Circuit principle
As shown, the input stage of this amplifier is a pair of field effectors, the advantage is that the input impedance is high, the dynamic range is large and the noise is low. VT3, VT4 constitutes the second level amplification, VT5 provides 1.8mA current to VT1, VT2; VT6 provides 9.5mA constant current to VT3, VT4. Due to the conduction of VD1, the current voltage does not affect the reference voltage of VT5 and VT6 even if there is fluctuations, and it is guaranteed to stabilize the VT1 and VT2 two tube currents. The VT9 ~ VT14 constitutes a complementary output stage, and its output current is controlled by VT8. VT8 is a constant voltage circuit, the dynamic internal resistance is small, the voltage of VCE8 is constant, and the regulation of RP2 can adjust the static operating current size. VT7, C3, R13, VD2 form a dynamic bias circuit, which is that the VT7 causes the base current of the VT8 to decrease when the signal input is increased, and the VCE of the VT8 is elevated, and the current of the VT9 to VT14 is correspondingly elevated. The highest peak can reach 4 to 5A, so that the push-pull output is automatically operated in the matrix, not only reduces the distortion and harmonic distortion, but also improves the utilization of the power supply, but the dynamic bias control is determined by the RL3. C5 is a large loop negative feedback capacitor to accelerate high frequency signals, with phase advanced compensation. Reverture transient distortion and intermodulation distortion and can be protected. The input stage also adds a high frequency filtering network to filter out some unnecessary high frequency noise and improve the signal-to-noise ratio. VT9, VT10, replaced with TIP41 and TIP42. VT11, VT12, VT13, VT14 can be replaced with TIP35C and TIP36C. Working voltages do not exceed + 50V and -50V, you can get 80W output.
Component selection
The BVCEO requirements of the transistors used in the circuit are greater than or equal to twice the supply voltage. Field effect tubes VDS> / = 50V, GM must also be consistent. The complementary output stage should measure the direct flow amplification of small current to large currents, require linearity, no more than 1% per bit, input level? T can be selected, output level? T should be lower. The resistance is preferably a metal membrane resistance for performance. C3 uses the CA series, the rest are the CBB capacitance. At the time of charge, the transistor selection of the promoter must be paid attention to. In fact, the working status of the promotion grade determines whether the unlailized output work can be easily free. The power supply voltage used in today's high-power amplifier is relatively high, and the output tube drive current is also relatively large, and the power consumption of the promotion level is considerable. This is the reason why the advanced power amplifier is used in the promotion of the heat and high power tube well.
Installation and commissioning]
Since the levels of the circuit are directly coupled. One of one of the components may affect the working state, or even damage the transistor. Therefore, the quality quality should be checked carefully. When installing, the two potentiometers that adjust the zero potential and the static current can be close to the vicinity of VT1, VT2, and VT7. The outer casing must be well grounded, and the high power tube and the emitter resistor are all under a large-scale radiator to avoid Unnecessary parasitic coupling, VT8 is installed in the middle of four transistors, and the VT8 is mounted in the middle of the four transistors to stabilize the static operating current. The debugging of this amplifier is very simple. First, a machine tool working lamp for a 40 to 60W, 27OHM or voltage 36V, and a power 60W of a machine tool work bulb is subjected to a fake load. The mid-point voltage should be around 0V with a multimeter measuring, and if the request is not required, it may be a differential tube, or the component tube foot weld is wrong. For debugging of quiescent current: adjusting RP2, so that the voltage across the emitter resistor of the output tube is reduced to about 100 mV, then the quiescent current is 200mA.
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