Transistor Sensor Amplifier

Introduction: Transistor Sensor Amplifier

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In this Instructable you will be building a BJT transistor sensor amplifier.

You will need a few NPN/PNP BJT (Bipolar Junction Transistor) transistors in case you burn them.


- Veroboard, NPN and PNP BJT transistors - 7 of each, 500 kohm - 2 Megohm potentiometer/variable resistor(I used 25 kohm potentionmeter/variable resistor), 1 Megohm resistor, 100 kohm resistors - 2, 10 kohm, 1 kohm resistor - 2, 100 ohm resistor (high power) - 3, 47 uF capacitor (bipolar), LED/Bright LED - 3.

Optional: NPN Power Transistor, Soldering Iron, Solder, Infra-red transmitter diodes, Wire stripper, Speaker, Remote Control, signal generator or 555 timer oscillator, a heat sink for the power transistor, heat transfer paste.

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Step 1: Build the Circuit

I used the solder and soldering iron but it is not necessary. The entire circuit can be assembled without soldering.

I used a 25 kohm potentiometer because I did not have a 500 kohm potentiometer.

The resistor values in the PSpice circuit simulation are different from what you see in the photo. I decided to build the circuit with the components that I already had. LED current = (Vs - Vled) / Rled = (3 V - 2 V) / 100 ohms = 10 mA. For two LEDs you need to reduce the Rled value in half to double the LED current. You can use 47 ohms. Do not raise to power supply voltage above 3 V or you will burn the LEDs. RledPower= Iled * Rled = 10 mA * 100 ohms = 1 Watt.

If you are using a speaker or a relay then you really need a power transistor which I did not have for this circuit. The three transistors are used for redundancy and reducing the power burden for each transistor. The Drelay diode is used to stop the discharging relay current from burning the three transistors. You need the Rc4 1 kohm resistor because when the LEDs are off the BJT transistor collectors is not connected to the power supply. However, you can still hear a quiet sound in your speaker without this resistor. I tried and it worked. However, avoiding the 1 kohm resistor will reduce the circuit cost but might not be necessary because you do not need a high power 1 kohm resistor.

The resistance of the speaker is usually 8 ohms. However, if this circuit is connected to an audio amplifier the input impedance could be very high. This is why I included the 10 kohm resistor.

I used 47 uF capacitor. If the output is connected to operational amplifier than the lower cut-off frequency is equal to: 1/(2*pi*(Rc4+Rl)*Cl) = 1/(2*pi*(1000 ohms + 10000 ohms)*(47*10^-6 F)) = 0.3078 Hz. If 8 ohm speaker is connected in paraller with 10 kohm resistor than the lower cut-off frequency is equal to: 1/(2*pi*(Rc4+Rspeaker)*Cl) = 1/(2*pi*(1000 ohms + 1008 ohms)*(47*10^-6 F)) = 3.3594 Hz. Thus the Cl capacitor value can be reduced to 22 uF, almost doubling cut-off frequency to 7.1769 Hz. The minimum cut-off frequency must be 10 Hz. Small capacitors usually have a lower maximum:

- power, voltage, and current ratings.

Bipolar capacitors can conduct in both directions without failing. Thus you need to use a bipolar 47 uF/22 uFcapacitor placed in series with speaker, althought it is not likely that the speaker voltage will exceed the collector voltage of Q4C transistor, thus reversing the polarity of the capacitor.

The connection of Q4C and (Q4A and Q4B) BJT NPN transistors will affect the linearity and frequency response of this circuit (from sensors to speaker). Thus you might want to disconnect the collector of Q4C from Q4A and Q4B pair. However, this will put eliminate the redundancy and power protection feature of this circuits. A good option is to use more transistors in parallel (additional transistor for Q4C and anotehr for Q4A and Q4B pair).

Step 2: Test the Circuit

On the photo above you can see the difference between the bright LED and photodiode.

I used an oscillator connected to an Infra-red transmitter diode connected in series with 1 kohm resistor to signal generator. I set the frequency to 100 Hz.

You need to vary the potentiometer value until the LEDs are just on.

In the second photo, you see the two photodiodes that I used. If the diodes are not receiving the signal then you need to switch the polarity. The photodiodes need to be biased in reverse because forward biasing will cause a 0.7 voltage drop. The cathode (negative terminal) is shorter than the anode (positive terminal) and was thus connected to the power supply as you see in the photo. The anode was connected to 100 kohm resistor, Rbias1 in the circuit. (

The Infra-red signal is reflecting from my hand and turning off the LEDs. The circuit will not work well outside in the sunlight.

If the LED is oscillating than you can use a low pass RC (resistance capacitance) filter ( Connect the power supply to filter input and the circuit to filter output. The risks are:

- blowing up the capacitor,

- damaing the power supply,

- or blowing up the battery.

If the resistance value is 10 ohms and the capacitor value of the RC filter is 1000 uF than the cut-off frequency will be 1/(2*pi*10*(1000*10^-6 F)) = 15.9154 Hz. Thus you are filtering the 50 Hz power supply ripple very well with those values. 1000 uF capacitor is charging from zero volts. Thus the maximum current across RC filter resistor will be 3 V / 10 = 0.3 A and maximum resistor power rating will be = 3 V * (3 V / 10 ohms) = 0.9 W. Do not use higher than 10 ohm resistor values for the RC filter. The circuit is drawing at least 10 mA from power supply (for each LED). Thus if you use 100 ohm resistor than the voltage drop across this resistor will be 1 V.

You can use this circuit for other current sensors. It might not work for ultrasound because of low bandwidth and might not work well for microphones because of possible poor linearity. You can use high-frequency transistors for the ultrasound receivers. However, there could be issues with the stray capacitance of high-value resistors that I used.

Step 3:

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