Design Method of Small Power Amplifier

The table below shows the design specifications of the power amplifier for a Walkman. The maximum output of a Walkman is 1V. About 1. If the voltage amplification of the circuit is 10 times, it can make a small speaker sound at a certain volume. In this case, an output power of 0.5W is sufficient.

This is a circuit diagram of the designed power amplifier. This circuit is monophonic. To play stereo sound, a circuit for another channel is required.

As a whole circuit structure, the input signal is amplified by the common emitter amplifier circuit. The bias circuit inserted in the collector of the common emitter circuit generates the bias voltage of the emitter follower, and the current is amplified by the push-pull emitter follower.

Determine the supply voltage

The supply voltage is determined by the output power.

The maximum output power PO is 0.5W for an 8Ω load (the impedance of the speaker). So the output voltage Vo at this time is:

Vo=√Po=_Z√0.5Wx8Ω

=2Vms(4.4)

Z = load impedance

This value is an effective value. If the input signal is a sine wave, the peak-to-peak value of the output waveform is 5.7V (≈2Vms×√2×2).

For an output voltage of 5.7V, the power supply voltage Vcc is set to a value above the several volts of loss generated by the circuit, including the voltage drop across the emitter resistor of the common emitter circuit, the voltage drop across the emitter resistor of the emitter follower, and the saturation voltage between the collector and emitter of the transistor, etc. Here, Vcc = 15V (single power supply).

The operating point of the common emitter amplifier circuit

The collector current of the common emitter amplifier circuit is set to a large value, which is much larger than the base current of the emitter follower supplied to the lower stage.

When the load is 8fl and the output power is 0.5W, the output voltage VO is 2Vms (assuming the waveform is a sine wave).

Its peak value is 2.8V (≈2V×√2). At this time, the load current (=collector current of Tr3 or Tr4) is 350mA (-2.8V/8Ω) (also the peak value).

Here, the hFE of the transistor used in the emitter follower is assumed to be 100, and the base current provided by the common emitter circuit is 3.5 mA (-350 mA, All00). The current flow is shown in the simulation.

Assume that the collector current of the common emitter circuit is much larger than the base current 3.5mA, that is, 20mA.

For Tr1, select a device with a collector current of 20mA or more, a collector-base voltage and a collector-emitter voltage of 15V (power supply voltage) or more. (Select a transistor model of 2SC2458)

If the emitter potential of Tr1 is too high, a large collector amplitude cannot be obtained; if it is too low, the collector current increases with temperature. Considering all factors, it is taken as 2V here.

In order to set the collector current (emitter current) to 20mA, the resistor Rs+R6 between the emitter of Tr1 and GND is set to 100Ω (-2V/20mA).

The part that determines the magnification

If the collector potential of Tr1 is set to 8.5V, the maximum amplitude can be obtained (the emitter follower bias potential generated by Tr2 is ignored here).

In order to make the collector potential 8.5V, the voltage drop on R3 is 6.5V (=15V-8.5V), so

R3=6.5V/20mA≈330Ω(4.5)

Also, divide R5+R6-100fl into two parts, let R5=22Ω, R6=75Ω, and connect R6 to ground with C3, the AC voltage amplification factor of the circuit is

Av=R3/R5=330Ω/22Ω(≈24dB)(4.6)

Since the actual gain is smaller than the value obtained by formula (4.6) and the loss on the emitter resistor of the emitter follower stage (described later), the setting value of A should be set to a value slightly larger than the design specification (the design specification is 10 times).

In addition, C3 is a capacitor that bypasses R6 to increase the amplification. R5, R6 and C3 form a high-pass filter. In order to meet the frequency characteristics of the design specifications, C3 is set to C3 = 330μF.

R1 and R2 play a role in determining the base potential. In order to make the emitter potential 2V, the base potential is set to 2.6V (=2V+VBE). Here, assume that the current flowing through R1 and R2 is 0.5mA, R1=24kΩ, R2=5.6kΩ. Therefore, the input impedance of the circuit is 4.5kΩ (=R1∥R2).

The cut-off frequency of the high-pass filter formed by the coupling capacitor Cl on the input side and the input impedance of the common emitter circuit is below 20Hz (according to the design specifications), which determines the value of C. Here, C=10μF (cut-off frequency is 3.5Hz).

VR is a variable resistor for adjusting the input level (volume) and is set to 10 kΩ.

Emitter follower bias circuit

In order to omit the coupling capacitor, an emitter follower circuit is inserted between the collector of the transistor Tri of the common emitter circuit and the load resistor R3.

It shows the relationship between voltage and current in various parts of the bias circuit.

The transistor Tr2 selected here can be of any type as long as it meets the conditions that the maximum collector current is above 20mA and the maximum rated values ​​VCBO and VCEO between collector-base and collector-emitter are above 1.2V (two VBE).

However, considering the thermal coupling problem between Tr3 and Tr4, the low-frequency medium-power amplifier transistor 2SC3423 is generally considered. It is mounted in the full mold package of T0126 (insulated mold package with no exposed metal part).

The current flowing on the base side of this circuit (VR2 and R4) is determined by R4, here R4=300Ω. The current flowing through VR2 and R4 is 2mA (=0.6V/300Ω). On the other hand, the collector current of Tr1 is 20mA, and the collector current of Tr2 is 18mA (-20mA-2mA).

Even in such a circuit (same as the amplifier circuit), the current flowing on the base side is set to 1/10 of the collector current (in order to ignore the base current).

In order to make the collector-base voltage of Trz 2VBE (VBE of Tr3 and Tr.), the value of VR2 can be made the same as R4. Therefore, VR2=470Ω (500Ω is also acceptable) is used, so that the resistance of the semi-fixed resistor is 300Ω when the sliding head position is near the center.

C2 bypasses the bias circuit to make the impedance "seen" by the base of Tr3 and Tr4 equal. Due to the insertion of C2, the high-frequency distortion rate is improved.

The larger the value of C2 is, the lower the base-to-base impedance of Tr3 and Tr4 will be. However, if the value of C2 is too large, it will be meaningless. Here, Cz=3.3μF is taken.

6 Emitter Follower Power Loss

This circuit sets the power supply voltage to 15V and the collector potential of Tr1 to 8.5V. Therefore, if the bias voltage caused by Tr2 is ignored, the emitter follower circuit is the same as the common emitter circuit, and can output a signal with a peak voltage of 6.5V.

When this output signal drives an 8Ω load, a peak current of about 800mA (-6.5V/8Ω) flows as the collector current through Tr3 and Tr4.

On the other hand, when the output voltage reaches the positive and negative peaks, the power supply voltage (15V) is directly applied between the collector and emitter of Tr3 and Tr4.

Generally, when the output waveform is a sinusoidal wave (as shown in this circuit), in a push-pull emitter follower operating in class B, the maximum collector loss Pc of each transistor is 1/5 of the maximum output power (for details, see reference [1]).

Assuming the output waveform is a sine wave, the maximum output voltage of the circuit is an effective value of 4.6V. (-6.5V/√2), so the maximum output power is 2.65W (≈4.6V2/8Ω), and the maximum value of Pc of Tr3 and Tr4 is 1/5 of it, that is, 0.53W.

Therefore, Tr3 and Tr4 select transistors with collector current above 800mA, collector-base voltage and collector-emitter voltage above 15V, and Pc above 0.53W.

Here, Tr3 and Tr4 use the complementary pair 2SD1406 and 2SB1015 for low-frequency power amplification.

The total collector loss of Tr3 and Tr4 is 1.06W, so a heat sink is necessary. In this circuit, a heat sink that can sufficiently dissipate 1W of heat is used (MC24-L20, ryo-san, any heat sink of the same level is acceptable).

In order to thermally couple Tr2, Tr3 and Tr4, the three transistors are mounted on the same heat sink.

It should be noted that the heat generated by 1.06W only occurs when the output is the maximum output of 115. If the sound is not too loud often, and the tube shell of Tr3 and Tr4 is large enough (T0220 full plastic mold), there is no need to install a heat sink.

At this time, Tr2 to Tr4 are thermally coupled, and Tr2 is clamped by Tr3 and Tr4 and fixed with screws.

Components around the output circuit

The emitter resistors R1 and R8 of Tr3 and Tr4 play the role of limiting the output current and absorbing the temperature change of the VBE value of Tr3 and Tr4. However, when the emitter resistance value is small, as in this circuit, we cannot have too high expectations for the absorption of temperature changes. If the value of R7 and R8 is large, because the load current flows through R7 and R8, a large power loss will be generated in the resistor.

For example, when power is supplied to an 8Ω load, assuming R7=R7=16Ω, the power that can be supplied to the load is 1/2 of the original output power (because the output impedance of the circuit is R7∥R8=8Ω), so the voltage amplification is also estimated to be 1/2.

Therefore, the emitter resistor should be set to a value smaller than the connected load resistance, that is, less than 1/10. In this circuit, the load is 8Ω (speaker), so R7=R8=0.5Ω (power and voltage amplification have 3% loss).

Even if the maximum load current of 800mApek continues to flow through R7 and R8, the power consumption is only 0.16W (≈(800mA/√2)x0.5Ω). Therefore, it is sufficient to use resistors with a rated power of 1/4W for R7 and R8.

However, it is difficult to buy a 1/4W, 0.5Ω resistor. Therefore, in the circuit making, two 1/4W, 1Ω resistors are connected in parallel as a substitute (see Photo 4.1).

C4 is a DC blocking capacitor. C4 = 1000μF, and the high-pass filter cutoff frequency formed by the speaker impedance 8Ω is 19.9Hz (meeting the design specification of 20Hz).

The load resistance of 8Ω is very low. When you want to lower the cutoff frequency, you must increase the value of Ct anyway.

When there is no load, R9 is used to discharge C4 (so that even if a speaker is connected after the power is turned on, there will be no vibration noise). A value that is too large is meaningless, and a value that is too small will cause power loss. Here, R9=1kΩ is used.

C5 is a decoupling capacitor for the power supply. In a single-supply power amplifier like this circuit, the impedance of the collector side (i.e., power supply) of Tr3 and Tr4 seen from the GND (i.e., OV) at the output end is very low at the frequency of the output signal. When a large amount of output current flows, the output waveform will be distorted.

The collector of Tr4 is connected to GND, and the impedance to GND is 0. However, the collector of Tr3 is connected to the power supply, so it has a certain value. Therefore, the value of C5 is made very large to reduce the low-frequency impedance to GND. Here, C5=470μF.