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.

[mkiijam]

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.

Right. I should know my own circuit, except it's mostly ingredients I've been grabbing to make my OWN pie!

The choice on 3 discrete relays was, at least in part, to facilitate lighting up LEDs on the input switches... I thought up that stuff about 2 years ago, so I'm kinda not remembering why I went with 3. The idea of not causing a short is intriguing. Let me think on that for a bit before you give it away.

[Alex Nikitin]

Over 20 years ago I've published (in Ukrainian magazine "RadioHobby" #2 2002), a circuit for the constant input impedance logarithmic attenuator. Here is the circuit, from that printed article, set for binary 1-2-4-8-16-32-64dB steps, giving the maximum of 127dB attenuation in 1dB steps, with only seven switchover contacts (preferably good signal relays) . For a 20dB attenuator step and a 10K load the resistors values should be 9K and 1.11K (two 18K in parallel and 1K1 should do nicely).

Cheers

Alex

[mkiijam]

So this could / would be controlled by a microcontroller or encoder or something?

[Alex Nikitin]

So this could / would be controlled by a microcontroller or encoder or something?

Yes, that was the idea. A microcontroller also allows for a timing trick to make the relay switching less noticeable, the worst example is if you go from -64dB to -63 or back from -63 to -64 and you switch all relays at the same time, you might release the lower attenuation bits before you engage the high attenuation bit, and get a very noticeable signal breakthrough up to 0dB, so you have to adjust the timing accordingly. If it is done right the switching is almost imperceptible on an audio signal. On the photo is the two channel stereo (or a single channel balanced) module from 2001 (I've worked as Chief Engineer at Creek Audio at the time), control was arranged in BCD format, 1-2-4-8-10-20-40 dB, 85dB max in seven stages plus a full mute (8 relays in total) , using laser trimmed resistor arrays.

Cheers

Alex

[Zero999]

Or use analogue switches. There are devices available with a resistance of under an Ohm. If it's significant, reduce the lower resistor values slightly.

[Alex Nikitin]

Or use analogue switches. There are devices available with a resistance of under an Ohm. If it's significant, reduce the lower resistor values slightly.

With very few exceptions solid state analogue switches are not good for audio applications, they more or less destroy the sound quality (at least practically all CMOS switches do). Also a limited headroom can be a problem. The volume control module I've posted above could take 50V RMS continuous signal input, try that with a solid state switch  .

The whole point of designing this circuit was to provide a better sound quality than solid state switches/controls and better precision and longer life than a motorized potentiometer, for a reasonable cost with as few relays as possible.

Cheers

Alex

[mkiijam]

Is there an example of the MCU that would control this or would i need to design it myself?

[Zero999]

Is there an example of the MCU that would control this or would i need to design it myself?

Any microcontroller will do. The relays are more important. Reed relays would be ideal.

Something like this would be great, as it can be driven directly from a microcontroller output pin.
https://www.pickeringrelay.com/product/101-1-c-5-3/

So this could / would be controlled by a microcontroller or encoder or something?

Yes, that was the idea. A microcontroller also allows for a timing trick to make the relay switching less noticeable, the worst example is if you go from -64dB to -63 or back from -63 to -64 and you switch all relays at the same time, you might release the lower attenuation bits before you engage the high attenuation bit, and get a very noticeable signal breakthrough up to 0dB, so you have to adjust the timing accordingly. If it is done right the switching is almost imperceptible on an audio signal. On the photo is the two channel stereo (or a single channel balanced) module from 2001 (I've worked as Chief Engineer at Creek Audio at the time), control was arranged in BCD format, 1-2-4-8-10-20-40 dB, 85dB max in seven stages plus a full mute (8 relays in total) , using laser trimmed resistor arrays.

Cheers

Alex

Of course, 126dB of attenuation isn't much good at low voltages.

The some non-standard resistor values in that circuit. Did you round to the nearest E96/E192 or go for two in series or parallel to get the exact value?

[Alex Nikitin]

The some non-standard resistor values in that circuit. Did you round to the nearest E96/E192 or go for two in series or parallel to get the exact value?

The custom resistor array trimmed to better than 0.1% was used in the original design, so no problem with non-standard values and at least 60dB CMRR in a balanced connection. In practice, one could use standard values (i.e. 1K1/81K for 1dB, 2K0/39K for 2dB, using two resistors combinations for higher attenuation), losing some accuracy but keeping the difference between channels to a minimum.

Any microcontroller will do. The relays are more important. Reed relays would be ideal.

Reed relays (as much as I like Pickering, use them a lot) have one particular problem in audio applications - a direct pickup from the coil (a closed contact makes it into a small transformer). A "normal" sealed signal relay with gold contacts is easier to use.

Cheers

Alex

[mkiijam]

Of course, 126dB of attenuation isn't much good at low voltages.

I've never understood this. My feed to the output amps / line drivers is from the output of the volume control which can be at ground? How is this anything but infinite (in theory) attenuation?

[Zero999]

Of course, 126dB of attenuation isn't much good at low voltages.

I've never understood this. My feed to the output amps / line drivers is from the output of the volume control which can be at ground? How is this anything but infinite (in theory) attenuation?

I was talking about Alex's circuit which has an atenuation range of 0 to 127dB in 1dB steps. 127dB is over 2.2 million, which is pretty pointless at low level signals. One might as well just disconnect the signal, using a switch. At releatively high voltages, such high levels of attenuation make more sense.

[magic]

The point of 127dB is that it costs only two resistors and a relay more than the obvious alternative of 63dB, which OTOH may be not enough

[mkiijam]

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.

It seems like I read someplace that the two signals combined will be louder, so this is why there are larger resistors?

[magic]

Two signals mixed by means of equal resistors become simply the average of the two signals.
For instance, if the signals are equal, the output is also equal to each of them.

The output of such passive "mixer" is X = (A + B)/2 = A/2 + B/2.
So each signal is attenuated by half and then they are summed together.

[mkiijam]

What I meant to say was: https://www.goldpt.com/mode.html

I just used this for MONO.

Is that wrong in my case? Better to use an active buffer for the mono mode?

[magic]

I believe the issue was that switching to mono increases output resistance of the mono/stereo module and causes further attenuation?

Several options to deal with it.

1. Simply live with it. An equal signal on L and R will be become quieter in mono mode, even though nothing really changes. You may want to ensure that the attenuator/volume module has constant input impedance so that the attenuator still makes exactly 20dB difference in mono mode too.

2. Insert a buffer between the mono/stereo module and the attenuator/volume module. It may be switched into the signal path for mono only or simply always connected.

3. Maybe you could find a way to modify mono/stereo to always have 22kΩ output impedance per channel, then the same attenuation will happen in stereo too. Compensate for it by adding gain or reducing attenuation somewhere else. Watch out for noise/clipping/dynamic range issues.

4. Reduce mono mixdown resistor values to reduce output impedance and the amount of attenuation in mono mode. This increases loading on the preceding stage when L and R aren't equal.

[Bltzbrg]

Hello,

I couldn't find any help with google or chatgpt and this forum is the closest to a possible solution.

I am designing a PCB basically to merge the signals from two 3.5mm audio jacks to a headphone and into a recorder. I want to enable an about -15db attenuation with a simple 2 position switch. I tried to understand which resistors I need but I couldn't figure it out. All I know is that the source that is too loud is usually connected to headphones with 64 ohm and a driver sensitivity at 1kHz/1mW (db) - 88db.
The signal is splitted. It should go unchanged into the headphones and with -15db into the audio output jack to my audio recorder.

I couldn't understand why some people calculate with 10kOhm.

Any help would be much appreciated

Thank you!

[magic]

10kΩ is a common input impedance of line level inputs (i.e. each channel is loaded to ground by 10kΩ), although certainly not guaranteed - it may be higher or sometimes lower.

A simple passive mixer (two equal resistors between the two inputs to be mixed) has high output impedance and won't drive headphones directly, so this device will need some active electronics: a buffer/amplifier and a power supply for it.

[Bltzbrg]

I don't know which numbers I should look for.

I want to put the signal of the Rode mic on the left audio channel in my gopro and the signal of my mp3 player on the right channel. The headphones of the mp3 player have got following specs:

Driver sensitivity at 1kHz/1mW (db) - 88db
Impedance - 64 Ohm
Dynamic Frequency range - 20-20.000 Hz


The microphone's technicals specs (Rode Smartlav+) are:

Acoustic Principle: Permanently Polarised Condenser
Active Electronics JFET
Polar Pattern Omni-directional
Frequency Range 20Hz - 20kHz
Output Impedance 3k Ω Typical
Signal To Noise Ratio 67 dB
Equivalent Noise 27 dB Typical
Maximum SPL 110 dB
Sensitivity -35dB (17.8mV @94dB SPL)
Dynamic Range 83dB Typical
Power Requirements Powered from iPhone TRRS
socket (2.7V)

Unfortunately I cannot find any technicals specs to the mp3 player.

Hearing the microphone on the left channel and the mp3 player on my gopro footag already works. I just want to make the mp3 player more quiet (but I need the Volume on the headphones). Its just too much for the gopro.

I want to lower the volume of the mp3 player just for my gopro by about -15db. I don't know with which impedance I should calculate and how many Watts the resistors should be able to handle. Also is Pi or T attenuator better?

Many thanks!

[Bltzbrg]

I found some more information about the GoPro Mic-Adapter:
Bias-Voltage: 2,5 V

Input impedance 2,2 kOhm in setting "standard-microphone" and "standard-microphone+"
Input impedance 47 kOhm in setting  "condenser microphone" and "condenser microphone+"
Input impedance of the GoPro AUX-Input 8 kOhm
unbalanced 3.5mm stereo

[magic]

I want to put the signal of the Rode mic on the left audio channel in my gopro and the signal of my mp3 player on the right channel.

Hearing the microphone on the left channel and the mp3 player on my gopro footag already works. I just want to make the mp3 player more quiet (but I need the Volume on the headphones). Its just too much for the gopro.

I want to lower the volume of the mp3 player just for my gopro by about -15db. I don't know with which impedance I should calculate and how many Watts the resistors should be able to handle. Also is Pi or T attenuator better?

So no need for actually mixing mp3 and mic signals onto a single channel and you don't care about hearing mic signal in the headphones?

This sounds fairly straightforward now.
First of all, connect the headphones directly to the mp3 player and use a passive splitter to get a second output to the GoPro.

To attenuate the mp3 signal, use an ordinary resistor divider. You already have 8kΩ resistors to ground on each input pin inside the GoPro (if aux input is the one you are using for mp3), so simply add 36kΩ in series with the cable for 0.18x attenuation and that's it. Power dissipation in these resistors is negligible - output voltage of mp3 players is no more than a volt or two RMS at full volume, which makes fractions of mW when applied to such high resistance.

[Bltzbrg]

I checked the settings of my GoPro and it uses the "standard-microphone" as default. I want to stitch on that to stay end-user friendly.

I will not hear the mic signal in the headphones because it is routed differently.

So 2.2k Ohm it is.

So how exactly should I place the resistors? So the standard 65 mW is enough? 100mW does cost 1 cent or even less more so I could also go for that.

I just mentioned all the other devices and their specs because I read something about T and Pi Attenuator and Input and Outpur Impedance and it confused me. I thought having some context may prevent some errors in the answer.
Do I need a T or Pi Attenuator?

A small drawing would be nice, if it doesn't bother you I guess I just need the part: Ground, Signal wires from the switch -> Resistors -> GoPro TRS

Thank you very much already for your fast replies and willing to help

[magic]

This should work. Power rating really doesn't matter. I would try it with ordinary 0.25W THT resistors, and if it works as intended put the resistors and cables in any random plastic box with panel mount jack connectors for inputs/outputs.

Code: [Select]

       +---------------- R HP out
       | +-------------- L HP out
       | | +------------ HP GND
       | | |
       | | |
R in --+------ 22k --+
         | |         +--- R out
L in ----+---- 22k --+
           |
GND -------+-----+------- GND out
                 |
GND -------------+
Mic in ------------------ L out
This assumes the receiving device has a stereo electret mic input with separate 2k2 loads on each channel - one for the divider (R) and one for the electret mic (L).

It is important that HP ground is connected close to the input connector and MIC ground close to the output connector, as drawn. Don't allow HP return current to flow through the wire between output GND and mic input GND.

[Bltzbrg]

Let's start with the fact that your code drawing looks really good

I don't have a stereo signal. I merged L and R right after the TRS Socket to make things easier. For the headphones I just split the audio signal to L R again. This is the "end recording device" : https://gopro.com/en/us/shop/mounts-accessories/pro-3.5mm-mic-adapter/AAMIC-001.html

I hope it has two seperate loads. I think I have to assume it.
I connect all the GND with a Copper pour. So everything that is not a signal line is ground (classic PCB design). How did you calculate the 22k resistor? The online calculator (https://www.everythingrf.com/rf-calculators/attenuator-calculator) is probably not suitable for me right? So for my case I just need one resistor with 22k Ohm right before the output?

It would be nice if you could explain how you calculated it. Maybe I have to make adjustments to -20db or -10db in the future.

Thank you so much. You already helped me a lot!

[magic]

For mono use two paralleled 22k or simply 10k, doesn't matter for this kind of accuracy.

But how are you "merging" L and R after the TRS socket? You shouldn't simply short them together, the two output amplifiers of the mp3 player will "fight" and one will source lots of current while the other will sink it, both attempting to set a different level on the same line... They are probably protected enough to survive undamaged, but battery life will be impaired if nothing else.

These Π and T attenuators are for RF systems with 50Ω (or similar) input and output impedance everywhere, that's not how line level audio signals are transmitted and you don't need these circuits.

The mono circuit I suggest is a simple voltage divider consisting of 10k and 2k2 (inside your mic adapter) giving (10+2.2)/2.2 = ~5.5 attenuation ratio. The stereo version replaces 10k with two 22k in order to mix two input signals; if both signals are equal then it behaves the same as mono. For other ratio, the usual voltage divider rule applies.

[Bltzbrg]

Hi Magic,

So now I just added a 10kOhm 100mW resistor in the signal line before my output trs.

Oh, yes I just shortened the two signals of the mp3 player together and it seemed to work. These websites say with some resistors I could "merge" them:
https://www.instructables.com/Simple-Way-to-Convert-Stereo-to-Mono/
https://forum.arduino.cc/t/audio-input-switching-turning-stereo-to-mono-etc/242588/3

Could I do it like that? Which resistors do I need? The mp3 player is supposed to have equal left and right channels but I am not sure if it really is like that and there is no public information on that.

The cable solution I had before was like:
V-Adapter TRS Stereo to 2 TRS Stereo
reverse V-Adapter Left TRS and Right TRS to TRS "Stereo" (https://a.co/d/8TKS2M9)

Split the mp3 player signal with v-adapter1 to go into headphones and the other one goes into the reverse v-adapter e.g. left channel into the GoPro Adapter.
The mic goes directly into the reverse v-adapter e.g. right channel.

For simplicity I could obviously just use one channel but I don't want to risk it to loose sound that was just sent on the other mp3-player channel. Thats why I ask how I can convert the stereo signal to a mono signal.

We are almost there

Thanks again!

[magic]

They are not equal if there are different signals on L and R.
They also aren't equal if they have different DC offsets and there are no series capacitors on the outputs.

The resistors thing should work, but with a few dB loss of volume.

[Bltzbrg]

I just edited my previous post to avoid double posting. Sorry!

[Bltzbrg]

I would really appreciate an answer to my edited post 
Then I could continue working on the project.
Thank you

[magic]

Not sure what the question is at this point?

Yes, you should be able to mix the outputs of a stereo MP3 player to mono by means of those 33Ω resistors. As I said, it will be a few dB quieter than driving the headphones directly. Or you can simply connect the headphones in stereo and only send mono to the recording device, not sure what's the problem with that?

[Bltzbrg]

So 33 Ohm resistors are good for my usecase? How many dB would that be?
If a song plays "left, right, left, right" I want to hear that in the gopro. Not just "left,...,left,..."
I wired the headphones directly to the mp3-player and I guess I am going to use the two 33 Ohm resistors to connect the L and R channel together to get a mono signal.
You said that simply shorting the channels together would damage the mp3-player and drain the battery? Why do these small resistors prevent that?

Sorry, if I am being annoying but you are my only helpful source

[magic]

By Ohm's law.
Current flowing from left to right output (or vice versa) is voltage between these channels divided by total resistance (2x33Ω = 66Ω).

[Bltzbrg]

I bought some audio jacks and a resistor kit.

I connected the L and R channel of the mp3 player as following:
L-in->33Ohm---------->Mono-Out
R-in->33Ohm-----^
GND--------------------->

The signal is now about -30db quieter which is too much. I want to convert a stereo signal to a mono signal without making it quieter.

And I want (with a switch) to make the signal -10 or -15db quieter.

I also connected the 1k Ohm Resistor in just 1 channel to see how much quieter it makes the sound. Barely nothing. Sometimes maybe 3db.

Any tips or explanation why 1k Ohm does nothing but two 33Ohm do so much? If you are tired of explaining I get it. then just provide new Resistor sizes if you can

Thank you very much for your time

[magic]

This is the headphones mono adapter, correct?
Are you saying that the headphones are 30dB down now?

If not then what is connected to where and what are you measuring?

[Bltzbrg]

Oh ok, I thought I made it understandable, my bad.

First I tried the stereo signal to mono signal solution with the 2x 33Ohm Resistors. I connected everythin on a breadboard.
Mp3-player stereo signal TRS in -> 33 ohm resistor (1 per channel) -> connect both channels -> TRS out into my Gopro. Obviously I just had audio on the left channel (as intended) but it was too quiet (-30db). At this point I don't want to make it quieter. Just stereo to mono.


The second test I did, was the 1k Ohm Attenuator.
Mp3-player stereo signal TRS in -> on one channel 1k ohm -> TRS out into my gopro
I thought like that I could hear the about -15db difference between the channels. Or is that the issue that I connected the "louder" channel as well? In this scenario I almost had no attenuation. Maybe about -3db but not more.

I hope it is understandable now.

[magic]

Inserting 1kΩ between the source and a 2k2 input is expected to produce about -3dB, so nothing wrong with that. It's basic voltage divider math, use more resistance for more attenuation.

33Ω should produce much less attenuation, so something must have been wrong. Maybe used 33k instead?

[Bltzbrg]

The resistor-set I used was a cheap one from amazon. So that could be possible. Thank you for pointing that out. I found some useful calculators and I calculate it with 2.2 V.
I am a little bit embarassed. I made a mistake by taking 1k Ohm and not 10k Ohm and I didn't notice my mistake until now. I will try to find the correct 33 Ohm resistor tomorrow.
I feel like now I have all the missing numbers I need to use these online calculators
Do you have got a good source that I could learn why shorting the two channels with 2x 33Ohm is okay but not using the resistors is bad? I mean 33Ohm is not much, isn't it? Some other sources use two 1k Ohm Resistors. That would do an attenuation of about -2db. What would be the difference between 33Ohm and 1k Ohm in my use case?

Thank you!