Thread Starter

Box-O-Rocks

Joined Jul 14, 2022
12
Hello all,
Disclosure, I am an electrical engineer but I deal with designing power system for construction projects. I decided to go back for Masters in Power Electronics. I'm rusty with DC circuits.

My homework is the same as a previous thread... https://forum.allaboutcircuits.com/threads/buck-boost-converter-duty-cycle.128700/

I am not comfortable with the MOSFET. Could someone help explain how the current changes as it transitions through the MOSFET? my values are the same as in the link...
(Given) Vg= 12V, V= -5, R=2.5, f= 200kH,
(Solved) L=117uH, C=20uF, D=0.2941

I am looking to solve for the MOSFET peak current iQ1(t) with the ripple current.
 

Papabravo

Joined Feb 24, 2006
21,029
Hello all,
Disclosure, I am an electrical engineer but I deal with designing power system for construction projects. I decided to go back for Masters in Power Electronics. I'm rusty with DC circuits.

My homework is the same as a previous thread... https://forum.allaboutcircuits.com/threads/buck-boost-converter-duty-cycle.128700/

I am not comfortable with the MOSFET. Could someone help explain how the current changes as it transitions through the MOSFET? my values are the same as in the link...
(Given) Vg= 12V, V= -5, R=2.5, f= 200kH,
(Solved) L=117uH, C=20uF, D=0.2941

I am looking to solve for the MOSFET peak current iQ1(t) with the ripple current.
There are two conditions to worry about:
  1. The MOSFET is turned on. In this state the resistance of the channel from drain to sources is very small, on the order of tens of milliohms. That is the current flowing into the drain is the same as the current coming out of the source. The voltage drop across the channel will be by Ohm's Law, the drain current times rds(on), the channel resistance. If rds(on) is 0.010 Ω, then the voltage drop will be millivolts/Ampere.
  2. the MOSFET is turned off. In this state the resistance of the channel is large, perhaps a couple of MegOhms. Very small leakage currents may flow. Onside of the MOSFET is at the supply voltage, but the other side is at a voltage determined by the inductance and the time rate of chage of the current.
In between those two states the device experiences switching losses as the device transition from ON to OFF and form OFF to ON. These times are usually on the order of fractions of a microsecond.
 

Thread Starter

Box-O-Rocks

Joined Jul 14, 2022
12
In between those two states the device experiences switching losses as the device transition from ON to OFF and form OFF to ON. These times are usually on the order of fractions of a microsecond.
Therefore,
1. would i need to account for the resistance values in the on and off state?
2. (if no) would the current through the MOSFET be equal to current values found if there were just a switch as in figure 1 (see attached.)

No information was provided about the MOSFET, so I have to assume ideal conditions. The textbook I am using doesn't cover MOSFETs in this Chapter, so again I assume the book expects the reader to have basic knowledge of their ideal conditions. would you disagree?
 

Attachments

Papabravo

Joined Feb 24, 2006
21,029
1. No - not really
2. You can certainly model a MOSFET as a voltage-controlled switch. I do that all the time.

One more thing. You should swap the polarity on the output resistor. Your output voltage will be negative.
 

Thread Starter

Box-O-Rocks

Joined Jul 14, 2022
12
1. No - not really
2. You can certainly model a MOSFET as a voltage-controlled switch. I do that all the time.

One more thing. You should swap the polarity on the output resistor. Your output voltage will be negative.
Thank you. I treated the problem as if it was just a switch and considering that when the MOSET is on, the current will flow through in the Inductor and because they are in series, the current would be the load current from the Inductor.

I just kicked myself a bunch. Looking at these circuits, sometimes it just feels like there is a hidden resistance or a magically induced voltage, or some stray current just hiding in the components when in reality the answer is as simple as it appears.
 
Top