Transcript m-lagrande

No really, what is it?
• It is an Octave/Fuzz pedal for electric
guitar.
• It uses three BJTs along the way.
• It was first made in the early ‘70s.
Things To Consider
• First of all, this is a “positive ground” circuit. Basically
this means that most of the voltages are of lower potential
than the ground.
• A few of the parts were not available through normal
means, so I substituted. It worked out, so I’m satisfied.
• For simplicity, I modeled the output of an electric guitar as
a .1 V sin wave. This isn’t completely correct, as there are
harmonics thrown in there initially, but it illustrates the
point well.
The first transistor utilizes the collector voltage off the third
transistor for its emitter voltage. The input signal goes into
the base.
What does this do?
• The current that goes from the first
transistor’s collector to the base of the
second transistor depends on the
potentiometer.
• The spikes have different slopes to them…
this will be important later on.
For 1k resistance due to potentiometer.
10uA
5uA
0A
0s
10ms
20ms
30ms
40ms
50ms
60ms
-Ib(Q2)
Time
For .1k resistance due to potentiometer.
50uA
0A
-50uA
0s
10ms
20ms
30ms
-Ib(Q2)
Time
40ms
50ms
60ms
The Second Transistor
• This transistor is there for current amplification.
• PNP (2N3906)
The Third Transistor
• PNP (2N3906)
• Due to the amplification from the
second transistor, this transistor
flips in and out of saturation.
• The result is a clipped signal.
• It also amplifies the signal a little.
The result of the third transistor is that the output current is
clipped. It makes fuzz. Fuzz is good in this case, as this
pedal is supposed to provide fuzz.
The following is the –1 x the current from the collector of Q3 when
the potentiometer caused .5k ohms of resistance:
20mA
10mA
0A
0s
10ms
20ms
30ms
-Ic(Q3)
Time
40ms
50ms
60ms
Now that all is said and done with the transistors, we have most
of the ac component being able to pass the capacitor and go to
the transformer.
This is the resulting voltage on the primary side of the transformer
for a .5k potentiometer value:
5.0V
0V
-5.0V
0s
10ms
20ms
30ms
V(C7:2)
Time
40ms
50ms
60ms
That last, clipped signal will now go through
this transformer setup:
The transformer steps it down 1:3, and the
diodes then rectify the signal, taking off the .7V
off the voltages.
The following is the resulting voltage of the
rectification with the .5k ohm pot resistance:
1000mV
500mV
0V
0s
10ms
20ms
30ms
V(D5:2)
Time
40ms
50ms
60ms
This is after going through a little volume control.
There is actually a potentiometer instead of the two
resistors in series, but it doesn’t affect the circuit other
than amplitude.
1.0V
0.5V
0V
0s
10ms
20ms
30ms
V(C9:2)
Time
40ms
50ms
60ms
500mV
250mV
0V
0Hz
0.4KHz
0.8KHz
1.2KHz
1.6KHz
2.0KHz
V(C9:2)
Frequency
The fourier analysis shows that there is actually more of the
second harmonic (the octave) present than the first.
What About The Potentiometer?
Here are the resulting output graphs for two other values of the pot:
For 1k Ohms
1.0V
0.5V
0V
0s
10ms
20ms
30ms
40ms
50ms
1.6KHz
2.0KHz
60ms
V(C9:2)
Time
500mV
250mV
0V
0Hz
0.4KHz
0.8KHz
1.2KHz
V(C9:2)
Frequency
2.4KHz
For .1k Ohms
1.0V
0.5V
0V
0s
10ms
20ms
30ms
40ms
50ms
60ms
V(C9:2)
Time
200mV
100mV
0V
0Hz
1.0KHz
2.0KHz
3.0KHz
V(C9:2)
Frequency
4.0KHz
5.0KHz
Results
• As you can see, a greater resistance due to the
potentiometer causes higher octave amplification.
• A low resistance causes that steep signal we saw earlier to
blend into itself, distort, and the octave element is not
significant.
• It was constructed, and after some finagling, it worked.
• Before actually soldering the circuit I should work on the
biasing and optimize it for the desired affects.
• A true bypassing switch should be added, because
otherwise this of little use and a pain to use.
That Is All