# Serious: Switching Mode Power Supply

#### Sherif Amer

Joined Jul 6, 2010
8
Dear all,

I'm working on some project which requires 100v converted from 9v, so i called the basic idea of Boost converter.

The circuit is quiet simple, i use IRF540NS switched from PWM (Pulse Value 3.3v, Pulse Width 278us, Period 300us) I used TI application note http://www.ti.com/lit/an/slva372b/slva372b.pdf to calculate the values of the coil and the output capacitor using the equations 5,6 and 12, 13.
The inputs are: Vout = 100, Iout(max) = 5mA
After all, L = 148mH and Cout = 511uF

The simulation output is 10v not 100v!!! #### Papabravo

Joined Feb 24, 2006
12,407
You might need a "load" to provide a DC path when the switch is open.

#### Sherif Amer

Joined Jul 6, 2010
8
second it doesn't need a load, i figured out the problem, it's in the IRF, it doesn't turn on 3.3v, changed the pulse into 5v, it settled finally at 100v.

#### crutschow

Joined Mar 14, 2008
23,378
The IRF540NS is a standard MOSFET which requires 10V Vgs to fully turn on (it's only partially on at 5V). If you want one that turns on at a lower voltage you need to use a "logic level" type MOSFET.

#### Sherif Amer

Joined Jul 6, 2010
8
The graph in the datasheet VDS vs ID is drawn with VGS varies from 4.5v to 15v. I will need another MOSFET anyway. Thanks a lot guys!

#### ifixit

Joined Nov 20, 2008
650
What is you peak inductor current? I'll assume 2 Amp for now.

If the inductor max peak current is 2 amp or less then Vds with a Vgs of 5V is only 0.15 Volts. That is 0.15 X 2 Amp = 0.3 Watts and Rs = 0.15V / 2A = 75mΩ. Also, the temperature has very little effect at Vgs=5 so Rs won't change much either. A 5 V pulse is okay for this situation.

Regards,
Wayne