about boost converter

[Ahmed Kh]

hello
please i need help my input around 9v
i am doing the hardware for boost converter
my problem when i put load resister around 22 ohm going be too hot as a heater
if i put around 1k ohm the voltage will be increase too much and also if i remove the resister the voltage around be 50v and the normal output voltage have to be 20
what is the load have to be put on output (also with resister or with remove resister )
my best regards

 

[amayilsamy]

How much output voltage and current you require?

post your present circuit

 

[Ahmed Kh]

How much output voltage and current you require?

post your present circuit

the hardware as the same of this software
the duty cycle is 50%
i design for current 2 a
thanks

 

[chuckey]

What is happening is that during the time the FET is on you are charging the magnetic field of the inductor and when the FET is off this energy is discharged into the load via the diode. The capacitor smooths out the voltage pulses.
To get a stabilised output you have to sample the output voltage and feed this back to modify the mark space ratio or frequency, to get the correct voltage. If for simplicity you want it to be unregulated then you have to change the frequency or the size of the L or the amount of current it is taking (changing input voltage), but the voltage will only be correct at one load current.
Frank

 

[amayilsamy]

Try to do with simple switching regulator MC34063.

Refer the datasheet of MC34063.

 

[arunkumarshanti]

Hi,

You are missing the feedback loop which regulates the line/load in your breadboard implementation. You can google for the control techniques for the boost converter. There are voltage & current control techniques also, PWM & PFM techniques. However, if you don't need the regulation (meaning line & load are constant). Calculate the output voltage & load current. Then place your load resistor according