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FETs source voltage


c3r14l.k1l4
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Greetings,

I'm new here and this is my first post. All my electronics knowledge has been built from experience and books and I intend to help whenever I can. But as I'm not God I've got some doubts.

One of them wich is bugging me is how to get a FETs source voltage value. For instance, we know that in a BJT the emmitter is always the base - junction drop voltage (~0.7V), but in the books I can't find a similar relationship for the FETs. Only the drain current in function of the gate voltage (the parabola function).

Can anyone explain this to me?
And... sorry for my english, as it's not my native language...
Cheers!

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BJT's are current devices, they use the flow of electronics through the PN junction to create a cascade effect allowing a larger current to flow from Vcc to Vee.

Because the flow is proportional, a small current at the Base can control a large current at the collector. (And Emitter if you look at it that way)


FETS (Field effect Transistors) use VOLTAGE as the controlling agent.
As a result the actual Current flow is limited because in essence you are putting current into the fet at the PN junction backwards. It acts like a backwards diode so only a tiny current will leak through.

The MOS (Metal Oxide Silicon) FET enhances this by adding a layer of GLASS to the mix.  The glass isolates the gate from the Source and Drain.
In effect there is (Ideally) No current flow at the gate, Only voltage.

Just like a capacitor pulls a charge across an insulator (Dielectric) the Fet causes a static charge around the Gate. Because the current in the device needs to flow past the gate the static charge on the gate acts like a clogged pipe and restricts the flow.

As this charge builds it begins to PINCH OFF the current flow through the Drain-Source pathway. 

It is a lot like stepping on a garden hose.  While you are not directly adding any water to the flow, you can 'pinch off the flow' through the insulator of the hose.

The voltage required to totally shut off a FET is called the Pinch Off voltage.

Some FETs (Enhancement and Depletion mode) are OFF until a voltage is applied to the gate.  They work on the same concept but there are extra layers to the FET making it more complex,  Without getting into the would junction theory of how it works, Just think of them as working BACKWARDS.

The More VOLTS applied, the more OPEN the channel is.

does that Help ?

-Mike

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Nop. Sorry...

You explained me it's functioning... and I already knew that... I just want to know how to calculate the source voltage regarding the one applied to it's gate.

Check this:
123132.png
In the BJT circuit the emmitter is 5-0.7V so aprox 4.3V. In the FET one the source is 5-X=1.431 so X=3.569V? Where does this value come from? Is it the pinch voltage? I think so... just to make shure...

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