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According my book, which I think it might be an error, Q1 is a D-MOSFET, and the behavior of the circuit is shown in the following picture. But it seems to me that according to the behavior this should be an E-MOSFET. Because if it was a D-MOSFET it would be normally on and an increase in duty cycle would mean a shorter T off and thus shorter time the MOSFET is on and thus less V out. But the Waveforms show and say that a larger duty cycle with more T on produce an increase in V out

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  • \$\begingroup\$ note the polarity of the on/off signal. \$\endgroup\$ – Jasen May 26 '18 at 5:47
  • \$\begingroup\$ @Jasen exactly, the on/off signal has a larger on time and it says this will correlate with more Vout, unless by "on time" high signal they mean "driving it low" \$\endgroup\$ – Edwin Fairchild May 26 '18 at 5:50
  • \$\begingroup\$ @Jasen because I read that signal as HIGH as in applied Voltage means "on" , but a D-Mosfet would need no voltage to be "on" \$\endgroup\$ – Edwin Fairchild May 26 '18 at 5:52
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With an N-channel MOSFET, a positive change in gate voltage means an increase in conduction in the FET D-S channel, whether you are using a D-MOSFET (Depletion mode) or an E-MOSFET (Enhancement mode).

The difference between D- and E- is the level at which the change takes place. For instance an E-MOSFET is off with zero gate volts, and (depending on the specific model) on with 2v to 10v on the gate. A D-MOSFET will be on with zero gate volts. To turn it off you take the gate to a negative voltage, just like with a vacuum tube grid. In both cases, a positive change increases conduction.

As you see on the waveforms picture, on is a more positive voltage then off. The difference between using a D- and E- device would be where you draw the zero on the control voltage graph. With E- it would be the lower voltage, with D- it would be the higher voltage.

With a P-channel MOSFET, the polarities are opposite, and the states for zero gate voltage are the same as for N-channel.

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