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DC load line and operating point of diode

Short answer

The DC load line is ID = (VDD – VD)/R. Graphically, draw a line from (0, VDD/R) to (VDD, 0) on the diode I–V plane; its intersection with the diode characteristic gives the Q-point (VDQ, IDQ). For a 5 V supply and 1 kΩ resistor, IDQ ≈ 4.25 mA, VDQ ≈ 0.75 V.

What is Dc load line? Find operating point for the diode circuit graphically using load line method.

The dc load line is the straight line on the diode's i–v plane that shows every (VD, ID) pair allowed by the external circuit (source and resistor) under dc conditions. The operating point (Q-point) is where this line crosses the diode's actual characteristic curve.

Circuit

  VDD ---/\/\/---+---|>|---+
          R       |   D     |
                  ID ->    GND

Applying KVL to the loop:

VDD = ID·R + VD   →   ID = (VDD - VD)/R

This is a straight line in the VD–ID plane (the load line) with:

  • ID-intercept (VD = 0): ID = VDD/R
  • VD-intercept (ID = 0): VD = VDD
  • slope = -1/R

Graphical procedure

  1. Draw the diode characteristic ID versus VD (given or measured).
  2. Mark the two points (0, VDD/R) and (VDD, 0), and join them with a straight line.
  3. The intersection of the line and the curve is the Q-point (VDQ, IDQ). Read VDQ and IDQ from the axes.
 ID
 VDD/R |\
       | \        diode curve
       |  \     _.-
  IDQ  |   \ _.Q
       |    \'|
       |     \|
       +------+----\---- VD
              VDQ    VDD

Example

VDD = 5 V, R = 1 kΩ, silicon diode: the intercepts are 5 mA and 5 V. The line cuts the curve at about VD ≈ 0.75 V, ID ≈ (5 - 0.75)/1 kΩ ≈ 4.25 mA.

The Q-point is both on the diode curve and on the load line, so it satisfies the diode and the circuit equations at the same time.

✓ Answer checked by Claude Opus 5.5 (high effort)

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