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#1 |
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diyAudio Member
Join Date: Aug 2002
Location: TX
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Even though this is my first post, I have been reading this site for some time now (great place!). I have been seriously motivated to attempt BOSOZ and have read several posts - I now know where to get PCB (audioXpress) and case (par-metal). One missing thing has been an adequate potetiometer. I have been thinking about that when I stuck this idea (comp sc background?).
Put following 8 resitors in series with a switch in parallel with each of them (8 switches): ---R1---R2---R3---R4---R5---R6---R7---R8--- |_\__|_\__|_\__|_\__|_\__|_\__|_\__|_\__| S1 S2 S3 S4 S5 S6 S7 S8 The resistor values (in ohms) are: R1 = 20 R2 = 40.2 R3 = 80.6 R4 = 160 R5 = 320 R6 = 642 R7 = 1290 R8 = 2550 Now, if 0 is off and 1 is on then S1 S2 S3 S4 S5 S6 S7 S8 1 1 1 1 1 1 1 1 = 0 ohm 1 1 1 1 1 1 1 0 = 20 ohm 1 1 1 1 1 1 0 1 = 40.2 ohm 1 1 1 1 1 1 0 0 = 60.2 ohm 1 1 1 1 1 0 1 1 = 80.6 ohm . . 0 0 0 0 0 0 0 0 = 5102.8 ohm Hence with 8 switches on the panel we get 256 steps in increments of 20 ohms all the way upto 5kohms. With 6 switches we can get 64 steps, with 9 switches 512 steps, and so on (the resistor values will be different for acheiving 5k total). For 4 pots we can use ganged switches or even separate switches with separate set of resistors. Also, I think 8 switches on the panel will look more cool than a knob (I know my wife will have a serious challange adjusting volume with this scheme, but it comes almost intuitively to me). Does this look like a good idea, or am I smoking crack here? |
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#2 |
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diyAudio Member
Join Date: Jul 2002
Location: Netherlands
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What a fantasic idea, my girlfriend will go screaming nuts since she's already having trouble selecting the right input.
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#3 |
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diyAudio Member
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Well that could work except that you need a log characteristic for the volume attenuator. This is linear. On the other hand if you connect a resistor in parallell with this thing and then make it the shunt element in a shunt attenuator it does approach logaritmic and working state.
/UrSv |
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#4 |
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diyAudio Member
Join Date: Aug 2002
Location: TX
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UrSv:
Thanks for the tip. I was trying to achieve log response by adjusting resitor values but gave up after my head started hurting (doing the calculations - I am not even sure if that can be done). So, in order to achieve 5 k total, should it be shunt 10k and fixed 10k or some other values (higher or lower shunt as compared to fixed) are better? Also, pl note that I mentioned the resistors in reverse order. Actually they should be: R8 = 20 R7 = 40.2 R6 = 80.6 R5 = 160 R4 = 320 R3 = 642 R2 = 1290 R1 = 2550 |
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