#23 JFET self-bias — solve both roots, verify the physical one

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

1. 1. The goal: bias without a second supply

   Goal: I_D and terminal voltages before adding a signal

   One supply; source resistor creates the bias

   Narration transcript

   Imagine preparing this transistor to amplify a small signal. Before adding that signal, we need to know where the device rests: its current, and the voltages at its three terminals. Today we will find that operating point using one supply. Keep the whole circuit in view. The important new element is the resistor underneath the source, which will create the negative gate-source bias for us.

2. 2. The goal: bias without a second supply

   $\displaystyle V_{\mathrm{DD}}=20 V \cdot R_{D}=3.3 k\Omega \cdot R_{S}=1 k\Omega$

   $\displaystyle R_{G}=1 M\Omega \cdot I_{\mathrm{DSS}}=8 \left[\mathrm{mA}\right] \cdot V_{P}=-6 V$

   No separate negative gate supply

   Narration transcript

   Read the original circuit values before calculating. The supply is twenty volts. The drain resistor is three point three kilohms, and the source resistor is one kilohm. A one-megohm resistor connects the gate to ground. The device parameters are eight milliamps for I D S S and negative six volts for V P. Notice that the separate negative gate supply from fixed bias has disappeared.

3. 3. Grounded gate, positive source, negative bias

   $\displaystyle I_{G} \approx 0 \to I_{G} R_{G} \approx 0$

   Gate resistor to ground → V_G ≈ 0 V

   Source is not grounded: V_GS need not be zero

   Narration transcript

   Start with the gate branch. We neglect the tiny reverse gate leakage in this ideal DC analysis. The gate current is approximately zero, so the voltage drop across the gate resistor is approximately zero as well. The bottom of that resistor is grounded. Therefore the gate potential is zero volts. This does not yet tell us the gate-source voltage, because the source is no longer connected directly to ground.

4. 4. Grounded gate, positive source, negative bias

   $\displaystyle I_{S} \approx I_{D}$

   $\displaystyle V_{S} = I_{D} R_{S} > 0$

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

   Grounded gate, raised source → negative V_GS

   Narration transcript

   With negligible gate current, essentially the same current flows through the drain and source resistors. Current through the source resistor creates a drop from the source down to ground, so the source becomes positive relative to ground. Now subtract source potential from gate potential. Zero minus I D times R S gives negative drain current times source resistance. A grounded gate can therefore have a negative gate-source voltage. The reference terminal matters.

5. 5. Grounded gate, positive source, negative bias

   I_D ↑ → V_S ↑ → V_GS more negative

   Current tends to fall: negative feedback

   Improved stability ≠ perfect constant current

   Narration transcript

   This connection also explains the name self-bias. Suppose the current tries to increase. The source-resistor drop increases, the source rises, and the gate-source voltage becomes more negative. In the saturation model, that more-negative control voltage tends to reduce the current, opposing the original change. This is negative feedback. It improves stability, but it does not eliminate device variation or make a perfectly constant current source.

6. 6. Turn the feedback into one equation

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

   V_GS = −I_D R_S; both relations must hold

   Assume saturation now; verify it after solving

   Narration transcript

   We now have two descriptions of the same operating point. Shockley's saturation equation links drain current to gate-source voltage. The source resistor links that same voltage back to drain current. We must satisfy both relations together. For now, assume saturation so that we can form a candidate solution. After finding the current and drain voltage, we will return to this assumption and check it explicitly.

7. 7. Turn the feedback into one equation

   x = I_D/(1 [mA]): dimensionless; I_D=x [mA]

   $\displaystyle R_{S}=1 k\Omega \to V_{S}=x V$

   V_GS=−x V; units remain explicit

   Narration transcript

   Make the units explicit before expanding the equation. Define x equals drain current divided by one milliamp. The variable x is a dimensionless numerical value, so the current is x milliamps. Multiplying by the one-kilohm source resistor gives x volts at the source. Milli and kilo cancel. Therefore gate-source voltage is negative x volts. We are not dropping the units; we are choosing a clearly defined scaled variable.

8. 8. Turn the feedback into one equation

   $\displaystyle x = 8 \left[1 -\frac{-x}{-6}\right]^{2}$

   $\displaystyle x = 8 \left(1 -\frac{ x}{6}\right)^{2}$

   The ratio is positive; the outer subtraction stays

   Narration transcript

   Substitute the device parameters and the source-resistor relation into Shockley's equation. Divide the current equation by one milliamp. We obtain x equals eight times the square of one minus negative x divided by negative six. The two negative signs inside that ratio cancel. What remains is one minus x divided by six, all squared, and multiplied by eight. Keeping these signs visible prevents a common error in self-bias calculations.

9. 9. Turn the feedback into one equation

   $\displaystyle \left(1 -\frac{ x}{6}\right)^{2} = 1 -\frac{ 2x}{6 }+\frac{ x^{2}}{36}$

   $\displaystyle = 1 -\frac{ x}{3 }+\frac{ x^{2}}{36}$

   $\displaystyle x = 8 - \left(\frac{8}{3}\right)x + \left(\frac{2}{9}\right)x^{2}$

   Do not omit the cross term

   Narration transcript

   Expand the square carefully. One minus x over six, squared, is one minus twice x over six plus x squared over thirty-six. Simplify the middle term to get one minus x over three plus x squared over thirty-six. Now multiply every term by eight. The right side becomes eight minus eight thirds times x, plus two ninths times x squared. The middle term is essential; squaring two terms separately would give the wrong equation.

10. 10. Turn the feedback into one equation

    $\displaystyle 9x = 72 - 24x + 2x^{2}$

    $\displaystyle -24x - 9x = -33x$

    $\displaystyle 2x^{2} - 33x + 72 = 0$

    Narration transcript

    Clear the denominators by multiplying the entire equation by nine. The left side becomes nine x. On the right we have seventy-two minus twenty-four x plus two x squared. Subtract nine x from both sides. Negative twenty-four x minus nine x becomes negative thirty-three x. Rearrange into standard quadratic form: two x squared minus thirty-three x plus seventy-two equals zero. Each operation must affect both sides consistently.

11. 11. Solve both roots; test their meaning

    $\displaystyle a=2 \cdot b=-33 \cdot c=72$

    $\displaystyle \Delta = 1089 - 576 = 513$

    $\displaystyle x =\frac{33 \pm \surd 513}{4}$

    Narration transcript

    Apply the quadratic formula with a equals two, b equals negative thirty-three, and c equals seventy-two. The discriminant is b squared minus four a c. Thirty-three squared is one thousand eighty-nine; four times two times seventy-two is five hundred seventy-six. Their difference is five hundred thirteen. Negative b is positive thirty-three, and two a is four. So x equals thirty-three plus or minus the square root of five hundred thirteen, divided by four.

12. 12. Solve both roots; test their meaning

    x₁ ≈ 2.588; x₂ ≈ 13.912 → currents in [mA]

    Keep both algebraic roots; use unrounded values

    Physical domain decides, not “pick the smaller root”

    Narration transcript

    Evaluate both signs. Using the minus sign gives approximately two point five eight eight, while using the plus sign gives approximately thirteen point nine one two. Since x was defined relative to one milliamp, these correspond to currents in milliamps. Both satisfy the expanded quadratic algebraically. We keep the unrounded values in the remaining calculations. Now the physical gate-voltage range, not a habit of choosing the smaller number, must decide which candidate is usable.

13. 13. Solve both roots; test their meaning

    Required:

    $\displaystyle -6 V \le V_{\mathrm{GS}} \le 0$

    Large root → V_GS ≈ −13.912 V: reject

    Small root → V_GS ≈ −2.588 V: gate check passes

    Narration transcript

    For this n-channel saturation transfer model, gate-source voltage must lie between negative six volts and zero. The larger root would make gate-source voltage about negative thirteen point nine one two volts, outside the allowed interval. Squaring the expression extended the algebra beyond the valid device domain. Reject that root. The smaller root gives approximately negative two point five eight eight volts, which passes the gate check. We still need to verify the drain voltage.

14. 14. Recover every node voltage

    $\displaystyle V_{S} = I_{D} R_{S} \approx 2.588 V$

    $\displaystyle V_{G}=0 \to V_{\mathrm{GS}} \approx -2.588 V$

    Source rises; the gate does not fall below ground

    Narration transcript

    Recover the source voltage first. Multiply the accepted candidate current by the one-kilohm source resistor. The source voltage is approximately two point five eight eight volts. The gate stays at zero volts, so zero minus that positive source voltage gives negative two point five eight eight volts from gate to source. This is the central picture of self-bias: the gate does not move below ground; the source rises above it.

15. 15. Recover every node voltage

    $\displaystyle V_{\mathrm{RD}} = I_{D} \times 3.3 k\Omega \approx 8.539 V$

    $\displaystyle V_{D} = 20 - V_{\mathrm{RD}} \approx 11.461 V$

    V_D is relative to ground; it is not V_DS

    Narration transcript

    Next calculate the drain-resistor drop. Multiply the unrounded current in milliamps by three point three kilohms. The drop is approximately eight point five three nine volts. Starting at the twenty-volt supply, subtract that drop to reach the drain node. The drain potential is approximately eleven point four six one volts. This is a voltage relative to ground. It is not yet the drain-source voltage, because our source is above ground.

16. 16. Recover every node voltage

    $\displaystyle V_{\mathrm{DS}} = V_{D} - V_{S} \approx 8.873 V$

    $\displaystyle V_{\mathrm{DS}} = V_{\mathrm{DD}} - I_{D} \left(R_{D} + R_{S}\right)$

    8.872 V in the original: early-rounding difference

    Narration transcript

    Drain-source voltage means drain potential minus source potential. Subtract the source rise from the drain potential using the unrounded values. The result is approximately eight point eight seven three volts. Equivalently, subtract the current times the sum of both resistors from the supply. The original notes rounded the current early and obtained eight point eight seven two volts. That tiny difference is rounding, not a different circuit or a different operating principle.

17. 17. Verify the region and the voltage sum

    Saturation requires V_DS ≥ V_GS − V_P

    $\displaystyle V_{\mathrm{GS}} - V_{P} \approx -2.588 - \left(-6\right) = 3.412 V$

    8.873 V > 3.412 V → saturation check passes

    Narration transcript

    Now check the assumption that allowed us to use Shockley's equation. Saturation requires drain-source voltage to be at least gate-source voltage minus V P. Here the threshold is negative two point five eight eight minus negative six, or approximately three point four one two volts. Our drain-source voltage, eight point eight seven three volts, is greater than the required threshold. Unlike the previous lesson's original fixed-bias circuit, this candidate passes the saturation check.

18. 18. Verify the region and the voltage sum

    

    $\displaystyle 8.539 + 8.873 + 2.588 \approx 20 V$

    Shockley substitution returns the same I_D

    Ideal model: adequate ratings assumed; real parameters vary

    Narration transcript

    Perform two final consistency checks. Along the drain-source loop, the drain-resistor drop, transistor voltage and source-resistor drop add back to twenty volts. Then substitute the unrounded current-derived gate voltage into Shockley's equation; it returns the same current. These checks support the ideal operating point. They do not certify an arbitrary physical part: we still assume adequate breakdown and power ratings, and real J FET parameters can vary significantly between devices.

19. 19. A self-consistent operating point

    Blue: Shockley curve · Purple: source-resistor relation

    Intersection satisfies both; drain-region check is separate

    $\displaystyle Q \approx \left(V_{\mathrm{DS}}=8.873 V, I_{D}=2.588 \left[\mathrm{mA}\right]\right)$

    Narration transcript

    The graph gives a useful final view of the same calculation. The blue curve is the saturation transfer relation. The purple line comes from the grounded gate and source resistor. Their intersection selects the current and gate-source voltage that satisfy both equations. The drain circuit then passes the separate saturation test. Our operating point is about two point five eight eight milliamps and eight point eight seven three volts from drain to source.

20. 20. A self-consistent operating point

    Reference → bias relation → device equation

    All roots → physical domain → voltages → region check

    Next: common gate, with a smaller saturation margin

    Narration transcript

    Remember the method, not just the final numbers. Establish the gate reference, express source voltage in terms of current, and combine the circuit relation with the device equation. Solve all mathematical candidates, then reject those outside the physical domain. Recover the remaining voltages and verify the operating region. In the next lesson, a common-gate DC circuit will use the same reasoning, but its available saturation margin will be much smaller.

Source video: Electronics Basics #23 | JFET Self-Bias — Step-by-Step DC Analysis (11:18)