Inverter is also crazy! Field effect power supply production process and circuit analysis

The inverter introduced in this article (see Figure 1) is mainly composed of MOS field effect tubes and ordinary power transformers. Its output power depends on the power of the MOS field effect tubes and the power transformer, eliminating the tedious transformer winding and suitable for amateur production by electronics enthusiasts. The following introduces the working principle and production process of the inverter .

1. Circuit Diagram

Figure 1 Inverter system circuit diagram

2. Working Principle

Here we will introduce the working principle of this inverter in detail.

2.1. Square wave signal generator

Figure 2 Square wave signal generator

Here, six inverters CD4069 are used to form a square wave signal generator. R1 in the circuit is a compensation resistor, which is used to improve the unstable oscillation frequency caused by changes in the power supply voltage. The oscillation of the circuit is completed by charging and discharging the capacitor C1. Its oscillation frequency is f=1/2.2RC. The maximum frequency of the circuit shown in the figure is: fmax=1/2.2×3.3×103×2.2×10-6=62.6Hz; the minimum frequency fmin=1/2.2×4.3×103×2.2×10-6=48.0Hz. Due to component errors, the actual values ​​will be slightly different. For other redundant inverters, the input terminals are grounded to avoid affecting other circuits . 2.2 Field effect tube drive circuit

Figure 3 Field effect transistor drive circuit

Since the maximum amplitude of the oscillation signal voltage output by the square wave signal generator is 0~5V, in order to fully drive the power switch circuit, TR1 and TR2 are used here to amplify the oscillation signal voltage to 0~12V, as shown in Figure 3.

2.3 MOS field effect tube power switch circuit

The following is a brief description of the working process of the application circuit composed of C-MOS field effect tube (enhancement MOS field effect tube) (see Figure 4). The circuit combines an enhancement P-channel MOS field effect tube and an enhancement N-channel MOS field effect tube. When the input end is low, the P-channel MOS field effect tube is turned on, and the output end is connected to the positive pole of the power supply. When the input end is high, the N-channel MOS field effect tube is turned on, and the output end is connected to the power ground. In this circuit, the P-channel MOS field effect tube and the N-channel MOS field effect tube always work in opposite states, and their phase input and output ends are opposite. Through this working mode, we can obtain a larger current output. At the same time, due to the influence of leakage current, the gate voltage has not reached 0V, usually when the gate voltage is less than 1 to 2V, the MOS field effect tube is turned off. Different field effect tubes have slightly different turn-off voltages. Because of this, the circuit will not cause a power short circuit due to the simultaneous conduction of the two tubes.

Figure 4 MOS field effect tube power switch circuit

Based on the above analysis, we can draw the working process of the MOS field effect tube circuit part in the schematic diagram (see Figure 5). The working principle is the same as described above. When this low voltage, high current, 50Hz frequency alternating signal passes through the low voltage winding of the transformer, it will induce a high voltage AC voltage on the high voltage side of the transformer, completing the conversion from DC to AC. It should be noted here that in some cases, such as when the oscillation part stops working, sometimes a large current will pass through the low voltage side of the transformer, so the fuse of this circuit cannot be omitted or short-circuited.

Working process of MOS field effect tube circuit part 3. Production points

The circuit board is shown in Figure 6. The components used can be referred to Figure 7. The transformer used in the inverter uses a finished power transformer with a secondary voltage of 12V, a current of 10A, and a primary voltage of 220V. The maximum drain current of the P-channel MOS field effect tube (2SJ471) is 30A. When the field effect tube is turned on, the drain-source resistance is 25 milliohms. At this time, if a current of 10A passes through, there will be a power consumption of 2.5W. The maximum drain current of the N-channel MOS field effect tube (2SK2956) is 50A. When the field effect tube is turned on, the drain-source resistance is 7 milliohms. At this time, if a current of 10A passes through, the power consumed is 0.7W. From this, we can also know that under the same working current, the heat generated by 2SJ471 is about 4 times that of 2SK2956. So this point should be taken into account when considering the heat sink. Figure 8 shows the position distribution and connection method of the inverter field effect tube introduced in this article on the heat sink (100mm×100mm×17mm). Although the heat generated by the field effect tube when working in the switching state is not very large, the heat sink selected here is slightly larger for safety reasons.

Inverter location distribution and connection method

4. Inverter performance test

The test circuit is shown in Figure 9. The input power used for the test here is a 12V car battery with low internal resistance and large discharge current (generally greater than 100A), which can provide sufficient input power for the circuit. The test load is an ordinary light bulb. The test method is to change the load size and measure the input current, voltage and output voltage at this time. The output voltage decreases with the increase of load, and the power consumption of the light bulb changes with the voltage. We can also find out the relationship between output voltage and power through calculation. But in fact, since the resistance of the light bulb will change with the voltage applied at both ends, and the output voltage and current are not sine waves, this kind of calculation can only be regarded as an estimate.

Take a 60W light bulb as an example: Assume that the resistance of the light bulb does not change with the voltage. Because R = V2/W = 2102/60 = 735Ω, when the voltage is 208V, W = V2/R = 2082/735 = 58.9W. From this, the relationship between voltage and power can be calculated. Through testing, we found that when the output power is about 100W, the input current is 10A, and the output voltage is 200V.

Inverter test circuit