Fender Active Midboost Kit ("Clapton" midboost)

[geo]

So I got this kit a while back and I haven't had a chance to install it, but I took a glance at the wiring diagram and it's got me a little confused. Can anyone tell what the hell they're doing with the tone-end of the switch?

www.guitarcenter.com/Fender-Pre-Amp-Mid-Boost-Kit-25-DB-102019098-i1384771.gc?esid=midboost+kit



I just automatically assumed there'd be two leads you put a pot between to control the midboost, two leads for V+/ground, and two leads for signal, but after staring at this for a couple hours I'm pretty puzzled. It looks like the last two leads (6,7) are for tweaking the amount of midboost by connecting (what I'm assuming are) outputs for the regular signal and the mid-boosted signal and mixing them together through a pot, then pushing that to the volume pot as well to control total volume. If someone with a brain for electronics could hash this out for me a little more intuitively I'd greatly appreciate it.

-Geo

EDIT: This isn't exactly the diagram that came with mine, and it's actually elucidated the tone-end of the Fender switch a little bit, looks like they've introduced the tone to the regular signal chain and just jumped all of the tone connections. Why do these have to be jumped? (It seems to be the same circuit-wise, just without extra lines put in there to confuse people, and with fewer unexplained numbers.)

[asmith]

Nov 21, 2011 5:26:01 GMT -5 geo said:

Can anyone tell what the hell they're doing with the tone-end of the switch?

They're doing nothing with it. You don't need to.

I mocked up this diagram of the controls. It might prompt some "clicking:"



The selector switch's only function in the circuit is to select which pickup you're hearing. Then you go through the high-cut tone control, which uses a dual-gang pot.

Then you hit the active Mid-Boost, which'll work work it's mojo, then the volume control. Then the mid-boost control, which fades between "clean" and "mid-boosted" signal, then out.

Does that clear anything up?

[geo]

Shouldn't the volume only cut the signal coming from the brown wire from the midboost, since you've got some length of the midboost volume control's resistor between the output and the ground?

[asmith]

I'm not sure what you're asking. The mid-boost resistor isn't connected to ground - which is represented by the bottom line of the diagram.

What I think is happening here is - and this is a harzarded guess:

• Signal goes in Yellow wire.
  
• Clapton 'copies' the signal, let's call it "B-Signal," and boosts the mids by a finite amount.
  
• Clean signal exits Green wire.
  
• Clean signal heads through volume control, goes back in Brown wire.
  
• Clapton compares the volume level between Yellow and Brown wire.
  
• Clapton adjusts the volume of "B-Signal" accordingly.
  
• Clapton outputs the boosted, vol-adjusted signal out the Violet wire.
  

That way all the mid control is doing is fading the output between an adjusted boost-signal, and a clean signal.

I would guess that the design works that way because the circuitry is actually providing a geniune 'boost,' i.e. adding signal strength. In turn, it makes it simpler to keep the volume pot attached to a signal strength that you know is going to go through it, instead of having to play with pot values for adjusted signals.

Hope that helps.

[geo]

Ah, gotcha. Seems kinda silly that it'd have to match the volume at the end, that was pretty counter-intuitive and really screwed me up trying to figure this puppy out.

(Not to mention it's going to have to subtract the B-signal passing through the resistance between brown & violet when it calculates how much volume to send out that way. Seems like a real headache to do it that way.)

[ashcatlt]

Don't honestly believe that's what's going on. More likely the output of the 5-way is buffered before going on to the pots. The signal is split at the one end of the blend pot, then the EQ thing is applied and pumped into the other end.

[geo]

Kit comes with a 250K and a 50K pot. I'm assuming the 250K is for volume and the 50k is to regulate the midboost?

[asmith]

Actually, I'd say that's more likely to uphold Ash's statement, that the signal post-tone-control is buffered. I would imagine the 50k pot is the volume, and the 250k pot is the blender.

If you wanted to use the TBX controls with this scheme, they would go where the current tone control is.

[geo]

I was actually thinking of doing this for tone control:
guitarnuts2.proboards.com/index.cgi?board=wiring&action=display&thread=6056

What does it mean that the signal is buffered? I only have a basic understanding of electronics.

[newey]

Here's wikipedia's take on it, for what it's worth. In the context of guitar circuitry, buffering is usually used to take the guitar's output impedance from instrument-level signal to a line (or audio) level signal input.

Voltage buffer

A voltage buffer amplifier is used to transfer a voltage from a first circuit, having a high output impedance level, to a second circuit with a low input impedance level. The interposed buffer amplifier prevents the second circuit from loading the first circuit unacceptably and interfering with its desired operation. In the ideal voltage buffer in the diagram, the input resistance is infinite, the output resistance zero (impedance of an ideal voltage source is zero). Other properties of the ideal buffer are: perfect linearity, regardless of signal amplitudes; and instant output response, regardless of the speed of the input signal.

If the voltage is transferred unchanged (the voltage gain Av is 1), the amplifier is a unity gain buffer; also known as a voltage follower because the output voltage follows or tracks the input voltage. Although the voltage gain of a voltage buffer amplifier may be (approximately) unity, it usually provides considerable current gain and thus power gain. However, it is commonplace to say that it has a gain of 1 (or the equivalent 0 dB), referring to the voltage gain.

As an example, consider a Thévenin source (voltage VA, series resistance RA) driving a resistor load RL. Because of voltage division (also referred to as "loading") the voltage across the load is only VA RL / ( RL + RA ). However, if the Thévenin source drives a unity gain buffer such as that in Figure 1 (top, with unity gain), the voltage input to the amplifier is VA, with no voltage division because the amplifier input resistance is infinite. At the output the dependent voltage source delivers voltage Av VA = VA to the load, again without voltage division because the output resistance of the buffer is zero. A Thévenin equivalent circuit of the combined original Thévenin source and the buffer is an ideal voltage source VA with zero Thévenin resistance.

Current buffer

Typically a current buffer amplifier is used to transfer a current from a first circuit, having a low output impedance level, to a second circuit with a high input impedance level. The interposed buffer amplifier prevents the second circuit from loading the first circuit unacceptably and interfering with its desired operation. In the ideal current buffer in the diagram, the input impedance is infinite and the output impedance is zero (impedance of an ideal current source is infinite). Again, other properties of the ideal buffer are: perfect linearity, regardless of signal amplitudes; and instant output response, regardless of the speed of the input signal.

For a current buffer, if the current is transferred unchanged (the current gain βi is 1), the amplifier is again a unity gain buffer; this time known as a current follower because the output current follows or tracks the input current.

As an example, consider a Norton source (current IA, parallel resistance RA) driving a resistor load RL. Because of current division (also referred to as "loading") the current delivered to the load is only IA RA / ( RL + RA ). However, if the Norton source drives a unity gain buffer such as that in Figure 1 (bottom, with unity gain), the current input to the amplifier is IA, with no current division because the amplifier input resistance is zero. At the output the dependent current source delivers current βi IA = IA to the load, again without current division because the output resistance of the buffer is infinite. A Norton equivalent circuit of the combined original Norton source and the buffer is an ideal current source IA with infinite Norton resistance.

I know, clear as mud, right?

[sumgai]

Nov 28, 2011 23:58:16 GMT -5 newey said:

Here's wikipedia's take on it......

...... clear as mud, right?

Was to me.

Believe it or don't, some of us Engineer types actually do use Thevenin's and Norton's Theorems much more often than might appear necessary. They really do cut down on the time spent analyzing things, once you get the hang of it.

HTH




sumgai

[geo]

So here's the deal: I've got the switches in place, but I apparently can't measure as well as I thought I could, 'cause there's no place below my two 5-way switches for the mid-boost kit.

I have two options:
(1) Route a compartment like on the Clapton Strat
(2) Block the tremolo and sneak it in the back compartment.

I rarely use the tremolo, but it's nice to have the option. I guess what I do depends on:

How much worse will (1) it sound versus how much better (2) will make it sound.

Feedback?

[newey]

'cause there's no place below my two 5-way switches for the mid-boost kit.

Two 5-way switches? I think this is the first you've mentioned that. I can see where that would cause fitment problems for any added bits.

I don't think the slight loss of wood will have a deleterious effect on the tone. I've never heard anyone complain about that on the Clapton Strat.

The decision, in my way of thinking, should instead center on whether the guitar would lose value if not able to be converted back to "stock"- in other words, does it have any vintage value? And then, your use of the trem. And also, your comfort level with applying a router to your guitar!

Before you do any routing, I'd suggest a "test fit" with the trem block idea, you can at least see if that's a realistic option. You shouldn't need to actually block the trem permanently to be sure it will fit.

[geo]

Thanks a million, guys! It came out beautifully!
guitarnuts2.proboards.com/index.cgi?board=music&action=display&thread=6596

Wound up routing a compartment where the Clapton Strat does so I could keep the trem.