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How does this Tremolo work?

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  • How does this Tremolo work?

    Hey guys,

    I built this tremolo circuit (attached) from the Beavis Audio Research site. It sounds good and it's my first effects pedal. However, i dont know how it works and what the main parts of the circuit are doing to give the tremolo effect. I've also read something about LFO, but in this circuit im not sure which component or part is carrying out the oscillations. Can anyone help explain this please? Thank you.
    Attached Files

  • #2
    Hi Mark,
    AFAICS the circuit works as follows:

    The first 2N5457 acts as a follower / buffer, then the signal goes to the 2N3904, acting as a common emitter (variable ) gain stage. the gain is modulated ( giving you the "tremolo" effect ) by means of the second 2N5457, which, driven by the 2N5088 on the right, varies its Rds ( the resistance between its Drain and Source ) altering the degree of NFB in the 2N3904 gain stage, thus controlling the stage's gain. Quite obviously, the 2N5088 on the right is the heart of the LF OSC section.

    Hope this helps

    Best regards

    Bob
    Hoc unum scio: me nihil scire.

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    • #3
      - way more knowledgeable people than I on here, so in the spirit of testing my limited knowledge I will have a go.

      You have two Q2s, a fet and a bipolar.

      Q3 is the oscillator. It takes time to charge those 1uF caps, but when they reach a certain level of charge the increased voltage on the base of Q3 reaches the level at which it conducts, and discharges the caps, then the cycle starts again.

      Q1 is a buffer, it gives a high-impedance input and a uniform output for the next stage, which is the output transistor Q2-the-bipolar.

      Q2-the-fet applies the oscillating voltage to the emitter of Q2, varying the amount to which it amplifies the signal.

      Do tell me if I'm wrong, I'm a lifelong learner, me.

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      • #4
        Oops cross posted with someone who knows way more than me .

        What's that 22uF do (C4)?

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        • #5
          Thanks Robert & Alex,

          I think I understand the circuit a little more now
          What does NFB mean? I suppose when the resistance between drain and source increases, the "depth" or NFB increases too?

          Also, what are functions of R1, R2, R4, R6? Why are R1 and R2 1M ohms?

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          • #6
            @Mark,

            In a nutshell,
            R1 and R2 are part of the first FET biasing network, ensuring that the gate is placed half-way between 9 V and GND (guess what....4.5V). They form a potential divider, and their value is high to keep the stage's input impedance high (1MOhm in this case). The first FET is a source follower (impedance buffer/adapter, gain slightly less than 1, very high input impedance, low output impedance). R4 is the follower's "main" load resistor (the actual load is formed by the parallel of R4 and the following stage's input impedance). R6 is the second "leg" of the Q2 base biasing network (voltage divider) , and the base voltage of Q2 is 9V*(R6/(R5+R6))= 1.9V.

            NFB stands for Negative FeedBack. In a BJT gain stage, increasing the emitter resistor increases the NFB, because the more current flows in the resistor, the higher the voltage across it, and, since the emitter voltage gets higher, Vbe decreases by the same amount, resulting in an "opposition" to the Ice increase. In our case, when the second FET is not being driven by the LF OSC, it doesn't "bypass" the lower emitter resistor.

            @Alex,

            That 22uF cap in series with the second FET's source (as you correctly noted that FET is erroneously labeled as Q2, it should be labeled Q3, since the 2N3904 right before it uses the same label already, and the OSC BJT should be labeled Q4) is the gain stage's bypass cap, and it plays a role in the circuit only when the second FET is being driven (and its Rds becomes very low) by the OSC, thus increasing the stage's gain.

            Oh, and you don't need to stand corrected, as your circuit analysis is correct IMHO.

            BTW, I'm NOT all that knowledgeable, only terribly curious!
            ( and I've got still a LOT to learn )

            Luckily, this forum is populated with people way more knowledgeable than I am!

            Cheers

            Bob
            Last edited by Robert M. Martinelli; 03-11-2010, 01:11 PM.
            Hoc unum scio: me nihil scire.

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