14 views Yuda Electronic (HK) Technology Co.,Limited. 2018-07-13
Phone USB charger production design
In the figure, the resistor F1 of 1 ohm acts as a fuse, and the rectification is completed by a diode D1. After the power is turned on, C1 will have a DC voltage of about 300V, and supply current to Q1 through R2. The emitter of Q1 has R1 current detecting resistor R1. After Q1 is energized, the collector current will be generated through T1 (3, 4). At the same time, the induced voltage is generated on (5, 6) (1, 2) of T1, wherein the T1 (1, 2) output is rectified by D7, and C5 is supplied to the load through the USB socket; wherein T1 (5, 6) is rectified by D6 , C2 through IC1 (actually 4.3V voltage regulator), Q2 constitutes a sampling comparison circuit to detect the output voltage level;
Among them, T1 (5, 6), C3, and R4 also form the feedback circuit of the Q1 triode, which allows Q1 to operate at high frequency and continuously supply power to the T1 (3, 4) switch. When the output voltage rises due to light load or high power supply voltage, T1 (5, 6), IC1 sampling comparison leads to Q2 conduction, Q1 current decreases, electrode current decreases, load capacity becomes small, resulting in output The voltage is reduced; when the output voltage is reduced, Q2 will be cut off again, the load capacity of Q1 will become stronger, and the output voltage will rise again; this will automatically regulate the voltage.
Although the circuit has few components, it also designed overcurrent overload and short circuit protection. When the load is overloaded or short-circuited, the electrode current of Q1 increases greatly, and the emitter resistance R1 of Q1 will generate a higher voltage drop. The high voltage generated by this overload or short circuit will turn on Q2 through R3, so that Q1 will stop. To prevent overload damage. Therefore, changing the size of R1 can change the load capacity. If the required output current is small, for example, only need to output 5V100MA, the resistance of R1 can be increased. Of course, if you need to output 5V500MA, you need to reduce R1.
What is the role of C4, R5, D5? The T1 transformer is an inductive component. When Q1 is turned off, it will induce a very high voltage at the electrode. This voltage may be as high as 1000 volts or more, which will damage Q1. Now there is switch D5, this voltage can charge C4, C4 Immediately after charging, it can be discharged through R5, so Q1 will not be damaged by the high voltage of the electrode. Therefore, if these three components are damaged, Q1 is very dangerous and may be damaged.
1N4007 is a low frequency diode, FR107 is a high frequency high voltage diode, and 1N5819 is a low voltage high frequency Schottky diode. (FR107 can replace 1N4007, but not vice versa; while 1N5819 can’t be replaced by other diodes. The turn-on voltage of 1N5819 is very low. Therefore, the low-voltage rectification efficiency is very high. If other diodes must be used instead, the efficiency will be low.)
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