#21 JFET and Shockley’s equation — calculate, then check

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. The goal: control current with voltage

   Sensor: small voltage, limited current

   Goal: voltage controls a larger current

   Control input ≠ drain-to-source current path

   Narration transcript

   Imagine a sensor that gives us a small voltage, but cannot supply much current. Our goal is to let that voltage control a larger current without drawing much current from the sensor itself. Look at the two paths on the screen: a control input at the side, and a current path from drain to source. They do different jobs.

2. 2. The goal: control current with voltage

   

   Junction field-effect transistor (JFET)

   Picture → equation → calculation → physical check

   Energy comes from the external supply

   Narration transcript

   Today we meet the junction field-effect transistor, or J FET. By the end, you will be able to find its drain current from a gate voltage, and decide whether your answer makes physical sense. We will begin with the picture, then build the equation, then calculate a simple example. The energy in the controlled current comes from the external supply, not from the control signal.

3. 3. Three terminals, a different control mechanism

   G: gate · D: drain · S: source

   N-channel: conceptual cutaway, not to scale

   Electric field changes depletion; no mechanical squeezing

   Narration transcript

   There are three terminals. G means gate, D means drain, and S means source. In the n-channel device shown here, the gate controls a conducting channel. Our cutaway is a conceptual model, not a drawing of a particular commercial part. The gate material does not mechanically squeeze the channel; an electric field changes the depleted region inside the semiconductor.

4. 4. Three terminals, a different control mechanism

   Conventional current: D → S

   Electron motion: S → D

   Blue: current · Purple: controlling gate

   Narration transcript

   The conventional drain current enters at D and leaves at S. Electrons in this n-channel device travel the other way, from source to drain. We will show conventional current with the blue arrows, so we do not switch directions halfway through the lesson. The purple region belongs to the gate, which controls the channel from the side.

5. 5. Three terminals, a different control mechanism

   

   Earlier BJT approximation:

   $\displaystyle I_{C} = \beta I_{B}$

   JFET first approximation:

   $\displaystyle I_{G} \approx 0$

   I_G ≈ 0 does not imply I_D ≈ 0

   Narration transcript

   In our earlier BJT bias calculations, we used collector current equal to beta times base current. A base current was part of the calculation. For the J FET, the first approximation is different: the gate current is approximately zero. Do not conclude that the drain current must also be zero. The gate controls a separate current path; it does not have to supply that current.

6. 6. Why almost no gate current?

   Gate-channel PN junction is reverse-biased

   Real leakage: approximately zero, not exactly zero

   JFET junction gate ≠ MOSFET oxide gate

   Narration transcript

   Why is the gate current so small? The gate forms a P N junction with the channel. In normal n-channel operation we reverse-bias that junction. There is still a small leakage current in a real device, so approximately zero is more honest than exactly zero. This is a junction gate, not the insulated oxide gate of a MOSFET.

7. 7. Why almost no gate current?

   $\displaystyle I_{S} = I_{D} + I_{G}$

   $\displaystyle I_{G} \approx 0 \to I_{S} \approx I_{D}$

   Current conservation is unchanged

   Narration transcript

   Now apply current conservation. The current leaving the source equals the drain current plus the gate current entering the device. Replace the gate current by approximately zero, and the source current is approximately equal to the drain current. That is one less unknown to worry about. This is the same conservation law you already used with a BJT; only the size of the control-terminal current has changed.

8. 8. Why almost no gate current?

   

   $\displaystyle V_{\mathrm{GS}} = V_{G} - V_{S}$

   $\displaystyle V_{\mathrm{GS}} = 0 - 2 = -2 V$

   V_GS is variable; do not substitute 0.7 V

   Narration transcript

   Next, read the voltage correctly. V G S means gate potential minus source potential: first letter minus second letter. If the gate is at zero volts and the source is at two volts, zero minus two gives negative two volts. Unlike the fixed zero-point-seven-volt approximation we used for a silicon BJT junction, the J FET gate-source voltage is not a fixed constant.

9. 9. Two landmarks and Shockley’s equation

   I_DSS: I_D at V_GS = 0, with enough V_DS for saturation

   Signed n-channel cutoff parameter: V_P = V_GS(off) < 0

   Narration transcript

   Two device parameters give us our landmarks. I D S S is the drain current at zero gate-source voltage, with enough drain-source voltage for saturation operation. It is not the current at every possible drain voltage. The other parameter is V G S off: the negative gate-source voltage that brings an n-channel device close to cutoff. In our source notes, this signed voltage is called V P.

10. 10. Two landmarks and Shockley’s equation

    $\displaystyle I_{D} = I_{\mathrm{DSS}} \left(1 -\frac{ V_{\mathrm{GS}}}{V_{P}}\right)^{2}$

    Ratio → subtract from 1 → square → multiply by I_DSS

    V/V has no unit

    Narration transcript

    Shockley's equation connects the two landmarks with a curve. Drain current equals I D S S, multiplied by the square of one minus the ratio of gate-source voltage to V P. Do not let the subscripts distract you. We are taking a voltage ratio, subtracting it from one, squaring the result, and multiplying by a current. The ratio has no unit because volts divide by volts.

11. 11. Two landmarks and Shockley’s equation

    $\displaystyle V_{\mathrm{GS}} = 0 \to \frac{ V_{\mathrm{GS}}}{V_{P} }= 0$

    $\displaystyle \left(1 - 0\right)^{2} = 1 \to I_{D} = I_{\mathrm{DSS}}$

    $\displaystyle V_{\mathrm{GS}} = V_{P} \to \left(1 - 1\right)^{2} = 0$

    Ideal cutoff:

    $\displaystyle I_{D} = 0$

    Narration transcript

    Test the endpoints before trusting a formula. At zero gate-source voltage, zero divided by V P is zero. One minus zero is one, and one squared is one. So the drain current equals I D S S. At cutoff, gate-source voltage equals V P. Their ratio is one; one minus one is zero; zero squared is zero. The ideal model then gives zero drain current.

12. 12. Work it out: every step

    Teaching example; not a device specification

    $\displaystyle I_{\mathrm{DSS}} = 8 \left[\mathrm{mA}\right] \cdot V_{P} = -4 V \cdot V_{\mathrm{GS}} = -2 V$

    V_DS = 5 V · Find I_D, then check the region

    Narration transcript

    Let us add a numerical example to make that concrete. These are teaching values, not a specification for a particular part. I D S S is eight milliamps, V P is negative four volts, and gate-source voltage is negative two volts. We also choose five volts from drain to source. Our target is the drain current, and afterward we will check the operating region.

13. 13. Work it out: every step

    $\displaystyle \frac{V_{\mathrm{GS}}}{V_{P} }=\frac{-2 V}{-4 V}= 0.5$

    Both minus signs matter; the ratio is positive

    Narration transcript

    Begin with the ratio inside the brackets. Negative two volts divided by negative four volts is positive one half, or zero point five. The two negative signs cancel. This is where a sign error can quietly make your answer much too large. Keep both minus signs visible until you have actually performed the division.

14. 14. Work it out: every step

    $\displaystyle 1 - 0.5 = 0.5$

    0.5 × 0.5

    $\displaystyle 0.5^{2} = 0.25 =\frac{ 1}{4}$

    Half the voltage range does not mean half the current

    Narration transcript

    Now do the subtraction: one minus zero point five equals zero point five. We are not finished. The brackets are squared, so multiply zero point five by zero point five. That gives zero point two five, one quarter. Halfway across the allowed voltage range does not mean half the current, because this relationship is curved, not a straight line.

15. 15. Work it out: every step

    $\displaystyle I_{D} = 8 \left[\mathrm{mA}\right] \times 0.25 = 2 \left[\mathrm{mA}\right]$

    The current unit comes from I_DSS

    $\displaystyle 2 \left[\mathrm{mA}\right] = 0.002 A$

    $\displaystyle I_{S} \approx 2 \left[\mathrm{mA}\right]$

    Narration transcript

    Finally multiply eight milliamps by one quarter. Eight times zero point two five equals two, so the drain current is two milliamps. The bracket had no unit; the milliamps came from I D S S. In amperes this is zero point zero zero two. Because gate current is negligible, source current is also approximately two milliamps.

16. 16. Work it out: every step

    

    Sanity check:

    $\displaystyle 0 \le 2 \left[\mathrm{mA}\right] \le 8 \left[\mathrm{mA}\right]$

    18 [mA] would signal a likely sign error

    Magnitude alone does not verify the operating region

    Narration transcript

    Does that size make sense? With the gate moved from zero toward cutoff, current should decrease from eight milliamps toward zero. Two milliamps lies between those limits. If you found eighteen milliamps for these inputs, stop and inspect the signs before moving on. A reasonable magnitude is a useful check, but we still have to check the region.

17. 17. Pinch-off is not the same as cutoff

    Pinch-off near drain: current continues on a plateau

    Cutoff: sufficiently negative gate voltage suppresses current

    Check the book’s voltage definition and sign convention

    Narration transcript

    Here is an important language trap. Pinch-off near the drain end marks the start of approximately constant-current operation. It does not mean the drain current has stopped. Cutoff is the different situation where a sufficiently negative gate voltage suppresses the current nearly to zero. Textbooks sometimes use the name pinch-off voltage for the cutoff parameter too, so always check which voltage and sign convention they mean.

18. 18. Pinch-off is not the same as cutoff

    

    Model: V_P ≤ V_GS ≤ 0; below breakdown

    Saturation threshold:

    $\displaystyle V_{\mathrm{DS}} \ge V_{\mathrm{GS}} - V_{P}$

    $\displaystyle V_{\mathrm{GS}} - V_{P} = -2 - \left(-4\right)$

    $\displaystyle -2 + 4 = 2 V$

    Narration transcript

    Our simple Shockley calculation describes the saturation region, below breakdown, with gate voltage between V P and zero. For this signed n-channel convention, saturation begins approximately when drain-source voltage reaches gate-source voltage minus V P. In our example that is negative two minus negative four. Subtracting a negative adds four: negative two plus four equals positive two volts.

19. 19. Pinch-off is not the same as cutoff

    5 V > 2 V: saturation check passes, below breakdown assumed

    Operating point:

    $\displaystyle \left(V_{\mathrm{DS}}, I_{D}\right) = \left(5 V, 2 \left[\mathrm{mA}\right]\right)$

    Small V_DS: ohmic region; saturation formula is not valid

    Narration transcript

    We chose five volts across drain and source. Five is greater than two, so the ideal saturation check passes, assuming five volts is safely below the actual device's breakdown rating. Look at the graph: the point is on the plateau at two milliamps. At a very small drain voltage, the device would be in the ohmic region and the same saturation-current formula would not give its actual drain current.

20. 20. The calculator does not know the device

    Bias relation + squared Shockley equation → two algebraic roots

    A calculator does not check device physics

    Narration transcript

    Our original notes finish with a warning about quadratic equations. In later bias circuits, gate-source voltage can depend on the unknown drain current. Substituting that circuit relation into the squared Shockley equation can produce a quadratic, with two mathematical roots. A calculator can find both, but it cannot tell you which root describes a working transistor.

21. 21. The calculator does not know the device

    x = (−b ± √(b² − 4ac))/(2a)

    For each root: gate voltage → allowed interval → drain region

    Positive or smaller is not a physical selection rule

    Narration transcript

    The general roots are negative b, plus or minus the square root of b squared minus four a c, all divided by two a. We do not need to solve a full bias circuit today. The habit we need now is this: for each candidate current, calculate its corresponding gate voltage, check the permitted gate-voltage interval, and then check the drain-voltage region. Do not select a root merely because it is positive, or because it is the smaller one.

22. 22. The calculator does not know the device

    Try V_GS = −6 V: outside the V_P = −4 V cutoff bound

    Bare algebra:

    $\displaystyle \frac{-6}{-4}=1.5; \left(1-1.5\right)^{2}=0.25$

    Algebra predicts 2 [mA] again — not a valid operating result

    Reject extrapolation: the device is near cutoff

    Narration transcript

    For a quick demonstration of the danger, take our same parameters but try a gate voltage of negative six volts. That is beyond the negative-four-volt cutoff. The bare algebra gives negative six over negative four equals one point five; one minus one point five equals negative zero point five; squaring gives one quarter. It then predicts two milliamps again. The algebra ran, but its operating assumptions failed. The physical device is near cutoff, not back on the same two-milliamp point.

23. 23. Your JFET analysis checklist

    Goal answered: voltage controls current with I_G ≈ 0

    Know G/D/S; I_S ≈ I_D; V_GS is not fixed

    8 [mA], −4 V, −2 V → I_D = 2 [mA] in saturation

    Narration transcript

    Let us return to our goal. A gate-source voltage controls the channel current while drawing very little steady gate current. We named gate, drain and source; used gate current approximately zero to get source current approximately equal to drain current; and replaced the fixed-junction-voltage habit with a variable gate-source voltage. For eight milliamps and negative four volts as device parameters, a negative-two-volt control gave two milliamps in saturation.

24. 24. Your JFET analysis checklist

    Signs → Shockley → physical checks

    Next: find V_GS from the fixed-bias circuit

    Narration transcript

    Keep this order: voltage signs first, Shockley's equation second, physical checks last. The number is not finished until its assumptions pass. Next time we put the J FET into a fixed-bias circuit and find the gate voltage from the actual connections, instead of receiving it as a given. That is where this small set of rules becomes a complete circuit-analysis method.

Source video: Electronics Basics #21 | JFET & Shockley’s Equation — Step by Step (11:25)