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According to this topic ,how to get this formula, I use RC circuit and can't get this formula enter image description hereenter image description here enter image description here

This is my answer, and I don't know where is wrong

Power JJ
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    It's rare that you need to calculate this given that most times (in critical timing applications) you would use a push-pull driver. If you are not using a push-pull driver then what are you hoping to force home with the value of the gate-source pull-down resistor i.e. why do you need to calculate it? What are you trying to achieve? – Andy aka Dec 05 '20 at 12:38
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    I take issue with this calculation from gsills' answer as it makes a lot of assumptions (like that \$C_{gs}\$ and \$C_{gd}\$ are constant, which they're not, they vary when with the state of the transistor (on or off). So in my opinion, you get a **bogus answer** from this calculation. The rest of the answer is better but I agree most with Olin's answer: *"1 kΩ, 10 kΩ, or 100 kΩ ought to work."* and what actually works depends on the **rest of the circuit**. – Bimpelrekkie Dec 05 '20 at 12:58
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    That gate resistor serves one purpose: discharge \$C_{gs}\$ and almost any resistor can do that. The question is how fast does the discharge need to be (simple R * C calculation) and what value of \$R_{gate}\$ will not interfere with the other functions. So TLDR: **forget** about this calculation as **in my opinion**, it makes little sense. Also: not every value of every component **can** always be calculated or even **needs** to be calculated (precisely). **Very often** an experienced designer will just suck his/her thumb and choose a value that will do the job. – Bimpelrekkie Dec 05 '20 at 13:00
  • @Andyaka Because I see the mosfet put a pull down resistor, and I want to know how to calculate the resistor, and I can't get the same formula with him – Power JJ Dec 05 '20 at 13:01
  • @Bimpelrekkie you mean I don't need to calculate right? – Power JJ Dec 05 '20 at 13:10
  • @Jason why do this when you can use a push-pull driver. Like, what are your trying to achieve? Take note of what bimpelrekkie says also. It's not an exact science and, if you need it to be then use a push-pull driver. Calculating a pull-down is far from meaningful in many applications. – Andy aka Dec 05 '20 at 13:12
  • @Andyaka Using the Rgs resistor is to prevent the mosfet false turn on using the push-pull driver, and the we don't need the Rgs resistor? – Power JJ Dec 05 '20 at 13:21
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    You don't need a pull-down resistor when driving push-pull. This is why I was asking you these questions! I suspected you didn't. – Andy aka Dec 05 '20 at 13:25
  • @Andyaka I got it thanks – Power JJ Dec 05 '20 at 13:39
  • @Jason then maybe you should make your own answer here? – Andy aka Dec 05 '20 at 13:41
  • @Andyaka you mean I write the answer in there? – Power JJ Dec 05 '20 at 13:42
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    Sure I do. There's nothing wrong in doing that. You can explain why you initially got confused and how comments helped you to understand that using a pull-down resistor is ineffective when you are using a push-pull driver. It'll get my upvote if that's of any interest to you. But, to do your original question some justice, you should explain that you believe the original linked question is not useful in the light of the two capacitances varying so much with applied gate voltage. If you feel this is beyond your depth then just leave it. – Andy aka Dec 05 '20 at 13:44
  • Can you tell me how to do this? because I am not familiar with this – Power JJ Dec 05 '20 at 13:48
  • @Andyaka how to set your answer is best answer – Power JJ Dec 05 '20 at 14:20
  • I've done that below @Jason – Andy aka Dec 05 '20 at 14:23

1 Answers1

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The original information linked in your question said this important piece of information that you may have missed: -

how to calculate the correct value for a pulldown resistor for a MOSFET's floating gate.

And, from a comment conversation with the OP, it transpires that he is driving the MOSFET with a push-pull driver: -

Using the Rgs resistor is to prevent the mosfet false turn on using the push-pull driver

This effectively means that the MOSFET doesn't have a floating gate because it is driven by a push-pull driver. This means that a pull-down resistor is not required for normal operation of the MOSFET because the push-pull driver performs that function.

However, there are circumstances when it might be prudent to add one. For instance, if the driver didn't activate until several volts of supply were applied, the MOSFET might be partially turned on by driver circuit leakage currents. This might damage the MOSFET if a "strong" load were connected.

But, no calculations are needed for this because simply choosing a 10 kΩ resistor in most cases will do the trick. The 10 kΩ resistor would barely affect normal circuit functionality or performance but, if there is a suspicion that it does, then choose a 100 kΩ pull-down resistor.

No need for lengthy calculations or formulas.

Andy aka
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