Check the listing for details. High Power 100W*2 OCL Module Two Channel Amplifier Board Electronic DIY Kits. Color: Green, Condition: New. Listed at 13.88 USD. ☆ Welcome☆ to ☆ my ☆ shop ☆ OCL power amplifier board 100W high power two-channel pure post-level DIY audio power amplifier board DIY Kits (board color random: green board, red board) Features: Fully separated components, four high-power triodes, two stereos, each with a power of up to 100W. This power amplifier adopts the typical OCL circuit, which has the advantages of high stability, wide frequency response range and good fidelity. It is often used in high-fidelity sound reproduction equipment. This OCL stereo amplifier is suitable for use by a wide range of electronic enthusiasts and audiophiles. First, the working principle of the circuit As can be seen from the figure below, the output of the speaker and the amplifier is directly coupled, and the output capacitor for DC blocking is omitted. In order to make the DC potential of the output of the circuit zero volt, a positive and negative symmetrical power supply is used, and the differential amplifier input is used. And other measures. In Figure 1, vt i, vt2 are differential amplification input stages, vt3 is the excitation stage, and vt4~vt7 are composite complementary output stages. The audio signal is sent to the base of vt i through the coupling capacitors c 1 and ri. After amplification, it is output from the collector of vt i and sent to vt3 to further increase. The excitation signal of the collector output of vt3 pushes the power output stage vt4~ Vt7 works so that the amplified audio signal can drive the speaker to work. To facilitate further analysis, Figure 1 can be simplified to the form of Figure 2. Vt4 and vt6 are combined to be equivalent to one npn transistor, and vt5 and vt7 are equivalent to one pnp transistor. It can be seen from the vt4, 6 and vt5, 7 of the circuit of Fig. 3 and the power supply filter capacitors c9, c10 that they are equivalent to a bridge. When vt4, 6, vt5, and 7 are identical, c9 and c10 are also identical, the bridge arms are balanced, and the speaker does not pass through. If the positive and negative power supplies are completely symmetrical, the output potential can be guaranteed to be 0V. Since the circuits are all directly coupled, any changes in ambient temperature and component parameters will affect the potential at the output (point a, Figure 2). To this end, vt 1, vt2 constitute a differential amplifier to overcome the zero drift, DC negative feedback is also applied in the circuit, that is, the output is applied to the base of vt2 through r6, so that the potential of the output (point a) is zero. Volt. The feedback process is: a point potential ↑—ube2 ↑—ie2 ↑—ur4 ↑—ube l ↓一ic l↓一uc 1 ↑—ube3 ↓一ie2 ↓—ur7↓一ube4,6 ↓(ube5,7 ↑) A vt4, 6 internal resistance v (vt5, 7 internal resistance ↓) a point potential ↓. On the other hand, if the potential at point a is ↓, the potential at point a will be paralyzed by the opposite change process. Second, the installation and production The printed circuit diagram of this circuit is shown in Figure 3. It includes a two-way OCL power amplifier circuit and a DC power supply circuit. A heat sink is required between the body of the 4 high power tube 2N3055 and the printed circuit board, and is fixed by screws. The resistors are installed horizontally. The capacitors and triodes are installed vertically and close to the circuit board. The soldering requirements are firm and reliable. There are two jumpers on the circuit board, which can be soldered with iron wires. Third, debugging The power transformer used in this circuit needs to be prepared by itself. It adopts a center-tapped dual output transformer (AC: 1 5VX2), the power is not less than 40W, and is connected to the AC~ and the upper part of the printed circuit board. After being energized, it is generated at both ends of C9 and C10. With a DC voltage of ±18V, the voltage across the speaker is zero volts. First adjust the current of the differential amplifiers vt l, vt2, in order to avoid the large current flowing through the power tube, first short the base of vt4, vt5 with a wire, so that vt4 ~ vt7 cut off. Then solder one end of the resistor r6 to the ground and solder it to ground. The total emitter current of the differential amplifier stage is determined by r4. The R4 is adjusted so that the total emitter current of vt 1 and vt2 is 1 mA. After the resistor r6 is restored, the voltage across the speaker should be o. If there is an offset, adjust r3. 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