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Can someone explain "Buffering"?

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  • Can someone explain "Buffering"?

    I'm trying to teach myself a few concepts, and one of the ones i've come across a few times is "Buffering". Its mentioned in terms of a signal either from a guitar or bass.. Such as "The signal from the pickup goes through a buffer"..

    Can someone explain what this means, and the benefits/cons of it?

    bel

  • #2
    When you buffer a pickup, the pickup "sees" a high impedance load. Normally you have the pot resistance and cable capacitance loading the pickup. This acts like a second order low pass filter, and dulls the top end and lowers the resonant peak. Some people hear this as "warm", "organic" or "vintage". A buffered pickup will sound brighter and more hi-fi, or "modern".

    I've used simple JFET buffers in basses that didn't add any gain or change the EQ, and when engaged you would hear the high end extend higher, and the low end sounded tighter and deeper. For guitars this is often not the tone you want. My Tele was way too strident with a buffer, and I couldn't play it without turning down the tone knob a little, or sticking my fingers in my ears! That made it hard to play. So I removed it.

    But... a buffer with a lower input impedance would give a different tone. I have an old Bartolini Hi-A TC-1 preamp, and its input impedance is 200K, so it doesn't accentuate the top end as much as the JFET buffers. It sounded more like the pickup used passive, but with a slightly crisper tone.

    The Audere bass preamp has impedance switches to change the loading.

    Now some bass players like the active tone, and some don't. So it's common to have bypass switches for the preamps. But this takes away some of the benefits you can have with an onboard bass preamp, such as active blending of the pickups. And some pickups, like some of the old Lane Poors, needed to have their level boosted because they were very under wound for a cleaner tone. Also some pickups are overly dark without the buffer, like with Wals.
    It would be possible to describe everything scientifically, but it would make no sense; it would be without meaning, as if you described a Beethoven symphony as a variation of wave pressure. — Albert Einstein


    http://coneyislandguitars.com
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    • #3
      I think of the buffer as a way to "lock in" the tone of a pickup close to the source. Nothing downstream of the buffer (volume pots, cable capacitance, impedance loading and inductive loading from other pickups) will be able to affect the tone very much.
      It can be a good thing -or not...

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      • #4
        Thanks very much for the information .. So are buffers that are typically in guitars/basses Voltage buffers or Current buffers? My guess is that they are current buffers because they are commonly used on low impedance pickups...

        Comment


        • #5
          Originally posted by belwar View Post
          Thanks very much for the information .. So are buffers that are typically in guitars/basses Voltage buffers or Current buffers? My guess is that they are current buffers because they are commonly used on low impedance pickups...
          Belwar,

          Buffers have one primary role; isolating the pickup impedance from external loads that tends to change the tonal characteristics of the typical passive pickup. Buffers provide no gain but because the loading on the pickup is reduced, some higher frequencies may sound a slight bit louder when switching between the passive pickup and switching in the buffer circuit.

          Buffers, by their nature, convert a high impedance to a low impedance. This low impedance is less affected by the capacitance added by normal guitar cables in the range of 300 pf to 500 pf for a 10 ft cable.

          Passive guitar pickups have a resonant peak that is somewhat loaded by the on-board volume and tone controls. With a buffer mounted right after the pickup or right after the selector switch, the natural resonant frequency of any guitar pickup is accentuated because the pickup loading is much less.

          Buffered pickups can drive long guitar cables with no noticable high frequency losses.

          Piezo pickups are high impedance and need very high buffer input impedances in the range of 5 to 10 meg ohms to have their full response.

          One clever design puts a FET semiconductor inside the guitar-end of the guitar cord jack so that any guitar using this device has the benefit of a partially buffered load on the pickups. However, I use the word partially because the volume controls and tone controls still present some loading on the pickup. What is eliminated is the 300 pf to 500 pf coax loading that can slightly dull the very high harmonics.

          Aphex makes a line of products called the "Xciter" for acoustic, electric and bass guitars. These devices typically have an input impedance of 10 meg ohms and provides active gain through the use of multiple tone controls.

          To answer you question, the buffers work as voltage buffers to allow the pickup to output it's maximum voltage. During the process of buffering the low output impedance actually increases the drive current to drive lower impedance devices without these devices reflecting their low impeance load back on to the pickup where the pickup's tonal quality would be affected.

          I hope I have answered your question?

          Joseph Rogowski

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          • #6
            Buffering can make a pickup sound less bright. This happens when the removal of the cable capacitance puts the resonant peak up too high. Then one no longer has the resonant boost in the few KHz range. When you buffer a pickup from the cable capacitance you can and should adjust the resonant peak with capacitance before the buffer or use active tone circuits in order to get the best sound for the application.

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            • #7
              I still cant figure out how to measure impedance.. I get the concept that it is the AC equivilant of DC Resistance, but how does one measure it?

              What kind of impedance (example measurement) would one expect to see after a buffer?

              matt

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              • #8
                Originally posted by belwar View Post
                I still cant figure out how to measure impedance.. I get the concept that it is the AC equivilant of DC Resistance, but how does one measure it?

                What kind of impedance (example measurement) would one expect to see after a buffer?

                matt
                Matt,

                Impedance is only present in AC circuits and varies with frequency.

                The guitar has a primary frequency range of 82Hz to about 1300Hz, however the harmonics, which contain lesser energy than the fundamental, and may not reach the ear if they are masked by the stronger lower harmonics or if the high frequency response of the pickup falls off at too low a frequency. Ultimately, what sound like a good guitar pickup is not a scientific question, but what many people say "does sound good" can be scientifically analyzed, discussed and replicated.

                All pickups, (magentic (inductive) and piezo (capacitive)) have a source impedance that is mostly inductive or capacitive. Example. The windings on a pickup have a thin insulation causing each wire wind to lay very close to other winds. With the many thousands of AWG 42 or 43 wire turns the coil has some capacitance, called phantom capacitance, typically between about 50 pf to about 200 pf. This circuit would look like a pure inductor (the pickup) in parallel with this phantom capacitance and would exibit a resonant frequency in the mid (3Khz to 6KHz) frequency range. This is what gives different pickups their characteristic sound.

                The pickup typically sees the following loads on it which affect it's tonal quality, volume pot value, tone pot value, tone capacitor value, guitar cable capacitance and amplifier input impedance. The buffer also sees these loads but with a much lower output impedance so the effects of these loads have less effect on the sound quality and do not reflect back to the pickup. EMG pickups have built in buffers and typically use 25K ohm pots which is 10 time lower than traditional 250K pots and 20 times less than 500K pots.

                Connect your guitar to an amp with a very short cable (3 ft) and then use a longer cable (15 to 20 ft) and you can hear the effect of the additional cable capacitance on the upper tonal range of the guitar harmonics. The impedance of a guitar cable changes with frequency. Example: a 10 ft coax selected at random, exhibits a measured (with an Extech LCR meter) AC resistance of 2.8K ohms, Q of 162, capacitance of 356pf at 1KHz with an impedance of about 453K ohms and about an AC resistance of 6.7K ohms, Q of about 300 at 120Hz with an impedance of about 2 meg ohms. This cable capacitance acts in parallel with the phantom capacitance of the pickup wire to lower the resonant peak of the pickup. From a tonal perspective the guitar cable had the same effect as putting a 356pf capacitor across the pickup's output.

                Using the Extech meter you can measure impedance by multiplying the AC resistance reading by the "Q" reading. This gives the reactance which will be very close to the impedance at two frequencies, 120Hz and 1000Hz.

                When the inductive reactance of the pickup (based on it's inductance) is equal to the capacitive reactance of the phantom capacitance and the guitar cable capacitance, the pickup will exhibit a resonant peak. This resonant peak has an amplitude called "Q". Pot values across the pickup tend to lower the effective Q of the resonant peak. Up to the resonant peak, the impedance is mostly goverened by inductive reactance. Beyond the resonant point the impedance is mostly governed by capacitive reactance. Change guitar cable lengths or guitar cable diameter (quality or brand) and you will vary the capacitive load on the pickup and in some situations, like my example above, hear the difference.

                Because the resonant peak of most pickups is well above the fundamental frequency of the played notes, only the harmonics are affected by the resonant frequency. This resonant frequency will tend to accentuate some harmonics over others, giving single coil, humbuckers or over wound pickups their characteristic sound.

                The rule of thumb is to use a load, such as a pot value that is 10 times the impedance of the source (the pickup). That is why if you want a telecaster to have a snappy sound, some people will change the 250K pot to a 1Meg pot to reduce the loading and make the resonant peak a little more pronounced. However the effective load of the pot is also in parallel with guitar cable and the effective input impedance of the amplifier input circuit impedance which is typically around 1Meg.

                One way to fine tune your pickup, but only your ears can do this, is to add a high impedance buffer which isolates the pickup from the guitar cable capacitance and amplifier input impedance and then manually load the input to the buffer with a resistance and/or a small capacitance (100pf to 400pf). to suite your ear. This way the sound of your guitar will be the same with short or long guitar cables or different amp input impedances. This is a personal and subjective evaluation but can be described as I am doing.

                Pickups are complex sound generators because their individual designs causes them to act a resonant filters, accentuating certain frequencies and attenuating a particular range of harmonic frequencies. Buffers are one way to keep the response of a given pickup rather fixed, independent of what follows it.

                Joseph Rogowski
                Last edited by bbsailor; 09-13-2009, 06:48 PM.

                Comment


                • #9
                  Originally posted by belwar View Post
                  I'm trying to teach myself a few concepts, and one of the ones i've come across a few times is "Buffering". Its mentioned in terms of a signal either from a guitar or bass.. Such as "The signal from the pickup goes through a buffer"..

                  Can someone explain what this means, and the benefits/cons of it?
                  Having read the other explanations of impedance, which are perfectly clear to me yet I fear may not be sufficient, I think it's time to resort to the hydraulic analogies traditionally used to teach electricity.

                  Voltage is equivalent to hydraulic pressure (pounds per square inch).

                  Current is equivalent to fluid flow (gallons per minute).

                  Resistance (the DC form of impedance) in ohms is equal to the ratio of the voltage (in volts) across a component to the resulting current through that same component (in amps).

                  In hydraulics there is a parallel to resistance, being the pressure (in psi) across a hydraulic component or assembly divided by the resulting flow (in gpm) through that component.

                  Let's assume that the hydraulic component is a bit of rigid metal pipe of specified length and internal diameter. For a given pressure, the longer the pipe and/or the smaller the pipe, the lower the flow, so a long thin pipe would have a high resistance to flow, and a short large pipe would have a low resistance to flow.

                  If we suddenly raise or lower the pressure, the flow will suddenly increase or decrease. (I'm assuming that the speed of sound is too fast to measure in this test.)

                  Now for some fun. What happens if the rigid metal pipe is replaced with a length of rubber tubing? While a sudden increase in pressure causes a sudden increase in fluid entering the tube, it will be some time before the flow leaving the tube increases to match, the difference being taken up by expansion of the tube. And later when the pressure is suddenly decreased, the flow out of the tube continues for awhile as the tube shrinks back to its original size.

                  What is the resistance of this rubber tube?

                  What electrical component does the rubber tube resemble in behavior?

                  Comment


                  • #10
                    Originally posted by Joe Gwinn View Post
                    Having read the other explanations of impedance, which are perfectly clear to me yet I fear may not be sufficient, I think it's time to resort to the hydraulic analogies traditionally used to teach electricity.

                    Voltage is equivalent to hydraulic pressure (pounds per square inch).

                    Current is equivalent to fluid flow (gallons per minute).

                    Resistance (the DC form of impedance) in ohms is equal to the ratio of the voltage (in volts) across a component to the resulting current through that same component (in amps).
                    Im with you so far.. Basic stuff from Ohms Law. R=E/I

                    Originally posted by Joe Gwinn View Post
                    In hydraulics there is a parallel to resistance, being the pressure (in psi) across a hydraulic component or assembly divided by the resulting flow (in gpm) through that component.
                    Ok you've lost me - That sounds exactly like resistance based on your example above. If you voltage is your PSI, and your current is your GPM, then the formula is the same... PSI/GPM = Parallel resistance? or R = E/I.. What am I missing?

                    Originally posted by Joe Gwinn View Post
                    Let's assume that the hydraulic component is a bit of rigid metal pipe of specified length and internal diameter. For a given pressure, the longer the pipe and/or the smaller the pipe, the lower the flow, so a long thin pipe would have a high resistance to flow, and a short large pipe would have a low resistance to flow.

                    If we suddenly raise or lower the pressure, the flow will suddenly increase or decrease. (I'm assuming that the speed of sound is too fast to measure in this test.)

                    Now for some fun. What happens if the rigid metal pipe is replaced with a length of rubber tubing? While a sudden increase in pressure causes a sudden increase in fluid entering the tube, it will be some time before the flow leaving the tube increases to match, the difference being taken up by expansion of the tube. And later when the pressure is suddenly decreased, the flow out of the tube continues for awhile as the tube shrinks back to its original size.

                    What is the resistance of this rubber tube?

                    What electrical component does the rubber tube resemble in behavior?
                    The rubber tube acts like a capacitor, storing energy..

                    Beyond that im lost as to what the above example tells me about impedance...

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                    • #11
                      Originally posted by belwar View Post
                      Ok you've lost me - That sounds exactly like resistance based on your example above. If your voltage is your PSI, and your current is your GPM, then the formula is the same... PSI/GPM = Parallel resistance? or R = E/I.. What am I missing?
                      Nothing. The parallel is exact, although with one resistor, parallel versus series doesn't apply.

                      The rubber tube acts like a capacitor, storing energy.
                      Exactly.

                      What's the hydraulic equivalent to inductance?

                      Beyond that I'm lost as to what the above example tells me about impedance...
                      Note that the flow is now out of time phase with the drive voltage. What effect does this have on the definition of resistance or impedance?

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                      • #12
                        OK I'm going to jump right in and say that it's frequency dependent.
                        Do I get a prize?

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