The constant-voltage-drop (CVD) model approximates the Diode by saying: when it is on, the voltage across it is a fixed value V; when it is off, no current flows (open circuit). It is by far the most widely used diode model in this course because it captures the one fact that matters for hand analysis — a conducting silicon diode drops about 0.7 V — without dragging you into transcendental algebra.
Why 0.7 V is a good constant
The Diode equation is exponential, so every decade of current costs only about 60 mV of extra . Across the whole practical current range a small-signal silicon diode sits between roughly 0.6 V and 0.8 V. The CVD model just pins that to a single representative number, 0.7 V, and treats it as a fixed source. So a conducting diode becomes a 0.7 V battery (with the + terminal at the anode), and a non-conducting diode becomes an open circuit. Exactly which value you use is a convention — 0.7 V is standard; some texts use 0.6 V or 0.65 V.
The diode becomes a 0.7 V battery when on, an open circuit when off.
As with the Ideal diode model, you must assume a state, solve, and check it. Assuming “on”, a valid result has (current really flowing anode→cathode). Assuming “off”, a valid result has the diode reverse biased ().
Worked example
A 5 V supply, a 1 kΩ resistor, and a diode in series. Assume the diode is forward biased, so the CVD model replaces it with a 0.7 V drop. Apply Kirchhoff’s voltage law around the loop:
Solve for the diode current:
That is the entire analysis. Two relations — one KVL equation and the model’s V — give the Operating point directly, no iteration. Check the assumption: mA is positive, so the diode really is forward biased and the assumption holds.
Worked example: V, kΩ, V ⇒ mA via KVL.
Compare this with the Exponential diode model, which would give a slightly different (because the true at 4.3 mA is not exactly 0.7 V) at the cost of Iterative diode analysis. For most circuits the CVD answer is close enough that the extra work is not worth it.