Understanding Fender inout jack values and effects

[jukelemon]

Hi all.

I have a question.

Can someone please explain the details/specifics of the voltage divider circuit/impedance theory behind the 1meg/2 x 68k input resistors on a typical Fender input.  Why 68k and what happens if you reduce/increase the 1 meg resistor or the 2 68k resistors?

Thanks.

Jason

[sluckey]

http://home.comcast.net/~seluckey/amps/misc/Amp_Scrapbook.pdf

see page 6.

[HotBluePlates]

The 68k resistors also perform a grid stopper function, even though the ideal location of a grid stopper is right at the tube grid pin.

Miller capacitance reduces the high frequency response of tube stages; this is used for a positive along with series grid resistance to form a low-pass filter to reduce RF interference in guitar amps.

Miller capacitance is the tube's grid-plate capacitance time the stage amplification (plus 1), plus the grid-cathode capacitance. An average 12AX7 has:
Cg-k = 1.6pF
Cg-p = 1.7pF
Gain = ~60

So total Miller capacitance is Cg-p(A+1) + Cg-k = 1.7pF * (60+1) + 1.6pF = ~105pF

The -3dB point of the filter formed by the series resistance and the Miller capacitance is 1/(2*pi*R*C), so
= 1/(2*3.14*68kΩ*105pF) = ~22.3kHz [for the low gain jack] or
= 1/(2*3.14*34kΩ*105pF) = ~44.6kHz

This isn't a brick-wall, so highs beyond these numbers still get through, though they're attenuated.

Also, there is some stray capacitance in the wiring from jack to tube, which will add a little to the Miller capacitance and bring that -3dB point lower.

By themselves, neither of these should be audible, except in the lack of RF interference. But other treble roll-offs in the amp (and in your guitar cord, etc) add on and can bring the net -3dB point at the treble end of the frequency spectrum lower.

Oh yeah... the 1MΩ is a practical upper limit, as there is a maximum recommended value of resistance from any tube's grid to ground. If you get larger, you run into issues with grid current (which are put to use in grid-leak biasing by using a series grid cap).

[jukelemon]

Perfect and makes sense.

Now, let's say I am using 2 amps.  I pug guitar into the HI channel and then send the LO to the other amp's Hi channel.  The 1st amp's input is HI i.e. regular signal but the 2nd amp's signal is cut.

Is the 2nd amp's signal essentially coming from the LO channel i.e. in Sluckey's explanation it is 50% of the signal.

And if so, could you wire a pot so that you could control the amount of gain/signal loss in the LO channel?

In other words, could you sub in a pot instead of a 68k resistor and dial in/out the amount of signal loss?

Thanks.

Jason

[sluckey]

My explanation of the hi/lo jacks goes out the window when you plug something into both jacks at the same time.

The original intent of having two jacks on an amp was to be able to plug in two instruments, not daisy chain one instrument into two amps. Back in the day that Fender came up with that simple hi/lo jack circuit, it was desirable for two band members to use the same amp. When two identical instruments were plugged in, each could deliver the same amount of signal to the amp. The 68K resistors are really isolation resistors that provide a degree of isolation between the two instruments, thus preventing the volume control on one instrument from affecting the volume of the other instrument. The 'grid stopper' effect is just an added benefit and the fact that Fender put them on the input jacks makes me think he was not really concerned with RF interference at the time.

[HotBluePlates]

... Now, let's say I am using 2 amps.  I pug guitar into the HI channel and then send the LO to the other amp's Hi channel.  ...

Is the 2nd amp's signal essentially coming from the LO channel i.e. in Sluckey's explanation it is 50% of the signal. ...

Sluckey's notebook doesn't cover "plug in both inputs". So I'm gonna guess based on what I think I see in the schematic.

If you plug into the Lo input with nothing in the Hi input, signal is cut by half. But if you plug into both jacks, neither switched contact comes into play. With Type A, as Sluckey has drawn, each side gets its signal passed to the tube with little or no attenuation.

But you're asking about jumpering: From the hot connection of the high jack, the signal sees a series 68kΩ and then another 68kΩ until the signal exits through the Lo jack. The net 136kΩ forms a divider with whatever resistance the other amp has from grid to ground (probably a 1MΩ just like the original amp).

This is only down 10% from "full up" and would be like having a volume control on the 2nd amp set just a bit above 9. In other words, almost no attenuation. And my experience jumpering channels (within a single amp with 2 channels) suggests there's almost no volume difference between the two when doing this. That is, same volume settings for both channels seems to yield a roughly balanced volume level (enough that at most only a minor tweak of the volume control is needed).

So I'd suggest you don't need any extra trim control on the input of a "slave amp". You could add one if you want, to lower the volume level/drive in the slave amp, but inserting a volume control before any tube stage tends to degrade the signal to noise ratio (you could just as easily add the volume control after the 1st gain stage of the slave amp, where it probably already exists).

[HotBluePlates]

... The 68K resistors are really isolation resistors that provide a degree of isolation between the two instruments ... The 'grid stopper' effect is just an added benefit and the fact that Fender put them on the input jacks makes me think he was not really concerned with RF interference at the time.

I think we're both right.

In Dave Funk's book, he cited all 3 reasons for why the 68kΩ resistor were the way they were.

In an effort to double-check our explanations, I looked back at the earliest tweed Fender schematics. Originally, no series resistors of any kind on any amp. When they were added, they were to isolate jacks that fed a single tube grid from each other.

Around the 5D-series, Fender seemed to add 1MΩ resistors to ground with switching on the jacks. Before this, it looks like he relied on the guitar's volume control to be the grid reference for the 1st stage (unless grid leak bias was used).

Funk suggests that 68k-70kΩ was used because it is about 10 times bigger than the d.c. resistance of the typical Fender pickup, thereby making it an effective value for isolation. By the time Fender got to making the 68kΩ's do triple-duty, I think the value and the Miller capacitance effect happens to be a useful, if accidental, result. Especially as booster/distortion pedals start driving amp inputs (but the 68k arrangement predates that by at least 5-6 years).

[jojokeo]

Especially as booster/distortion pedals start driving amp inputs (but the 68k arrangement predates that by at least 5-6 years).

Which leads to the next logical point that in driving multiple amplifiers at the same time - one would imagine that there would be pedals involved & happening on the front end as in a performance situation (otherwise what's the point?)? Therefore, a preamp boost or line level signal would help minimize signal loading/degredation and coupled through a simple mixer circuit used in reverse as a "multi path distributor" could drive as many outputs (or amplifiers) as needed w/ complete isolation from one to another. This could be no frills/no controls or elaborate w/ separate volume and tone controls for each channel so the prior concerns become moot.

[jjasilli]

It seems to me that if you daisy-chain Fender inputs, then re Amp1:

*  if you plug the guitar into the Hi input, it will will see only a 68K grid stopper > Amp1;

*  the Lo input of Amp1 will serve as an "output" > Amp2; the 2X 68K input resistors of Amp1 will function as 136K of series resistance to the "output" of Amp1's lo input jack.

*  This 136K of resistance, at the very least, will serve as a voltage divider with the grid leak resistor of the input of Amp2.  This will get more complicated depending upon the specific circuitry of Amp2's input.   Amp2 will see a diminished signal but this can probably be compensated for with dial settings.

EDIT:  Whoops, l now see that Hotblue already covered this.  Also my experience with daisy chaining amps at their inputs matches his.

If you plug the guitar into the Lo jack of Amp1, then IMHO you don't deserve a second amp!  

[tubenit]

You can always mic one amp's speaker into the input of another amp by using a hi-lo Z  adapter.  I did this with a Dano Centurian and a Princeton Reverb. And it sounded good to me!

With respect, Tubenit

[jjasilli]

My question, therefore, do you select one input over the other based on the impedance of the signal source?

No, the input impedance is set by the 1M grid leak resistor.  The 68k resistors are so small by comparison, that they don't "change the equation".   (Note that the 1M resistor is in parallel with the guitar pots).  This input circuit wants to mate with hi impedance sources like passive guitar pickups, or old fashioned hi impedance mic's.

The difference I have found is the source voltage rather than the source impedance.  E.g., my Champ prefers that humbucker PU's -  higher voltage -- be plugged into the Lo input jack.  Note that humbuckers and single coils are both in the range of hi impedance devices.

[jukelemon]

Thanks all.

A very good explanation.

[PRR]

> do you select one input over the other based on the impedance of the signal source?

Hi-gain input is just 1Meg, as said.

Lo-gain input is near 136K. A bit low, but much higher gets into hiss problems. Low enough to take the zing off of single-coil, but SC pickups don't need low gain. Humbuckers do, but they don't have much zing, so no loss.

2-guitar use: input impedance is complicated, but not-less than 136K.

[loogie]

Richard Kuehnel devotes a large chunk of his book 'Guitar Amplifier Preamps' to the subject.  Well worth reading.  Pentode Press also has some great interactive amp tools and articles:

http://www.ampbooks.com/home/books/preamps/

Well worth checking out.

[jukelemon]

... Now, let's say I am using 2 amps.  I pug guitar into the HI channel and then send the LO to the other amp's Hi channel.  ...

Is the 2nd amp's signal essentially coming from the LO channel i.e. in Sluckey's explanation it is 50% of the signal. ...

Sluckey's notebook doesn't cover "plug in both inputs". So I'm gonna guess based on what I think I see in the schematic.

If you plug into the Lo input with nothing in the Hi input, signal is cut by half. But if you plug into both jacks, neither switched contact comes into play. With Type A, as Sluckey has drawn, each side gets its signal passed to the tube with little or no attenuation.

But you're asking about jumpering: From the hot connection of the high jack, the signal sees a series 68kΩ and then another 68kΩ until the signal exits through the Lo jack. The net 136kΩ forms a divider with whatever resistance the other amp has from grid to ground (probably a 1MΩ just like the original amp).

This is only down 10% from "full up" and would be like having a volume control on the 2nd amp set just a bit above 9. In other words, almost no attenuation. And my experience jumpering channels (within a single amp with 2 channels) suggests there's almost no volume difference between the two when doing this. That is, same volume settings for both channels seems to yield a roughly balanced volume level (enough that at most only a minor tweak of the volume control is needed).

So I'd suggest you don't need any extra trim control on the input of a "slave amp". You could add one if you want, to lower the volume level/drive in the slave amp, but inserting a volume control before any tube stage tends to degrade the signal to noise ratio (you could just as easily add the volume control after the 1st gain stage of the slave amp, where it probably already exists).

Hi. In your B example, the 2 68k resistors in Amp 1 (not the slave amp) would create a high freq cut.  Correct?  The signal would not see the 1 meg resistor in Amp 1 but would see it in Amp 2 (slave amp).  Correct?  How is that different from just going into the low jack of say amp1 (as far as signal)?  The LO jack has ~50% gain/signal right?  But the chaining of amps do not yet they have the same resistance.  I guess it is because how the signal sees the resistors.  Right?

If I wanted to Trim the highs/attenuate the highs between the chaining of the amps, what resistor(s) would I play with?  The 68k's?

So, the 1meg on Amp 1's HI input jack is never in the circuit?  Only the 2 68k resistors?

Thanks all

[sluckey]

In your B example, the 2 68k resistors in Amp 1 (not the slave amp) would create a high freq cut.  Correct?

The 68K resistors AND the interelectrode capacitance of the tube create the high freq cut filter.

The signal would not see the 1 meg resistor in Amp 1 but would see it in Amp 2 (slave amp).  Correct?

Wrong. When you plug something in both jacks of Amp 1, the 1M resistor will still be in the circuit.

How is that different from just going into the low jack of say amp1 (as far as signal)?  The LO jack has ~50% gain/signal right?  But the chaining of amps do not yet they have the same resistance.  I guess it is because how the signal sees the resistors.  Right?

Wrong. It's different because you now have a plug in each jack of amp 1. Doing that opens the switch of both jacks and the circuit is changed such that the 50/50 voltage divider is switched out. Study the circuit, perhaps even draw it out with both jack switches open and maybe it will make sense to you.

If I wanted to Trim the highs/attenuate the highs between the chaining of the amps, what resistor(s) would I play with?  The 68k's?

Diddling with any of the resistors on the input jacks is not gonna really affect the guitar signal. The 68Ks come into play with much higher frequencies, ie, rf. If you want to diddle the tone of the signal going to the slave amp, turn the tone control in the slave amp, or put a hi cut filter inside the slave amp.

So, the 1meg on Amp 1's HI input jack is never in the circuit?  Only the 2 68k resistors?

Wrong. I explained that in the pdf.

[HotBluePlates]

What Sluckey said.