Ok, I’ve sat down and tried to work out some of the math for this and have gotten somewhere, but not completely.
So with my starting stats of 150Ω/2000Ω I can follow example 1 from that detailed webpage for the post part.
- Convert 20dB to a ratio. The formula is: k = 10^(db/20). That’s 10 to the power (db/20). The calculator says 10. 20 dB is a ratio of 10:1. If you have a table of K factors, you’ll find this value in the column with K at the top.
- To achieve this value of loss, the resistors that make up the voltage divider need to have values that fulfill this relationship: k = 1 + (Rseries/Rshunt). But where do you start? There are a jillion values you could start with, but the one that makes sense is the shunt resistor, which effectively sets the value of the output impedance.
So, Rshunt = Zout = 150 ohms (the microphone imedance).
- Using algebra, manipulate the formula to solve for Rseries when we know Rshunt and k.
Rseries = Rhunt * (k - 1). Since k = 10, k - 1 = 9. In the K factor table, there is a column for the value K - 1.
- Continuing, Rseries = Rshunt * 9 = 1350 ohms.
- Since this is a U-pad, for a balanced system, we split the series arm into two equal parts. 1350 / 2 = 675 ohms.
- Resistors only come in particular values, so you need to convert these calculated values into standard ones. For 5% tolerance resistors, there are 24 values per decade (a decade of values is a range of 10:1, the first two digits are the same, and they get multiplied by factors of 10 to get the higher values). For 1% resistors, there are 96 values per decade. We’ll use 5% resistors here. 150-ohms is a standard value, and 680-ohms is the nearest standard value.
- Summarizing: get 2 680-ohm 5% resistors and 1 150-ohm 5% resistor. Connect as shown in the diagram for a U-pad (balanced-L). If you want to use 1% resistors for improved common mode rejection (CMR) and maybe sonics, the nearest 1% values are 681 ohms and 150 ohms.
I was able to follow that and more-or-less make sense of it. This gives me the values I would need if I’m wiring up a U-configuration pad, and will show up to the mic as an impedance of 1500Ω. Where it gets confusing is that my RME has an actual impedance of 2000Ω. I don’t know how fussy this kind of stuff is, but I guess I’d want the pad to show up as transparently as possible, so I want to try to solve for that.
On that webpage, the next step sort of covers this:
Find the values for the 30 dB pad using the table. The pad input impedance is 4550 ohms (verify this for yourself). What do we need to parallel this with to get 1500 ohms? This additional parallel resistor is R4 in the diagram.
Use a variation of the parallel resistor formula: Rtotal = 1/((1/R1)+(1/R2)). Solving for R1 we get 1/((1/Rtotal)-(1/R2)). The calculator says 2238 ohms. 2200-ohms is the closest 5% value.
I find that really confusing, but using my trusty Max was able to implement the formula and get the same values:
EXCEPT
In my context, my “total value” (Rtotal) is 1512Ω and the value I want at the end is 2000Ω.
I don’t know enough about maths to make proper sense of what’s going on, but it appears that no value of resistor will give a higher value than the input here:
I’m guessing because this is some kind of voltage divider, the impedance produced can’t be higher than what you start with.
SO
Is the difference between a preamp impedance of 1500Ω to 2000Ω a “big deal”, or is there a way to compute the values such that I can get a 20dB loss and still have an input impedance of 2000Ω?
