One thing I don't understand about jfets

C

circuit975

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Hello everyone


I understood and learned the mathematical expression of the q operating point very well. Now I am trying to learn how to find the operating point on the graph. But in the examples I researched, there is such a solution. (which I drew in red) Id is given values as 3 mA and 5mA. then answers were found by replacing them in the Vgs equation. What is the logic here? According to what values such as 3mA, 5mA are given. How do we know these values?
 
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circuit975 said:
How do we know these values?
Click to expand...
You look at the voltage-v-current graph(s) in the datasheet for the particular jfet you are using.
Welcome to ETO!
 
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It is a quadratic function with an intercept on the horizontal axis at the pinch-off voltage. That is the voltage at which no drain current flows because the channel has been eliminated. The vertical axis intercept is just Idss, or the drain current that flows when Vgs=0 that flows. From the graphs in the data sheet and your own diagram you can determine the empirical relationship.
 
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circuit975 said:
I think it can only be found from the given circuit and the table created
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No, you need the JFET's characteristics, as alec stated.
That's where the red curve you drew comes from.

You can measure those characteristics for the particular JFET you are using, if you like.
 
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Well, give Id 0 and find Vgs, then give Vgs 0 and find Id and find the intersection like this. Isn't that the right step? That's what I learned to do.
 
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circuit975 said:
Well, give Id 0 and find Vgs, then give Vgs 0 and find Id and find the intersection like this. Isn't that the right step? That's what I learned to do.
Click to expand...
Except that experimentally you can't set Id and measure Vgs, you can only set Vgs and measure Id.
 
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