Electrical & Computer
Diode I–V Lab
A diode is the textbook nonlinear component, and this lab makes its behaviour concrete. The current follows the Shockley equation I = Is·(e^(V/nVt) − 1): essentially zero (−Is) in reverse, then climbing exponentially in forward bias — which is why a silicon diode seems to 'switch on' so abruptly around 0.6–0.7 V. Tune the saturation current Is, the ideality factor n and the temperature (which sets the thermal voltage Vt = kT/q ≈ 25.85 mV at 300 K) and watch the knee shift. Switch on the resistor load line and the operating point is solved for you — the single (Vd, I) that satisfies both the diode curve AND Kirchhoff's voltage law Vs = I·R + Vd — the graphical method every circuits course teaches for analysing a nonlinear element in a linear circuit.
25.85 mV
Thermal voltage Vt
1.00 pA
Saturation Is
574.19 mV
Op. point Vd
4.43 mA
Op. point I
The method, with your numbers
Shockley equation + load line
- 1
Find the thermal voltage
Vt = k·T / q
Vt = k·300 K / q = 25.85 mV
Vt depends only on temperature — about 25.9 mV at room temperature (300 K). Every diode equation is scaled by it.
- 2
Write the diode law (Shockley equation)
I = Is·(e^(Vd/(n·Vt)) − 1)
I = 1 pA·(e^(Vd/(1·25.85 mV)) − 1)
Every 60 mV of extra forward bias multiplies the current ×10 — that exponential is why the diode "switches on" so sharply.
- 3
Write the load line (KVL around the loop)
Vs = I·R + Vd → I = (Vs − Vd)/R
I = (5 V − Vd)/1000 Ω (crosses the axes at Vd = 5 V and I = 5 mA)
A straight line: everything the resistor allows. The circuit must sit on this line AND on the diode curve.
- 4
Intersect the two: the operating point
Is·(e^(Vd/(n·Vt)) − 1) = (Vs − Vd)/R
Vd = 0.574 V, I = 4.43 mA
This equation is transcendental — no algebraic solution exists. The sim bisects on Vd (on paper you'd guess Vd ≈ 0.7 V and iterate).
- 5
Sanity-check with the 0.7 V constant-drop model
I ≈ (Vs − 0.7)/R
I ≈ (5 − 0.7)/1000 = 4.3 mA vs exact 4.43 mA
The shortcut most exam problems use. It works because Vd barely moves once the diode is on — raise R and watch Vd stay pinned near 0.7 V.
Drag any control above — every number here recalculates. Want this method for any problem? Step Sheets →
The Shockley diode obeys I = Is·(e^(V/nVt) − 1). Forward current rises exponentially, so a silicon diode appears to switch on near 0.6–0.7 V. Overlay a resistor load line (I = (Vs − V)/R) and the operating point sits where it crosses the curve — graphical circuit analysis in one picture.
How to use this simulation
A diode is the textbook nonlinear component, and this lab makes its behaviour concrete. The current follows the Shockley equation I = Is·(e^(V/nVt) − 1): essentially zero (−Is) in reverse, then climbing exponentially in forward bias — which is why a silicon diode seems to 'switch on' so abruptly around 0.6–0.7 V. Tune the saturation current Is, the ideality factor n and the temperature (which sets the thermal voltage Vt = kT/q ≈ 25.85 mV at 300 K) and watch the knee shift. Switch on the resistor load line and the operating point is solved for you — the single (Vd, I) that satisfies both the diode curve AND Kirchhoff's voltage law Vs = I·R + Vd — the graphical method every circuits course teaches for analysing a nonlinear element in a linear circuit.
Everything runs in your browser — no sign-up, no download. Change a value and the result updates instantly, so you can build a feel for how each input shapes the outcome. It pairs with Crameleon's practice exams and step sheets when you want to go from intuition to working the problems.