Amplifier Circuit Diagrams

100W Subwoofer Amplifier Circuit

A subwoofer is a loudspeaker that delivers low-frequency audio signals. Ken Kreisler invented the first subwoofer amplifier in 1970. Its primary purpose is to enhance the bass quality of audio signals. We create a subwoofer amplifier that produces audio signals at low frequencies between 20 and 200 Hz and has a 100W output power to drive a 4 ohm load.

Outline

  • Subwoofer Amplifier Circuit Principle
  • Circuit Diagram of 100W Subwoofer Amplifier
  • Circuit Components:
  • Subwoofer Amplifier Circuit Design:
    • Audio Filter Design:
    • Pre Amplifier Design:
  • Power Amplifier Design:
  • Subwoofer Amplifier Circuit Operation:
  • Applications of Subwoofer Amplifier Circuit:
    • Limitations of the Circuit:

Subwoofer Amplifier Circuit Principle

Audio Signal is first filtered to remove the high frequency signals and allow only the low frequency signals to pass through it. This low frequency signal is then amplified using a voltage amplifier.  This low power signal is then amplified using a transistor driven class AB power amplifier.

Circuit Diagram of 100W Subwoofer Amplifier

Subwoofer Amplifier

Circuit Components:

COMPONENTVALUE
R16K
R26K
R3130K
R422K
R515K
R63.2K
R7300 Ohms
R830 Ohms
R9, R103 K
C1, C20.1uF, electrolyte
C3,C5,C610uF, electrolyte
C41uF, electrolyte
Q12N222A
Q2TIP41
Q3TIP41
Q4TIP147, PNP
D1, D21N4007
Dual Supply+/-30V

Subwoofer Amplifier Circuit Design:

Audio Filter Design:

Using the OPAMP LM7332, we created a Sallen Key low pass filter. The quality factor was supposed to be 0.707, and the cutoff frequency was assumed to be 200Hz. Similarly, if the number of poles is 1, and the value of C1 is 0.1uF, the value of C2 can be determined to be 0.1uF. The value can be obtained by replacing known values in the equation, assuming R1 and R2 are the same.

Q/(2pifc*C2) R1 = R2

This gives each resistor a value of 5.6K. R1 and R2 are 6K resistors in this case. We don’t need resistors at the non inverting terminal, which is shorted to the output terminal, because we want a closed loop gain filter.

Pre Amplifier Design:

The preamplifier is based on class A operation of transistor 2N222A.  Since the required output power is 100W and load resistor is 4 Ohms, here we require a supply voltage of 30V.

 Assuming the collector quiescent current to be 1mA and collector quiescent voltage to be half of supply voltage, i.e.15V, the value of load resistor is calculated to be equal to 15K.

R5 = (Vcc/2Icq)

Base current is given by, Ib = Icq/hfe

Substituting the values, hfe or AC current gain , we get the base current to be equal to 0.02mA. The bias current, Ibias is assumed to be ten times the base current, i.e. 0.2mA.

The emitter voltage is assumed to be 12% of the supply voltage, i.e. 3.6V. The base voltage, Vb is then equal to Ve +0.7, i.e. 4.3V.

Values of R3 and R4 are then calculated as given:

R3 = (Vcc – Vb)/ Ibias  and R4 = Vb/Ibias

Substituting the values, we get R3 to be equal to 130 K and R4 to be equal to 22K

The emitter resistor (Ve/Ie) is calculated to be 3.6K. However, this resistance is shared by two resistors, R6 and R7, with R7 serving as a feedback resistor to mitigate C4’s decoupling impact. The value of R7 is discovered to be 300Ohms when the values of R5 and gain are added together. The value of R6 is then 3.2K.

We determine the value of C4 to be 1uF since the capacitive reactance of C4 should be less than the emitter resistance.

Power Amplifier Design:

The power amplifier is designed using Darlington transistors TIP142 and TIP147 in class AB mode. The biasing diodes are selected such that their thermal properties are equal to that of the Darlington transistors. Here select 1N4007.

Since a large value of bias resistor is required for a low bias current, we select R9 to be equal to 3K.

The driver stage is used to provide the power amplifier with a high-impedance signal. In this case, we’re using a TIP41 power transistor in class A mode. The emitter resistor, R8, is equal to 28.6 Ohms and is determined by the values of emitter voltage, Ve (1/2Vcc-0.7) and emitter current, Ie (equal to collector current, i.e. 0.5A). We’ll use a 30 Ohm resistor in this case.

The value of the bootstrap resistor R10 should be such that the Darlington transistors have a high impedance. R10 is set to 3K in this case.

Subwoofer Amplifier Circuit Operation:

The Sallen Key low pass filter is used to filter the audio signal, allowing only frequencies below and equal to 200Hz to pass while staying filtered. Through the coupling capacitor C3, this low frequency signal is fed into the transistor Q1’s input. In class A mode, the transistor creates an amplified version of the input signal at its output. Q2 converts the amplified signal into a high impedance signal, which is then sent to the class AB power amplifier. The two Darlington transistors work together to produce a full cycle of output signal, with one conducting for the positive half cycle and the other for the negative half cycle.R11 and R13 are utilised as emitter resistors to reduce any differences between the matching transistors. The diodes are employed to guarantee that cross over distortion is kept to a minimum. This high-power output signal is then utilised to drive a low-impedance loudspeaker or subwoofer (approximately 4 Ohms). Note that for testing purposes, we’ve utilised an 8 Ohm resistor.

Applications of Subwoofer Amplifier Circuit:

  1. This circuit can be used at home theatre systems to drive subwoofers to produce a high quality, high bass music.
  2. This circuit can also be used as a power amplifier for low frequency signals.

Limitations of the Circuit:

  1. The filter circuit tends to increase the DC level of the audio signal, causing a disruption in the biasing.
  2. The use of linear devices causes power dissipation, thus reducing the efficiency of the circuit.
  3. It is a theoretical circuit and output contains distortion.
  4. The circuit doesn’t provide any provision to remove noise signal and thus the output may contain noisy disturbance.

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