Microphone pre-amp for microcontrollers


This microphone amplifier is based arond the TLC272 double opamp.
The first opamp is  used as an amplifier. The voltage offset on the positive input is set to half Vcc with R2 and R3. The Closedloop gain is set at 100 by R5 and R4. There is an AC coupling/high pass filter to the 2nd opamp by C4 and R6. The second opamp is used as comparator: R7 and R8 determine a level sensitivity of 71 mV (with 5 Volt Vcc).

SPL is the sound pressure level. There is also an audio output for the amplified microphone signal. As the audio signal is DC coupled, the signal will swing around 1/2 Vcc

Microphone pre-amp for Arduino
Note: In the picture you may see a TL072 chip rather than a TS272 (TLC272). That is because I briefly tried that one. The TL072 is a double opamp especially for audio purposes. My results with it however were not so convincing. Best use the TS272 (TLC272)

Versatile microcontroller PSU


Though 5 Volt is commonly used for microcontrollers, not seldomly higher voltages are required as well, such as 9 or 12 volt for relays or lamps.

The presented psu offers that and it came to life because i had an old 2×6 print transformer laying around that i finally decided to put to some use.

The rectification is a classic one to get a double voltage from 2 transformer coils, although one could of course use 2 bridge rectifiers. This gives a rough voltage of 8.5 and 17 Volt.

The rectified voltages are then stabilized to  5 and 12 volt respectively and we end up with 4 voltages as shown in the figure.


Electret microphone amplifier


Amplifier for an electret microphone that can be used to detect sound with an Arduino or other microcontroller. The op-amp used is a TLC272 that is pin compatible with a number of other dual op-amps such as the LM393 the LM358 or the OPA2344 and OPA2345.

IC1a is used as a non-inverting amplifier. It offers a higher impedance as the inverting amplifier. The gain of the non-inverting amplifier is determined by 1+R4/R1 and is 40dB (100x). If desired a trim pot can be used for R4.

In most cases it is possible to DC couple the circuit. However in this case it is necessary to ensure that the non-inverting has a DC path to earth for the very small input current that is needed. This can be achieved by resistor, R2 in the diagram. The value of this may typically be 100 kOhms or more. If this resistor is not inserted the output of the operational amplifier will be driven into one of the voltage rails.

When adding a resistor in this manner the capacitor-resistor combination forms a high pass filter with a cut-off frequency. The cut off point occurs at a frequency where the capacitive reactance is equal to the resistance. This is expressed in the Formula 1/2πRC. For the used combination it is =16Hz.

The output on pin 1 is at best a sinus wave and though microcontrollers do not really like those or any other odd shaped wave, when fed to their ADC input that does not make a difference. However, if  the output of IC1a is fed to an ADC there will be very strong fluctuations in the reading and as such the signal is not really representing the total sound pressure. We do that  with op-amp IC1b that is used as a comparator with a reference voltage on the inverting entrance. When fed with  5Volt, the Voltage on the inverting entrance will be 72mV ((5000/69)*1)
As long as V- < V+, Vo will be low, but when the sound level rises above 72mV, IC1b will make the output go high.
With a gain of 100 in IC1a, a voltage off 720uV from the microphone would be enough.  The Voltage build up over C3 is depending on the number of times the Output goes high and thus is an indication for the sound pressure.


The PCB is quite simple. The convoluted area above the double op-amp is to either insert a resistor, or an upright trim pot that can be inserted facing, or facing away from the IC.
There are 4 extra components in comparison with the circuit. Those are 2 LED’s and 2 resistors. These give an indication of the Output voltage as well as the supply voltage.  They are optional however:

There are many commercially available electret breakout boards for the Arduino (or other micro controllers). One of them is the Sparkfun breakout board:
electret-sparkfun Sparkfun had a bit of a rocky start in this area with their BOB-08669 and they replaced it with the BOB-09964, the difference between the two being some changes in resistor and capacitor values.

The Sparkfun board is centered around 1 op-amp, the OPA344 that is used as an inverting amplifier. The gain is determined by R5/R4 and is 40dB (100x). The input high pass RC filter cuts off frequencies below 10Hz. The RC network with C3 and R5 cut off above 13 kHz. The original BOB-08669 would cut off too high and it’s impedance was too low (as can be expected from an inverting op amp). The cut-off  for the low pass filter can be calculated from xc=1/2πRC, in which xc being the reactive capacitance. With the given values of 12pF and 1Mhz the low pass is determined at 13.2kHz. The high pass around C1 and R4 gives a cut off of 3.3Hz (so not 10Hz as stated).

R2 and R3 give an off-set adjustment centered around half voltage (2.5 Volts).

Though the new board is an improvement over the old one, still a lot of people experienced problems with it. It did not really change in design, just a few R and C values were adapted.

Dealextreme  offers a board based around a 393. It has an audio out and a digital out for if the signal comes above a certain threshold. It is the KY-037 from Keyes and has a 12 turn potentiometer to set the threshold. It is rumoured to have low sensitivity.

Freetronics board
The Freetronics board is also based upon the TLC272. It offers an audio output and a sound pressure output with visual indication. Here the peak detection on pin 6 of the IC is set to 71 mV and the gain of the first opamp is set to 100

Using a Single Opamp

Using a single op-amp of course is also possible and this time we will use the inverting pin of an LM741. The gain = R4/R3 =1000=60dB.
D1 and D2 form a voltage doubler, rectifier and peak detector.
C3 and R5 form an RC delay for the analog entrance of the Arduino


Using the LM386
Although the LM386 is primarily aimed to drive a speaker, it can be used as a microphone pre-amp. results with it however have been less than satisfying.

Using the LM358
The LM358 gives better results than the LM386.

The following code can be used to easily test the workings of the amplifier in combination with an Arduuino. The code reads the input 10.000 tmes and then prints the min, the max and the spread

 * Measures the voltage on an analog pin over a ~1s period
 * of time and sends the Min, Max and Diff (Spread) values 
 * over Serial.
 * Author: 
 * Dimitar Kovachev, http://lowvoltage.wordpress.com/
const int analogPin = 0;

void setup() {

void loop() {
  int mn = 1024;     // mn only decreases
  int mx = 0;        // mx only increases

  // Perform 10000 reads. Update mn and mx for each one.
  for (int i = 0; i < 10000; ++i) {
    int val = analogRead(analogPin);
    mn = min(mn, val);
    mx = max(mx, val);

  // Send min, max and delta over Serial
  Serial.print(" M=");
  Serial.print(" D=");

Cheap PIR hack

P1040700c I picked up a motion sensitive nightlight at a thrift store. It runs on 3 AA batteries and lights up 3 pretty bright LED’s if an object approaches the sensor. The one I have goes under different names sysch as the YL-253 (YL stands for Yilin). In the Netherlands it is sold at the ACTION shops for 2.59 euro

As it goes through its batteries pretty fast and one of the dead cells had leaked in the battery comportment, I wondered if I could perhaps use it as a motion sensor for my Arduino.

The case opens with 4 screws and then reveals a circuit board with an LP0001 chip on it and a big fat photosensitive element that probably is a PIS209S or a D203s. The D, S and G of this sensor do not stand for Drain.Source and Gate, but for Drain, Source and Ground. The signal is picked up from the Source. It is a Dual Element detector with an openings angle of 120 degrees. It operates between 3 and 15 Volts and has a load resistance of 47k.

The DIL16 chip is an LP0001. That is a PIR (passive infra-red controller) and supposedly one of the most stable that is available. Sometimes the chip is called TL0001 or BISS0001.

A datasheet gives the following circuit:
Screenshot from 2014-04-10 08:58:05

This is not exactly the circuit in the night light, but it comes close. The Nightlight  e.g. does not have a relay but 3 bright LED’s are driven. It also does not have a an LDR but a photodiode. The function of the  diode (or LDR) is to ensure the night light only works at night. The board also has  switch for On-Off and Auto. On using the circuit, it becomes clear that the LED’s stay on for a  specified amount of time, that is called the Trigger time. The Trigger time is determined by 24576 xR10 x C6. Once triggered, the chip cannot be triggered again for a certain amount of time that is called the Inhibition time. This inhibition time or Ti is defined  by Ti≈24 xR9xC7  (watch out, the Chinese data sheet has switched these two values, the English data sheet is correct). With the current values that wild give a Trigger time of 2.5 seconds and  an inhibition time of 0.036 secs. (These values are for the circuit of the datasheet. in the night lamp the on time is abt 5 seconds)
To get rid of the light/dark function, it suffices to  replace the photo diode with a resistor in the region of 220kOhm.

The data sheet of the LP0001 shows 3 outputs: Output 1 (on pin 16) and Output 2 (on pin 12) as wel as Vo (on pin 2).
Vo is the detector output that is used to drive a transistor (that drives a relay) and it can be used as such. The other two outputs deliver an analog signal.
Output 1 is the output of an opamp (with + and – input). That has a gain of  R7/R8 which is equal to 2M/47K= 40 Output 1 is fed to input 2. That opamp has a gain of R6/R5 = 100. So it seems most logic to pick up the final PIR signal from either opamp 1 or opamp 2. The best seems to be from the Output of opamp 2 on pin 12


What one then only needs to do is to solder a wire to PIN 12 of the chip to pick up the PIR signal and to lead that to an analog input on the arduino.

P1040698c After you have soldered the wire connect that to e.g. analog pin 0 on your arduino and connect the Ground as well.

Copy the following program in your Arduino:

int pirPin=0; 
  Void setup() { 
  pinMode(pirPin, INPUT);
  void loop() { 

Move your hand to and fro the sensor

theoretically this will give readings between 0 and 1023.

If that works and the sensor readings seem to have at least some correlation with your hand movements that means the sensor works.

One can do several things with this output, e.g. use it to drive an LED or a lamp with analogWrite, or use it for a threshold value above or below which e.g. a lamp or relay is switched on or off.

Whatever one choses, it is generally a good idea to take the average of a few readings and one can do that as follows:

int pirPin=0; 
int value=0; 
void setup() 
  Serial.begin(9600); //just to check
  pinMode(pirPin, INPUT); 
void loop() 

int sample(int z) 
/* This function will read the Pin 'z' 5 times and 
   take an average. 
   Afterwards it will be mapped by dividing by 4 
   Could of course immediately divide by 20 but this 
   way it is easier to follow the program 
  int i; 
  int sval = 0; 
  for (i = 0; i < 5; i++){
  sval = sval + analogRead(z);// sensor on analog pin 'z' 
  sval = sval / 5; // average 
  sval = sval / 4; // scale to 8 bits (0 - 255) 
  return sval; 


The digital signal (from Vo) ofcourse can be easily incorporated in a program that reads a digital pin on the Arduino.

EEPROMs on the Arduino with I2C

eeprom2 Attaching an EEPROM to the Arduino is quite simple and the easiest way to do that is via the I2C bus. EEPROMs come in many forms but the 24 LS256 is a good choice as it is easy to use and pretty cheap (85 euro cents at my supplier). It can hold 256 kilobits of data (that’s 32 kilobytes).

The 24LS256 can also run on 3.3V which is handy if one is using a Lilypad or Pro Mini 3.3V. The 24LS256 uses 3 pins for selection of its address, so you can use up to eight at once on the same bus.
The above figure shows the base address register of the 24L256. It is addressed as follows: 1010A2A1A0. If you are only working with 1 EEPROM, the easiest is to connect A2-A0 with Ground. This gives the address of 1010000 which is 0x50 Hex. In reading and writing to the EEPROM one needs to realize that it has 32 kB (actually 32767) and therefore one byte is not enough to address all the memory.

So when one wants to send read and/or write requests, one needs to send two bytes – one for the MSB or higher end of the address (the 8 bits from left to right), and the next one for the LSB or lower end of the address (the final 8 bits from left to right).

If for example one wants to use address 21000, that goes as follows: In binary, 21000 is 0101001000001000. Split that up into 01010010 and 00001000, then convert the binary values back to numerical bytes to send with Wire.send().

That sounds more complicated than it is, as there are in fact two operands to help with that. This first one is >>, or bitshift right. This will take the MSB of the byte and drop off the lower end, leaving only the first 8 bits. To get the  LSB of the address, one can use operator &, or bitwise and. This operand, when used with 0xFF will give the lower bits.

Writing data to the 24LC256

Writing data is quite easy. First initialize the I2C bus with:

Wire.beginTransmission(0x50); // for pins A0~A2 set to GND

then send some data. The first data to send are the two bytes for the address (25000) were one wants to write to the memory.

Wire.send(21000 >> 8);  // send the MSB of the address
Wire.send(21000 & 0xFF); // send the LSB of the address

Subsequently send the byte to store at address 21000 and  then close the connection:

Wire.send(15); //just sending ‘15’  as example

That concludes the writing. now for reading:

Reading data from the 24LC256

Reading goes similar. First initialize the connection and provide the address of the data to read:

Wire.beginTransmission(0x50); // Chosen base address
Wire.send(21000 >> 8);  // send MSB of the address
Wire.send(21000 & 0xFF); // send LSB of the address

Then, ask for the number of bytes of data starting at the current address:

Wire.beginTransmission(0x50); // base address Wire.requestFrom(0x50,1); // We want one byte

Here, inbyte is a byte variable chosen to store the data retrieved from the EEPROM.

The Power of the I2C bus is of course that various devices can be connected to the same lines. The top figure shows such a set up with two EEPROMs. It is key of course that they each have their own address. In the figure  I have chosen to use addresses 0x50 and 0x51. One gets that by connecting A0-A2 to ground for  one chip, but connecting A0 to Vcc (‘ High’)  for the second chip. The resulting address is then 1010001.

A print layout i given below


Print design can be found here

Power LED drivers

led-driver4R2 determines the current.
– LED current is approximately equal to: 0.5 / R2
– R3 power: the power dissipated by the resistor is approximately: 0.25 / R2.
choose a resistor value at least 2x the power calculated so the resistor does not get burning hot. so for 700mA LED current: R2 = 0.5 / 0.7 = 0.71 ohms. closest standard resistor is 0.75 ohms. R2 power = 0.25 / 0.71 = 0.35 watts. we’ll need at least a 1/2 watt rated resistor.

The design of this LED driver is very simple and probably can be found in many variations on the interenet: T2 operates in linear mode and acts as a variable resistor that controls the current through the LED’s. R1 pulls-up the gate of T2 so that T2 starts up in its conductive state.
As the current begins to flow through the LED’s, T2, and R2, the voltage drop across R2 increases until it is high enough to switch on T1, and in so doing starts to pull the gate of T2 to ground.
This causes T2 to increase its resistance, which decreases current flow through itself, the LED’s, and R2, which decreases the voltage drop across R2, which causes T1 to let T2’s gate float back toward positive supply, thus increasing the current flow. Within a few milliseconds, the current flow stabilizes around a specific set point determined by the value of R2. The required supply voltage overhead is only about 0,6-1.2 volts, less than half of what for instance an LM317 regulator circuit requires.

As the value of R2 is based on the transition emitter-base voltage of T1, calculating it is easy. Most conventional silicon general-purpose NPN transistors switch fully at around 0.7VDC and start to transition from off to on at 0.56-0.58VDC. Assuming a transition voltage of 0.58VDC and a target load current of 320mA, the math is as follows (Ohms law):

R = V / I
R = 0.58VDC / 0.320A
R = 1.8 Ohms

Calculating the power dissipation for R2 under normal operation is done by Watt’s Law:

P = V * I
P = 0.58VDC * 0.320A
P = 0.185 Watt

In this case a quarter-watt resistor will work but a one-watt would have a safety margin of double. Using a design with an LM317 would lead to a much greater loss.

Choosing the input voltage is simple. In the example above I have used 3 LED’s, so you would have to total the forward voltage of those and add about 1.2 Volts:

For example:
Supply voltage for red LEDs=3*2.5V+1.2V=8.7V
Supply voltage for blue LEDs=3*3.8V+1.2V=12,6V

Perhaps you could even get away with an excess voltage of 0.6 instead of 1.2 Volts

The circuit has a disadvantage: It tries to push as much current through the load as it can as long as the voltage across R2 doesn’t exceed what starts to turn T1 on. So, for increased fault tolerance, the power rating for R2 could be selected based on the worst-case scenario of the maximum amount of available current flow in the event of a failure: a short across the LED’s as well as T2. For that, simply calculate dropping the supply power (voltage times current) across R1. It’s generally cheaper to simply use a fuse rated for just above the desired current set point, though, as a power supply providing 1A @ 12VDC would require a 15-watt resistor for R2 and this would be far more costly than a simple fuse.

Dissipation through the MOSFET under operating conditions is calculated the same way, only the voltage used in the calculation should be the supply voltage so that the heatsink selection will account for the worst-case scenario of a dead-shorted load at the target current. In the above example, assuming a 12VDC supply and 320mA current limit, the math works out as follows:

P = V * I
P = 12VDC * 0.750A
P = 3.84 watts

The IRF520 can handle 9.2A at 100VDC, and can dissipate 60 watts, so only a modest heatsink might be necessary.
To calculate dissipation through T2 with an accidentally shorted load, simply calculate the maximum supply power that T2 will see. If the supply is unregulated and capable of large current flows in short time spans (e.g., batteries), again a fuse would be wise as a last-ditch protective measure against catastrophic failures by blowing the circuit open so it cannot try to short the supply. If the set current is 320mA, a 500mA fuse in series with this circuit would be a wise, and relatively inexpensive, addition – all the more so for portable projects using lithium batteries, given how these like to catch fire (and for nonrechargeable lithium primaries, explode) when shorted.
The only limiting factors on load current and supply voltage relate to the components – using a suitable resistor for R2, this circuit can easily regulate up to a few amps of current in as little as one square inch of board space. R1 is not too critical: 1Megaohm will work and I also have seen 10k for R1 and 33Ohm for R2 driving 3 LED’s at a 12 V supply. A good way to calculate the value of R1 is: R1= (Vcc – 1.4)/(I/20)

If you need precision, the easiest way to get it is to build a test circuit with a fixed supply (12VDC is a great value, but use the desired or intended supply voltage if there’s a need to be precise), a dead short for a load, and a 100 ohm resistor for R2. This will set the current limit to about 6mA. Once the circuit is powered up, measure the drop across R2 to determine the exact transition voltage for T2, and calculate a specific value for R2 for your current requirements based on the math described earlier.

The extra parts for PWM support include T3, R3. Since T3 is a PNP transistor and R4 biases its base to ground, T3 starts up turned on and pulls T1’s gate to ground regardless of what T2 is doing, and this forces both T1 and T2 to turn off. R3 limits current draw to the PWM signal source. Provide a positive voltage greater than ~0.7VDC to T3 through R3, and T3 turns off, which allows the rest of the regulator to function as described above. In this manner a PWM signal can be used to vary the brightness of a load of power LEDs, or a simple on-off switch effect can be implemented by merely pulling T3’s base to positive supply with a few milliamperes of current. In this manner it would be possible to use the circuit as a self-current-limiting switch that only needs a tiny little low-current button as its actuator even though the load could be an amp or more.

For LEDs, this circuit is  very simple and wastes very little power if the source and load voltages are kept close to each other. Ideally this circuit works most efficiently with the supply being right at 1.5VDC above the ideal load voltage at the set current. The power source should be able to supply the required amount of current.

If the circuit is used for other loads than LED’s there will be some ripple depending on the type of load being powered, especially during power-up, but decoupling capacitors could be used to compensate for this. The circuit can be driven directly from any logic circuit (even at 3.3VDC).

The circuit can drive resistive, inductive, capacitive, or mixed loads. Capacitive loads may well experience slower charge times, as the circuit will clamp the charge current. Inductive loads should be paralleled with a reverse-polarity diode as is common with e.g. relais (and in some cases, also a proper snubber capacitor) to suppress any back-EMF from the load when power is disconnected so that the voltage spike won’t damage the MOSFET. (Most power MOSFETs include integrated protection diodes, but this should never be relied upon as the sole protection mechanism.)

A simple LED driver with a 2N3055

Even simpler
If you do not have a MOSFET, but you happen to maybe still have an old trusty 2N3055 (or similar like the TIP series) workhorse laying around, the following circuit is for you:

The current is determined by R2 and is:
I =0.8/R2

The maximum value of I in this driver is 10A. .
DC-power supply 9V-15V DC

R2 for 1W LED -2.7ohms 1W
R2 for 3W LED -1.5-ohms 1W
R2 for 5W LED -0.6-ohms or 2 x 1.2-ohms/1W in parallel

Using an LED
stroombron2 Instead of 2 diodes it is also possible to use an LED to set the reference voltage. Ofcourse for this one needs to know the exact voltage of the LED.
Calculations of the resistors is as follows:
R2=(Vs-Vd)/(Id*KB) in which Vd= the LED voltage, Id the current through the circuit LED and where K = 1.2 to 2 (so that R1 is low enough to ensure adequate IB)