guitar preamp schematic theory

Discussion in 'General Electronics Chat' started by fez, Jan 6, 2013.

  1. fez

    Thread Starter Member

    Dec 6, 2009
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    I posted on this topic a while ago, but it was more general, so I'm opening a new thread so nobody has to be distracted by the older replies, as I've a specific question this time around.

    I've been pondering over the schematic for the MXR Microamp found here, and that damn R3 resister is driving me nuts. What is its purpose? I was told previously that its helps in the frequency-gain aspect, but how and why? I did some monkeywork using the open & short capacitor tests, and these are the time constants I got for each individual capacitor:-

    Short-circuit capacitor test:-
    C1's tau: C1(R2)
    C3's tau: C3(R6+R5)
    C4's tau: C4(R8//R7//R2)
    C5's tau: C5(R9+R10+R6+R5+R4)

    Open-circuit capacitor test:-
    C2's tau: C2[R4//(R5+R6+R10+R9)]

    As you can see, R3 plays no part in any of them! Please help.
     
  2. Brownout

    Well-Known Member

    Jan 10, 2012
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    It looks superfluous to me. It might work with the input capacitance of IC1 as a crude high frequency filter. I can think of no other reason for its existance in that circuit.
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    R3 will limit the current into the op amp +input in the event of overvoltage or static discharge on J1.

    Regarding your time constants: In actual circuit operation,

    C5's tau is just C5*(R9+R10), because the op amp output impedance is essentially zero.
    C2's tau is just C2*R4. This is because pin 2 is a virtual ground from an AC standpoint. This is somewhat complicated by the fact that, assuming a infinite bandwidth op amp, the gain will never be less than 1 at high frequencies, but the frequency at which this occurs depends on the value of R5+R6. Another way of thinking of this is that the high frequency response always starts rolling off at Fc=1/(2*pi*R4*C2), but starts to flatten out when F=1/(2*pi*(R5+R6)). In the actual circuit, the op amp's bandwidth will cause the high frequency response to continue to roll off.
     
    Last edited: Jan 6, 2013
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  4. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    R3 into a non-inverting jfet input is like putting a sheet of paper on a brick wall for added protection. It is not needed from my point of view.
     
  5. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    That's what I would guess as well. Inputs that go out to the world sometimes conduct large voltages in by accident.

    Also: when pulled above or below the supply voltage, they sometimes turn on internal parasitic diodes and you always want to limit current under that condition or you can blow the chip.
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Yeah, if it's a one-off that you're making for yourself, you might never think of static discharge protection. When you're selling thousands of them, it becomes important to make the design robust, so it can withstand abuse. Another example of this is D1, which protects against power supply reversal.
     
  7. fez

    Thread Starter Member

    Dec 6, 2009
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    Hmm, the over voltage protection does make sense for R3. But how is C2's time constant C2*R4? The way I see it, the input, being an independent source, is at ground potential for the open-circuit test. Which means the + input is at ground potential. Which, given its an opamp, means the - input is at gnd potential too and thus so is the opamp output terminal too. Given all that, the time constant seems to be zero!
     
  8. fez

    Thread Starter Member

    Dec 6, 2009
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    1
    I just remembered something about the tests. When you're finding the effective resistance seen by a capacitor, not do you only 'zero' all independent sources, but you also replace the capacitor itself by a test source. Now, replacing C2 by a test voltage source 'vt', thus the - input is at ground potential but the output is at a voltage of (-vt). Using this, I found the effective resistance seen by C2 (vt divided by I-through-vt) to be R4*(R9+R10)/(R9+R10+R4), which is approximately (R9+R10) because of R4's larger value. So is the time-constant actually C2*(R9+R10)?

    On a side note, how does one get better intuition in analyzing analog circuits? Any special tips? Mathematical monkeywork, while fun (because its simple), develops no intuition or depth whatsoever, so its also repulsive at the same time.
     
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