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QWE4562009
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Published on 2024-3-23 17:20
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qwqwqw2088
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Published on 2024-4-14 14:41
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Vi is connected to the IO port of the MCU to detect whether the adapter is inserted, but I don't understand how to detect it. Why not just use resistors to divide the voltage?
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Published on 2024-4-21 16:36
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Published on 2024-4-20 23:02
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Thank you very much for this brother's reply. It is very helpful. If USB is not connected, VI has low voltage drop, otherwise it has high voltage drop. How did you get this sentence? Please explain the principle. Thank you
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Published on 2024-4-21 16:35
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QWE4562009
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If VI is the IO port of the MCU, it can be configured as an external interrupt detection. If the USB is inserted, the middle of R34 and R37 is 2.5V, plus the voltage drop of D8, the voltage at DI is high level (3.3V MCU). If the USB is not inserted, the middle of R34 and R37 is 0V, and the voltage at DI is low level. The MCU only needs
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Published on 2024-4-21 16:53
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QWE4562009
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Published on 2024-4-21 16:53
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What does it mean? I don’t understand DI?
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Published on 2024-4-27 10:41
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Published on 2024-4-21 16:56
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8
Published on 2024-4-21 16:58
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VI cannot be 4V. What does this mean? Why does the diode need to be turned on? When the charger is plugged in, the cathode of the diode is now 2.5V. What do you mean? I don't understand. Please enlighten me.
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Published on 2024-4-27 10:43
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QWE4562009
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QWE4562009
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Published on 2024-4-28 09:12
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Published on 2024-4-28 09:16
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