# Cheap high quality volume control for BOSOZ

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#### jag

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?

#### rwagter

What a fantasic idea, my girlfriend will go screaming nuts since she's already having trouble selecting the right input.

#### UrSv

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

#### jag

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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This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.