I need to lower an audio signal by 20dB. *Schematic added*

[mkiijam]

I need to build a -20dB or (maybe 30) but can't seem to find any standard solution? I see lots of circuits for balanced audio, but this will be unbalanced. Is it possible with a DPDT switch?

[antenna]

I need to build a -20dB or (maybe 30) but can't seem to find any standard solution? I see lots of circuits for balanced audio, but this will be unbalanced. Is it possible with a DPDT switch?

What is the impedance of the audio circuit?

[mkiijam]

I'm really not sure. I plan to place the DIM switch right after a volume control that will have 10k to ground. The DIM switch will feed op amps as buffers into line drivers.

This brings up a question: Should the DIM and MUTE switches be before or after the volume control?

[antenna]

I would put the attenuator before the volume pot, simply because it represents a fairly known load impedance (assuming the opamp imputs don't load it down too much).  A PI attenuator can be designed for 10kΩ and -20dB (1/10th the voltage) with an online calc.  Attached is a screenshot of the result I got.
 

[antenna]

And yes, a DPDT switch will work.  It would look like this (I used standard values here, which gives 19.76dB attenuation)

[mkiijam]

Thanks! Sure appreciate it !

[radiolistener]

you're needs to take into account load impedance which is planned to connect on output of your circuit. This impedance affects resistors value for attenuator. If you don't know it, it will be impossible to calculate proper resistor values for your attenuator.

The attenuator circuit above with don't take this impedance into account, it shows values for 10 kΩ load... But I suspect you're planning to connect load with impedance different than 10 kΩ, in such case these values will be incorrect...

The same it expects 10 kΩ impedance before attenuator, but most of all it is different... So, calculations also incorrect.


You're needs to measure output impedance of your circuit (it can be done with test loads and AC voltmeter with bandwidth > 20 kHz) and measure input impedance of your device which you're planning to connect through attenuator... Then you can calculate attenuator.

[jonpaul]

5400 ohm R + 600 Ohm R

Or just buy a 20 db atten

j

[magic]

The simplest solution: add 9x the volume pot resistance in series with the input terminal of the volume pot.
Worst case output-referred volume pot noise will occur at maximum volume, 12dB 6dB above the usual worst case which is at -6dB volume.
Loading on the upstream circuit decreases 10x.

A lower noise solution: add 0.9x volume pot resistance in series with the input terminal of the volume pot, add 0.11x volume pot resistance from volume pot input to ground.
Noise and loading similar to "bare" volume pot.

[Zero999]

A lower noise solution: add 0.9x volume pot resistance in series with the input terminal of the volume pot, add 0.11x volume pot resistance from volume pot input to ground.
Noise and loading similar to "bare" volume pot.

That's what I would do. It also makes it less dependent on the tolerance of the potentiometer, which is normally quite poor.

[tszaboo]

If you are not sure of the load and source resistance, then buffer the signal on the input and the output. A simple voltage follower opamp. So opamp-> divider-> opamp. It's not excessive, and done quite often, you have hundreds of opamps in a mixer table. I would run the Opamps on +/- 15V, that's plenty of headroom usually for RCA connections.

[antenna]

you're needs to take into account load impedance which is planned to connect on output of your circuit. This impedance affects resistors value for attenuator. If you don't know it, it will be impossible to calculate proper resistor values for your attenuator.

The attenuator circuit above with don't take this impedance into account, it shows values for 10 kΩ load... But I suspect you're planning to connect load with impedance different than 10 kΩ, in such case these values will be incorrect...

The same it expects 10 kΩ impedance before attenuator, but most of all it is different... So, calculations also incorrect.


You're needs to measure output impedance of your circuit (it can be done with test loads and AC voltmeter with bandwidth > 20 kHz) and measure input impedance of your device which you're planning to connect through attenuator... Then you can calculate attenuator.

If the opamps being driven by the wiper signal on the pot have a reasonably high input impedance and don't load down the wiper tap, the pot becomes the load for whatever signal is driving that pot.  So we know the load.  If the signal going into that pot is driving 10k, is it not safe to assume that the source, wherever it is coming from to drive that pot, is also 10k (or equally capable of driving a 10k attenuator in place of the 10k pot)?  Unless no thought was originally put into impedance matching, cutting into a 10k line and inserting a 10k input and output attenuator should be just fine. I think of it like going outside and cutting into my radio coax.  Its going to be 50Ω on both sides.  There would be no reason for me to go recheck my coax impedance as the matching was done prior to me tapping into it.   This is why I thought it would be better to put the attenuator BEFORE the pot.  What did I miss?

[antenna]

Could there be a 200Ω signal driving that 10k pot, sure, but if we then go and change the attenuator input from 10kΩ to 200Ω, now that source has been loaded differently than it is now and the level of attenuation needed will likely change.  He may need less than 20dB attenuation if he tries to match the source currently driving that 10k pot.

[magic]

Nobody "impedance matches" audio circuits like it's RF. Usual practice is "low impedance drives high impedance" everywhere, with little loading loss.

You don't need the attenuator to have 10kΩ input impedance. More likely, there is a specification like "10kΩ minimum". You don't want the attenuator to have 10kΩ output impedance either, because it's unnecessary Johnson noise. A pot can be driven by a lower impedance just fine, and noise of the attenuator will be lower.

Really all it takes is a two resistor voltage divider, remembering to treat the volume pot as part of the lower resistor in calculations.

[antenna]

Nobody "impedance matches" audio circuits like it's RF. Usual practice is "low impedance drives high impedance" everywhere, with little loading loss.

You don't need the attenuator to have 10kΩ input impedance. More likely, there is a specification like "10kΩ minimum". You don't want the attenuator to have 10kΩ output impedance either, because it's unnecessary Johnson noise. A pot can be driven by a lower impedance just fine, and noise of the attenuator will be lower.

Really all it takes is a two resistor voltage divider, remembering to treat the volume pot as part of the lower resistor in calculations.

Thank you for explaining that.

[rsjsouza]

Indeed. In audio there are no concerns about standing waves/reflections that require such careful impedance matching. The proposed solution with the two resistors is more than enough.

[tszaboo]

Nobody "impedance matches" audio circuits like it's RF. Usual practice is "low impedance drives high impedance" everywhere, with little loading loss.

You don't need the attenuator to have 10kΩ input impedance. More likely, there is a specification like "10kΩ minimum". You don't want the attenuator to have 10kΩ output impedance either, because it's unnecessary Johnson noise. A pot can be driven by a lower impedance just fine, and noise of the attenuator will be lower.

Really all it takes is a two resistor voltage divider, remembering to treat the volume pot as part of the lower resistor in calculations.

But without context it's not that simple. What you are writing is true for a box that has RCA inputs and outputs and it's all properly designed audio circuits.  But we don't know if that's the case. The input can be a DAC with variable output resistance based on the output code, and the output a headphone amplifier end stage with 1K input resistance.

[radiolistener]

is it not safe to assume that the source, wherever it is coming from to drive that pot, is also 10k

Usually audio output has low impedance, something about 10-100 Ω. Note that source impedance and load impedance will be connected in parallel to your attenuator input and output, so the resistance will be different and can significantly affect attenuation ratio...

[MathWizard]

After trying a little bit of RLC matching networks, now I get why attenuation boxes are so nice. And not impossible to build.

[mkiijam]

Here is the relevant portion of the schematic. There may be some issues or errors as this is a work in progress.

[magic]

The MUTE/DIM module is doing the exact thing I suggested, with different resistor values which also give ~20dB attenuation but variable input impedance and variable noise.

Johnson noise of the attenuator may or may not be a problem for you. Good news is that if it is, you only need to reduce resistor values.
Variable input resistance of the MUTE/DIM module shouldn't be a problem, as it seems to be driven by some line receiver chip through nothing more than a bunch of relays, unless I missed something.

Excellent use of a single switch pole per channel to control attenuation. I'm absolutely stealing this trick next time I will need this.

[magic]

Variable input resistance of the MUTE/DIM module shouldn't be a problem, as it seems to be driven by some line receiver chip through nothing more than a bunch of relays, unless I missed something.

I missed the mono mixdown mode involving R54 and R55.

This circuit has 11kΩ output impedance and is loaded by MUTE/DIM. In passthrough mode, MUTE/DIM input impedance is 5kΩ (simply the two volume pots) resulting in 10dB extra (maybe unexpected/unwanted?) attenuation. In 20dB mode, the loading is 38kΩ and attenuation is much lower (of course the 20dB attenuator still provides its own 20dB on top of that).

If this is a problem, maybe R54 and R55 could be reduced (also better for noise), or the output of the mono circuit buffered. You can also modify the attenuator to have constant 10kΩ input impedance (my initial suggestion) so that enabling mono mixdown introduces constant 10dB loss, regardless of 20dB attenuation or not.


I don't know whether this is your intent or not, but enabling multiple bus selection relays shorts the input buses together.

[mkiijam]

I don't know whether this is your intent or not, but enabling multiple bus selection relays shorts the input buses together.

The control of those relays is driven by an electronic version of the mechanical interlock switch. i.e. when you press one button it deselects the others.

[mkiijam]

In passthrough mode, MUTE/DIM input impedance is 5kΩ (simply the two volume pots)

Are the two volume controls in parallel in someway? I'm not sure how you are getting 5k instead of 10k?

[magic]

RL6 shorts the pots together. Also R54 with R55. The outcome is Thevenin-equivalent to a 11k/5k voltage divider fed by the average voltage of left and right channel.

As for the relays, "it will never happen" are famous last words before something happens
I think they could be rearranged to prevent any possibility of a short, and only two would suffice to choose from three inputs.