You will create a small passive mixer. It will be extremely bare bones. It will cover the basics of routing to an output and hearing that output. Some of the drawbacks of such a mixer will become apparent and set the stage for the improved active mixer.
What is a passive mixer?
In electronics a passive component is one which does not require a power source to work, or in other words, you don’t have to plug it in. An active device does require a power source. It could be a battery or wall power but an active device will not work without a power source.
Passive sources are very convenient but have major limitations making them much less common. One such limitation is that passive devices have to way to provide gain.
Passive sources are very convenient but have major limitations making them much less common. One such limitation is that passive devices have no way to provide gain.
We will construct a passive mixer using some simple 3.5mm jacks, many wires, some potentiometers, an active source and a receiver, namely your phone and a speaker.
The mixer that we are building will not require any power, but this also means the mixer will not be able to amplify signals, it will only be able to turn them down. This is just one of the downsides to such a device. Because of this we must have signals that are already amplified and a source that can handle these signals as well. For this job we will use a phone as the input (which has a built in amplifier) and a speaker (many of these also have their own amplifiers built in) as the output.
Experiment #1: The breadboard as a patch bay:
Before we jump to a basic mono mixer we will first see how versatile a breadboard can be. Namely, it functions as a very cheap patch bay.
Step 1: Attach the audio jacks to the breadboard like so:
Place these carefully, each pin should be connected. We need room in the back to get the audio from the jack via wires. The 2 left most pins will have one channel of audio, the middle pin (pin 3) will have ground, and the two rightmost pins will have another channel. This is true for TRS inputs.
|TRS Connector||Pin Number||Channel of Audio|
Note: the black rings are insulation, they are not the “ring” in TRS, it’s the metal band that is the ring. These jacks require TRS. Many cables these days are TRRS, tip ring ring sleeve.
On the left is TRS and on the right is TRRS. We must have TRS for the connectors to line up correctly to the jack.
Now we will check the continuity of the board to the cable. Continuity simply shows that there is no break in the wire and that the connection is good. It can be very handy to figure out how things are connected or to check if a component is working. To do so we will use the continuity check on a multimeter. Set the multimeter to the following setting:
Some multimeters do not support such a check so you will need to check yours. Ensure the red lead is in the correctly marked slot (if it’s not marked then place it in the voltage slot) and the black lead should be connected to COM slot (black). Some meters have one setting do multiple things and you may need to find a “function” button or something similar to cycle through them. Now when you touch the leads together you should hear a beep indicating continuity, meaning they are electrically connected.
Note: Some multimeters have very poor quality leads, this may cause you some difficulty when checking continuity.
If your meter doesn’t have a continuity check you can use the ohm setting to check continuity. If the meter reads a low value close to 0 then it means there is very little resistance and the connection is good. If the meter reads a high value or says OL (overload) it means the circuit has a high resistance and is open. The little meter in the picture for example shows a 1 when the resistance is too high to measure indicating an open circuit.
We specifically want to see what parts of the TRS input are connected to the jack so we can figure out how to connect things. To do so we connect one end of the cable to the jack and attach a jumper to the other end. We can move the jumper around now and check the continuity each time to see what part of the jack is connected to that particular pin.
To show this is the case you can plug in an auxiliary cable. One end connects the breadboard and the other is not connected to anything. This will eventually connect to your phone, but for now it just hangs unconnected to anything. Importantly this means the tip, ring and sleeve of the connector are now connected to the breadboard and we should be able to probe the hanging part of the cable and the breadboard to verify continuity. Now take a jumper wire and connect it to the breadboard on one of the outputs.
For the pins on the audio jack use the outer pins and middle pin only. Treat it as a 3 pin device. Ignore the middle inner pins. In other words if the jack has 1,2,3,4,5 pins on it just use pins 1, 3, and 5. If it only has three pins then just use the three pins.
Recall the chart:
|TRS Connector||Pin Number||Channel of Audio|
Lab Continuity Verification
In your lab report include a picture of each measurement. Have the jumper, breadboard, end the wire, meter, and its measurement clearly visible.
Now that we are 100% sure of how the connections are made we can proceed to create a bus. A bus will take the voltage and send it to another part of the breadboard. We can then take this bus and connect it somewhere else in the circuit later.
We can pick anywhere to make a bus, I like to make them in the middle on the back part of the breadboard so that there is plenty of room for more inputs and outputs.
NOTE: All grounds must be connected.
These buses have 2 important properties.
- Voltages that arrive at the same point will add together.
- We can take outputs from this bus to send them somewhere else.
There is also a very important electronic principle we must understand:
Any points of lower potential will act as receivers. (This will come up later as we add more inputs)
Now that we have our bus, let's suppose we have our phone connected to the input via an auxiliary cable. At this point an input will be present and the input must be amplified already for this to work since the mixer is passive. This signal will appear at the busses but currently it doesn’t go anywhere. Let’s add an output.
With the output connected we now have a basic set up. The input will come in, flow down the channels through the bus and then to the output, which can be headphones or a speaker. All we have done is pass the signal along. What we have made is a basic patch bay. Not super useful yet. But we now know what parts of the jack connect to the TRS cable and how to get signal into and out of the board. Go ahead and plug in your headphones and listen to hear it for yourself.
These jacks are stereo. This means it has 2 channels of audio. The pins on the left are one channel and the pins on the right are another channel. Note these are separate channels of audio, nothing is forcing you to pair them. Say we want only the right audio to come in and get connected.
Here for example only the right audio channel is connected. In this case you will only hear the right audio channel. Now consider the following:
Now on the input we have only the right as an input but on the output we have two right channels, meaning in your headphones you would hear the right channel in both ears.
We could have a similar scenario for the left channel. The point is these jacks can carry two signals, how we choose to connect these signals up is entirely your decision!
Say I wanted 3 in and 3 out. Then technically I only need 2 jacks for the input (they offer up to 4 connections) and 2 jacks for the output. Something like this:
While this technically works it's very unintuitive for TRS jacks because we use TRS jacks when 2 related signals are needed. For example, a left and right audio signal to form 1 stereo image. Balanced audio is another example of related signals where 2 channels are needed.
Instead if our intention is 3 separate signals, say they are microphones and we want each to have its own output then we can wire up the following:
Now it makes more sense, however if we intend to listen to each channel then this circuit would only cause 1 side of the headphones to work the other would be silent! We can copy the signal to both headphone outputs so that both sides of the headphone work despite there being only mono information.
This set up would copy the signal to the other side of the headphones so it works.
The important point is that the 2 connections are arbitrary. It's up to you how the input and output are used.
- In the above schematic the signal for the mics is connected to the left most input. What part of the jack is this connected to? What would changing it to the right side do?
- These jacks are for TRS, how would a TS jack be different?
Now let’s say we need to copy the input to more than one output. We can do so by simply adding another output and taking it from the main bus. Main bus in this case being the stereo left right busses combined.
Note: I am choosing wiring to make it visible on the breadboard. This is not the ideal wiring I would use in practice. I don’t want wires directly covering other wires, so some of them will be at angles when it’s deemed necessary.
We now have the same thing as a headphone splitter! This is where some of the downfalls of a passive system begin to show however.
First adding headphones in parallel is like adding resistances in parallel. This can lead to issues but for our setup won’t be much of a problem.
Second, the passive mixer here will have everyone who listens at the same volume level! If someone wants a lower volume there is no way to do that.
An active mixer can handle these problems. Despite the issues with this way of splitting things, passive splitting is often extremely useful and easy to implement and can sound quite good in the right circumstances!
Adding an Input
Just like how we added an output, we can add an input.
Now when we place 2 sources at the inputs.
Consider the following:
The first input will have a source, a phone lets say. The second input will not have an input. The first output will have a pair of headphones. Now answer the following knowing this information:
- Will any audio appear at the second output? Explain.
At this point we have a rudimentary patch bay. We can add more inputs and outputs, create busses and route signals as we like. Please create a schematic for the following:
Create a circuit that has 2 inputs and routes them to unique outputs. Meaning input 1 to output 1 and input 2 to output 2. There is no bus. Your answer should be presented as a schematic. Build this circuit and include a picture of it on the breadboard.
Experiment #2 Mixing:
Having a patch bay is very nice but currently the board has no way to control the volume of the sources coming in. We could use a volume control on the phone (our input) or whatever stage is before the mixer but this is very inconvenient. Instead we can simply add a potentiometer to control the volume.
Let's take a moment to consider why resistance would cause a volume level drop. Remember that in the circuit the signal is represented by a voltage. If this voltage is small then the volume in your headphones will likewise be small.
When we take a signal from a circuit we do so “across a resistance”, or a “load”. In our case the load is the headphones, however we have no control over the headphones resistance directly, and we want to create something we can control so that we can create a volume knob, therefore we introduce another resistance that we can control to do so.
Let's take a very basic case and consider a DC voltage. (Forget about the mixer for a moment and just consider what a resistor's behavior is like.)
Let's say we have
In the simulation the green tint shows how the voltage drops and the V with a circle in it represents a voltmeter. See how the entire voltage source appears across the resistor?
Remember we want a voltage not a current. Increasing the resistance will reduce the current but will do nothing to the voltage.
See how in this simulation the resistor is much greater, yet the voltage is the same! The current is far less but the voltage remains the same. How then are we to create a volume knob with a resistor?
The solution here is to use 2 resistors.
Here is the idea, if a resistor is large compared to another one in series (they must form a loop because of KVL), then a large voltage drop will occur over the big resistor and a small one will occur over the small resistor.
Here we see 2 voltmeters measuring the voltage across each resistance. One resistor is small compared to the other and only has 49.505mV! This voltage is much smaller compared to the input, hence a "low volume"! The input was 5 volts and across that resistance it's only about .05 volts! The large resistor however gets a big voltage drop at 4.95V, meaning there is basically no volume drop across the second resistor, 5V came in and nearly 5V went out. Using this we now have a way to reduce the voltage and hence reduce the volume!
Now let's flip it.
We see the opposite, there is now a large voltage drop across the first resistor and a small one across the second! If the resistors were the same the voltage drop across each would be ½ the input or 2.5 volts.
When we use resistors in this way it’s called voltage division.
This voltage division allows us to create a volume control, by varying how big one resistor is compared to the other one we can control the voltage drop across it, and since the volume level is controlled by the voltage level we therefore have also created a volume control. We simply take the output voltage from across one of these resistors since we can control that voltage by simply picking the resistor.
In this situation the values of the resistor are not directly important, it's the ratio of one resistor to the other that is important! You will see this is a common property we take advantage of.
A potentiometer is a device that is just like 2 resistors in series:
Inside a potentiometer is a strip of resistive material. A wiper moves across this material splitting it into 2 sections. There is a section from pin 1 to the wiper (pin 2), and from the wiper to pin 3.
If the wiper is at pin 1, then it is like there is no resistance between pins 1 and 2! It’s like the small resistor because very little to no resistive material is between pins 1 and 2. The resistance between pins 2 and 3 however will be very great! This is because the entire resistive strip is between pins two and 3! As the wiper moves across the strip this relationship will reverse. In this way we can control the ratio of two resistors.
So now across each channel we can add a potentiometer. We want to configure the potentiometer to act like a voltage divider. We also want the far left position to be a low volume and the far right position to be a high volume so that it acts like a volume control we are familiar with. If we wire it wrong it could be backwards! An easy mistake to make.
Since a low voltage means a low volume, and since ground is 0V, we should make one side ground to be our “low” voltage. Lets pick pin 1 to be ground so that when the wiper is turned far left it will connect pin 2 to ground acting like the little resistance. Therefore we want the output to be across pins 1 and 2 which will have a low resistance when the wiper is turned to the far left. As the wiper moves the resistance between pins 1 and 2 will grow and likewise the voltage drop will grow as well!
In other words, we connect pin 1 to ground because now the resistance between pins 1 and 2 has the ability to go all the way to ground which is 0 volts. Pin 2 will be our output since that is what we are able to move and hence is the node that “changes voltage”, meaning we can use it to change volume. Pin 3 must receive the input so that our signal can drop across the potentiometer.
Here is a simulation showing this. Try changing the slider that controls the potentiometer.
- In a circuit with only one resistor will the voltage change with the size of the resistor? Explain.
- In a circuit with 2 resistors in parallel will the voltage change with the size of the resistor? Explain.
- What would happen if the output is taken across pins 2 and 3 instead of pins 1 and 2? Explain.
For now let's simplify the board to have only 1 input and 1 output. Also, the potentiometer is 1 channel so lets only process the left channel and duplicate it on the output.
We now have a basic 1 channel volume control. Try listening to it! Stereo potentiometers do exist that have 2 rows of pins. We will use them in later labs.
Design Problem #2:
Add a second potentiometer for the right channel and make the output full stereo. The balance of these 2 channels becomes a pan control. What interaural cue would such a setup rely on for placing something in a stereo field? With your answer also submit a schematic and a picture of the breadboard.
Design Problem #3:
Add a second input and a second output to your circuit from design problem #2. Create a bus with both channels present and have volume controls on both channels. You should only need two potentiometers to do this. Submit a schematic and a picture of the breadboard.
Design Problem #4:
Design: Create a 2I2O (2 in 2 out) circuit using 3 jacks (one for the input two for the output). Keep in mind these jacks are stereo, so they can contan 2 channels. You need 1 jack for the input and two for the output. The input jack will carry a left and right audio channel, our “2 in”. Each input should have a volume control. The output should have 2 jacks, one for the left and one for the right, our “2 out”. On the output clone the signal so that both sides of the headphones work. Submit a schematic and a picture of the breadboard.