Period 0<time<0.5 the MOSFET is OFF

[Matt007]

Hi all,

I have been looking at this question and I am having trouble sketching the wave forms.

I know that during the period 0<time<0.5
the MOSFET is OFF
the diode is ON
i = 4amps
v =0
id=0
VDS=?

Then during 0.5<time<1 period
the MOSFET is ON
the diode is OFF.
i=0
v=?
id=?
VDS=0

How do I go about working out what values are v and id are doing during 0.5<time<1 period so i then can draw the wave forms.

If you can help that would be appreciated

thanks

Matt

 

[Zryn]

Where is 'v' on the attached diagram?

Is this a CCM buck converter?

 

[Matt007]

v is across the diode from cathode to anode.

Thanks for pointing that out

 

[Zryn]

It can be useful in these circumstances to have two different pictures, one for each phase, similar to what I have attached. I have highlighted equi-potential areas on each diagram.

For 0p < t < 0.5p;

When the diode is on, it is conducting (i = 4A) and has a forward voltage drop across it, according to the pictures information. Thus you are given v.

The MOSFET drain source voltage is considered an open circuit when its turned off, thus no current flows through the drain side, since there's no voltage.

For 0.5p < t < 1p;

The MOSFET drain source voltage is considered a short circuit when its turned on, and the diode is considered an open circuit in this case as it is off, but you still want the voltage across it. Looking at the lines of equi-potential, can you see what the voltage might be?

The drain current in this case is the circuit series current returning from the load resistance to complete the loop from the source. Can you see anywhere what the circuit series current might be?

 

[Matt007]

Thanks for the reply I have worked though it now and have solved it