Buck Converter Problem(PLEASE HELP ME!!!)

It's possible that you're not driving your MOS hard enough, you want it to be very very on, on hard, ohmic ... could that be it.

Are you able to generate low voltages, ie < 5v fairly easily?

Added after 4 minutes:

OOps, I forgot to add this ... is the current mos you're using an NMOS ... have you tried a PMOS type.

Additionally, what is the load?
What is the "Ron" for the MOS you're using currently?

I hope I'm understanding your question correctly. ^-^

Firstly, thank you for your interest...

rhesusmonkeyboy said:

It's possible that you're not driving your MOS hard enough, you want it to be very very on, on hard, ohmic ... could that be it.

Are you able to generate low voltages, ie < 5v fairly easily?

Click to expand...

Yes i can generate voltages smaller than 5V.if i make the load resistance be higher, than i can generate higher voltages(i can get the 12v output which is my input when the load is in megaohms)

Added after 4 minutes:

OOps, I forgot to add this ... is the current mos you're using an NMOS ... have you tried a PMOS type.

Click to expand...

it is N-channel MOSFET. i didn't tried any P-type.

Additionally, what is the load?

Click to expand...

the Load is 2.2ohms-11watts resistance.i am sure that it is not related to the load's power.

What is the "Ron" for the MOS you're using currently?

Click to expand...

in the data sheet, it says:drain to source on resistance, rDS(on)=0.4ohm. but i am not sure that u asked for it.. i attach that page from the datasheet.

I hope I'm understanding your question correctly. ^-^

Click to expand...

thanx a lot!!! i must solve the problem. it is my project..thank u...

1. u got to make the Vgs of IRF510 to be 10V to obtain Rdson=0.4 ohm!
but in ur circuit, ur Vg=12V, but Vs=12-Vds=12-Rds*I < 10V, and from
ur statement, Vout=5V @duty=100%, I think at that time ur Vs maybe around 6V,
So ur real Vgs=12-6=6V only and ur Vin is not 12V, it is 6V,. if ur Vout is 5V,ur duty is 5/6=83%.
2. to get Vgs > 10V, one common way is to use bootstrap technique (diode+Cap), u can refer to ISL6522 from Intersil.

The problem is of course that you cannot drive the MOSFET into saturation.

The eaiest thing to do is use a P-ch device and reverse the signal of the PWM.

Else, use a charge pump to generate a voltage about 8-10V higher than the input voltage and power your controller from that voltage. That way you will be able to always drive the MOSFET into saturation.
You can actually use the same 555 to build the charge pump.

Go PMOS!

Find a comparable PMOS, definitely.

I made that self-same mistake once.

To reverse the control signal, just connect your ramp to where your control voltage was, vice versa.

If you don't do this, what'll happen is that it'll look like (1-D) – ie D' – is controlling the output, rather than D.

Enjoy!

^-^

thank u so much. i will try to do that by using a p-type mosfet. i wish it works. if it does not, i will be here again!!! ))
take care..

hello people first i thank people who helped me to solve my problem.i overcome the problem by replacing a P-type mosfet instead of an N-type. so why did you suggest me to do that?what is the reason?for the N-type, we control the source voltage which is our the output.but for the P-type, source is connected to +12 volt. our output is connected drain. so i cannot saturate the N-type because of this?Vgs is always changing(Vg and Vs changes together) can you explain it?please tell why it becomes!.. thank u very much!.....

To saturate an n-ch MOSFET you need to apply a voltage between gate and source, such that the gate is more positive than the source.
With an n-ch MOSFET in the buck regulator, when the transistor saturates its source will be at the same potential as the drain, that is equal to the input voltage.
But if the source is at the same potential as the input voltage, then to keep the transistor saturated you need to have its gate at something higher than the input voltage. That did not happen in your circuit. You need an extra voltage, usually supplied by a charge pump, that delivers power to the MOSFET driver.

With the p-ch, the source is at the input voltage potential, but driving a p-ch MOSFET requires that the gate be more negative than its source. That is simply accomplished by pulling the gate to the gound. So as long as the input voltage is high enough for the MOSFET to stay saturated, this will work.

thank u very much!................................