How to determine the load output voltage of the fluorescent lamp electronic ballast?

Answer:

1 to C4 to Q1 to the R6, L1c to L2, and the filament 2 to the C5, the current is going to be the discharge circuit of C5. This cycle forms the oscillating square wave (D6, D7). About within 0.3 s cause of L2, the C5, C4 LC series resonant circuit, the formation of a lot of harmonic current through the filament, inside hydrogen ionization, and then make the mercury mercury vapor, high pressure and mercury vapor at the ends of the C5 form arc discharge, inspire wall phosphor luminescence.

Electronic ballast (EB) of the no-load U don't need to test, to test with the oscilloscope, EB and switch power supply some similar but different features, always can't afford to EB no-load wanted to be a switch power supply, its frequency in 20 ~ 100 KHZ, ordinary multimeter does not have the high frequency voltage, is connected with the instant is U up circuit, then without loop is stopped

Can't test. Because there is no voltage output when the load is empty. As shown in figure. When the load is empty, both sides of A and B are positive. Test, must load, load resistance is big, it wants to output high voltage.

2 to L2, L1c to Q2, R5 to GND, current to C5 charge; Q2 when conduction, Q1 will be because of the feedback effect of L1 as, Vc2 by D5 discharge current through the L1b decline, decrease, causing L1b ends on a negative with positive feedback voltage. According to the principle of the same end, L1a have left negative feedback voltage is right, so that Q1 quickly saturated conductivity, and the positive feedback effect of the L1 and make Q2 as quickly. When Q2