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At the moment I am trying to create an audio circuit, in which I need to reduce an audio signal from 2 V into a variable output of 0,030 V to 0,350 V. For that I think a Logarithmic Pot will do the trick, but I don't know how to get it to 0,350 V first.

But here another thing. I plan to enclosed the circuit in a box and I don't want to have potentiometer inside of it which I can precisely control, i.e. let's say the box has a hole so you can reach the potentiometer inside a rotate accordingly with a screwdriver, there's no real way to say: "ok I move it a quarter to the right or to the left. Therefore I would like to have a switch, digital, a hex switch I don't know I have no idea, that I can externally select let's say letter B and that means that the voltage will be decrease 20 dB, or when I select letter G the voltage will decrease 40 dB.

I hope I have explained myself, I'm a little new to this so sorry if I am mistaken or misunderstood something.

Thanks in advance for your help.

=)

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  • \$\begingroup\$ Just us a switch or rotary switch for multiple attenuation levels. There are rotary encoder trimmers that are adjusted with little plastic screw drivers that are binary (not gray code, actual binary) that could in theory be wired to switch resistors in various ways that yield stepped attenuation (eg 3 wires + common is 2*2*2 = 8 steps). Although they are designed for digital signals so they could only handle low voltage / low current signals. \$\endgroup\$ – squarewav Nov 6 '14 at 2:30
  • \$\begingroup\$ Hello ioplex thanks for your help, I stil have a question I don't if anybody can answer it, maybe it's a stupid question but I don't know the answer. How do I lower the Voltage from 2 volts to 0,030 V before starting to "control" the range from 0,030V to 0,350V . Thanks. \$\endgroup\$ – MeMe2 Nov 6 '14 at 7:37
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All a volume control is, is a variable voltage divider.

You don't have to have it as a smooth track - you can do it with switches and fixed resistors.

For instance, the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

With no switches closed, Vout == Vin.

With SW1 closed you get \$V_{OUT} = \frac{R2}{R1 + R2}×V_{IN} = \frac{100k}{10k+100k} = 0.9×V_{IN}.\$.

With SW2 closed you get \$V_{OUT} = \frac{R2}{R1 + R2}×V_{IN} = \frac{10k}{10k+10k} = 0.5×V_{IN}.\$.

With SW3 closed you get \$V_{OUT} = \frac{R2}{R1 + R2}×V_{IN} = \frac{1k}{10k+1k} = 0.09×V_{IN}.\$.

You can also combine switches. With SW1 and SW2 closed the "lower" resistor becomes \$\frac{R2×R3}{R2+R3} = \frac{100k × 10k}{100k + 10k} = 9090\Omega\$ so the output voltage would be \$\frac{9090}{9090 + 10000} = 0.476×V_{IN}\$

You can have as many switches as you like, and whatever combination of resistors suits your needs.

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  • \$\begingroup\$ Hello ioplex thanks for your help, I stil have a question if there anyway to simulate a logarithmic pot with this switched you mentioned? Thanks. \$\endgroup\$ – MeMe2 Nov 6 '14 at 7:38
  • \$\begingroup\$ Just choose resistor values that suit how you want to work. \$\endgroup\$ – Majenko Nov 6 '14 at 10:01
  • \$\begingroup\$ Vout 0,3556 0,25 0,1782 0,125 0,0892 0,0625 0,0446 0,03126 in dB’s -15dB -18dB -21dB -24dB -27dB -30dB -33dB -36dB These are the Voltage ouput that I need to obtain the corresponding dB attenuation, my input is 2 V. But I dont know how to manage all the switch/resistors relation. Thanks. \$\endgroup\$ – MeMe2 Nov 6 '14 at 10:06
  • \$\begingroup\$ Majenk what I mean is, I already have the relationship between Vout/Vin I need, I have R1, how am I supposed to isolate the R2, R3, and so on in order to know those resistor values? \$\endgroup\$ – MeMe2 Nov 6 '14 at 14:01
  • \$\begingroup\$ If you know the input voltage, and you know the output voltage, and you know R1, you can very easily calculate R2. \$R2 = R1\frac{1}{\frac{V_{IN}}{V_{OUT}}-1}\$. More detail on: en.wikipedia.org/wiki/Voltage_divider \$\endgroup\$ – Majenko Nov 6 '14 at 14:25
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I think you might be able to use one of the step type variable attenuators. Copal has advertised them as T2114. It is specified from DC to RF and standard at 600 Ohm I think, looks like it is available in balanced and unbalanced versions. They are binary devices that have a maximum of 15.5 or 31 dB attenuation.

I only ever saw one (same form factor, supplier unknown) in a dead facsimile machine to set the TX levels for regulatory compliance to match country PSTN standards. I de-soldered it of course and briefly did use it to calibrate an anemometer to a random moving coil meter successfully.

If you can find these within budget, the ratings (max attenuation, resistance, power) are suitable, the 16 steps are enough (you can stack two for 256 steps) and you do not need to change the setting regularly (will wear out) then you are sorted.

copal-electronics.info/en/CopalElectronics/attenuators_en/rf-components_en.pdf

(edit) to reduce the voltage from 2000 to 350 mV you need to reduce it with a input attenuator. Something like the one below. You will have to calculate he values yourself for your chosen attenuation and impedance.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Hello KalleMP Thanks for your help. Looks like a nice component, I just took a lot at the link but I still have an answer, how would I wire the Vin and Vout to this attenuator? and also how do I know the voltage for each attenuation. Thanks \$\endgroup\$ – MeMe2 Nov 6 '14 at 7:40
  • \$\begingroup\$ Hi KalleMP, I guess I figured out the output Voltage for each attenuation (just for my information). It seems like this component will do what I need, that is, to attenuate according to the little switch position to the combination I want. Now my question is, to get the proper ouput I need to reduce the Voltage to 0,350 before getting to this component, how do I do that? Thanks \$\endgroup\$ – MeMe2 Nov 6 '14 at 10:52

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