The Core Method: KVL and the Constant Voltage Drop Model

When you encounter examples of light emitting diode circuits on university exams or in practical design reviews, the underlying physics relies on the Shockley diode equation. However, solving the transcendental Shockley equation by hand is impractical for timed exams. Instead, we apply the Constant Voltage Drop (CVD) model combined with Kirchhoff’s Voltage Law (KVL).

The CVD model treats the LED as a fixed voltage source (its forward voltage, $V_F$) in series with an ideal closed switch once the threshold is reached. According to Georgia State University's HyperPhysics, this $V_F$ is dictated by the semiconductor's bandgap energy—typically 1.8V for red (AlGaAs) and 2.2V to 3.3V for green/blue (InGaN) variants. For precise component-level analysis, we reference manufacturer datasheets, such as the All About Circuits semiconductor guide, which emphasizes checking $V_F$ at your specific target forward current ($I_F$), as $V_F$ shifts slightly along the I-V curve.

Walkthrough 1: Sizing a Current-Limiting Resistor for a Standard 5mm LED

Problem Statement: Design a bias circuit for a Kingbright WP7113SRD (Super Red) 5mm LED powered by a 9V alkaline battery. The target forward current is $I_F = 20\text{ mA}$. The datasheet specifies a typical $V_F = 1.85\text{ V}$ at $20\text{ mA}$. Calculate the required series resistor ($R_S$) and select the nearest standard E12 resistor value. What is the actual current with the selected resistor?

Step-by-Step Algebraic Solution

  1. Define the KVL Loop: Starting from the negative terminal of the battery and moving clockwise: $-V_S + I_F R_S + V_F = 0$.
  2. Isolate the Resistor Term: $I_F R_S = V_S - V_F$.
  3. Solve for $R_S$: $R_S = \frac{V_S - V_F}{I_F}$.
  4. Substitute Known Values: $R_S = \frac{9\text{ V} - 1.85\text{ V}}{0.020\text{ A}}$.
  5. Calculate: $R_S = \frac{7.15\text{ V}}{0.020\text{ A}} = 357.5\ \Omega$.
  6. Select Standard Value: The nearest E12 series value above 357.5 $\Omega$ (to ensure we don't exceed the 20mA absolute maximum rating) is 390 $\Omega$.
  7. Recalculate Actual Current: $I_{actual} = \frac{7.15\text{ V}}{390\ \Omega} = 0.01833\text{ A}$, or 18.3 mA.

Sanity Check & Verification

  • Units: Volts divided by Amps yields Ohms. The order of magnitude ($\sim 10\text{V} / \sim 0.02\text{A} = 500\ \Omega$) aligns with our 357.5 $\Omega$ result.
  • The Trap: The most common exam mistake is treating the LED as a short circuit and calculating $R = 9\text{V} / 0.02\text{A} = 450\ \Omega$. This ignores the bandgap voltage drop, resulting in an under-driven, dim LED.
  • Independent Verification: On the bench, power the circuit and measure the voltage directly across the 390 $\Omega$ resistor with a multimeter. It must read exactly 7.15V. If it reads 9V, your LED is dead (open circuit). If it reads ~0V, the LED is shorted or wired backward.

Walkthrough 2: Parallel LED Strings and the Current Hogging Trap

Problem Statement: A student connects two different LEDs in parallel: a Green LED ($V_F = 2.2\text{ V}$) and a Red LED ($V_F = 1.8\text{ V}$). They share a single $330\ \Omega$ current-limiting resistor connected to a 5V DC logic rail. Calculate the current through each LED and describe the physical outcome.

Step-by-Step Algebraic Solution

  1. Identify the Node Voltage: In a parallel configuration, the voltage across both LEDs must be identical. The node voltage ($V_{node}$) will be clamped by the LED with the lowest forward voltage threshold.
  2. Determine the Clamp: The Red LED turns on at 1.8V. Therefore, $V_{node} = 1.8\text{ V}$.
  3. Analyze the Green LED: The Green LED requires 2.2V to conduct. Since it only sees 1.8V, it remains below its conduction threshold. Thus, $I_{green} \approx 0\text{ mA}$.
  4. Analyze the Red LED (KVL): The voltage drop across the shared resistor is $V_R = V_S - V_{node} = 5\text{ V} - 1.8\text{ V} = 3.2\text{ V}$.
  5. Calculate Red Current: $I_{red} = \frac{V_R}{R} = \frac{3.2\text{ V}}{330\ \Omega} = 0.00969\text{ A}$, or 9.7 mA.

Sanity Check & Verification

  • Units & Magnitude: 9.7 mA is a safe, standard operating current for a 5mm indicator LED. Total circuit current is conserved ($I_{total} = I_{red} + I_{green}$).
  • The Trap: Students frequently calculate the current for each LED independently as if they had their own resistors, or they average the $V_F$ values. In reality, parallel LEDs with mismatched $V_F$ curves result in "current hogging," where the lowest $V_F$ component steals all the current. As noted in Texas Instruments' LED driver fundamentals, parallel strings without individual ballasting resistors lead to severe current imbalance and potential thermal runaway.
  • Independent Verification: Visually inspect the board; the green LED will be completely dark. Measure the voltage across the parallel LED pair; your meter will read ~1.8V, proving the red LED is clamping the node.

Frequently Asked Questions: Examples of Light Emitting Diode Applications

What are common examples of light emitting diode types used in university lab exams?

University labs typically rely on standard 5mm through-hole T-1¾ packages because they are cheap and robust against minor wiring faults. The most common examples of light emitting diode components include the Kingbright WP7113 series (Red/Green/Yellow) and the Lite-On LTL-307 series. Exams favor these because their $V_F$ values (1.8V to 2.2V) map cleanly to standard 5V and 9V lab power supplies, keeping the arithmetic straightforward while testing core KVL concepts.

How do high-power examples of light emitting diode circuits change the thermal calculations?

When moving from 5mm indicators to high-power illumination (like the Cree XLamp XP-G3 series), the CVD model is no longer sufficient on its own. High-power examples of light emitting diode circuits require thermal derating. The $V_F$ drops by approximately -2mV/°C as the junction temperature rises. Exam problems involving these components will force you to calculate the thermal resistance ($\theta_{JA}$) and adjust the $V_F$ variable in your KVL equation based on the ambient temperature and power dissipation ($P_D = V_F \times I_F$). Failing to account for this negative temperature coefficient results in an under-estimated current and potential catastrophic thermal runaway.

Why do infrared examples of light emitting diode require different bias resistor values than visible LEDs?

Infrared examples of light emitting diode components, such as the Vishay TSAL6200 used in remote controls, are fabricated from Gallium Arsenide (GaAs) rather than Gallium Arsenide Phosphide (GaAsP). This material difference results in a much lower bandgap energy, yielding a typical $V_F$ of just 1.35V (compared to 1.8V+ for visible light). If a student blindly uses a standard 1.8V assumption for an IR LED in an exam problem, the calculated resistor will be too small, pushing the IR LED past its absolute maximum continuous forward current (usually 100mA for standard packages) and destroying the component. Always verify the specific chemistry and $V_F$ from the datasheet.