Lateral Mosfets

Saltwater

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Hi,

I have two ECW20N20 Lateral mosfets. https://www.exicon.info/PDFs/ecw20n20.pdf
And I have set them up as a common emitter amplifier. With a resistor before the load and the transistor acting as a varistor to ground.

But I don't get these parts very well, or I am doing something wrong. Or is there something wrong?

I have them set up at 20V and the current they draw through the resistor is about 1.33A.
Looking for the bias point, the bias point is below two volts. And the voltage swing is about a volt.
Where the transistor is only operating at a few milivolts. Which is hardly an amplifier.


The data sheet seems a bit out on this as well, in being brief.
Is this OK? does this mean I have to supply loads of current to one of these to get a decent amplification going?
 

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Please show your schematic. From your description it isn't clear how you have it connected, there is no such thing as a "common emitter" MOSFET amplifier but I think I know what you mean.

Brian.
 

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I had to draw it, but that's the gist of it. Edit: I take the measurement from above the transistor after the resistor.
 
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Bias and load have to tango. Too much or too little kills gain.
Your peak gm is found at lowish current, around VT, and gain
is maximized by high Z load (at cost of speed). The best stage
gain you can get is gm*Rload. So gm and R both high.
 
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I don,t know the math too well.
dick_freebird said:
The best stage gain you can get is gm*Rload. So gm and R both high
Click to expand...
If I infer correctly, if the load line is matched. To create a higher voltage at the load, is the damping at the source. Which is reliant on the transistors internal resistance.


With a "regular" mosfet this will put the minimum bias point at about or over 3V. And the voltage swing will be larger.


But the results with these ECW20n20 are different. So much so I doubt even if the parts are not fake.

Here you can see they must be very sensitive, so much so a fixed bias instead of a pre-amplifier seems the better choice.
Distortion is low.. But the output is only 100mV. If U=I*R and R = 8 thats terrible. The output is there since the current is there. But that's not very strong.
 
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