The prototype main control IC for this debugging is a CR6842S, which is a highly integrated current control PWM controller, which can be used for medium to large off-line power converter. The main features are: low starting current, built-in soft start, green mode unique frequency conversion and frequency hopping mode. Mainly used in power adapters and battery chargers.
[Photo of prototype]
[Output specification]67V2A
We all know that the switching MOS tube is the most critical device in the switching power supply in addition to the main control IC, if the MOS tube is damaged, then the entire switching power supply will completely fail, and may lead to AC short circuit, which is very dangerous, MOS breakdown will also lead to a large part of the surrounding device damage, troubleshooting is also very troublesome. Therefore, the prevention of MOS damage is very important, so have you ever encountered the phenomenon that the MOS waveform will be overrushed when the no-load is cut to full load? Below I will share a method to solve this phenomenon.
It can be seen that before the MOS waveform is not changed, the MOS will overshoot when the no-load cutting full load, and then slowly down, although only 1 to 2 seconds, but if the MOS positive benefit and relatively critical state, there will be a risk of exceeding the MOS voltage value. Generally, the MOS voltage value we commonly use is 650V. The following waveform is the MOS waveform of 220VAC input, the highest value has been as high as 610V, then it is likely to exceed the MOS withstand voltage value at 264VAC input.
So how to solve this problem? In fact, it is likely that the optocoupler pin 1, that is, the resistance of the power supply foot (R17 in the figure below) is not caused by the wrong value
This resistance for the optocoupler power supply resistance, you can see that it is directly connected to the output, but also bear the role of voltage division, the greater the output voltage, the greater the value of the resistance, otherwise the above situation will occur, the general output voltage is low machine (such as 12V, 24V) value is generally 3-5KΩ, then large (such as 48V, 72V), It is necessary to take more than 10K to ensure that the power supply of the optocoupler is in the normal range (the general optocoupler 1 pin voltage is 36V), and the output voltage can also be connected in series to regulate the voltage of a voltage regulator below 36V. The original machine used the resistance is 1K, obviously too small, according to experience I took a 15K resistor, can also calculate a rough value according to the formula:
R= VO-optocoupler positive guide voltage drop -Vka/1.5/2, where the optocoupler positive guide voltage drop is 1.2-1.4V, Vka is 431, the comparison voltage between V and K, generally Vka>=2.5V.
(67-1.4-2.5)/1.5=42K, in order to retain the margin this value should be divided by 2, so R=21K, the 67V output power supply optocoupler power supply resistance value can be up to 21K, leave the margin because the optocoupler is a light emitting diode, there will be a certain light decay.
The modified MOS waveform is as follows:
It can be seen that the modified MOS waveform has no obvious overshoot, and the MOS peak is only about 540V for 220VAC input, even for 264VAC input, it will be within the safe range.
At this point, the switching power supply of a bomb risk has been eliminated, you can see that the optimization of the method is still very large, you can also try this method if you encounter similar situations.
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Post time: Nov-06-2024