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  • Negative feedback, and resonance question?

    Hi everyone!

    I'm trying here to understand the negative feedback circuit of my Marshall dsl 15(see schematic bellow).

    My first question is when i engage the deep switch the signal of the negative feedback passes trough R91 then R94? or goes from R91 to the R93 in parallel with C69 path ?

    How can i now what frequencies are boosted with the resistor R93 in parallel with the C69 capacitor? there is somekind of graphic calculator for this?

    What is the purpose of the R91 resistor, is just to reduce the signal, or it has effect on the deep switch?

    Also alot of amp schematics dont have a R58 resistor, in the negative circuit this resistir R58 is just to reduce the signal strenght?


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  • #2
    When the switch is open it's in 'deep' mode.

    The deep corner (-3dB) frequency is 1/2/pi/R91/C69 = 1/2/3.14/22k/22nF = 330Hz. The gain is highest at low frequencies and decreases at -6dB /octave to level out after 330Hz. R91 sets both the upper corner frequency and the gain above that frequency.

    R93 stops the gain from rising below 33Hz.

    IMHO R58 is superfluous. It would only do anything if CON2 was disconnected and even then nothing especially useful.
    Experience is something you get, just after you really needed it.

    Comment


    • #3
      1/(2pi*220k*22n) =33Hz, not 330Hz.

      What do you mean with "R91 sets both the upper corner frequency and the gain above that frequency"?

      Comment


      • #4
        Originally posted by Rod View Post
        1/(2pi*220k*22n) =33Hz, not 330Hz.

        What do you mean with "R91 sets both the upper corner frequency and the gain above that frequency"?
        It's 22k not 220k so 330Hz is correct

        Upper corner freq depends on R91 * C69, HF gain depends on R91/R57 thus R91 affects both.
        Experience is something you get, just after you really needed it.

        Comment


        • #5
          R58 should have no affect on audio, but it does keep C44 charged, possibly kills potential pops when the deep switch is pushed or at least when that connector is opened.
          Education is what you're left with after you have forgotten what you have learned.

          Comment


          • #6
            Thanks guy's for your answers and help.

            In the schematic bellow, in the negative feedback circuit we have a resistor R50 in series after the resistor R51 in parallel with the capacitor C36.

            Putting the resistor R50 i series after or before the resistor R51 in parallel with C36 it's the same thing? If it is diferent, it's diferent in what?

            Thanks.

            Click image for larger version

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            • #7
              Originally posted by Rod View Post
              Thanks guy's for your answers and help.

              In the schematic bellow, in the negative feedback circuit we have a resistor R50 in series after the resistor R51 in parallel with the capacitor C36.

              Putting the resistor R50 i series after or before the resistor R51 in parallel with C36 it's the same thing? If it is diferent, it's diferent in what?

              Thanks.

              [ATTACH=CONFIG]42434[/ATTACH]
              The order of the R50 and R51//C36 make no difference.
              Experience is something you get, just after you really needed it.

              Comment


              • #8
                Originally posted by nickb View Post
                The order of the R50 and R51//C36 make no difference.
                So putting R50 before or after R51//C36 the result is the same?

                Comment


                • #9
                  Originally posted by Rod View Post
                  So putting R50 before or after R51//C36 the result is the same?
                  Yes. It depends on the total feedback impedance. In this case R50 + Zx; where Zx is the impedance of R51 and C36.

                  Hopefully you will agree than 2+4 = 4+2 so, R50 + Zx = Zx + R50

                  For completeness Zx = R51*Zc/(R51 + Zc); Zc= 1/2/3.14/f/C36
                  Experience is something you get, just after you really needed it.

                  Comment


                  • #10
                    Originally posted by nickb View Post
                    Yes. It depends on the total feedback impedance. In this case R50 + Zx; where Zx is the impedance of R51 and C36.

                    Hopefully you will agree than 2+4 = 4+2 so, R50 + Zx = Zx + R50

                    For completeness Zx = R51*Zc/(R51 + Zc); Zc= 1/2/3.14/f/C36

                    Thanks nickb for the help, it's still confusing for me but it's a little more clear.

                    If i understand it right so if increase the resistor R50 for exemple to 100k, it will reduce the high freaquency and the gain in the high frequencys right?

                    What is the caculation method to see at what frequencys the gain will lower?

                    Comment


                    • #11
                      Originally posted by Rod View Post
                      If i understand it right so if increase the resistor R50 for exemple to 100k, it will reduce the high freaquency and the gain in the high frequencys right?
                      R50 and R51 are the negative feedback resistors. If you increase the resistance the negative feedback is reduced and the gain will therefore increase. At high frequency C36 is a short circuit so the gain is set by R50. Doubling the value of R50 will double the gain (if you ignore the finite open loop gain).

                      Comment


                      • #12
                        If you increase R50 the gain at all frequencies will increase, but the effect is greater at high frequencies.

                        To get a feel of what is going on, look at the circuits at the extremes of zero and infinite frequency. The impedance of a capacitor is 1/2/pi/f/c. So at zero the capacitor has infinite impedance and the gain is (p to (R50 + R51)/VR9. At infinite frequency the capacitor is a short and the gain is R50/VR9.

                        So lets think about inbetween. Start at zero and imagine the frequency is increasing. The capacitor's impedance is big compared to R51 so the gain is steady. Soon the impedance of C36 will equal R51 and this is our first "corner" frequency i.e. when 1/2/pi/f/C36 = R51 (or rearranging f= 1/2/pi/R51/C36). After that the gain will be reducing at the capacitor's lower impedance dominates R51//C36 ( i.e. R51 in parallel with C36). Go higher in frequency and we reach a point where C36//R51 is small enough compared to R50 that we have a second, upper, corner frequency. After that the gain levels out again.
                        The upper corner frequency is f= 1/2/pi/c/[R50.R51/(R50+R51)]

                        I plotted a graph for you - green is R50 = 47k and blue is =100K

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                        BTW, I am pretending the amplifier has infinite gain to keep it simple. The truth is a long way from that but I don't think that helps your understanding of the concept at this stage.


                        The gain equation, with the infinite amplifier gain assumption is

                        (R51+R50)/VR9 * SQRT ( 1+ (2 * pi * f * C36* R50 * R51 /( R50 + R51))^2)/(1 + (2 * pi * f * C36* R51)^2 ( if I did it right )

                        If am afraid that if I give you the step by step calculation it will only serve to confuse you. You really need to read up on reactive circuits theory, know about imaginary numbers and negative feedback theory.
                        Last edited by nickb; 02-19-2017, 09:04 PM.
                        Experience is something you get, just after you really needed it.

                        Comment


                        • #13
                          Thanks guys for your answers and help.

                          Nickb did you use some program to plot the graph or online calculator, a calculator like that will be very helpful

                          Just one more question about a mod i'm thinking in the negative feedback, going back to the schematic attached of the Marshall.

                          If i remove c44 and put a wire in it's place, then remove R94 and in it's place, put the capacitor C44 in series with the R94 resistor, will this mod have the same result of leaving everything has it is, or not doing this mod?

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                          Comment


                          • #14
                            Originally posted by Rod View Post
                            Thanks guys for your answers and help.

                            Nickb did you use some program to plot the graph or online calculator, a calculator like that will be very helpful

                            Just one more question about a mod i'm thinking in the negative feedback, going back to the schematic attached of the Marshall.

                            If i remove c44 and put a wire in it's place, then remove R94 and in it's place, put the capacitor C44 in series with the R94 resistor, will this mod have the same result of leaving everything has it is, or not doing this mod?

                            [ATTACH=CONFIG]42445[/ATTACH]
                            It won't work. C44 is needed to prevent DC from flowing.


                            I used LtSpice, a free circuit simulator, to do the plots.

                            PS: Instead of 50 questions how about you tell us what you are trying to do?
                            Experience is something you get, just after you really needed it.

                            Comment


                            • #15
                              But there is negative feedback loops or circuits schematics, that dont have a capacitor in the circuit.

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