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Panning for fun
A1
When pot is all the way toward B, then voltage at A1 is
A
R
R
A1 = A X
0.707R
R+(0.707R||R)
=Ax
(0.707Rx R)
(0.707)
R+(0.707R)
1+(0.707)
R+
R
B
(0.707R||R)
R
(0.707Rx R)
=Ax
1+
R+(0.707R)
= A x 0.29289
(0.707)
1+(0.707)
B1 is of course zero - it’s shorted to ground
B1
When pot is all the way toward A, then voltage at B1 is likewise B x 0.29289
When pot is in the middle, then voltage at A1 is
(0.35355Rx R)
A1 = A X
(0.35355R||R)
R+(0.35355R||R)
(0.35355)
1+(0.35355)
R+(0.35355R)
=Ax
R+
=Ax
(0353557Rx R)
1+
R+(0.35355R)
(0.35355
= A x 0.2071
1+(0.35355)
And the voltage at B1 is B x 0.2071 as well.
Since the middle of the pot is grounded, A and A1 do not interact with B and B1.
A
R
-
R
Out
+
0.707R
R
B
A
-
R
R
Out1
+
0.707R
R
R
R
-
If we take note that the inverting input of an opamp is at a virtual
ground, we can connect both of the output R’s that were grounded
and get the same result, but the opamp mixes the signals, and at the
same time, can supply some gain. If we make the gain setting
resistor in the feedback loop be just enough to make either A or B
have no loss when the pot is turned all the way to one side, then
we have a constant power cross fader. The value of resistor to do
this is one divided by the loss or 3.41 times R. Convenient values
of resistors are R=15K, 0.707R = 10K, and the feedback resistor
is 51K.
Out2
+
If we want to pan one signal to two differnent places, we can
do it by using two opamps, each with a gain of 3.41, but connect
them to the two different sides. Now A is panned from one output
to the other, in a constant power manner.
+
-
R
A
R
+
Rf
R
R
A
0.707R
R
-
Out1
R
R
Rf
3.41R
0.707R
R
-
R
Out1
R
R
2.41R
Out2
+
We can do the same thing without inverting the signal if we take the
signal at A1 and B1 to two separate opamps, and correct the gain of
the opamp. Now we need another resistor from the inverting input
to ground, and Rf needs to be 2.41R. If R=15K, 0.707R=10K and
2.41R = 36K.
Out2
+
We can even get one output inverted, the other not - just use the right
opamp circuit on the side you want true/inverted. Notice that we could
use two signal sources, too. It doesn’t matter what the signal is. The
circuit pans the input(s) with constant power.
Copyright 2003 R.G. Keen. All rights reserved. No permission for local copies or display from web sites other than http://www.geofex.com.
To Effect Input
From Effect Output
Pan in and Pan Out
51K
15K
100K
1M
+9
1M
0.22 to 1uF
-
+
Output
+
+
10K
100K
Input
15K
0.22 to 1uF
Vb
-
1M
+
0.22 to 1uF
+
15K
LED
Vb
+9
4.7K
15K
Here’s a panning application - an effect blender. The input signal is inverted by the first opamp and drives the panner circuit but
also goes out to an effect-loop out jack. The return from the effect loop goes into the other side of the panner circuit. The “bypass”
switch does not bypass. What it does is in one position, it grounds the effect side of the panner circuit, and opens the wiper lug to
ground. This is the same as forcing the pot to be rotated all that way to the effect return side, so the effect-only side of the panner
has signal through it as though the pot were turned to full signal. In the blend position, the switch grounds the lug of the panner pot
and the other section of the switch turns on an indicator LED.
To Effect Input
From Effect Output
The input impedance is not particularly high, only 100K. This is because the input opamp inverts the signal to keep the final output
in true phase with the input. If this is not a high enough impedance, an emitter follower input buffer can be added.
-- or you could use the high impedance version shown below. The input stage has been reworked to have an input impedance of
about 1M, ignoring the 2.2M at the input cap. This version will give no treble loss with a raw guitar input.
22pF
100K
cw
51K
100K
15K
1K
Input
100K
2.2M
Vb
+
+
10K
0.1uF 1M
-
0.22 to 1uF
+
1M
+9
Vb
15K
1M
Output
+
0.22 to 1uF
Vb
1M
+
0.22 to 1uF
15K
LED
+9
4.7K
15K
Copyright 2003-20063 R.G. Keen. All rights reserved. No permission for local copies or display from web sites other than http://www.geofex.com.
Pan between outputs
51K
100K
15K
100K
15K
-
+
+
+
Vb
Output
0.22 to 1uF
Vb
1M
100K
10K
+9
51K
0.1uF
Input
+9
+9
+
15K
10K
1M
15K
-
Vb
+
Output
10K
+
510K
10K
+
+
22uF
22uF
0.22 to 1uF
Vb
1M
Another panning application - panning one source to two outputs. You might use this to pan one guitar between two effects chains. You
may be able to pan between two amps; but remember, that application is fraught with hum problems, and this circuit will not make the
hum issues better or worse.
I have only shown generic opamp pinouts. You will have to match the pins on your actual opamps with the functions shown, like the
non-inverting input (+), the inverting input (-) and outputs. You will also have to select opamps. I would use two dual opamps like the
TL072 or LM833. But a single quad opamp like the TL074 will work too.
See the trick with the bias voltage? I did a normal bias voltage for the opamp inputs, but separated the “ground” connection ot the panning
pot. This ensures that the signal from the panning pot does not pollute the bias voltage and cause oscillation problems.
Copyright 2003-20063 R.G. Keen. All rights reserved. No permission for local copies or display from web sites other than http://www.geofex.com.