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Schoeps dc/dc converter was GenChinaMic dc converter question

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  • Goran Finnberg
    ... You can only get _accurate_ results when using an electrostatic measuring meter. Even using a 1 Gohm resistor in series with my 100 Mohm multimeter shows
    Message 1 of 15 , Nov 4, 2010
      Unknown aka "taperchuck2112" <millers2112@...> :

      > Are you guys getting 60v on the stock circuit?
      > I'm getting 44v on one and 38v on the other of my CAD GXL1200's.

      You can only get _accurate_ results when using an electrostatic measuring
      meter.

      Even using a 1 Gohm resistor in series with my 100 Mohm multimeter shows an
      error of nearly 20 percent compared to an electrostatic meter.

      This type of DC/DC converter must be unloaded and that is what it is when
      feeding a dc biased condenser capsule whose insulation properties is nearly
      perfect.


      --

      Best regards,

      Goran Finnberg
      The Mastering Room AB
      Goteborg
      Sweden

      E-mail: mastering@...

      Learn from the mistakes of others, you can never live long enough to
      make them all yourself. - John Luther
    • Mark Fitzsimons
      This seems oversimplified to me - why must it be unloaded? it has a very low current capability , not none at all. mine seems to work perfectly well with a
      Message 2 of 15 , Nov 5, 2010
        This seems oversimplified to me - why must it be unloaded? it has a very low current capability , not none at all.

        mine seems to work perfectly well with a 1M+8.2M voltage divider on the output, this has the added advantage of setting a known maximum output impedance for measuring the voltage with a DMM.

        is the 20% error you mention purely due to loading, or is there some other factor?

        My understanding of how a condenser mic works is that the capsule is discharged and recharged periodically in response to its changing capacitance, so in fact there is only no load when there is no sound. As soon as there is a AC signal voltage then work is being done so some current, however small, must be flowing. Is this not correct?

        It follows from this that if the current available of the polarisation voltage source is too low then the capsule will be slow to recharge which will limit slew rate? all in theory of course, I imagine the current required is very very small - but still theoretically non-zero.




        On 5 Nov 2010, at 00:16, Goran Finnberg wrote:

        > Unknown aka "taperchuck2112" <millers2112@...> :
        >
        > > Are you guys getting 60v on the stock circuit?
        > > I'm getting 44v on one and 38v on the other of my CAD GXL1200's.
        >
        > You can only get _accurate_ results when using an electrostatic measuring
        > meter.
        >
        > Even using a 1 Gohm resistor in series with my 100 Mohm multimeter shows an
        > error of nearly 20 percent compared to an electrostatic meter.
        >
        > This type of DC/DC converter must be unloaded and that is what it is when
        > feeding a dc biased condenser capsule whose insulation properties is nearly
        > perfect.


        [Non-text portions of this message have been removed]
      • Mark Fitzsimons
        oops - I take back what I said about getting 75v out of this circuit. It seems that my 20yr old multimeter for which I have lost the manual may have an input
        Message 3 of 15 , Nov 5, 2010
          oops - I take back what I said about getting 75v out of this circuit.

          It seems that my 20yr old multimeter for which I have lost the manual may have an input impedance of 20M or more, so i may be in fact only getting 40V or so... the 75 v was assuming a 10M input impedance

          it measures 38.6V loaded only with the multimeter. but i don't know the actual input impedance of the multimeter.

          loaded with a cheap 1M input multimeter it reads 19.7v
        • userno232000
          ... No, If you think about an electret you will see. The capsule is operated in constant charge mode and the net is no DC current. The work is all mechanical.
          Message 4 of 15 , Nov 5, 2010
            --- In micbuilders@yahoogroups.com, Mark Fitzsimons <mf@...> wrote:
            >
            > This seems oversimplified to me - why must it be unloaded? it has a very low current capability , not none at all.
            >
            > mine seems to work perfectly well with a 1M+8.2M voltage divider on the output, this has the added advantage of setting a known maximum output impedance for measuring the voltage with a DMM.
            >
            > is the 20% error you mention purely due to loading, or is there some other factor?
            >
            > My understanding of how a condenser mic works is that the capsule is discharged and recharged periodically in response to its changing capacitance, so in fact there is only no load when there is no sound. As soon as there is a AC signal voltage then work is being done so some current, however small, must be flowing. Is this not correct?
            >

            No, If you think about an electret you will see. The capsule is operated in constant charge mode and the net is no DC current. The work is all mechanical.
          • Mark Fitzsimons
            OK i don t fully understand, but i see your point, if the polarisation voltage was doing any work then the electret backplate would discharge, and it doesn t.
            Message 5 of 15 , Nov 5, 2010
              OK i don't fully understand, but i see your point, if the polarisation voltage was doing any work then the electret backplate would discharge, and it doesn't.

              so my understanding was wrong - a common enough occurrence.

              I'll read up again...

              still not necessarily a reason to not load the voltage converter though...


              On 5 Nov 2010, at 14:41, userno232000 wrote:

              >
              >
              > --- In micbuilders@yahoogroups.com, Mark Fitzsimons <mf@...> wrote:
              > >
              > > This seems oversimplified to me - why must it be unloaded? it has a very low current capability , not none at all.
              > >
              > > mine seems to work perfectly well with a 1M+8.2M voltage divider on the output, this has the added advantage of setting a known maximum output impedance for measuring the voltage with a DMM.
              > >
              > > is the 20% error you mention purely due to loading, or is there some other factor?
              > >
              > > My understanding of how a condenser mic works is that the capsule is discharged and recharged periodically in response to its changing capacitance, so in fact there is only no load when there is no sound. As soon as there is a AC signal voltage then work is being done so some current, however small, must be flowing. Is this not correct?
              > >
              >
              > No, If you think about an electret you will see. The capsule is operated in constant charge mode and the net is no DC current. The work is all mechanical.
              >
              > _


              [Non-text portions of this message have been removed]
            • Mark Fitzsimons
              please check this is correct.... My DMM has an input impedance of 20M (checked with a 20M resistor and a 9V battery & reading was half with the resistor) I
              Message 6 of 15 , Nov 5, 2010
                please check this is correct....

                My DMM has an input impedance of 20M (checked with a 20M resistor and a 9V battery & reading was half with the resistor)

                I read the output of my voltage converter at approx 0.8v through a 1G resistor, and as approx 6.8v through a 100M resistor

                0.8v x 51 = 40.8V
                6.8v x 6 = 40.8V

                reading with the meter directly about 38V

                I conclude that since there was no difference between the 1G and 100M readings that the voltage output really is around 40.8v

                if the load was dragging the voltage down then the 1G measurement should have shown a higher voltage, but it didn't.

                margin of error of course... but close enough.

                looks like i might need to install a trimmer or change the input zener to get the output up to 60v
              • Jerry Lee Marcel
                You were right in saying that the capacitor gets slightly discharged when positive pressure is applied, but the next half-period of signal, it will charge
                Message 7 of 15 , Nov 5, 2010
                  You were right in saying that the capacitor gets slightly discharged
                  when positive pressure is applied, but the next half-period of signal,
                  it will charge again. The net result is constant charge and no current
                  consumption at all.
                  The only time the bias supply has to produce some current is during the
                  initial charging period. So yes, the bias supply must have some current
                  capability. Very tiny though; a 60nA charge current can 99% charge a
                  40pF capsule in less than 300msec.

                  Le 05/11/2010 16:02, Mark Fitzsimons a écrit :
                  >
                  > OK i don't fully understand, but i see your point, if the polarisation
                  > voltage was doing any work then the electret backplate would
                  > discharge, and it doesn't.
                  >
                  > so my understanding was wrong - a common enough occurrence.
                  >
                  > I'll read up again...
                  >
                  > still not necessarily a reason to not load the voltage converter though...
                  >
                  > On 5 Nov 2010, at 14:41, userno232000 wrote:
                  >
                  > >
                  > >
                  > > --- In micbuilders@yahoogroups.com
                  > <mailto:micbuilders%40yahoogroups.com>, Mark Fitzsimons <mf@...> wrote:
                  > > >
                  > > > This seems oversimplified to me - why must it be unloaded? it has
                  > a very low current capability , not none at all.
                  > > >
                  > > > mine seems to work perfectly well with a 1M+8.2M voltage divider
                  > on the output, this has the added advantage of setting a known maximum
                  > output impedance for measuring the voltage with a DMM.
                  > > >
                  > > > is the 20% error you mention purely due to loading, or is there
                  > some other factor?
                  > > >
                  > > > My understanding of how a condenser mic works is that the capsule
                  > is discharged and recharged periodically in response to its changing
                  > capacitance, so in fact there is only no load when there is no sound.
                  > As soon as there is a AC signal voltage then work is being done so
                  > some current, however small, must be flowing. Is this not correct?
                  > > >
                  > >
                  > > No, If you think about an electret you will see. The capsule is
                  > operated in constant charge mode and the net is no DC current. The
                  > work is all mechanical.
                  > >
                  > > _
                  >
                  > [Non-text portions of this message have been removed]
                  >
                  >


                  [Non-text portions of this message have been removed]
                • Nick Zuccaro
                  why don t you put a high valued load resistor on that voltage supply so you can have a known, fixed amount of current draw? it only would have to draw a
                  Message 8 of 15 , Nov 5, 2010
                    why don't you put a high valued load resistor on that voltage supply so you can
                    have a known, fixed amount of current draw? it only would have to draw a trickle
                    of current but it will sort of tie things down load wise so you don't have an
                    'unknown' in the oscillator/stepup load. it would also make it less sensitive
                    to DMM input impedances, etc. etc...



                    ________________________________
                    From: Mark Fitzsimons <mf@...>
                    To: micbuilders@yahoogroups.com
                    Sent: Fri, November 5, 2010 11:39:33 AM
                    Subject: Re: [micbuilders] Re: Schoeps dc/dc converter was GenChinaMic dc
                    converter question


                    please check this is correct....

                    My DMM has an input impedance of 20M (checked with a 20M resistor and a 9V
                    battery & reading was half with the resistor)

                    I read the output of my voltage converter at approx 0.8v through a 1G resistor,
                    and as approx 6.8v through a 100M resistor

                    0.8v x 51 = 40.8V
                    6.8v x 6 = 40.8V

                    reading with the meter directly about 38V

                    I conclude that since there was no difference between the 1G and 100M readings
                    that the voltage output really is around 40.8v

                    if the load was dragging the voltage down then the 1G measurement should have
                    shown a higher voltage, but it didn't.

                    margin of error of course... but close enough.

                    looks like i might need to install a trimmer or change the input zener to get
                    the output up to 60v




                    [Non-text portions of this message have been removed]
                  • Mark Fitzsimons
                    Do you mean the total charge in the capsule is the same all the time, but the voltage of that charge changes as the capacitance of the capsule changes? Like
                    Message 9 of 15 , Nov 5, 2010
                      Do you mean the total charge in the capsule is the same all the time, but the voltage of that charge changes as the capacitance of the capsule changes?

                      Like pressure in a piston? you make the volume larger the pressure decreases, smaller and it increases, but the total amount of air is the same?


                      On 5 Nov 2010, at 16:39, Jerry Lee Marcel wrote:

                      > You were right in saying that the capacitor gets slightly discharged
                      > when positive pressure is applied, but the next half-period of signal,
                      > it will charge again. The net result is constant charge and no current
                      > consumption at all.
                      > The only time the bias supply has to produce some current is during the
                      > initial charging period. So yes, the bias supply must have some current
                      > capability. Very tiny though; a 60nA charge current can 99% charge a
                      > 40pF capsule in less than 300msec.
                      >
                      > L


                      [Non-text portions of this message have been removed]
                    • Mark Fitzsimons
                      thats essentially what i m doing when measuring with a 100M resistor + 20M meter, total load = 120M i did have a 1M + 8.2M voltage divider on it, total load
                      Message 10 of 15 , Nov 5, 2010
                        thats essentially what i'm doing when measuring with a 100M resistor + 20M meter, total load = 120M

                        i did have a 1M + 8.2M voltage divider on it, total load 9.2M, but that was when I thought I needed to reduce the voltage due to mistakenly thinking my DMM input was 10M when in fact it is 20M.

                        with any load over about 40M the difference is minimal. - conjecture - output impedance of the circuit is approx 5M?


                        On 5 Nov 2010, at 17:47, Nick Zuccaro wrote:

                        > why don't you put a high valued load resistor on that voltage supply so you can
                        > have a known, fixed amount of current draw? it only would have to draw a trickle
                        > of current but it will sort of tie things down load wise so you don't have an
                        > 'unknown' in the oscillator/stepup load. it would also make it less sensitive
                        > to DMM input impedances, etc. etc...
                        >
                        > _____________


                        [Non-text portions of this message have been removed]
                      • Nick Zuccaro
                        i meant to leave the resistor in the circuit, to provide a fixed, permanent termination for the output of the oscillator that defines the load impedance. it s
                        Message 11 of 15 , Nov 5, 2010
                          i meant to leave the resistor in the circuit, to provide a fixed, permanent
                          termination for the output of the oscillator that defines the load impedance.
                          it's only a step in the direction of goodness ;]



                          ________________________________
                          From: Mark Fitzsimons <mf@...>
                          To: micbuilders@yahoogroups.com
                          Sent: Fri, November 5, 2010 3:37:19 PM
                          Subject: Re: [micbuilders] Re: Schoeps dc/dc converter was GenChinaMic dc
                          converter question


                          thats essentially what i'm doing when measuring with a 100M resistor + 20M
                          meter, total load = 120M

                          i did have a 1M + 8.2M voltage divider on it, total load 9.2M, but that was when
                          I thought I needed to reduce the voltage due to mistakenly thinking my DMM input
                          was 10M when in fact it is 20M.

                          with any load over about 40M the difference is minimal. - conjecture - output
                          impedance of the circuit is approx 5M?

                          On 5 Nov 2010, at 17:47, Nick Zuccaro wrote:

                          > why don't you put a high valued load resistor on that voltage supply so you can
                          >
                          > have a known, fixed amount of current draw? it only would have to draw a
                          >trickle
                          >
                          > of current but it will sort of tie things down load wise so you don't have an
                          > 'unknown' in the oscillator/stepup load. it would also make it less sensitive
                          > to DMM input impedances, etc. etc...
                          >
                          > _____________

                          [Non-text portions of this message have been removed]




                          [Non-text portions of this message have been removed]
                        • Mark Fitzsimons
                          yes, I see. I was thinking basically the same thing. If I leave a 20M load in circuit then at least i d know the actual voltage is the same as the voltage I
                          Message 12 of 15 , Nov 5, 2010
                            yes, I see. I was thinking basically the same thing. If I leave a 20M load in circuit then at least i'd know the actual voltage is the same as the voltage I measure with my 20M DMM, (as long as I don't have both at once of course), or if I don't want to drag it down that much I could make it 100M load and measure with an 80M + 20M DMM

                            i'm pretty sure of what the real voltage is now though. with or without the load.

                            if varying the load above a certain threshold doesn't change the result then thats the true figure.


                            On 5 Nov 2010, at 20:33, Nick Zuccaro wrote:

                            > i meant to leave the resistor in the circuit, to provide a fixed, permanent
                            > termination for the output of the oscillator that defines the load impedance.
                            > it's only a step in the direction of goodness ;]
                            >
                            > ____________


                            [Non-text portions of this message have been removed]
                          • Jerry Lee Marcel
                            That s exactly it! ... [Non-text portions of this message have been removed]
                            Message 13 of 15 , Nov 5, 2010
                              That's exactly it!

                              Le 05/11/2010 19:25, Mark Fitzsimons a écrit :
                              >
                              > Do you mean the total charge in the capsule is the same all the time,
                              > but the voltage of that charge changes as the capacitance of the
                              > capsule changes?
                              >
                              > Like pressure in a piston? you make the volume larger the pressure
                              > decreases, smaller and it increases, but the total amount of air is
                              > the same?
                              >
                              > On 5 Nov 2010, at 16:39, Jerry Lee Marcel wrote:
                              >
                              > > You were right in saying that the capacitor gets slightly discharged
                              > > when positive pressure is applied, but the next half-period of signal,
                              > > it will charge again. The net result is constant charge and no current
                              > > consumption at all.
                              > > The only time the bias supply has to produce some current is during the
                              > > initial charging period. So yes, the bias supply must have some current
                              > > capability. Very tiny though; a 60nA charge current can 99% charge a
                              > > 40pF capsule in less than 300msec.
                              > >
                              > > L
                              >
                              > [Non-text portions of this message have been removed]
                              >
                              >


                              [Non-text portions of this message have been removed]
                            • Goran Finnberg
                              ... When unloaded the ripple is close to zero. As you load it down the ripple increases markedly. The original Schoeps CMC 5 amp has extensive filtering to
                              Message 14 of 15 , Nov 11, 2010
                                Mark Fitzsimons:

                                > This seems oversimplified to me - why must it be unloaded?
                                > it has a very low current capability , not none at all.

                                When unloaded the ripple is close to zero.

                                As you load it down the ripple increases markedly.

                                The original Schoeps CMC 5 amp has extensive filtering to assure that the
                                near 2 MHz dc/dc oscillator will not found its ways into the capsule buffer
                                amp nor be fed out of the mic into any external mic preamp.

                                > I read the output of my voltage converter at approx
                                > 0.8v through a 1G resistor, and as approx 6.8v
                                > through a 100M resistor
                                >
                                > 0.8v x 51 = 40.8V
                                > 6.8v x 6 = 40.8V
                                >
                                > reading with the meter directly about 38V
                                >
                                > I conclude that since there was no difference between
                                > the 1G and 100M readings that the voltage output
                                > really is around 40.8v

                                Measurements of a Schoeps CMC5 amplifier body S/N 14109

                                10 MOhm load = 54.85 V DC
                                100 MOhm load = 59 V DC
                                1G1Ohm load = 59.4 V DC
                                Electrometer = 62.5 V DC

                                The 1GOhm used in series with my 100 MOhm multimeter was selected to be 1
                                GOhm within 0.2 percent.

                                My Keithley 177 microvolt DMM was used as the calibration standard.

                                I wrote:

                                > Even using a 1 Gohm resistor in series with my
                                > 100 Mohm multimeter shows an error of nearly 20
                                > percent compared to an electrostatic meter.

                                This is from measurements done on a CMC 5 body in 1977.

                                My dim memory should have recalled that the loading was 10 MOhm not 1G1.

                                Sorry for tha error.

                                The above S/N 14109 CMC 5 body shows a near 13 percent loss when loaded with
                                10 MOhm compared to the electrometer measurement.




                                --

                                Best regards,

                                Goran Finnberg
                                The Mastering Room AB
                                Goteborg
                                Sweden

                                E-mail: mastering@...

                                Learn from the mistakes of others, you can never live long enough to
                                make them all yourself. - John Luther
                              • Mark Fitzsimons
                                If the signal is taken from the opposite side of the capsule from the polarisation, and the polarised side is AC grounded through a cap, then does that take
                                Message 15 of 15 , Nov 12, 2010
                                  If the signal is taken from the opposite side of the capsule from the polarisation, and the polarised side is AC grounded through a cap, then does that take care of the ripple? or would it still be an issue?

                                  the voltages and loads resistances you give here are much more in line with my experience.

                                  I can live with 3v error. I was concerned it might be 10v out.


                                  On 11 Nov 2010, at 23:39, Goran Finnberg wrote:

                                  > Mark Fitzsimons:
                                  >
                                  > > This seems oversimplified to me - why must it be unloaded?
                                  > > it has a very low current capability , not none at all.
                                  >
                                  > When unloaded the ripple is close to zero.
                                  >
                                  > As you load it down the ripple increases markedly.
                                  >
                                  > The original Schoeps CMC 5 amp has extensive filtering to assure that the
                                  > near 2 MHz dc/dc oscillator will not found its ways into the capsule buffer
                                  > amp nor be fed out of the mic into any external mic preamp.
                                  >
                                  > > I read the output of my voltage converter at approx
                                  > > 0.8v through a 1G resistor, and as approx 6.8v
                                  > > through a 100M resistor
                                  > >
                                  > > 0.8v x 51 = 40.8V
                                  > > 6.8v x 6 = 40.8V
                                  > >
                                  > > reading with the meter directly about 38V
                                  > >
                                  > > I conclude that since there was no difference between
                                  > > the 1G and 100M readings that the voltage output
                                  > > really is around 40.8v
                                  >
                                  > Measurements of a Schoeps CMC5 amplifier body S/N 14109
                                  >
                                  > 10 MOhm load = 54.85 V DC
                                  > 100 MOhm load = 59 V DC
                                  > 1G1Ohm load = 59.4 V DC
                                  > Electrometer = 62.5 V DC
                                  >
                                  > The 1GOhm used in series with my 100 MOhm multimeter was selected to be 1
                                  > GOhm within 0.2 percent.
                                  >
                                  > My Keithley 177 microvolt DMM was used as the calibration standard.
                                  >
                                  > I wrote:
                                  >
                                  > > Even using a 1 Gohm resistor in series with my
                                  > > 100 Mohm multimeter shows an error of nearly 20
                                  > > percent compared to an electrostatic meter.
                                  >
                                  > This is from measurements done on a CMC 5 body in 1977.
                                  >
                                  > My dim memory should have recalled that the loading was 10 MOhm not 1G1.
                                  >
                                  > Sorry for tha error.
                                  >
                                  > The above S/N 14109 CMC 5 body shows a near 13 percent loss when loaded with
                                  > 10 MOhm compared to the electrometer measurement.
                                  >
                                  > --
                                  >
                                  > Best regards,
                                  >
                                  > Goran Finnberg
                                  > The Mastering Room AB
                                  > Goteborg
                                  > Sweden
                                  >
                                  > E-mail: mastering@...
                                  >
                                  > Learn from the mistakes of others, you can never live long enough to
                                  > make them all yourself. - John Luther
                                  >
                                  >



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