Why does the high-side switch turn ON with the Buck converter in the figure below?

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Ken Inoue on 16 Sep 2017
Commented: Ken Inoue on 19 Sep 2017
Thank you for your patronage. Why does the high-side switch turn ON with the Buck converter in the figure below? Since the threshold value of the high side switch is 0.5 V, to turn on the high side switch, for example, VG = 10 + 1 V is required as shown below. However, in the simulation circuit, the high side switch is turned on with VG = 1 V. Please give me a hint. The simulation circuit can be obtained from the following link. Thanks. https://www.dropbox.com/s/oeuho5yebzn27tx/Question.slx?dl=0   Abhi Sundararaman on 18 Sep 2017
Edited: Abhi Sundararaman on 18 Sep 2017
It looks like the ideal semiconductor block is intended to be in the "on" state when the gate-cathode voltage exceeds the threshold voltage. In your model's case, the cathode is the right side of the semiconductor, and the gate voltage is the signal that is coming from the "constant-PWM generator" part of the model.
Therefore, if the output of the "Constant - PWM generator" portion is 1 volt, then the gate-cathode voltage is 1V, and therefore exceeds the threshold value of 0.5 V.
To demonstrate this, you could attach a voltage sensor across the ideal semiconductor switch, and remove the PWM generator block, like so: Now, run the model once with "Constant2" equal to greater than the threshold voltage, and and once with "Constant2" less than the threshold voltage.
You will see that in the first case, the voltage measured is very small, because the switch is "on", and thus, has a low on resistance.
In the second case, the voltage will be roughly 10, because the switch is off. I believe the above is intended behavior.
In summary, I think that the switch is intended to turn on when the gate voltage is greater than the threshold voltage, independent of the source voltage. So in this case, since Gate Voltage = 1V, which is > 0.5 V, the switch will be on, regardless of the value of the voltage source's voltage (10V).
Ken Inoue on 19 Sep 2017
Thank you for your reply. I am not good at English.Therefore, my question was difficult to understand. But, you answered in an easy-to-understand way. Thanks.