Electronics Basics #32 | BJT + JFET Bias: Follow the Emitter Current

Instructor: Dr. Süleyman Burak ÇELİK

Find the JFET drain voltage in the original circuit with a BJT above a self-biased JFET. Here the shared current is drain current = BJT emitter current, not collector current. Begin with the whole circuit, solve the JFET equation, select the physical root, then use the BJT base loop. Original lesson 47: 16 V supply, RC 3.6 kΩ, RB 470 kΩ, RS 2.4 kΩ, grounded JFET gate, beta = 80, assumed VBE = 0.7 V, IDSS = 8 mA and VP = −4 V. The three source PDF versions describe the same problem. Checked ideal-model result: ID ≈ 1.060 mA, VGS ≈ −2.544 V, IB ≈ 13.086 µA, VD ≈ 9.149 V. JFET VDS ≈ 6.605 V exceeds its 1.456 V saturation boundary; BJT VC ≈ 12.231 V is above VB ≈ 9.849 V. Both operating regions are verified. The source’s 9.15 V and 9.19 V answers differ because rounding IB early to 13 µA is amplified by the 470 kΩ base resistor. We retain guard digits and explain the approximately 41 mV difference. Idealized DC analysis, negligible gate current, fixed VBE and simplified device laws; not an exact real-device prediction. Chapters: 0:00 Locate the requested drain voltage 1:02 Let the grounded gate set the relation 2:06 Solve the current and choose the physical root 4:07 Use the emitter current to find the drain node 5:07 Verify the regions and keep enough precision 6:51 Read the operating point from both models