soulsonic wrote:I see what you mean. I have certainly gotten some good results with it being lower before. The trend with "selling" a buffer seems to be for it to be high, though...
I have a question: there is 2M2 going to ground at the input, and this is in parallel with the 2M2 going to virtual ground after the 100n capacitor. So the AC impedance for the input then is about 1M, I assume? And what difference does it make to how the pickup interacts that part of the load is DC coupled and part is AC coupled? Would there be an appreciable difference in the sound if instead of using equal values, have perhaps the first one be 10M and AC coupled one be 1M? Would this sound very much different, I wonder?
Hi,
Let’s assume you have a guitar connected to the input and further only take into account that the output of an electric guitar is a choke of relatively low Q with a peak caused by wiring capacitances. We’ll for the sake of below simplicity model the output as a choke with a capacitive load and henceforth call it the guitar pick up.
What the pick ups sees is the composite load and if you draw an AC schematic of the input circuit you’ll see that Volume and Tone controls ( assumed to be on full) appear as parallell loads and in parallell with those is also mainly the capacitance of the guitar cable used ( for the sake of this model let’s ignore other ghost
components that appear in the wiring-this because most of those form filters way above the audio band) and further in parallell you get as parallell load a resistor that holds input capacitor at ground potential; let’s now assume that the reactance of the input capacitance is low enough to be ignored for audio frequencies
( when in doubt compute corner frequency) as in this example corner frequency computes at roughly 1,6 Hz; then the bias return resistor at the non inverting input appears as parallell load and lastly also in parallell the input impedance of the OP amp appears- this is computed as the Ri of the OP amp multiplied with the gain
at the point in bandwidth and for practical purposes here this value can be ignored with the OP amp used as this load will be more thousands times larger than any of the other loads.
So what do we have with values shown and just looking at the input of the circuit AC load computes roughly at DC reference resistor//OP reference resistor and with values 10M//2M2≈ total load
Hence the composite load that the input circuit forms for AC becomes the parallell of the loads ( regardless if they are connected to ground or a DC-potential that for AC is considered ground)
Depending on OP amp used the resistor to non inverting input may be selected to give the smallest DC offset at output-some fine bipolar OP amps draw input current and for instance the RI of a 741 is on the order of 100K while that is to be multiplied with the gain at the frequency that impedance is calculated and thus for
the most part on the order of a few Mega Ohms.
Techniques to freely adjust input impedance and keep DC shift low includes ’Bootstrapping’ and the point of ’Bootstrapping is to reduce current through the OP reference resistor to near zero and thus the impedance it forms for AC will be on the order 100 Mega Ohms or more.
Summary is that if you wish to achieve a particular input impedance to work as load on a guitar pick up selection of resistor values can be done freely.
Further assume a guitar like a Stratocaster and using the bridge pick up:
Internal load in the guitar is the volume pot, let’s say 250K Ohms
If you make a model of this ( and for this purpose ignore the ghost components created) you get a choke output that is loaded by 250K and it is when connected with a cable further loaded with the capacitance of the cable say it’s a 6 meter standard cable and internal capacitance of that is 750pF.
Further assume you have connected this to a buffer that has an input impedance of 2M2.
Simplifying and jast taking the major influences gives
Choke output is loaded by 250K//2200K//750pF ( to solve you first compute the resistance)
What the capacitance does is form a 2 pole LP filter with the choke output meaning that slope will be about 12dB/Octave ( but this filter will have resonance)
Compare again the same but now with buffer input at 500K
Choke output is now loaded with 250K//500K// 750pF ( to solve you first compute the resistances)
If you draw a frequency/impedance graph of the strat pick up ( i.e. a typical Fender stratocaster pick up in bridge position) in this example we can assume that a low passfilter with a resonant peak is formed and such a filter will mainly have its resonance affected by the composite load it is exposed to.
One can show that if this pick up is loaded by 100pF and a varying load of 2M2 to 10K the sharp resonance at 2M2 is flattened to a slope without resonance at 10K and in between values adjust the height of the resonance
It is sufficient to accept this to see what the load does and now there is no correct answer to a perfect load as what sounds perfect is firstly not scientifically defined in any model and further pick up loading is just a small part of the composite system string to speaker cloth
However a fun experiment is to make the DC reference resistor (as mentioned above variable from 2M to 10K and make the buffer have large enough impedance not to affect too much.
You should be able to hear how higher loads allow more of the pick up resonance and you may also find a point where you think resonance just feels good.
Anyway just adjusting from say composite load of 1M to 10K will do two things partly affecting the resonance of the pick up and also affect the taper of the volume control when set at other positions that full or zero.
At your service
BJ
BJF Electronics
sab-buffer1.pdf
