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What is the function of this diode? Isolation? Is it necessary? When the voltage is 3.3V, Vi is 4V? [Copy link]

 

What is the function of this diode? Isolation? Is it necessary? When the voltage is 3.3V, Vi is 4V?

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Wrong, VI was written as DI. The pin of the MCU corresponding to the VI is configured as the external interrupt acquisition mode, and the logic level at the VI is collected.   Details Published on 2024-4-28 09:16
 
 

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Charger detection? It should be related to the Vi pin control. The diode is used to prevent reverse voltage inflow.

Which voltage is 3.3V?

How can Vi be 4V?

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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?  Details Published on 2024-4-21 16:36
 
 
 

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Vi probably means USB voltage input monitor, which is used to detect whether the USB is connected (whether there is voltage). If the USB is not connected, VI will have a low voltage drop, otherwise it will have a high voltage drop. From the circuit, we can guess that the USB voltage is 5V and the MCU voltage is 3.3V.
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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  Details Published on 2024-4-21 16:35
 
 
 

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beyond_笑谈Published on 2024-4-20 23:02 Vi probably means USB voltage input monitor, which is used to detect whether the USB is connected (whether there is voltage). If the USB is not connected, VI low voltage...

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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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  Details Published on 2024-4-21 16:53
 
 
 

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qwqwqw2088 posted on 2024-4-14 14:41 Charger detection? It should be related to the Vi pin control. The diode is used to prevent reverse voltage inflow. The voltage division is 3.3V. Which voltage is Vi? How to...

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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You can also use resistor voltage division or the circuit in the picture. Anyway, the ground signals of MCU and USB are not isolated.  Details Published on 2024-4-21 16:56
 
 
 

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QWE4562009 posted on 2024-4-21 16:35 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-----this sentence...

If VI is the IO port of MCU, just configure it as external interrupt detection.

If the USB is plugged in, the middle of R34 and R37 is 2.5V, plus the voltage drop of D8, the voltage at DI is high (3.3V MCU).

If the USB is not plugged in, the middle of R34 and R37 is 0V, and the voltage at DI is low level.

The MCU only needs to detect the IO level of this VI to know whether the USB is inserted

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What does it mean? I don’t understand DI?  Details Published on 2024-4-27 10:41
 
 
 

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QWE4562009 posted on 2024-4-21 16:36 Vi is connected to the MCU's IO port to detect whether the adapter is inserted, but how to detect it I don't understand...

You can also use resistor voltage division or the circuit in the picture. Anyway, the ground signals of MCU and USB are not isolated.

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According to the voltage divider resistor value, the voltage divider value is 2.5V, and VI cannot be 4V. The diode does not have such a high forward conduction voltage drop. In order to reduce power consumption, the resistance value of the voltage divider resistor can be increased.

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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.  Details Published on 2024-4-27 10:43
 
 
 

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beyond_笑谈Published on 2024-4-21 16:53 If VI is the IO port of MCU, just configure it as external interrupt detection. If USB is inserted, the middle of R34 and R37 is 2.5V, plus the voltage drop of D8, DI...

What does it mean? I don’t understand DI?

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Wrong, VI is written as DI. The pin of MCU corresponding to VI is configured as external interrupt acquisition mode, and the logic level at VI can be collected.  Details Published on 2024-4-28 09:16
 
 
 

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beyond_笑谈Published on 2024-4-21 16:58 According to the voltage divider resistor value, the voltage divider value is 2.5V, and VI cannot be 4V. The diode does not have such a high forward conduction voltage drop, and in order to reduce power consumption, the voltage divider...

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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VI cannot be 4V, it should be the red circle in the picture below. [attachimg]805199[/attachimg]   Details Published on 2024-4-28 09:12
 
 
 

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QWE4562009 posted on 2024-4-27 10:43 VI cannot be 4V. What does this mean? Why does the diode need to be turned on? When the charger is plugged in, the negative side of the diode is...

VI cannot be 4V, it should be the red circle in the figure below.

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QWE4562009 posted on 2024-4-27 10:41 What does it mean? I don’t understand DI?

Wrong, VI was written as DI.

The pin of the MCU corresponding to the VI is configured as the external interrupt acquisition mode, and the logic level at the VI is collected.

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