saturate a mosfet

Thread Starter

baby_1

Joined Jun 3, 2011
39
i have a IRF3205 and i want to saturate in 45 amper,could you explain how can i do it? with equation and datasheet?

VDS=24 volt?
 

miguel cool

Joined Mar 15, 2010
9
First VTO=3.73234
And Saturation condition is: Vds>Vgs-Vto
from figure 1 of the datasheet and for ID = 45Amp
In the Id draw a line horizontal for 45A and meet the curve for 75°C this gives you 5.2=Vgs

If you wish calculate:

M1 9 7 8 8 MM L=100u W=100u
.MODEL MM NMOS LEVEL=1 IS=1e-32
+VTO=3.73234 LAMBDA=0 KP=95.6501

ID = 1/2 Kp W/L (Vgs - Vto)^2 for saturation region
 

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Wendy

Joined Mar 24, 2008
22,164
Welcome to AAC!

A rule of thumb is simply bias the gate and source with 10V. That will saturate the vast majority of MOSFETs.

A problem you will run into with 45A is the leads are simply not big enough. They will get hot too!. Dumping heat away from a part like this is your biggest problem, long term temperature will degrade, then fry your parts.

Simply turning a MOSFET on/off is as simple as applying a minimum voltage to the gate. Where it gets complex is if you are switching it. A gate resistor is used to prevent ringing (which in turns causes bad switching and heat), and the resistor is very low value, anywhere from 1Ω to 100Ω. The gate to drain looks like a capacitor, which can go as high as 0.1µF. Driving this capacitance is the real problem.
 

Thread Starter

baby_1

Joined Jun 3, 2011
39
my big problem is driving an element that takes more current,now i want to use it in switching mode Ferquency=1Khz,how can do it with mosfet? with little price?
 

Thread Starter

baby_1

Joined Jun 3, 2011
39
Thanks bill
i have another question its about mosfet power dissipation
some where i read that Pd=r*r*Rds
or
PD=VDS*ID
and some where else pd=f*(ton)*cn*vds

for pwm wave and switching

all of them get me different value

could you tell me the difreences?
 
Last edited:

Wendy

Joined Mar 24, 2008
22,164
I won't be able to help as much as I would like on this aspect. I have a technicians point of view, plus some practical experience with using these devices. Most cases for really high currents you can parallel MOSFETs to spread the heat around, but eventually they will fail. I used to maintain a in house piece of equipment that used high currents where this was the case.

Ron contributes to some of the heat, but only in a minor way. The real problem is switching speeds. The less time spent switching, the less heat generated because of the MOSFET being in between on and off. A resistor generates heat, a switch doesn't. Frequency only contributes because it dictates how many times / second the device turns on and off.

This is the real reason drivers are so important. If I was using a MOSFET as a simple on/off I can get by with slow switching speeds, because the number of times / second is so low. When you are switching several thousand times / second you must keep the MOSFET out of it's linear mode as little as possible.

Another aspect I understand about MOSFETs but don't know how to calculate is the gate resistor. A gate looks like a large capacitance. Add a long length of wire to that and you have an inductor. An LC circuit that is being pumped with a square wave will ring, and ringing will keep the MOSFET in the analog region we are trying to avoid to keep from generating heat. The resistor is meant to damp ringing, and should be as close to the gate as possible. It should be low value, as there is a trade off between the resistor slowing the switching speed (it is an RC circuit) and preventing ringing.

I'll ask and see if one of the engineers can explain the math better, and step aside at this point.
 

shortbus

Joined Sep 30, 2009
7,942
While not a engineer - At the bottom of page number 10 of this link; http://focus.ti.com/lit/ml/slup169/slup169.pdf there is a formula for figuring the gate resistor.

In another link that I can't find right know, it said as a rule of thumb for gate resistors- minimum 4.7Ω ,maximum 20Ω. This is ohms NOT Kohms. A lot of schematics on the web show the values as KΩ and they are wrong.
 
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