I think you're best getting the correct lpad, it'll work but with what drawbacks is out of my experience.
Alright, I really tend to get sucked into these things too deep. I have two 8 Ohm L-pads lying around; I haven’t been able to find 16 Ohm L-pads so far, and if I do I have a feeling they will be too expensive.
So here’s my thoughts about the L-pad being 8 Ohm for a 16 Ohm rated speaker:
The maths part: Assuming I have understood how a 8 Ohm L-pad works with a 8 Ohm speaker, the following should be true:
1) at zero attenuation, the series R in the L-pad is close to 0 and the parallel R close to infinite; reflected impedance is 8 Ohm (for those frequencies the speaker impedance is actually 8 Ohm)
2) at full attenuation, the series R is 8 Ohm, parallel R is close to 0; reflected impedance is the 8 Ohms to ground
3) at half attenuation, the series R is 4 Ohm, parallel R is 8 Ohm; reflected impedance is 4 Ohm + the 4 Ohm from the 8 Ohm in parallel with 8 Ohm = 8 Ohm
Now if I apply the same values to a 16 Ohm speaker:
1) at zero attenuation: 0 + 16 Ohm = 16 Ohm
2) at full attenuation: 8 Ohms to ground
3) in the middle: 4 Ohm + (8 Ohm + 16 Ohm) / 2 = 16 Ohm
Considering the blanks between these 3 points fill in smoothly in a curve, does that mean that if I put a 8 Ohm L-pad in front of a 16 Ohm speaker, the reflected impedance stays at 16 Ohm from no attenuation to half attenuation and then falls off to 8 Ohm?
If yes, this wouldn't be all that bad – ”half” attenuation sounds plenty.
Or the other option: I get a few fixed resistors and try 3-5 different combinations until I find the one I like best.
Cheers!
