Inverse curves related to JFET characteristics, help

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The discussion focuses on understanding how to determine the small-signal drain-source resistance (rd) from the JFET characteristics curve. The key point is that rd is calculated as the inverse of the slope of the IDS versus VDS curve at a specific operating point, which in this case is VDS = 15 V and IDS = 10 mA. A participant clarifies that to find the slope, one can draw a tangent at the operating point, calculate the change in IDS over the change in VDS, and then take the inverse of that slope. This method provides an approximate value for rd, although accuracy may be limited due to the graph's scale. The discussion emphasizes the importance of understanding the relationship between the curve and the parameters involved in JFET characteristics.
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Inverse curves related to JFET characteristics, help!

Hi,
This is an example given in a lesson which is closely related to a coursework question I'm trying to complete, the problem is I can't understand how they have got the results they have.

Here is the statement giving the relationship

"The parameter rd is the small-signal drain-source resistance and can be found from the inverse of the slope of the curve of IDS against VDS at the operating point. From FIGURE 1, rd can be estimated as 12.4 kΩ at the operating point of VDS = 15 V and IDS = 10 mA."
(figure 1 is the attached JPEG)

I cannot understand how they have got 12.4kΩ as the value of Rd, how do I find the inverse of a curve if I have no relationship for x and y ?

We are looking at the curve Vgs = -1

Can anybody help, I'm sure this is simple but I've been looking at it for two days now and am getting desperate.
 

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Hi Trespaser5! :smile:
Trespaser5 said:
I cannot understand how they have got 12.4kΩ as the value of Rd, how do I find the inverse of a curve if I have no relationship for x and y ?

The inverse of the slope is simply 1/slope.

ie if you imagine the graph is a road, then where it goes through the point (15,10),

you'd see a road sign saying "Gradient 1:12.4" :wink:
 
Hi,
Thanks so much for the reply and...I think I may understand but I just want to check. I drew a tangent on the curve at that point, the took the difference in y (Ids) and divided it by x (Vds) and divided it into 1. It gave me a close answer, it also gave me a close answer to some other examples. Does this sound right to you ?
 
yup! :biggrin:

(but, as you've probably noticed, it's not very accurate on such a small graph! :wink:)
 

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