Transistor Power Amplifier

This article is about a simple transistor power amplifier.

Figure 1: Device.

You can see how my circuit works in this video:

Step 1: Design the Circuit

I have drawn the circuit via PSpice software:

Figure 2: Design the Circuit.

The optional capacitor low pass filter circuit is useful when the amplifier is connected to mains powered 8 V power supply. The two 10 mF capacitors are needed to filter the possible 60 Hz or 50 Hz ripple. You do not need this circuit if the amplifier is battery-powered.

I have chosen two 10,000 uF capacitors for low pass frequency of:

Flp1 = 1 / (2 * pi * (2 * Cs1a) * (Rs1a / 2))

= 1 / (2 * pi * (2 * 10 mF) * (4.7 ohms / 2))

= 3.3863 Hz

The maximum current across the Cs1a and Cs1b capacitors is when the voltage across the capacitors is zero:

IcsMax = Vs / (Rs1a / 2) / 2

(divide by 2 because there are two capacitors)

The maximum voltage is 12 V.

= 12 V / (4.7 ohms / 2) / 2

= 2.5532 Amps

The maximum power is at half supply voltage:

PcsMax = VcsHalfSupply * IcsHalfSupply / 2

(divide by 2 because there are two capacitors)

= (Vs / 2) * ((Vs / 2) / (Rs1a / 2)) / 2

= 7.6596 Watts

Calculate the low pass frequency of the second power supply low pass filter:

Flp2 = 1 / (2 * pi * (3 * Cs2a) * (Rs2a / 2))

= 1 / (2 * pi * (3 * 470 uF) * (100 ohms / 2))

= 2.2575 Hz

The average current across the LEDs equals to:

Iled = (Vs - 2 * Vled) / (Rd1 + Rd2)

 = (12 V - 2 * 2 V) / (470 ohms + 470 ohms)

= 8 V / 940

= 8.5106 mA

The maximum current across the LED is equal to:

Iled = (Vs - Vled - Vbe) / Rd1

= (12 V - 2 V - 0.7 V) / 470 ohms

= 19.7872 mA

The 1 Meg potentiometer allows a clipping sound effect.

Step 2: Simulations

Time Domain:

Figure 3: Simulations Transient.

Frequency:

Figure 4: Simulations Frequency.

Step 3: Make the Circuit

I made the circuit on a small matrix board:

Figure 5: Make the Circuit.

Step 4: Encasement

I used a cheap $2 gift box to save money:

Figure 6: Encasement.

I used big 10 mF capacitors and 10-watt 4.7 ohms resistors because I connected the circuit to a 12 V battery.

Step 5: Testing

I used a Hantek 6022BE USB Oscilloscope to test my circuit.

The yellow plot is the input and the green plot is the output.

Figure 7: Testing 100 Hz Sine Wave Input.

Figure 8: Testing 1 kHz Sine Wave Input.

Figure 9: Testing 10 kHz Sine Wave Input.

Figure 10: Testing 20 kHz Sine Wave Input.

Figure 11: Testing Voice Input.

Talking:

Conclusion

The circuit can be produced without a 1 Megohm potentiometer and without the LEDs if you bias the output transistor emitters at half supply voltage.

LM311 Touch Buzzer

This article is about LM311 comparator IC (integrated circuit) touch buzzer circuit:

You can see the circuit working in this video:

Step 1: Design the Circuit

I have drawn the circuit in PSpice version 9.1 student edition:

The output of LM311 comparator is an NPN transistor collector. The current in pin 7 starts flowing when the positive input (pin 2) is below the negative input (pin 3). The output is inverted because the output of the comparator is an inverter logic gate.

Comparator is designed to drive coils. A buzzer coil turns ON and OFF similarly to a relay turning ON and OFF. The current in the coil continues to flow in the same direction when the current to the coil is discontinued. This is why I connected the diode.

The buzzer current is 40 mA. The 9 V battery can supply a current of 100 mA. The remaining 60 mA current is for the comparator IC (integrated circuit) and the resistor circuits.

The circuit will work without the 10 Megohm resistor. However, it is bad practice to leave pin 2 disconnected.

Step 2: Simulations

Simulations show that when the human resistance falls before about 8 Megohms high current begins to flow.

Step 3: Make the Circuit

I used a wire wrap socket and wire wrap tool to connect the circuit. I tried Rtouch resistor of 1 Megohms, 4.7 Megohms. However, the circuit was not turning ON well with those resistors (1 Megohm and 4.7 Megohm) when compared to Rtouch of 10 Megohms.

Conclusion

This circuit is useful for toy applications.

Emergency Lights

This article shows the emergency lights system.

The current source in my circuit represents a photodiode. When lights are turned OFF the current source produces almost zero current and the circuit thus activates emergency lights by providing current to the light bulb. The light bulb remains ON for a few seconds in case the photodiode is damaged due to an emergency.

This circuit is suitable for a 1.5 V light bulb. If you are using bright LEDs then you need to increase the power supply to at least 3 volts. However, if you raise the voltage to 6 V or above then you should use MOSFET transistor instead of BJT transistors. MOSFETs also have a maximum gate-source voltage that should not be exceeded.

My circuit allows the driving of higher current LED lights or light bulbs.

Step 1: Design the Circuit

The circuit that I designed is able to drive high-current light bulbs. You can use fewer transistors if you are connecting a bright LED.

Q3 and Q4 should be power transistors. I used PSpice student edition simulation software that does not include power transistors.

Each photo diode has a dark current. Dark current is the current across diode in darkness (when no light enters the diode). We can now calculate the maximum dark current of the photodiode:

Idark = Ipd

= (Vs - Rb1 * Ib1 - Vbe1) / Rpd - Ib1

= (Vs - Rb1 * Ib1 - Vbe1) / Rpd - Ic1 / Beta

= (Vs  - Rb1 * Ib1 - Vbe1) / Rpd - (Vs - Vbe2 - VceSat1) / (Rc1 + Rb1) / Beta

Ib1 = (Vs - Vbe1 - VceSat1) / (Rc1 + Rb1) / Beta

= (1.5 V - 0.7 V - 0.2 V) / (1,000,000 ohms + 220,000 ohms) / 20 

= 133.5113nA

Idark = 637.1162nA

If the photodiode dark current is above this value, the circuit might not work.

This is the Matlab code:

clear all
Vs=1.5;
Beta=20
Vbe1=0.7;
VceSat1=0.2;
Vbe2=0.7;
Rc1=4700;
Rb1=220000;
Rpd=1000000;
Ib1=(Vs-Vbe1-VceSat1)/(Rc1+Rb1)/Beta;
disp(['Ib1 = ' num2str(Ib1*1000000000) ' nA'])
Idark=(Vs-Rb1*Ib1-Vbe1)/Rpd-(Vs-Vbe2-VceSat1)/(Rc1+Rb1)/Beta;
disp(['Idark = ' num2str(Idark*1000000000) ' nA'])

I used Octave software.

You can see in the circuit above that I assumed the photodiode dark current to be 100 nA.

Step 2: Simulations

When the Ct capacitor is charged the Vt voltage is almost zero (with reference to ground) and the lights are ON. When the Ct capacitor is discharged the Vt voltage is almost equal to the power supply voltage and the lights are OFF.