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Junction field-effect transistors (JFET)
43 questions By Tony R. Kuphaldt
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Question 13 of 43
Explain what cutoff voltage (VGS(off)) is for a field-effect transistor. Research the datasheets for some of the following field-effect transistors and determine what their respective cutoff voltages are:
- J110
- J308
- J309
- J310
- MPF 102
Reveal answerI’ll let you research the definition of VGS(off) and the parameters of these specific field-effect transistors.
Follow-up question: based on your research of these datasheets, how constant is VGS(off) between different transistors? In other words, is this a parameter you can accurately predict from the datasheet before purchasing a transistor, or does it vary significantly from transistor to transistor (of the same part number)?
Notes:Discuss with your students the significance of VGS(off), especially its stability (or instability, as the case may be) between transistors. How does this impact the design of FET circuits?
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Question 14 of 43
The equation solving for drain current through a JFET is as follows:
$$I_D = I_{DSS}(1-\frac{V_{GS}}{V_{GS(off)}})^2$$
Where,
ID = Drain current
IDSS = Drain current with the Gate terminal shorted to the Source terminal
VGS = Applied Gate-to-Source voltage
VGS(off) = Gate-to-Source voltage necessary to “cut off” the JFET
Algebraically manipulate this equation to solve for VGS, and explain why this new equation might be useful to us.
Reveal answer$$V_{GS} = V_{GS(off)} (1 - \sqrt{\frac{I_D}{I_{DSS}}})$$
Notes:This question is primarily an exercise in algebraic manipulation. Have your students show their work in front of the class, to show others the strategy involved to manipulate such an equation.
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Question 15 of 43
The power dissipation of a JFET may be calculated by the following formula:
$$P=V_{DS}I_D+V_{GS}I_G$$
For all practical purposes, though, this formula may be simplified and re-written as follows:
Explain why the second term of the original equation \((V_{GS}I_G)\) may be safely ignored for a junction field-effect transistor.
Reveal answerIG is zero for all practical purposes.
Notes:This question asks students to look beyond the equation to the device itself and think about the relative magnitudes of each variable. Many equations in electronics (and other sciences!) may be similarly simplified by recognizing the relative magnitudes of variables and eliminating those whose overall effect on the equation’s result will be negligible. Of course, what constitutes “negligible” will vary from context to context.
