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In the schematic diagram, the FD2 signal passes through P2 of D16 and DAN217, and then passes through P1 of D16. Will the P2 line of D16 in the middle short-circuit P1 of D16 above? I am a novice, please give me some advice, thank you!

微信图片_20180727140648.png (32.03 KB, downloads: 0)

2号引脚那块不会短路吗

2号引脚那块不会短路吗

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There is no diode between pins 7 and 8. The upper left diode and the lower left diode in the first post are photodiodes. The 8th pin of D16 is connected to the positive terminal of the power supply, and both photodiodes are in reverse. When the photodiode is in reverse and without light, there is no current. When there is light (the light emitted by the right LED), although the photodiode is in reverse, there will be a small amount of current (the direction is from the positive terminal of the power supply). The current in this photodiode flows into the base of the transistor, becoming the base current of the transistor, making the transistor conduct. The transistor conduction reduces the potential of the input terminals D1A and D1B.  Details Published on 2018-7-27 16:52
 
 

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"Won't the P2 wire of D16 in the middle short-circuit P1 of D16 above?" D16 was found, but "the P2 wire of D16" and P1 were not found.

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1 and 2  Details Published on 2018-7-27 16:05
 
 
 

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The way the OP reads the map is very special. Please make it clear whether it is 2 and 1 or P2 and P1 of D16.

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1 and 2, both said they were newbies, how embarrassing!  Details Published on 2018-7-27 16:04
 
 
 

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There will be no short circuit. FD2 should be a control signal. V39 is a clamping diode to prevent the FD2 signal from being too large. When FD2 is high, the upper right LED of D16 will not light up. When it is low, the two LEDs will be turned on. The optocoupler is turned on. When FD2 is too large, it will pass through V39 and enter P15DC.

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Then why doesn't it light up when the voltage level is high, but lights up when the voltage level is low?  Details Published on 2018-7-27 16:19
 
 
 

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qwqwqw2088 posted on 2018-7-27 14:30 The OP's method of reading diagrams is very special. Make it clear what P2 of D16 and P1 of D16 are. Are they 2 and 1 or P2 and P1?
1 and 2, everyone says they are newbies How embarrassing!
 
 
 

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maychang posted on 2018-7-27 14:24 "Won't the P2 line of the middle D16 short-circuit the P1 of the upper D16?" D16 was found, but "the P2 line of D16" was not found, nor was...
1 and 2

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Does “1 and 2” refer to pin 1 and pin 2 of HCPL-053K-500R?  Details Published on 2018-7-27 16:12
 
 
 

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Does “1 and 2” refer to pin 1 and pin 2 of HCPL-053K-500R?

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Yes, the P1 I mentioned is different from pin 1.  Details Published on 2018-7-27 16:15
 
 
 

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maychang posted on 2018-7-27 16:12 “1 and 2” refers to pin 1 and pin 2 of HCPL-053K-500R?
Yes, the P1 and pin 1 I mentioned are different.

Comments

Usually, it is Pin 1 and Pin 2. The FD2 signal usually does not exceed the power supply voltage (15V). Even if it exceeds, the two diodes in series in DAN217 are turned on, and the diode between pin 1 and pin 2 in D16 is reversed at this time, and will not be turned on, let alone short-circuited.  Details Published on 2018-7-27 16:28
Usually, it is Pin 1 and Pin 2. The FD2 signal usually does not exceed the power supply voltage (15V). Even if it exceeds, the two diodes in series in DAN217 are turned on, and the diode between pin 1 and pin 2 in D16 is reversed at this time, and will not be turned on, let alone short-circuited.  Details Published on 2018-7-27 16:24
 
 
 

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844121774 Published on 2018-7-27 15:06 There will be no short circuit. FD2 should be a control signal. V39 is a clamping diode to prevent the FD2 signal from being too large. When FD2 is high, the upper right light-emitting diode of D16 will not...
Then why is it not lit when it is at a high level, but will light up when it is at a low level?

Comments

"Why doesn't it light up when it is at a high level, but lights up when it is at a low level?" The LED will not light up until the voltage of FD2 and FD3 is close to 15V. The reason why it does not light up at this time is that there is no current passing through the LED.  Details Published on 2018-7-27 16:33
 
 
 

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This is the overall schematic diagram

微信图片_20180727162157.png (57.69 KB, downloads: 0)

整体原理图

整体原理图
 
 
 

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Liu123 posted on 2018-7-27 16:15 Yes, aren't the P1 and pin 1 different?
Usually we have to say Pin 1 and Pin 2. The FD2 signal usually does not exceed the power supply voltage (15V). Even if it exceeds, the two diodes in series in DAN217 are turned on. The diode between pin 1 and pin 2 in D16 is reversed at this time and will not be turned on, let alone short-circuited.

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Do the four LEDs in D16 affect each other? I think this is an integrated IC.  Details Published on 2018-7-27 16:26
 
 
 

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maychang posted on 2018-7-27 16:24 Usually we have to say Pin1 and Pin2. The FD2 signal usually does not exceed the power supply voltage (15V). Even if it exceeds, it is the two diodes in series in DAN217...
Do the four LEDs in D16 affect each other? I think this is an integrated IC.

Comments

There are not four LEDs in D16, but only two, one between pins 1 and 2, and one between pins 3 and 4. The other two are photodiodes.  Details Published on 2018-7-27 16:31
 
 
 

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Liu123 posted on 2018-7-27 16:15 Yes, P1 and pin 1 are different.
The diode between pin 1 and pin 2 of D16 is in reverse state when the voltage of FD2 is very high. However, since V39 is turned on, the reverse voltage will not exceed the tube voltage drop of the two series-connected diodes in V39, so it will not break down due to excessive reverse voltage. Pins 3 and 4 of D16 will not break down when subjected to reverse voltage.
 
 
 

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Liu123 posted on 2018-7-27 16:26 Do the four LEDs in D16 affect each other? I think this is an integrated IC
There are not four LEDs in D16, but only two, one between pin 1 and pin 2, and one between pin 3 and pin 4. The other two are photodiodes.
 
 
 

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Liu123 posted on 2018-7-27 16:19 Why is it not lit when it is at a high level, but it is lit when it is at a low level?
"Why is it not lit when it is at a high level, but it is lit when it is at a low level?" The FD2 and FD3 voltages must be close to 15V before the light-emitting tube will not light up. The reason why it does not light up at this time is that there is no current passing through the light-emitting tube.

Comments

Thank you, I understand now. I will study the circuit carefully.  Details Published on 2018-7-27 16:44
Thank you, I understand now. I will study the circuit carefully.  Details Published on 2018-7-27 16:36
 
 
 

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maychang posted on 2018-7-27 16:33 "Why doesn't it light up when it is at a high level, but lights up when it is at a low level?" The voltage of FD2 and FD3 must be close to 15V before the light tube will not light up. ...
Thank you, I understand now. I will study the circuit carefully
 
 
 

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maychang posted on 2018-7-27 16:33 "Why doesn't it light up when it is at a high level, but lights up when it is at a low level?" The voltage of FD2 and FD3 must be close to 15V before the light-emitting tube will not light up. ...
And how does the 7.8 pin work when it is turned on or off? Why is this diode reversed?

Comments

I understand. I looked up the working principle of photodiodes on the Internet, and found out that this is the case. . . . . . .  Details Published on 2018-7-27 16:52
I understand. I looked up the working principle of photodiodes on the Internet, and found out that this is the case. . . . . . .  Details Published on 2018-7-27 16:45
 
 
 

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Liu123 posted on 2018-7-27 16:44 Also, how do the 7.8 pins work when they are turned on or off, and why is this diode reversed?
I understand. I checked the working principle of the photodiode on the Internet, and it turns out to be like this. . . . . . . .
 
 
 

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Liu123 posted on 2018-7-27 16:44 Also, how do the 7.8 pins work when they are conducting or not conducting, and why is this diode in reverse?
There is no diode between pins 7 and 8. The upper left diode and the lower left diode in the first post are photodiodes. The 8th pin of D16 is connected to the positive terminal of the power supply, and both photodiodes are in reverse. When the photodiode is in reverse and without light, there is no current. When there is light (the light emitted by the right LED), although the photodiode is in reverse, there will be a small amount of current (the direction is from the positive terminal of the power supply). The current in this photodiode flows into the base of the transistor, becoming the base current of the transistor, making the transistor conduct. The transistor conduction reduces the potential of the input terminals D1A and D1B.
 
 
 

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