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|8/9/2000 1:50 AM|
||how to bias a mosfet?|
Could someone please explain how to bias an enhancement mosfet, such as the BS170 or 2n7000. I'm trying to make a simple mosfet booster and can't get it to work. I looked at Joe's DualMos and don't understand his biasing technique. Right now I'm going through a .1uF cap to the gate, with a 1M to ground and a 1M (tried up to 2.2M)to +9V. I've tried from a 100 ohm to 2.2K from source to ground. And I've tried a 1K to 15K R from Drain to +9V. With all these combinations I get no signal at all using a BS170.
Does anyone have pointers here, or know of a good internet tutorial, or book.
|8/9/2000 3:49 AM|
Biasing a MOSFET is just a tiny bit more complicated than biasing a bipolar. The trick is to realize that the range for Vgs from just barely on to fully absolutely saturated is limited.
For a small signal MOSFET like the BS170, the range is about 0.5V to 2V. So if you have a linear amplifier, the gate will always be 0.5 to 2V more positive than the source. Since you're doing barely-on currents (by the MOSFET's standards again), the likeliest Vgs is going to be just barely on, say 0.5 to 1V.
So, we put the gate wherever we want it with resistors. The two 1M's work. The gate will be at 4.5V with a 9V supply. The source absolutely must, has to be 0.5 to 1V lower, at 3.5 to 4V.
Knowing that, we just use ohm's law to set the current in the source - and drain - to whatever we want. Let's say we want 1ma. So 4V/1ma is 4K, and if we put a 4K resistor (OK, OK, 3.9K) in the source to ground, and hook the drain up to the supply with any resistance that does not saturate the MOSFET, we get the desired current in the source, and drain.
The drain is conducting the same current, since the gate is insulated, so we can set the drain voltage to anything within the range of 9V on down to 4V by just choosing the resistor value between the drain and 9V supply. Let's say we want 6V on the drain. The resistor that does this is dropping 3V (9V-6V) at a current of 1ma, so it's 3K.
Let's say you want a gain stage with a reasonably large voltage swing on the collector. We allocate 1V to the source resistor, 4V to the drain resistor, and 4V to the drain-source drop, and we go peg the gate to a voltage to make this come true.
Right away, we know that the drain resistor is 4 times the source resistor, because the same current is flowing through them. Let's make this 1ma just to make the math be easy, so the source resistor is 1V/1ma = 1K, and the drain resistor is 4V/1ma=4K.
But where do we set the gate? Easy - it's the source voltage plus the right Vgs. The source is at 1V, so the gate has to be at 1V plus 0.5 to 1V, or 1.5 to 2.0V, and this will depend on which MOSFET you get. So, we use a 1M to ground from the gate and a 3.3M to +9V, which puts the gate at (1/4.3)*9V or just a hair over 2V. The thing will now be biased close to where we wanted, depending on the device, and will have a gain of 4. To get the gain higher, you have to bypass the source resistor with a capacitor.
In a nutshell - figure out what DC voltages you want on the drain and source, make sure that those voltages leave some Vds for the device to amplifiy. Select a drain current, and that picks the drain and source resistances. Then use a voltage divider to put the gate voltage where it's one barely-on Vgs above the source. Done.
|8/9/2000 11:28 AM|
Thanks RG! That was *very* helpful. Now its back to work with those misbehaving little mosfets.. hehe.
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