Re: Attenuator calculations.
No, no, it comes out more easily as a tapped series string of resistors. You don't have to make 48 completely independent 2-resistor attenuators.
1) it only needs half as many resistors
2) it avoids tolerancing of resistor values causing exaggerated jumps at some steps.
So a string of resistors. top stud to input signal 12k resistor to next stud, from that 12k resistor to next stud
then10k
9.1k
7.5k
6.8k
5.6k
4.7k
4.2k
3.6k
3.3k
2.7k
2.2k
2k
1.8k
1.5k
1.2k
1.2k
1k
910
750
680
560
470
420
360
270
220
200
180
150
120
120
100
91
75
68
56
47
42
36
33
27
22
20
18
15
and the last step is a big jump to complete attenuation, 100 ohms to the bottom (grounded) stud.
This gives you a 98.89k pot! and steps of about 1.12 to 1.32dB
David
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Can't afford the volcanic island yet, but the plans for my monorail and the goons' uniforms are done
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