Fet driving question

[Staigen1]

What is the purpose of the resistor between the output from the 555 and the gate of the powerfet often seen in switching applications. See the picture below.

//Staigen

 

[Staigen1]

Me to have problem with uploading picture, how?

//Staigen

 

[audioguru2]

The resistor to the gate of a Mosfet stops the Mosfet from making very high frequency oscillation when it switches.

 

[Staigen1]

Ok, so its a stopper then. How large should it be?

 

[audioguru2]

I have seen 100 ohms in series with the gate of power Mosfets to keep them from oscillating at a very high frequency when switching. If the value is too high, the Mosfet switches slowly because the resistor limits current that is needed to charge the high gate-source capacitance. ;D

 

[SM2GXN]

Hi Staigen!

I knew I had an application note from Farchild about this.
I've repaired quite a big number of buck converters on pc motherboards that's why I downloaded this application note.
I think some of your questions will be answered if you read this app.note.
http://www.fairchildsemi.com/an/AB/AB-9.pdf

Bjorn

 

[audioguru2]

Thanks Bjorn,
Fairchild recommends only 4.7 ohms? Wow, my driver's output impedance is usually much more than that. Their very high switching frequency benefits efficiency with such a low series resistance. ;D

 

[Kevin Weddle]

Okay, what is going on with the FET? Isn't there even the smallest amount of gate current. Doesn't it have an input impedance and isn't it gm dependent. Rin is equal what?

Maybe I'll get it straight. Okay, the input at the source of a FET is 1/gm. I think that is right.

 

[audioguru2]

Kevin,
The gate of a Mosfet doesn't have any connection to the drain or source, and it doesn't draw a gate current because it is the plate of a capacitor. A power Mosfet consists of thousands of little Mosfet cells in parallel, so their tiny input capacitance adds up to about 2nF.

In order for a power Mosfet to switch very quickly so it doesn't spend much time "half-way" drawing current through the load but with a fairly high voltage across it (P = I X V), you need a driver with a fairly high current to charge and discharge the gate capacitance very quickly.

 

[Kevin Weddle]

As far as I have read, there is an input impedance to the FET. It seems obvious that one may want to divide down the signal. Something to be said about resistors. Not only do they reduce the signal, but they also reduce the noise. If for any reason your threshold of noise is too high, just resist it out.

 

[audioguru2]

As far as I have read, there is an input impedance to the FET.

Of course, the input of a Mosfet is a fairly high value capacitor.

It seems obvious that one may want to divide down the signal.

Divide the input voltage or the input frequency?
Why?
Why not just use a driver with a high output current rating so it can charge and discharge the Mosfet's input capacitance quickly?

Something to be said about resistors. Not only do they reduce the signal, but they also reduce the noise. If for any reason your threshold of noise is too high, just resist it out.

If you make a signal attenuator with resistors, then the output signal will be too low. You will need to amplify it back up to the proper level and of course the noise will be amplified too.
What do you gain by "resisting it out"?
Resistors cause thermal noise and amplifiers have noise. If the noisy signal is low-level then if you attenuate it and amplify it you end up with even more noise.