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EA Tremolo Effect

M

Thread Starter

MikeD_72

Joined Nov 11, 2008
46
I would like to build this circuit (a tremolo guitar effect) for use with my digital piano. I'm having trouble understanding the role of the highlighted components - especially the interaction between Q1 and Q2.

Also, why did the designer select the type of transistors the way he did - is there an advantage to using the MOSFET/JFET/BJT as shown in the schematic?
 

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JDT

JDT

Joined Feb 12, 2009
657
Q1 is used as an amplifier. It's gain depends on the resistance attached to it's source (the 180 ohm and the 1k2 ohm). Lower resistance = more gain.

The resistance of Q2 depends on the voltage on it's gate. Due to the 22uF capacitor, at audio frequencies Q2 is in parallel with the 1k2 resistor.

So, as the gate voltage of Q2 is varied by the low frequency oscillator, its resistance changes, changing the gain of the amplifier.

Q2 is a JFET. Important to use this here as its resistance is proportional to its reverse gate voltage (unlike a conventional transistor). Also, it doesn't matter which way the current flows through the device - important here as there is no DC across it, only audio (AC).

A conventional transistor (bipolar) transistor could be used as Q1 but a MOSFET transistor gives a much higher input impedance due to it's insulated gate. This allows the use of a much smaller blocking capacitor and a higher value of biasing resistor.

So:-
Q2 is the gain control.
The 1M is the input biasing.
The 100k trim sets the DC operating point.
The 47uF removes any noise on the 9V supply from getting into the input.
The 1M on the input charges the DC blocking capacitor and prevents any nasty pops when you plug in your input.
The 4k7 is the amplifier drain load that allows current to flow through Q1.
The 180 ohm sets the maximum gain when Q2 is at its lowest resistance.
 
M

Thread Starter

MikeD_72

Joined Nov 11, 2008
46
JDT said:
Q1 is used as an amplifier. It's gain depends on the resistance attached to it's source (the 180 ohm and the 1k2 ohm). Lower resistance = more gain.

The resistance of Q2 depends on the voltage on it's gate. Due to the 22uF capacitor, at audio frequencies Q2 is in parallel with the 1k2 resistor.

So, as the gate voltage of Q2 is varied by the low frequency oscillator, its resistance changes, changing the gain of the amplifier.

Q2 is a JFET. Important to use this here as its resistance is proportional to its reverse gate voltage (unlike a conventional transistor). Also, it doesn't matter which way the current flows through the device - important here as there is no DC across it, only audio (AC).

A conventional transistor (bipolar) transistor could be used as Q1 but a MOSFET transistor gives a much higher input impedance due to it's insulated gate. This allows the use of a much smaller blocking capacitor and a higher value of biasing resistor.

So:-
Q2 is the gain control.
The 1M is the input biasing.
The 100k trim sets the DC operating point.
The 47uF removes any noise on the 9V supply from getting into the input.
The 1M on the input charges the DC blocking capacitor and prevents any nasty pops when you plug in your input.
The 4k7 is the amplifier drain load that allows current to flow through Q1.
The 180 ohm sets the maximum gain when Q2 is at its lowest resistance.
Click to expand...
Thanks, that was very helpful. I didn't know the JFET acted as a resistance. How can I use the information in the datasheet to predict the resistance of the transistor? (datasheet attached)
 

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Audioguru

Audioguru

Joined Dec 20, 2007
11,248
Almost every tutorial on sine-wave oscillators uses a JFET as a variable resistor to hold the output level constant.
Its resistance is difficult to measure because the level of the signal must be very low to avoid distortion and each JFET has a different amount.
Many JFETS have a very wide range of threshold gate voltage.
 
JDT

JDT

Joined Feb 12, 2009
657
MikeD_72 said:
Thanks, that was very helpful. I didn't know the JFET acted as a resistance. How can I use the information in the datasheet to predict the resistance of the transistor? (datasheet attached)
Click to expand...
Don't know enough about this to give you an exact answer.
Important to remember that the device is "Depletion mode" and is fully conductive (has the lowest resistance) when the gate voltage is zero. As the gate voltage goes more negative (with respect to the source) the resistance rises until the device is fully cut-off at Vgs(off).

I looked in my reference book "The Art of Electronics" (Horowitz and Hill) p. 240 has some text on this. Although it gives some formulas I have to say that it didn't help me much as the data sheet does not have much information.

Basically, looking at the second graph on page 2 of the data sheet you can see that close to the 0,0 origin the curves approximate to straight lines. These actually extend both sides of the origin. The slope of the line is the resistance.

It's not easy to tell from this graph but extrapolating the Vgs = 0V line near the origin seems to give a resistance of about 66 ohms. When the gate goes more negative than Vgs(off) the resistance should be very high (leakage current only).
 
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