums/vie ... 7&t=109553
Note that the schematic I found and linked there has the tone and volume pots reversed in the rhythm circuit. It's the tone pot that is 50k.
When the tone is on full its out the circuit. Its not effected by the strangle switch. 0 and 56k in parallel is 0.
It means unlike the rhythm circuit the resistance between pins 1 and 2 when the tone is turned down, never goes above that resistor. whilst between 2 and 3 its available up to 1meg. Otherwise you'd have a ton of volume loss, and odd effects when the tone is low and strangle is engaged. It does resolve the mid hump when you turn down the tone to a degree.

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oid
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Re: Why 1 meg pots?

Post by oid » Sun Jul 07, 2019 1:22 pm

jorri wrote:
Sun Jul 07, 2019 5:58 am
disagree on both those counts.
As I said, I was going off of memory on the resistor, and memory was faulty, my write up in the thread timtam linked looks to be correct.

They are separate things, you do not analyze a filter by analyzing it input, you can analyze the effect the filter has on the input, which is what the plots you linked show. Analyzing the filters on their own with idealized source and load gives us the most useful information since it tells us what it does to the signal we feed it, which is an unknown given the staggering amount of possible guitar pickups on the market and the near infinite number of possible loads the guitar could be plugged into. The plot of the filter/volume controls is also closer to the actual output of the guitar than a plot of the pickup. The real issue with the pickup plots is that they give no useful information to someone wanting to apply a circuit to different pickups, the huge number of possible pickup/circuit combinations makes providing these sorts of plots for every combination unlikely to ever happen. So we break things down into a few separate blocks so people can see how they would affect their setup, anyone can compare the plots of the circuit from a LP and a JM and figure out how the JM circuit would change their sound if they stuck it in their LP, you can not do that with pickup plots, all pickup plots show you is the POSSIBLE output of the pickup not actual output, they completely ignore the magnetic circuit and strings. We have to draw lines somewhere.
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