hi everyone,
Please help me with my third year BEng project "Designing SEPIC converter whose output is 5V to 20V and 1A current and also use micro controller as PWM controller (regulator)".
Problems
a. mosfet is not turning off.(both in simulation and practical)
b. snubber circuit which one is the best for sepic converter (i used rate of rise of voltage)
c. calculation to find the value of inductor(i.e. I calculated the L1 first and then made L1=L2. what happens if we calculate L2 first and make L2=L1?)
d. ground plane?
e. Does MOSFET driver necessary for 94kHz frequency ?
i have done simulation and calculation. I would like to know whether my calculation and simulation are right. I have attached my calculation and simulation result please help me I am stuck. Please give me some ideas. thank you

 

MOSFET turn on and turn off are controlled by the gate voltage, as I'm sure you know. I couldn't find any mention of gate voltage in yourr PDF.

 

I used 5V 94kHz pulse to gate. in practical MOSFET (VDS)is like bouncing.

 

In your schematic, you show an IRF540 which is not a logic level MOSFET, but a standard level MOSFET; it is only considered fully turned ON when Vgs=10v.

You don't show what you are driving the gate with. An IRF540N has a total gate charge (Qg) of 71nC, which won't be trivial to drive at 94kHz. There are many more modern MOSFETs that you could have chosen, with a far lower Qg.

 

Thanx for the help. I simulated the circuit again using logic MOSFET (IRLI530N) and MOSFET Driver (TC4427). I have attached the circuit simulation and result please have a look and suggest me. that would be very helpful.
thank you

 

Thanx for the help. I simulated the circuit again using logic MOSFET (IRLI530N) and MOSFET Driver (TC4427). I have attached the circuit simulation and result please have a look and suggest me. that would be very helpful.
thank you

A 2kΩ gate resistor is way too high. 2kΩ*2.5nF=5usec time constant. Your period is only ≈10usec. With 80% duty cycle, you have minimum pulse width of about 2usec. Make the gate resistor 47Ω or less. 10Ω might be better. Mount the resistor as close to the gate as you can get it.

 

A 2kΩ gate resistor is way too high. 2kΩ*2.5nF=5usec time constant. Your period is only ≈10usec. With 80% duty cycle, you have minimum pulse width of about 2usec. Make the gate resistor 47Ω or less. 10Ω might be better. Mount the resistor as close to the gate as you can get it.

thanx. i understand what you did. but does it give me current of 1.7mA. in another words how do i select the resistor for driver to give me the current i want and also less time constant plz.
thankyou

 

Read this Application Note, even if your not using IRF components, the basic working is the same in all brands of drivers. http://www.irf.com/technical-info/appnotes/an-978.pdf Section #6 is the part about gate resistors and layout.

Other component manufactures also have similar App notes.

 

The gate resistor should not be used to determine current. Gate resistors are there only to prevent parasitic oscillations in the MOSFET. They need to be low value, so that rise and fall times are fast, which minimizes wasted power.
Where do you want 1.7mA to flow?

 

tell me if i am doing wrong. to find out the rg (driver resistor),
QG(total gate charge) =C×V
C=QG/V [V=VGS(gate to source voltage)]
C=(15×10^-9)/6 [found QG from data sheet of mosfet]
C=2.5nF
and then
I=C dv/dt
where dt= desired time to charge the capacitor for e.g.
if we want it to charge in 10 ns then
I=(2.5x10^-9)x(6/(10x10^-9))
I=1.5A. since VGS=6V,
Rg=6/1.5A
Rg=4 ohm
is my calculation is right? Please tell me
thank you

 

Calculating the switching time of a MOSFET is very complicated.
http://www.digikey.com/Web Export/S...VishaySiliconix_MOSFETBasics.pdf?redirected=1