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Comments on Soft-start circuit behaviour

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Soft-start circuit behaviour

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Let's take a soft-start circuit that looks like this: Soft-start circuit schematic

To my understanding, the function can be described as follows:

  • C1 charges to open the gate. Value of C1 will dictate the soft-start delay.
  • R2 and R1 form a divider that establishes a bias on the gate voltage once the FET opens.
  • R1 also dictates startup time and serves to discharge C1 after input voltage is removed.

This circuit works in simulation and in the real world. LTSpice simulation results

Here, $V_{in}$ in the input voltage, $V_{out}$ the output, and $V_{n001}$ in the gate voltage of the FET. $V_{n001}-V_{in}$ is then the gate-source voltage.

The capacitor starts charging and at a certain point a sufficient $V_{gs}$ is reached after which the FET will begin to open. This is clearly seen on the simulation. However, I am not certain why the R2/R1 divider is at all necessary since the soft-start fucntionality is achieved through the R1/C1 circuit.

Therefore the question: what is the point of using R2 in this circuit?

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Unknown phrase (1 comment)
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Andy aka‭ wrote over 1 year ago

What does this mean --> open the gate