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Car lock relay output (open drain output) circuit confusion [Copy link]

Today I saw a colleague using a relay control circuit. The relay is external, with a freewheeling diode inside, and the control circuit is open-drain output, as shown in the figure below. There is one thing I don't quite understand here. What is the purpose of connecting the source of Nmos to the ground through diode D5? If not connected, what will be the problem if it is directly connected to the ground? I hope you can give me some advice.

开漏输出.png (82.68 KB, downloads: 0)

开漏输出.png

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This post was last edited by maychang on 2020-8-20 15:47 It is estimated that this diode is set to prevent the power supply from being reversed. If there is no D5, when the power supply is reversed, Q7Q9 will not be damaged because they have large resistances in series, but Q6 will not be damaged. Since the MOS tube has a diode inside, the diode will be turned on, and the diode connected in parallel on the relay will also be turned on. The diode inside the MOS tube and the diode connected in parallel on the relay winding will both pass a large current.   Details Published on 2020-8-20 15:45
 
 

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Counterattack, right?

 
 
 

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This post was last edited by maychang on 2020-8-20 15:47

It is estimated that this diode is set to prevent the power supply from being reversed. If there is no D5, when the power supply is reversed, Q7Q9 will not be damaged because they have large resistances in series, but Q6 will not be damaged. Since the MOS tube has a diode inside, the diode will be turned on, and the diode connected in parallel on the relay will also be turned on. The diode inside the MOS tube and the diode connected in parallel on the relay winding will both pass a large current.

 
 
 

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