Pourquoi avons-nous besoin d'un diviseur de tension lorsque nous obtenons la même tension à la sortie qu'à l'entrée?

J'utilise un circuit diviseur de tension de résistances de 100 Ω et 10 kΩ, et j'utilise leur sortie pour l'entrée du transistor (IRF740).

J'essaie de comprendre pourquoi les résistances ont ces valeurs spécifiques. Il est logique que si j'utilise un diviseur de tension, j'obtiens 0,99x avec 100 Ω et 10 kΩ, ce qui donne la même tension de sortie que la sortie.

Mais si c'est vrai, alors pourquoi ai-je besoin de ces résistances en premier lieu, si j'obtiens la même tension en entrée. Je ne peux pas faire la même chose sans eux?

— Edi
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It does form a resistive divider, but that is not their role. Its a gate resistor and a pull down resistor. If you put it in another configuration, you can make it not be a resistive divider.

— Wesley Lee

"... I get 0.999x as a result with 100Ω and 10kΩ ..." - You might want to check your maths on that ;)

— marcelm

Réponses:

The 10k $\Omega$ resistor is there to pull down the gate when the input is floating (thus avoiding an undefined/uncontrolled gate voltage). On the other hand, the 100 $\Omega$ resistor is there to limit gate charging/discharging current (due to the presence of gate capacitance) and to prevent oscillations.

But, as you have already detected, both resistors form a voltage divider. Thus, we need to avoid dropping too much voltage in the 100 $\Omega$ resistor, otherwise we might not be able to achieve the gate voltage required to turn ON the MOSFET.

The solution is to let the pull down resistor be much, much higher than the gate resistor. 20x higher is often used; in your case it's 100x higher, which is fine too. The pull down resistor value is not critical and is usually something between 4.7k $\Omega$ and 100k $\Omega$ .

— Enric Blanco
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The OP may not be aware that the gate exhibits some capacitance and the 100R limits the inrush current to charge and discharge this.

— Transistor

Thank you @Transistor, you're probably right so I'll update my answer will your comment contents.

— Enric Blanco

That's quite helpful! Thank you so much, both of you! :)

— Edi

Another solution is to put the pull down resistor before the gate resistor, so there's no divider at all.

— Vladimir Cravero

Here that 10K Ohms acts as a pull down resistor only ie., to make the gate low normally. in otherwords , we are defining the FET to be in OFF condition normally.

— Preethi Rajagopal
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