automatic wah.....help wanted.

[RLBJR65]

Its here http://members.shaw.ca/roma/ look under projects, guitar effects, Wah.

[GREEN FUZ]

Cheers. I found that site yesterday. When I saw the Morley-Wah schem it didn`t occur to me to go on to the next page.

[cosy]

Hey . so I've got the parts. but witch way should the transistor 2n5484 be mounted? the flat side to the left or right.

[Mark Hammer]

I don't know enough about discrete oscillators to judge, but the filter portion will work.  The op-amp is configured as a fairly straightforward inverting bandpass filter.

It is a fairly standard practice, adopted by many companies, to produce a wah effect by having two caps in series between the op-amp's output and inverting input pin, with a resistance to ground (or Vref) coming from the junction between them.  The gain of that bandpass filter is given by R12/R11, or x4.7.  Because a bandpass filter will lose some overall amplitude when it "dumps" high and low end, the added gain tends to keep effect and bypass signal somewhere in the same level range. 

In this case, the resistance linking C6 and C7 to ground (Vref) is comprised of 3 elements: R9, in series with the combined parallel resistance of Q2 and R10.  When Q2 is at its maximum resistance (which we'll assume is in the megohms), then there is effectively a 104.5k resistance to ground.  As the drain-source resistance of Q2 drops closer to zero, the overall resistance is close to the value of R9 (which sets the minimum resistance and could easily be a 10k trimpot or pot to suit to taste).

How is this a bandpass?  Well, remember that op-amps love to go full speed ahead, and the way we use them is to constrain how close they get to full out.  Those constraints are how much negative feedback we apply, and what portions of the frequency spectrum we apply them to.  Normally, when we have a (negative) feedback path, we calculate the upper frequency rolloff by combining the feedback resistance (in megohms) and feedback capacitance (in microfarads) in the formula Freq = 1/[2*pi*R*C).  Since two caps in series have a lesser capacitance than either cap on its own, the feedback capacitance formed by C6-C7 is actually .0015uf.  In tandem with R12, that creates a high-end rolloff at 219hz.  Kinda low, huh?

But wait.  I said that if you apply more negative feedback you decrease gain, which means that if you apply less negative feedback, you get more gain.  Okey doke.  Let's look at the network formed by C7 in conjunction with R9/R10/Q2.  C7 and the resistance to ground actually form a simple highpass filter.  If R9/R10/Q2 were at its maximum resistance value (104.5k), in tandem with C7, that would provide a highpass filter that rolls off below 761hz (in theory).  As the R9/R10/Q2 resistance gets smaller, that low-frequency rolloff point starts to move upward.  Anything below that frequency essentially gets dumped to ground.  Wait a sec....low-frequency negative feedback dumped to ground?  That means less negative feedback which equals more gain for those frequencies, right?  Right.

So, without wishing to make it any more complicated than it already is, as the R9/R10/Q2 resistance changes, it allows different amounts of content between upper and lower limits to flow from the output back to the inverting input pin, yielding a shifting amount/emphasis in negative feedback applied over the frequency spectrum.  Voila.  Bandpass filter.  Everything to the left of R6 is involved in producing a repeating change in the Q2 resistance.

The range of where that emphasis lives can be controlled in a bunch of ways.  The most obvious is to change the value of C6 and/or C7.  In a great many designs using this type of filter topography, you'll see equal values for the C6 and C7 equivalents.  For instance, the Dr. Q uses a pair of 4700pf caps, and some Morley wahs use .01uf caps if I'm not mistaken.  I gather the use of unequal values here is intended to give a slightly different "body" to the filtered sound, quite probably because the automatic aspect means you can't control it and get the emphasis where and when you want like a wah so it adjusts the passband area via unequal cap values to get a sound that works for all speeds.

Although not shown in the schematic, the range of sweep can be adjusted quite easily.  Remember that the sweep depends on how much change in resistance is created by Q2.  If R10 is smaller (than 100k) then changes to Q2 resistance will have less impact on the parallel resistance.  If R9 is bigger then the filter will not be able to sweep as far in that direction.  So, widest sweep is given by small values of R9 in tandem with larger values of R10.  This gives the widest range of resistance-to-ground values.  Narrower sweep is given by combining a larger value of R9 with a smaller value of R10.  This gives the smallest range of changes in resistance-to-ground values.

Finally, note that capacitors placed in parallel can substitute for capacitor values you don't have.  So, a .0047uf plus a .0022uf cap in parallel essentially equals a .0068uf (6800pf) cap, although it may well be too bulky to fit on your board.  Note that caps generally do not have very tight tolerances, so if the combined value is a bit higher or lower, don't worry.  There's a good chance that even if you had the desired value of cap, it wouldn't measure exactly what is printed on it anyways.  Plus or minus 10% is certainly alright.

There.  Enough to get you going?

[GREEN FUZ]

Mark, if you don`t know the answer just say so. We won`t think any less of you.

Looking at the flat side of the 2N5484 the pins are, in order

1. Drain
2.Source
3.Gate

Or DSG

Looking at the schematic the order of the pins is DGS. I guess that means that, unless you errm..creatively bend the legs it wont work. I`ll let someone more able than me say whether this is the case. Ehh..Mark?

[Mark Hammer]

Not sure if its true for all brands, but my Fairchild datasheet for the 2N5484 also says D-S-G, looking at the flat side, with the pins pointing down.

If you ARE going to bend pins to adapt one pinout to a different one, it is a good practice to:
a) leave the pins/leads a little longer than you might otherwise, so that less stress is placed on the leads, and
b) strip a bit of insuation off some hookup wire and slip a little over the leads on the component side of the board so that they don't accidentally touch each other or something else.

[cosy]

So i have assembled the running wah..... and when I hook it up to my guitar with no power on the wah, the guitar signal runs through without problems.
but when I put the battery on, I get a lot of hum.... the pots rv1 and rv2 sounds like they work but there is no wah sound.
the guitar signal is not comming through....

Any idears?

[GREEN FUZ]

Have you tried reorienting the pins of Q2?

[cosy]

Yes acording to the schematic....?

[GREEN FUZ]

Yes acording to the schematic....?

You don`t seem too sure.

Look, here is your transistor. Notice the pinout.

D = drain
S = source
G = gate

These should correspond with the schematic...

and the layout.


You will have to twist the legs to make them fit. I would take up Mark`s suggestions on how to implement this.

I`m not saying this will cure the problem but it is a start.