I'm going to build a passive matrix mixer using the information on this site and this schematic:

passive matrix mixer

This is the first time I have built any circuits, so I've been doing some research trying to work out what everything does.

If I've understood correctly, the 100 kΩ potentiometer will act as a voltage divider and send some of the signal to ground.

Why is there a 10 kΩ resistor connected to the output of the potentiometer? Does it essentially ensure that, even when the potentiometer is set to 0 resistance, that there is a minimum (10 kΩ) resistance, and so some voltage still goes to ground? If it does, why would it do that? Is it so, isn't the sum of the outputs from each of the potentiometers too big and prevented from doing damage to whatever device receives it?

I plan to use this mixer with a guitar and effects pedals. If I want to plug my guitar directly into the mixer, should I choose one of the inputs that is expressly for this purpose and get rid of the potentiometer and resistor for each of the outputs, replacing it with a switch before the outputs instead? Like this:

alternate circuit for direct guitar

I'm asking because, if I've understood correctly, the output voltage from my guitar pickup will be very low, and so sending some of it off to ground could be a bad idea unless I were to plug it into some kind of boost/op-amp before it went into the matrix mixer.

  • \$\begingroup\$ So how does that mixer work? Are we supposed to guess if inputs are on top and outputs on left, or the other way around? Also I don't recommend this to be used with instrument or guitar pickup level signals at all, this loads it too much already. Even the wiring from pickuo to multiple amp inputs adds capacitance from the wires. \$\endgroup\$
    – Justme
    Commented Feb 17, 2023 at 9:38

1 Answer 1


With the right combination of impedances (resistor about one magnitude higher than potentiometer), the "mixing resistors" reduce the influence of any given pot's position on the contributions from other sources.

Without any resistors, imagine one pot at the upper stop and another one feeding the same output at the lower:
the first source is effectively shorted.

Given the cost of amplifier components this century, amplify to a decent level as close to the source as feasible - I remember 1 mW into 600 Ω.
And consider an active mixer.
And input capacitors. And clamping diodes…

  • \$\begingroup\$ 1 mW into 600 ohm is about 0.775V. It has very little use these days as a definition. Guitar instrument level voltage is likely less than that and 600 ohms is too much load for it. 600 ohms load is also too much load for line level audio, and modern line level voltages are 2 VRMS or 5.6 VPP into 10k load or so. \$\endgroup\$
    – Justme
    Commented Feb 17, 2023 at 9:44
  • \$\begingroup\$ (@Justme 1 mW into 600 Ω is just the definition of line level. There is a reason why they started stating dBu.) \$\endgroup\$
    – greybeard
    Commented Feb 17, 2023 at 9:46
  • \$\begingroup\$ That is not a definition for anything, just an arbitrary value, without defining in which conditions that applies. For example, it would require some context, like when the definition true and with what kind of audio signal, will it be nominal level for quiet speech, or nominal level for loud rock concert. The nominal level is just long term average for some material, it does not define the maximum for peak amplitudes that must fit without clipping. \$\endgroup\$
    – Justme
    Commented Feb 17, 2023 at 10:56
  • \$\begingroup\$ (@Justme en.wikipedia on line level: The decibel unloaded reference voltage, 0 dBu, is the AC voltage required to produce 1 mW of power across a 600 Ω impedance (approximately 0.7746 VRMS).[2] This awkward unit is a holdover from the early telephone standards, which used 600 Ω sources and loads, and measured dissipated power in decibel-milliwatts (dBm).) \$\endgroup\$
    – greybeard
    Commented Feb 17, 2023 at 11:06
  • \$\begingroup\$ Yes but that does not define how quietly you need to whisper or how loudly you must yell to the microphone to produce the exact line level reference voltage on wires. You need to have a fixed reference sound pressure level definition which will equal to a fixed reference voltage level. Plus an allowed maximum range. For example, you must agree that CD players have line level outputs. But can you define which digital amplitude outputs your definition of line level voltage? These days, 0dBFS digital maximum amplitude is 2 Vrms. The voltages were likely different 40 years ago. \$\endgroup\$
    – Justme
    Commented Feb 17, 2023 at 11:19

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