Why is the laptop power light off?

　　The power adapter was sent in for repair due to no output voltage.

　　When the AC power is turned on, the green indicator light does not light up. Check the AC input cable and find that there is no circuit breakage. It is determined that the internal components of the power supply are damaged. Use a "one" screwdriver to pry open the shell and check the circuit board. It is found that the fuse Fl in the above figure has melted. It is suspected that the primary component has a breakdown current phenomenon. So the focus is on checking D4-D7, C7, and Ql. No abnormality is found. Therefore, Fl is replaced and the power is turned on for testing, but the power supply still has no output. At this time, the two ends of the main filter capacitor C7 are measured +300V, which is normal. After the power is turned off, there is still high voltage at both ends of C7, indicating that the switching power supply has not started. According to the internal schematic diagram of the control chip in the figure below, it can be seen that the (7) pin of KA3842 is the starting terminal. When working normally, +300V passes through the starting resistor Rll to start the control chip initially. This voltage must be above 16V (the starting current is about 0.5mA) to make the chip start normally. When KA3842 starts successfully, the pulse voltage of the switching transformer (3) and (4) windings is rectified by D1, limited by R2, and filtered by C5 to obtain a DC voltage of about 12V to replace the 16V starting voltage provided by the starting resistor Rll to power KA3842. When the power is on, the voltage of the (7) pin of KA3842 is ov. For convenience, Rll is not removed for measurement (which causes unnecessary losses later). Only the printed circuit of R11 and the (7) pin of KA3842 is cut off. When the power is on, the voltage across C5 can rise to more than 16V, indicating that the starting resistor R11 and C5 are normal. After the power is turned off, the resistance of the multimeter is used to measure that KA3842 is suspended. The positive and negative resistance of the (7) pin to ground are both 75Ω. According to the internal schematic diagram of the device, this should be abnormal. It is very likely that the 35V voltage regulator between the internal (7) and (5) pins has broken down, causing the startup voltage of R11 to short-circuit with the ground. So KA3842 was directly replaced with UC3842, and the printed circuit board wiring of pin (7) was connected and powered on. The power indicator lighted up and then went out. Inspection found that the voltage regulator in UC3842 had broken down again, just like KA3842. This is strange. The startup voltage is limited by R11, and the normal working voltage is also limited by R2. Moreover, if the resistor is damaged, the resistance will only increase or open the circuit, and generally there will be no short circuit or resistance decrease. Then what caused the newly replaced IC internal voltage regulator to break down again? To break down the diode inside the IC, first, the voltage must exceed the voltage regulation value of the tube, and second, a large current. However, after checking the printed circuit board wiring design, no possibility of high voltage leakage to pin (7) was found. Could it be that the resistance of R11 is decreasing? Although it is impossible, no other factors can be thought of.

　　Pin Function Definition

　　1 The internal error amplifier output terminal and (2) pin external RC compensation network can determine the gain and frequency response characteristics of the error amplifier

　　2 Feedback voltage input terminal, grounded in this circuit, control (1) pin to achieve this function

　　3 Current detection terminal, detect Ql source current

　　4 Timing terminal, external timing resistor and capacitor, the operating frequency of the oscillator is f=1.8 (RTCT)

　　5 strands of land

　　6 output terminals, the driving capability is positive and negative 1A.

　　7 (Start) Power supply terminal, the start voltage is lower than 16V. KA3842 does not work. After working, the limit voltage is 30V. The start threshold is 16V. If it is lower than 10V, it will shut down.

　　8 reference voltage output terminals, output accurate positive 5V voltage, up to 5mA

　　When Rll was removed for inspection, it was found that although its resistance value was normal, a closer look revealed that there was a little burn mark on the first ring of the resistor color ring. When the printed board was checked, it was found that the components on the printed board were arranged compactly. Rll and R12 were both installed upright and close together. The pin of R12 touched the shell of Rll. When the reverse peak high voltage on R12 broke through the insulation layer of the RI1 shell, it was directly connected to the (7) pin of IC, causing its internal voltage regulator to break down and stop working. This is equivalent to the negative electrode of D1 in the reverse peak absorption network directly touching the (7) pin of IC. The reverse peak high voltage and the starting voltage are superimposed on each other, and the voltage regulator in IC can be broken down in an instant. Therefore, Rll was installed, and R12 was separated by an insulating material at the contact point, and KA3842 was installed. The power-on indicator lighted normally, the output point output +20V was stable, and the switch power supply returned to normal.

　　Summary: After analysis, the cause of the failure is the unreasonable layout of the components of the unknown brand power supply. The two high-power resistors R11 and R12 are too close to each other. When the power supply is accidentally vibrated or dropped, the two resistors collide due to inertia and cause a failure (this is also a common problem of this type of switching power supply). The brand model of this power supply is L1SHININTERNA-TIONALEN-TERPRISECORP, model: 0335A2065.