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Schematic diagram - how does DZ1 protect the MOS tube? [Copy link]

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I saw on the Internet that a voltage regulator tube is connected in parallel between the drain and gate of the MOS tube .

It is said that it can protect the MOS tube when an external inductive load is connected .

Because the most contact is the inductive load parallel to the freewheeling diode,

I am not very clear about the process of MOS parallel voltage regulator:

When it is greater than 27.7V, is the high voltage released through the path of DZ1, VD1 and DZ3 (breakdown voltage 12V)?

Please help me analyze this process, thank you!

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[attach]639295[/attach]   Details Published on 2022-9-6 10:03

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This is a Schottky diode, which is used to prevent sudden input surges.

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At the moment of shutdown, the reverse electromotive force of the inductive load causes the voltage regulator to form a loop to balance the gate voltage, so that the gate voltage does not jump directly to zero, thereby suppressing the voltage overshoot phenomenon.

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This post is from Analog electronics

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Thanks for sharing! That is to say, the TVS part is negative feedback. If the collector voltage is too large during the tube shutdown process, the TVS tube will break down and generate base current, causing the tube to continue to conduct for a period of time to discharge the overvoltage. So I think this circuit should not be used in fast switching applications, because this protection delay  Details Published on 2022-9-6 12:25
Thanks for sharing! That is to say, the TVS part is negative feedback. If the collector voltage is too large during the tube shutdown process, the TVS tube will break down and generate base current, causing the tube to continue to conduct for a period of time to discharge the overvoltage. So I think this circuit should not be used in fast switching applications, because this protection delay  Details Published on 2022-9-6 12:22
 
 
 
 

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Thanks for sharing! That is to say, the TVS part is negative feedback. If the collector voltage is too large during the tube shutdown process, the TVS tube will break down and generate base current, causing the tube to continue to conduct for a period of time, discharging the overvoltage. So I think this circuit should not be used in fast switching applications, because this protection prolongs the tube shutdown time, and can only be used in "one-stick sales" situations.

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Another question is: Why do we need to connect a diode in series with the TVS tube? Isn't a single TVS enough?

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