2. Working principle
Here we will introduce the working principle of this inverter in detail.
2. 1. Square wave signal generator (see Figure 2)
Fig. 2 Square wave signal generator
Here, six inverters CD4069 are used to form a square wave signal generator. In the circuit, R 1 is a compensation resistor, which is used to improve the oscillation frequency instability caused by the change of power supply voltage. The oscillation of the circuit is completed by charging and discharging the capacitor C 1. The oscillation frequency is f= 1/2.2RC. The maximum frequency of the circuit is fmax =1/2.2× 3.3×103× 2.2×10-6 = 62.6 Hz. The minimum frequency fmin =1/2.2× 4.3×103× 2.2×10-6 = 48.0 Hz. Due to the error of components, the actual value will be slightly different. For other redundant inverters, the input terminal is grounded to avoid affecting other circuits.
2.2 FET driving circuit
Fig. 3 FET driving 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, TR 1 and TR2 are used here to amplify the oscillation signal voltage to 0~ 12V, as shown in Figure 3.
4. Inverter performance test
The test circuit is shown in Figure 4. The input power supply used here is 12V automobile battery, which has low internal resistance and large discharge current (generally greater than 100A) and 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 bulb changes with the change of voltage. We can also find out the relationship between output voltage and power by calculation. But in fact, because the resistance of the bulb will change with the voltage applied at both ends, the output voltage and current are not sine waves, so this calculation can only be regarded as an estimate.
Figure 4 Test circuit
Take a bulb with a load of 60W as an example: assume that the resistance of the bulb does not change with the change of voltage. Because R lamp = V2/W = 2102/60 = 735Ω, when the voltage is 208V, W = V2/R = 2082/735 = 58.9W. In this way, the relationship between voltage and power can be converted. Through testing, we found that when the output power is about 100W, the input current is 10A and the output voltage is 200 V. 。