Inverse curves related to JFET characteristics, help

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Discussion Overview

The discussion revolves around understanding the calculation of the small-signal drain-source resistance (rd) from the characteristics of a JFET, specifically how to derive this value from a curve relating IDS (drain-source current) to VDS (drain-source voltage) at a given operating point. The context includes coursework-related queries and practical application of theoretical concepts.

Discussion Character

  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about how to derive the value of rd (12.4 kΩ) from the curve without a clear relationship between x and y.
  • Another participant clarifies that the inverse of the slope can be found as 1/slope, suggesting a graphical interpretation of the slope at the operating point.
  • A participant describes their method of drawing a tangent at the operating point and calculating the slope using differences in IDS and VDS, seeking validation for this approach.
  • Another participant agrees with the method but notes potential inaccuracies due to the scale of the graph.

Areas of Agreement / Disagreement

Participants generally agree on the method of calculating rd using the slope of the tangent, but there is acknowledgment of potential inaccuracies in the graphical representation.

Contextual Notes

The discussion does not resolve the accuracy of the method used due to the limitations of the graph's scale and the lack of a defined relationship for x and y.

Trespaser5
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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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