I'm trying to understand impedance matching (or more to the point what input impedance to give a preamp cct when fed by a guitar pickup).
As I understand it traditional single guitar pickups have a DCR of approx 7k-8k ....I often read that the minimum acceptable input imedance for a guitar preamplifier is 1Meg?
why?
I'm interested in making my own low impedance pickup of say 2k DCR... I'll feed the output of this preamp into a preamplifier for boosting the signal level - what would be the optimum input impedance for the preamp circuit? (if it helps, the guitar frequency band is about 70Hz through 5khz (which I've included 5th order harmonics)
As I understand it traditional single guitar pickups have a DCR of approx 7k-8k ....I often read that the minimum acceptable input imedance for a guitar preamplifier is 1Meg?
why?
I'm interested in making my own low impedance pickup of say 2k DCR... I'll feed the output of this preamp into a preamplifier for boosting the signal level - what would be the optimum input impedance for the preamp circuit? (if it helps, the guitar frequency band is about 70Hz through 5khz (which I've included 5th order harmonics)
The interface is the classic low Z feeding a high Z so it is not matched in any sense - it doesn't need to be. However, all those turns on the pickup provide a large inductive component. The model would be a voltage source in series with the DC resistance and the inductance of the pickup. A more advanced model might include inter-winding capacitance. The pickup probably has a self resonance frequency where the inductance resonates with the distributed capacitance, but it should be at a very high frequency out of the audio range.
The inductance of the source, if loaded with significant capacitance forms a low pass filter
and will roll off the high end. The resistive component forms a voltage divider with the R-L of the source. So we want high Z and low capacitance to get the most signal transfer. The 1 Meg is usually the grid leak resistor as seen looking into a tube amp. Some use higher values.
You could put the pickup in series with a signal generator, and then load it with different loads to measure the frequency response. The pickup's generator is zero in this test and the external one acts in its place.
Good question by the way.
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Ok, but I'm more after the calculation to derive what the preferred input impedance for a preamp to mate with a specific coil (as I say, I want to make my own low impedance pickup, so I don't think the norm would apply)
So for a 7K DCR pickup coil, let's say the normal input impedance for a good matching preamplifier is 100K. (I don't know, I'm just using an example)
So what about a 3.5K DCR coil - given the above example, would the lowest acceptable input impedance therefore be 50K?
Is there a general rule of thumb or specific calculation I can use?
So for a 7K DCR pickup coil, let's say the normal input impedance for a good matching preamplifier is 100K. (I don't know, I'm just using an example)
So what about a 3.5K DCR coil - given the above example, would the lowest acceptable input impedance therefore be 50K?
Is there a general rule of thumb or specific calculation I can use?
What is your goal in doing this? If you wind a coil with fewer turns it will
have lower inductance and therefore a more extended high end assuming
that the capacitance is loading them down. You have to measure old and
new to see what the difference is. If you are trying to clone the old sound
then adjust things to match FR, if you want something new - experiment
to get the sound you are looking for.
As I understand it most want "hot" pickups, the more signal the better.
You could even do a SPICE model if you like.
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Hi PB2, sorry, my last answer was directed at Nigel's reply (check the times...we almost posted follow ups at the same time).
My goal is to understand impedance matching - particularly "if a coil's dc resistance is R, the inductance L, the maximum frequency is C ....then how do I punch those number in to come up with an preamp input impedance, so that I don't get an awful match. I don';t want to just take an easy option ....a 1Meg input resistor (which seems common), I want to understand why a 1Meg resistor. Why not for example...just a 10k resistor etc, etc?
Re your first reply - that was a mighty fine reply you gave, but I still haven't got a way of deciding what a preamp's input impedance should be for a given coil - is there no rule of thumb?
For example
My goal is to understand impedance matching - particularly "if a coil's dc resistance is R, the inductance L, the maximum frequency is C ....then how do I punch those number in to come up with an preamp input impedance, so that I don't get an awful match. I don';t want to just take an easy option ....a 1Meg input resistor (which seems common), I want to understand why a 1Meg resistor. Why not for example...just a 10k resistor etc, etc?
Re your first reply - that was a mighty fine reply you gave, but I still haven't got a way of deciding what a preamp's input impedance should be for a given coil - is there no rule of thumb?
For example
If you look at the volume and tone control pots on a typical guitar they are round about 250k.
In order for the package to work correctly the impedances of the pickups must be in a comparable range, or lower.
The net impedance seen looking back into the guitar output will vary as the tone and volume controls are operated.
As a very rough approximation the value will be a peak of 250k, dependent on the exact layout and settings.
'Matching' of input and output impedances is not really meaningful in this context since at least 10 octaves are encompassed. The wires to the amplifier are not a transmission line. It's sufficient to have a resistive termination in the range 100k-1M to ensure that the output voltage does not all appear across the Thevenin resistance of the source.
1M has the benefit of being less prone to volume changes associated with changes of the tone controls on the guitar, but a high impedance means that any induced current results in a bigger voltage swing which is more audible.
I like 250k.
w
In order for the package to work correctly the impedances of the pickups must be in a comparable range, or lower.
The net impedance seen looking back into the guitar output will vary as the tone and volume controls are operated.
As a very rough approximation the value will be a peak of 250k, dependent on the exact layout and settings.
'Matching' of input and output impedances is not really meaningful in this context since at least 10 octaves are encompassed. The wires to the amplifier are not a transmission line. It's sufficient to have a resistive termination in the range 100k-1M to ensure that the output voltage does not all appear across the Thevenin resistance of the source.
1M has the benefit of being less prone to volume changes associated with changes of the tone controls on the guitar, but a high impedance means that any induced current results in a bigger voltage swing which is more audible.
I like 250k.
w
Quite a few pro guitarists told me that they always have the volume and tone control fully up. That is max volume and max high tone.
If you choose 1M input resistor then use switched jack sockets to short out the input when no guitar is connected or you will get loud hum.
If you choose 1M input resistor then use switched jack sockets to short out the input when no guitar is connected or you will get loud hum.
wakibaki
Not sure if your answer relates to a typical passive pickup->vol-> guitar amp chain or not?
I'm trying to work out the best input impedance (for an onboard active preamp) for a bespoke low impedance pickups I hope to wind (which will about 1000 - 1500 turns of thicker wire vs a standard pickup). It seems a cop out to just have the input stage with a 1M eg Input impedance...just becuase that's what they normally do with higher impedance pickups, so I'd like to have got behind the calculations involved.
Not sure if your answer relates to a typical passive pickup->vol-> guitar amp chain or not?
I'm trying to work out the best input impedance (for an onboard active preamp) for a bespoke low impedance pickups I hope to wind (which will about 1000 - 1500 turns of thicker wire vs a standard pickup). It seems a cop out to just have the input stage with a 1M eg Input impedance...just becuase that's what they normally do with higher impedance pickups, so I'd like to have got behind the calculations involved.
look at theory/data at:
UIUC Physics 498POM Guitar Pickup Measurements
with typical pickups you have Henrys of inductance so peak coil Z in audio band is near a MegaOhm
winding C puts pickup self-resonance in the audio output response band, practically any loading affects frequency response
if you want you can EQ the response with differing coil loading
the lower the input Z in relation to the source, the greater the interaction/attenuation, 2 KOhm Zin will kill your signal with a very low corner frequency rolloff
if you want your electronics to not interact with the pickup's "sound" you would need to make a very high Z input over the audio range - likely requiring active bootstrapping of cable C and op amp common mode impedance
a "classic" guitar pickup "sound" will be specific to a particular combination of pickup, cable and preamp Z "mismatch"
"impedance matching" is not usually important in audio electronics, matching maximizes enery transfer, minimizes reflections which is important in RF and makes response equally sensitive to both sides of the interface
in audio electronics we usually try to minimize interaction between source and load by making the impedances as different as possible on each side of the interface: pre amp Zin >> source Z
a 1 dB attenuation from impedance loading would require a ~10x higher preamp input Z than source Z
UIUC Physics 498POM Guitar Pickup Measurements
with typical pickups you have Henrys of inductance so peak coil Z in audio band is near a MegaOhm
winding C puts pickup self-resonance in the audio output response band, practically any loading affects frequency response
if you want you can EQ the response with differing coil loading
the lower the input Z in relation to the source, the greater the interaction/attenuation, 2 KOhm Zin will kill your signal with a very low corner frequency rolloff
if you want your electronics to not interact with the pickup's "sound" you would need to make a very high Z input over the audio range - likely requiring active bootstrapping of cable C and op amp common mode impedance
a "classic" guitar pickup "sound" will be specific to a particular combination of pickup, cable and preamp Z "mismatch"
"impedance matching" is not usually important in audio electronics, matching maximizes enery transfer, minimizes reflections which is important in RF and makes response equally sensitive to both sides of the interface
in audio electronics we usually try to minimize interaction between source and load by making the impedances as different as possible on each side of the interface: pre amp Zin >> source Z
a 1 dB attenuation from impedance loading would require a ~10x higher preamp input Z than source Z
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The usual maths behind it is for maximum transfer of power and that happens when the source and destination impedances are the same.
In the real world it doesnt usually matter too much as long as the source impedance is less than the destination impedance.
In the real world it doesnt usually matter too much as long as the source impedance is less than the destination impedance.
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