Hopefully this is the right place to post. While this is related to the Crack, my questions are of a more general. nature. I apologize up front of the length.
I am slowly trying to understand the workings of the Crack circuit. A problem myself and others have had is that there is very little useful range on the volume control. This resulted in a couple of solutions to get more range on the control, which led me to think about other options; in particular using discrete switches for a combined coarse and fine volume adjustment. This thread,
http://bottlehead.com/smf/index.php?topic=9273.0, adds a discrete switch for variable attenuation before the volume control in the Crack circuit.
My questions are how to calculate attenuation and impedance for different circuits. As I understand it, whether a step attenuator or potentiometer, attentuation in its simplest terms is a voltage divider and is illustrated in the first diagram. One solution to the Crack volume issue was to simply place a resistor between the input and potentiometer; which would look like the second diagram. However, it was said that this would alter the impedance. The solution offered in the Crack FAQ#3 use two resistors, one as in the first solution and a second across the input to ground lugs of the potentiometer. I believe this can be represented as diagram three. Which could also represents a coarse/fine setup or that in the thread in the above. But in its simplest form is just a voltage divider in front of a voltage divider.
________ -----R3------
| |
R1 R1
|------ |-------
R2 R2
________|____ __________|_____
______
|
R3
|____
| |
R4 R1
| |-------
| R2
______|____|_____
To simply things, I will assume that R1 and R2 represent a potentiometer or step switch where R1 + R2 = constant = Rpot..
If I have it right, attenuation in dB is calculated by dB = 20*log(Vout/Vin). For diagram 1, this can be calculated as dB = 20*log(R2/Rpot). Impedance is R1 + R2 or Rpot.
Attenuation in the second circuit would be dB = 20*log(R2/(R3+Rpot)). Impedance would be R3 + Rpot.
Hopefully I’m correct to this point. However, I get confused on the 3rd circuit. From an attenuation standpoint, can the two dividers be considered as discrete and additive? That is the attenuation of the first voltage divider is added to the attenuation of the second voltage divider and its attentuation is calculated independent of the second voltage divider. Or does it have to be looked at in its entirety. That is the second voltage divider is in parallel with R4.
The former would have total attenuation calculated as dB = 20*( log(R4/(R3+R4) + log(R2/Rpot) ) and attentuation of the first voltage divider is dB = 20*log(R4/(R3+R4))
The latter would have attenuation of the first voltage divider as dB = 20*log( R||/(R3+R|| ). Where R|| = R4*Rpot/(R4+Rpot). (Since R1 and R2 are in series, they reduce to R = R1 + R2, which is Rpot). Total attenuation would be ??
Or is it something entirely different?
For the calculation of impedance I am at a total loss. R3+R||?
For those relating this post to the thread mentioned earlier, I believe that position 9 for the Right channel in Paul Birkeland’s diagram changes the effective circuit from my diagram 3 to diagram 2 for that position.
