Chapter 4 · Visualization 3 of 6 · Sections 4.3 and 4.5 (Figs. 4.10 and 4.17)

The Current as an Area

Two graphical constructions of the same current. On the left the drift current is a trapezoid under the charge-balance line; on the right the whole current is the area under the Chapter 3 curve −Q′I(VCB) between VSB and VDB.

Fig. 4.10 view: −Q′I versus ψS; the shaded trapezoid between the end points is ∝ IDS1

Fig. 4.17 view: IDS = (W/L)µ ∫ (−Q′I) dVCB from VSB to VDB

What to look for

IDS1 = (W/L)µ ∫ψs0ψsL (−Q′I) dψS, IDS = (W/L)µ ∫VSBVDB (−Q′I) dVCB

  • Raise VDB: the drain point slides right on the line and the trapezoid grows, until ψsL reaches ψsa and the trapezoid becomes a triangle: saturation.
  • On the right, beyond VP the curve is practically zero, so extending the upper limit adds no area.
  • Swap VSB and VDB: the same area, opposite sign. The ideal transistor is symmetric.

Try this

  1. Set VGB = 1.5 V, VSB = 0 and raise VDB from 0 to 2 V. At which VDB does the shaded area stop growing?
  2. Compare the two current readouts. Why do they agree?
  3. Raise VSB to 0.5 V with VDB = 2 V. Which part of the area disappears, and why does the current fall?
  4. Lower VGB until the device is in weak inversion. What happens to the trapezoid, and which current component dominates?