1. How do Operational Amplifiers Work?

Operational amplifiers are a type of integrated circuit designed to amplify the difference in voltage between the two input signals. Mathematically, the output signal (V_o) of an op. amp is proportional to the difference between the voltages being supplied to the positive (V₊) input terminal and the negative (V_–) input terminal:

V_o =A_{vo}(V_+-V_-)

The open loop gain (A_vo) refers to the amount of amplification applied to the difference between the two inputs. One characteristic that makes op. amps so useful is that the open loop gain/amplification value is very large (often around x100,000); this allows the output voltage to be regulated down to any desired value using some simple external components. The table below lists the main characteristics of operational amplifiers, and compares the ideal versus the actual characteristics seen in practice:

In practice, while op. amps don’t exactly follow the ideal assumptions, they are close enough so that these mathematical simplifications can usually be used when designing modular synthesiser circuits and figuring out suitable component values. Some of the situations when you need to watch out for non-ideal properties of op. amps are:

• The resistance of the input signals is very high
• The output load from the circuit is very low
• High frequency operation is required
• The output voltage needs to be able to swing right up to the supply voltages

In applications such as these, there are many specialised types of operational amplifiers which can fulfil these needs (but often are more expensive or make a compromise with other properties of the amplifier).

2. Voltage Comparator (Threshold Detection)

A comparator is a type of circuit which compares the voltages of two inputs, outputting a high voltage value when input 1 is larger than input 2, and a low value when input 1 is smaller than input 2. If the voltage of the second input remains fixed, a comparator circuit can be used to determine if the first input signal has exceeded the voltage threshold of the second signal.

(a) Simple Comparator

An op. amp can be used as a simple comparator circuit by connecting the two signals you want to compare to the inverting (V₂) and non-inverting (V₁) input terminals. If the voltage at the non-inverting input is greater than the one at the inverting input, then the output voltage will be positive; otherwise the output will be negative.

Due to the high gain of the amplifier, the output will saturate almost immediately to the maximum voltage which the op. amp can supply — this is dictated by the supply voltage rails. As a result, the output appears to be almost digital, switching between a fixed low and high voltage value.

Mathematical Derivation

• Starting with the op. amp formula:

V_{out} = A_{vo}(V_1 - V_2)

• If the gain is very large (approaching infinity), this means the output should also approach positive or negative infinity; however, the maximum output voltage is limited by the supply voltages connected to the amplifier:

\textbf{If: } \; A_{vo} \to \infin \quad \textbf{ Then:}

V_{out}=\begin{cases}
V_{s+} -V_{d} & V_1 > V_2 \\
V_{s-} +V_{d} & V_1 < V_2  \\
\text{Unstable} & V_1 = V_2
\end{cases}

Depending on whether the difference between the two input terminals is positive or negative, the output will jump to equal the positive or negative supply rails. When both inputs are equal, the output is unstable and can quickly jump between both rails. Note that the voltage of the output is reduced by the term V_d since most op. amps (unless advertised) introduce a voltage drop of up to 2 to 3 volts with respect to the rails.

Circuit Diagram

(b) Inverting Comparator with Hysteresis

In the above simple comparator circuit, when both input voltages are equal the output voltage can jump rapidly between both input rails. This can be an issue when the input signal itself is noisy; in that case the output jumps back and forth multiple times as the input voltage approaches the reference voltage instead of just giving a single clean transition. The most common method of dealing with this issue is to use positive feedback to offset the reference voltage after a transition has occurred. As a result, the reference voltage as the signal is increasing is different than that when the signal is decreasing (this is known as hysteresis).

Mathematical Formula

The voltage at the non-inverting terminal changes depending on whether the output voltage is currently at the high or the low voltage level. This means that the voltage threshold changes depending on whether the input signal is rising or falling.

For rising input voltages we can calculate the voltage at the non-inverting input (the threshold voltage) using superposition:

V_{+rising} = \frac{V_{ref}(R_1)}{(R_1+R_2)}+\frac{V_{outLOW}(R_2)}{(R_1+R_2)}

\therefore V_{+rising} = \frac{(V_{ref}R_1+V_{outLOW}R_2)}{R_1+R_2}

Similarly, for falling input voltages the threshold can be calculated:

\therefore V_{+falling} = \frac{(V_{ref}R_1+V_{outHIGH}R_2)}{R_1+R_2}

So the total hysteresis voltage swing can be calculated by subtracting these two thresholds:

V_{hysteresis} = V_{-rising}-V_{+falling}

\therefore V_{hysteresis} = (V_{outHIGH}-V_{outLOW})\frac{R_2}{R_1+R_2}

Circuit Diagram

In this circuit configuration, the output voltage is high if the input voltage is below the lower “falling” threshold, and it switches to a low output voltage when the input voltage exceeds the upper “rising” threshold. When the input voltage is in-between those two thresholds, it stays at whatever level it was at before:

V_{out}=\begin{cases}
V_{s+} -V_{d} & V_{in} < V_{+falling} \\
V_{s-} +V_{d} & V_{in} > V_{+rising} \\
\text{Unchanged} & V_{+rising} \le V_{in} \le V_{+falling}
\end{cases}

3. Unity Gain Amplifier (Signal Buffer)

In a unity gain amplifier (also known as a signal buffer or a voltage follower), the output of the op. amp is connected back into the inverting input terminal. Using this feedback, the amplifier actively tries to reduce the difference between the input and output voltages, making them the same. This circuit is very common, as it can be used to electrically isolate (buffer) two parts of a circuit. The output of the op. amp acts as an ideal voltage source, meaning the load connected to the output won’t affect the signal at the input of the op. amp and vice versa.

Mathematical Derivation

• Starting with the op. amp formula:

V_{out}=A_{vo}(V_{in}-V_{out})

• Rearranging:

(A_{vo}+1)V_{out}=A_{vo}V_{in}

• Due to the large open-loop gain of the amplifier (which is much larger than one), the left side of the equation can be simplified to remove the “+ 1” term.

\textbf{If: } \; A_{vo} \gg 1  \quad \textbf{ Then: } \; A_{vo}+1 \approx A_{vo}

\therefore \; V_{out} = V_{in}

Circuit Diagram

4. Simple Amplifier (Signal Boost)

(a) Inverting Amplifier

An amplifier circuit is a system which takes an input signal and increases the magnitude (amplitude) of that signal. For example in modular synthesisers, this would increase the volume of the audio waveform. An inverting amplifier configuration applies this increase and inverts the signal, meaning that the positive sections of the signal become negative and the negative sections become positive.

Mathematical Derivation

• To start off, we will substitute in zero for the non-inverting input terminal which is connected to ground:

V_{out}=A_{vo}(V_+-V_-)

\textbf{Since: } \; V_+=0 \quad \Rightarrow \; V_{out} =  A_{vo}(-V_-)

• Due to the high open-loop gain, the amplifier changes its output to try to keep the voltages of the inverting and non-inverting input terminals the same. This creates a “virtual ground” at the negative input.

\textbf{If: } \; A_{vo} \to \infin \quad \textbf{ Then: } \; V_- \to 0

• Using superposition we can derive another formula for the voltage at the inverting input (see explanation in box below):

V_-=  \frac{V_{in}R_2}{R_1+R_2}+\frac{V_{out}R_1}{R_1+R_2} = 0

• This can be rearranged and simplified to give us the relationship between the input and output voltages:

V_{in}R_2+V_{out}R_1 = 0

\therefore \: V_{out}=-V_{in}\frac{R_2}{R_1}

Circuit Diagram

(b) Non-inverting Amplifier

The non-inverting amplifier configuration is almost identical to the inverting amplifier above, except that the ground and input voltages are swapped. By adjusting the ratios of the two feedback resistors R₁ and R₂ the amount of gain/amplification of the circuit can be controlled.

Mathematical Derivation

• If the open loop gain is infinitely large, then the voltages at the inverting and non-inverting inputs needs to be equal:

V_{out}=A_{vo}(V_{in}-V_-)

\textbf{If: } \; A_{vo} \to \infin \quad \textbf{ Then: } \; V_{in} - V_- \to 0

\therefore V_{in} = V_-

• Using the formula for a potential divider, the voltage at the inverting input terminal can be related to the output voltage:

V_- = \frac{V_{out}R_2}{R_1+R_2}

• Rearranging to isolate the output voltage:

\therefore V_{out}=V_{in}\left( \frac{R_1}{R_2} + 1 \right)

Circuit Diagram

5. Summing Amplifier (Signal Mixer)

(a) Inverting Summing Amplifier

In modular synths you often want to add multiple signals together to create a variety of effects and harmonies. This can be done very simply by adding additional inputs to the normal inverting amplifier I described in Section 3(a). The contribution of each input signal to the output is dependent on the ratio of the input and feedback resistors.

Mathematical Derivation

V_{out}=A_{vo}(V_+-V_-) = A_{vo}(-V_-)

\textbf{If: } \; A_{vo} \to \infin \quad \textbf{ Then: } \; V_- = 0

• Since almost no current flows into the inverting op. amp input, the sum of all the current coming into the inputs needs to flow through the feedback resistor (R). Note that as the conventional current flows from the 0V point at the inverting input across the feedback resistor, this can only be true if the output is at a lower voltage (meaning it is negative).

I_R=\sum_{i=1}^N\frac{V_i}{R_i} \quad  \textbf{ and } \quad V_{out}=-I_RR

V_{out}=-\left(V_1\frac{R}{R_1}+V_2\frac{R}{R_2}+...+V_N\frac{R}{R_N}\right)

• If all the resistors are equal, then all the resistor terms in the equation cancel each other out and the op. amp simply sums all the input voltages together and inverts it:

\textbf{If: } \; \{R_i = R \text{ }| \text{ } i = 1,...,N\}

\therefore V_{out} = -(V_1 + V_2 + ... + V_N)

Circuit Diagram

(b) Non-inverting Summing Amplifier

It is also possible to create a non-inverting signal mixer, where the output is the positive sum of the inputs. However, this circuit is not as common in synthesisers since it is not as easy to add additional inputs. If more than two inputs are added, the ratio of the two resistors connected to the inverting input of the op. amp needs to be changed to ensure the output gain factor remains the same.

Mathematical Derivation

V_{out}=A_{vo}(V_+-V_-)

\textbf{If: } \; A_{vo} \to \infin \quad \textbf{ Then: } \; V_+ = V_-

• Same as the non-inverting amplifier circuit (4b), the voltage of the inverting input can be calculated using the potential divider formula.

V_-=\frac{V_{out}R}{R+R}=\frac{V_{out}}{2}

• The voltage at the non-inverting input can be calculated using superposition:

V_+=\frac{V_1R}{R+R}+\frac{V_2R}{R+R}=\frac{V_1+V_2}{2}

• Since the voltage at the inverting and non-inverting inputs needs to be equal, the output voltage therefore is the sum of the two inputs:

\therefore V_{out}=V_1+V_2

Circuit Diagram

6. Dealing with Unused Op. Amps

When working with operational amplifiers on a breadboard or in a circuit, each IC chip usually contains multiple op. amps. Commonly chips contains either 2 or 4 amplifiers; however, your circuit may not need to use all of these amplifiers meaning that one or two of them will remain disconnected. In this situation it is good practice to connect the pins of the unused amplifiers to a known input and output so that they won’t be changing randomly. If these pins are left floating, the unused op. amps may introduce noise and interference which negatively impacts the performance of the op. amps that are being used. To prevent this, the amplifiers should be wired up in a “Unity Gain” configuration (see Section 3), with the non-inverting input connected to a voltage which is halfway between the positive and negative rails:

1. Series Battery Method

One of the simplest ways to create a dual power supply is by using two sets of batteries. The batteries are connected in series, so that the positive terminal of one battery is attached to the negative terminal of the second battery. When this middle connection is used as the ground reference for the circuit, you will be able to get a positive and negative voltage from the batteries, as shown in the circuit diagram below. For small and portable synthesisers, this is often done using two 9V batteries as I’ve demonstrated on a breadboard in the image below. Since the voltage of both batteries will drop as the power is drained, we also need to include voltage regulators which ensure a stable voltage is supplied to the synthesiser. In the image below, you can see that the batteries I am using are almost empty as the voltage measured by my multimeter is only -7.11V.

This method only works when one or both voltage sources are said to be “floating”. This means that the power source is not connected to any absolute reference voltage, such as a connection to Earth. All batteries are floating power sources, however wired power supplies often aren’t. For example, if the negative terminal of both voltage sources is connected to ground, then attaching the positive and negative terminals of both sources together will simply create a short-circuit; this is something I would encourage you to avoid!

• Benefits:
  • Very easy to implement and troubleshoot.
  • Relatively Portable.
  • Voltage can be increased by adding more batteries in series.
  • The battery life and maximum output current can be increased by adding more batteries in parallel.
• Disadvantages:
  • Batteries constantly need to be replaced!
  • The voltage of the batteries will drop as they run out (as seen in the image), so an additional power regulator IC will still be required.

2. Dual AC to DC Rectification

The electricity being supplied in the mains socket alternates from a positive to a negative voltage many times a second (230V 50Hz in Europe, 120V 60Hz in the US). What we want to do is reduce this voltage down to a lower and more manageable voltage, taking the positive half of the AC signal to supply the positive output and the negative half for the negative output. This process requires the following steps:

• Step down the high voltage being supplied by the mains to a lower voltage using a transformer.
• Rectify the AC signal into a positive and negative signal using diodes.
• Smooth out the voltage using capacitors.
• Generate a stable output voltage using power regulators.

a. Half-wave Rectifier Circuit

This is the power supply design I used in my synthesiser, and it is probably the most common design used by DIY synth builders. This design is often preferred to the Full-wave Rectifier as you can use a commercial wall plug transformer to convert the mains power down to 12V AC, which is used by the power supply. This means that your circuit does not directly come into contact with the mains power, making it a little bit safer to work with (but you still need to be careful!).

Important: You need to make sure that the wall plug transformer you use outputs 12V alternating current, and not 12V direct current. The 12V DC plugs are a lot more common, so it may take some searching to find the correct type of 12V AC plug. Also make sure that the plug you get is rated for a current of at least 1000mA or above, and that the mains voltage input rating is correct for the country you are in.

An example of a half-wave rectifier circuit is shown in Schematic 2 below. The circuit takes in a 12V AC signal from the wall plug, and converts it into a stable positive and negative 12V output. I have seen many variations of this circuit, using a wide variety of different capacitor values.

How does it work?

1. The circuit takes in a 12V alternating current signal from the wall plug transformer. 12V AC refers to the root mean square (RMS) value of the signal. This signal has a peak voltage of ±17V, as shown in the waveform diagram below.
2. The diode D1 only allows the positive half of the AC signal to pass through, while D2 lets negative voltages through. This process is known as half-wave or half-bridge rectification, since only half of the AC waveform is used to power each of the voltage outputs. As a result, each output can theoretically only output half of the power (and consequently current) delivered by the wall plug transformer. The peak voltage of the rectified signals is ±16.3V, since the diodes introduce a 0.7V drop into the circuit.
3. The capacitors smooth out the waveform, ensuring that a more continuous voltage is being supplied to the voltage regulators. The reasoning behind selecting this specific capacitance value is discussed in the next section.
4. The LM7812 and LM7912 voltage regulators ensure that the outputs of the power supply stay at a stable +12V and -12V respectively. If you want to get +15V and -15V outputs instead, you can use a 15V AC power plug and replace these with the LM7815 and LM7915 regulators. If you are putting together your own circuit, watch out as the input, output and ground pins are in a different order on the positive and negative voltage regulators.
5. Capacitors C3 and C4 are mainly included to improve the transient response of the power supply; the capacitor can provide brief bursts of high current when there are sudden changes in load being applied to the power supply. According to the data-sheet for the negative voltage regulator LM7912, for stability the capacitor C4 should be at least 1μF (using a tantalum capacitor) or 10μF (using an electrolytic capacitor). The higher value of 100μF was chosen to give an additional factor of safety over this minimum value.
6. The two LEDs are there to indicate that there is power at the outputs. Some negative power regulators also require a minimum load to be applied at the output before they start up, so the LEDs help to provide that load.
7. According to the data-sheet for LM7912, the diode D4 is required when large capacitors such as C10 are used at the input. The diode prevents momentary input short circuits, which can occur when the circuit is powered up or down. The LM7812 does not necessarily need this, but I put D6 in for good measure.
8. The data sheet for both LM7812 and LM7912 specify that D5 and D3 should be present to prevent latch-up problems. These components act as clamping diodes, helping to protect the regulators from reversed polarity on the outputs. If one regulator starts up before the other one, devices such as operational amplifiers (op amps) can latch up and cause a short circuit between both power rails. This can prevent the second regulator from starting up. The diodes (preferably Schottkey) prevent the positive output from going below -0.3V and the negative output going above 0.3V, allowing both regulators to start up and the latch up condition to stop.

How to choose the capacitor values?

Why are there two capacitors at the input of each power rail (C1 & C7, C2 & C10)? How were the values of these capacitors chosen? I’ve looked at several schematics for half-wave rectifiers and there seems to be a lot of variance in what the capacitance value should be.

Generally there is one small non-electrolytic capacitor close to the input of each power regulator, which helps to stabilise, filter and smooth the input (C1 and C2). Usually this is between 100nF to 1μF. Small capacitors (ceramic, polyester, tantalum etc.) tend to be better than larger electrolytic film capacitor at filtering out high-frequency noise from the signal.

Then there is a bank of large electrolytic capacitors connected in parallel (C7 and C10; more capacitors can be connected if required), ensuring that there is a relatively constant reservoir of power even when the AC input signal is in the opposite half of the wave and no new power is being supplied. These capacitors are good at removing low frequency noise and stabilising variances in the DC voltage. The total capacitance of this reservoir depends on the amount of load you expect to put on the power supply. Here is how to calculate how much capacitance you may need:

V_{DC}=\frac{16.3+14.5}{2}=15.4V

\rho_\%=\frac{15.4-14.5}{15.4}\times100=5.8\%

R_L=\frac{V_{DC}}{I_{DC}}=\frac{15.4}{1}=15.4\Omega

P_D = (V_{DC}-V_O )(I_{DC})\newline=(15.4-12)(1)=3.4W

C_s=\frac{1}{\rho _\%R_L}=\frac{1}{5.8\times 15.4}=0.011F

\text{If } \quad C_s=0.0047F \quad \text{then:}

R_L=\frac{1}{5.8\times 0.0047}=36.7\Omega

I_{DC}=\frac{15.4}{36.7}=0.42A

So to summarise… if we want to get a full 1A of output current from our power supply, the combined capacitance value at the input of the regulator needs to be at least 11,000µF.

b. Full-wave Rectifier Circuit

In a “full-bridge” or “full-wave” rectification circuit, both the positive and negative sections of the alternating waveform are used to power both outputs. This means that the circuit can theoretically drive twice the load compared to a half-bridge rectifier. As can be seen in Schematic 3, most of the circuit is identical to the half-bridge rectifier. The only differences are that two additional rectification diodes have been added, and a transformer with three outputs (called a “Center Tapped Transformer”) is used. The central output of the transformer is used as the ground reference, while the other two connections output an identical 12V AC signal, but out of phase by 180°. This means that when one of the outputs is in the positive section of the alternating waveform, the other is in the negative section and vice versa.

This type of circuit is often used in professional equipment, but is not used as much by DIY synthesiser builders. Center-tapped transformers are not available as a pre-packaged wall-plug, so you would need to wire your own. Since one end of the transformer is connected to mains power, building this circuit involves a bit more risk and should only be attempted if you have the right equipment and know what you are doing! When buying a transformer, make sure that the mains voltage input rating is correct for the country you are in.

How does it work?

1. The transformer takes the mains alternating signal and reduces the voltage, outputing two 12V alternating current signals which are out of phase by 180°.
2. The four diodes are used to separate the positive and negative sections of the alternating signal, directing the positive halves to the +12V regulator and the negative halves to the -12V regulator. Since both AC signals are out of phase, this results in a continuous supply of power for both polarities.
3. The rest of the circuit is identical to the “half-bridge rectifier”, so you can refer to my description above to see how it works and what each component is doing.

3. DC to DC Inverting Charge Pump

It is also possible to generate a dual 12V power supply from only one +12V DC power plug. This is useful since DC power plugs are a lot more common and therefore cheaper to buy. It is also easier to find 12V DC plugs which have a high current rating, allowing more synthesiser modules to be powered from the same supply. This type of power supply design is often seen in portable modular synthesiser kits, and small Eurorack-compatible power modules. Since the transformer and rectification circuitry (large capacitors) are all contained within the external plug, the footprint of the electronics used in this design can be made a lot smaller than in the Dual AC-DC rectification circuits.

a. How does it work?

In its most simple form, an inverting charge pump uses a “floating” capacitor to carry charge over from the +12V side to the -12V side of the circuit. The capacitor is charged up from the +12V input being provided by the wall plug. Once full, the capacitor is disconnected from +12V input and the positive lead is instead connected to ground. Since the charge (and therefore voltage drop) across the capacitor remains the same, this means the negative terminal of the capacitor is now at a voltage of -12V. The capacitor then begins to discharge and this is used to power the negative rail. In our power supply, this process of charging and discharging is repeated many times a second. Schematic 4 shows an equivalent circuit, demonstrating how this system works. In a real circuit, the switching of the capacitor is done using an IC chip.

1. Initially, switches S1 and S3 are closed while switches S2 and S4 are open. Capacitor C1 is connected to Vin and ground, causing the charge in the capacitor to increase.
2. After a certain interval, the switches S1 and S3 are opened up again while S2 and S4 are closed. The top leg of the capacitor is now connected to ground instead of Vin. Since the charge in the capacitor hasn’t changed, there is still the same voltage drop across the capacitor. As a result, a voltage of -Vin is present on the bottom leg of the capacitor.
3. This switching mechanism is continuously repeated, charging the capacitor C1 with the positive input voltage and de-charging it again on the inverted output. The capacitor is essentially pumping the charge from the positive input to the inverted output.
4. Capacitor C2 acts as a power buffer/storage, smoothing the voltage on the output and ensuring that a continuous supply is available at the inverted output.

b. Implementing it in practice

In the example circuit show in Schematic 5, we are using the LTC1144 chip made by Analog Devices to do the switching for out inverting charge pump. The capacitor C6 is used to invert the charge, while C5 acts as reservoir so that the negative output has a more stable output. The graphs show how the circuit reacts when it is started up. The current through the capacitor C6 alternates from positive to negative at regular intervals as it charges from the positive supply and de-charges into the negative output. The voltage of the negative output quickly decreases as the reservoir capacitor C5 is charged up, levelling out at -12V over time.

In the LTC1144 chip, the frequency of the switching signal can be increased or decreased by changing the value of the capacitor connected to the OSC input pin. Charge pumps can operate at a wide range of switching frequencies, usually ranging from 1kHz to as high as 200kHz.

Note: I haven’t had the chance to try this circuit out in practice, so the capacitor values in Schematic 5 will probably need to be altered to make it suitable for use as a synthesiser power supply. The circuit simulations were done in the free program “LTspice” made by Analog Devices.

Building a synthesiser rather than buying a pre-assembled one has several benefits; first of all, it is significantly cheaper! For example, a simple voltage-controlled oscillator (VCO) Eurorack module usually costs at least €100, while the components for a DIY version can often be sourced for less than €20. But more importantly, building a DIY synth helps you to understand exactly how each circuit works and how it influences the overall sound. Once you become familiar with how a circuit works, you can also begin adding your own extra features and quirks! There is quite a large DIY synth community where people show off and share their own designs. For me at least, I find that building my own synth is just as much fun as trying to make music with it once it is complete.

What is a Modular Synth?

Synthesisers are devices which can create sounds and music from electronic circuits. A sound can be created very simply by generating an oscillating electrical signal, which is varying at a frequency which we can hear (usually between 20 to 20,000Hz). By changing the rate at which this signal oscillates we control the pitch of the tone, while the volume can be controlled by adjusting the peak-to-peak voltage of the signal. Diagram 1 shows what changing the volume and pitch of an sinusoidal signal looks like. However, generating a basic signal is only the starting point when using a synth! By combining multiple signals and filtering them in weird and wonderful ways, it is possible to create some truly unique sounds and music.

Basic synthesisers have multiple functions hardwired together to give you a specific type of instrument and tone. Modular synths are special, because each unique musical operation is broken out into a separate device/module. For example, you could have one module to generate an initial waveform (oscillator), then one to change the tone of the sound (filter), followed by a module to control the volume (amplifier). Each module has an audio jack for every input, output and control signal, allowing you to automate and change the way in which the module operates. This allows you to hook up the modules in a large number of different ways! Diagram 2 shows one way in which 4 common modules can be linked together to produce music.

So what does the result sound like? I’ve attached an early video of my DIY modular synth below, showing a demo tune I put together to see if all of the modules were working. I have since added some extra modules to my synthesiser, which I will talk about in a later post. At the top of the box is an 8-step sequencer which is used to control the pitch and duration of each of the notes. On the left (blue) is a single oscillator which can produce three different types of waveforms (sinusoid, triangle, square), as demonstrated at the start of the video. Finally, next to this is an amplifier (grey) which can control the volume of each of these waveforms. It is interesting to hear all the different types of tones which this basic synthesiser can produce, without needing to introduce any filtering.

Up next in this Tutorial Series…

As I continue building and improving my own DIY synth, I hope to post several tutorials describing various electronic circuits commonly used to create music. Using a variety of diagrams and practical demonstrations, I will try to show how each synth module affects the audio signal. I will also try and model parts of the circuit mathematically, to illustrate the process and consideration made by the designers when developing the designs. Here is a list of the modules I have already built, and may discuss in my future tutorials:

• Dual +-12V Power Supply
• Voltage Controlled Oscillator (VCO)
• Voltage Controlled Amplifier (VCA)
• 8-step Sequencer
• Attack, Decay, Sustain and Release module (ADSR)
• Voltage Controlled Filter – Low Pass and High Pass (VCF)