LDO Dropout Calculator
Design linear voltage regulators by calculating dropout voltage, power dissipation, thermal effects, and heat sink requirements for stable power supplies.
Linear vs. Switching Regulators
Linear (LDO): Low cost, simple, low noise, low output ripple. Efficiency = V_out/V_in (e.g., 5V from 12V = 42%). Best for <500 mA loads with small voltage drops.
Switching (Buck): High efficiency (85–95%), compact, suitable for high-current applications. More complex, higher cost, switching noise. Best for >1 A, large voltage drops, or tight thermal constraints.
Common LDO Family Performance
| Family | Max Current | Dropout (typ) | θJA (typ) | Cost |
|---|---|---|---|---|
| LM7805 (legacy) | 1.5 A | 2 V | 50 °C/W | $ |
| AMS1117 (standard) | 1 A | 1.2 V | 60 °C/W | $ |
| LP2950 (LDO) | 100 mA | 380 mV | 80 °C/W | $ |
| AP7930 (low dropout) | 3 A | 200 mV | 30 °C/W | $$ |
Design Tips for Thermal Management
- Keep power dissipation <2 W for compact designs without a heat sink
- Use a larger PCB copper area (>500 mm²) to spread heat
- Mount the regulator away from other heat sources (ICs, transformers)
- Add thermal vias under the package for improved heat sinking
- Use a heatsink compound (thermal paste) between the regulator and heat sink for thermal contact
Frequently Asked Questions
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Dropout voltage is the minimum voltage difference (input minus output) required for a linear regulator to maintain regulation and deliver the rated output current. Below this voltage, the regulator cannot maintain the output voltage and enters dropout mode.
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If input voltage approaches output voltage too closely, the regulator cannot operate reliably. For example, a 5V-to-3.3V LDO with 200 mV dropout requires at least 3.5V input. If input drops to 3.4V, regulation fails.
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Power dissipation in a linear regulator is P = (V_in − V_out) × I_out. For example, a 12V input, 5V output at 500 mA load dissipates (12−5) × 0.5 = 3.5 W. This power becomes heat and requires thermal management.
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Efficiency = V_out ÷ V_in. A 5V output from 12V input has 5÷12 ≈ 42% efficiency. The remaining 58% is wasted as heat. Linear regulators are less efficient than buck converters but offer simplicity and low noise.
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Use LDOs for low-current applications (<500 mA), where low noise, low cost, and simplicity are priorities. Use buck converters for high-current, high-voltage-drop applications where efficiency matters more than noise.
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Thermal resistance (θJA) is measured in °C/W and describes how much the die temperature rises per watt dissipated. A regulator with θJA = 50 °C/W dissipating 1 W will have a 50 °C temperature rise above ambient.
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T_junction = T_ambient + (P_dissipated × θJA). For example, if ambient is 25 °C, dissipation is 2 W, and θJA is 50 °C/W, then T_junction = 25 + (2 × 50) = 125 °C. Most regulators are rated to 150 °C maximum.
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Most regulators have thermal shutdown protection at 140–160 °C. If junction temperature exceeds this limit, the regulator shuts down, stopping current flow. This protects the device but disrupts circuit operation. Use thermal management (heat sink, copper area) to prevent shutdown.
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Lower input voltage (smaller voltage drop), reduce output current, or switch to a switching regulator (buck converter). For fixed loads, smaller dropout voltage helps: a low-dropout (LDO) regulator typically has 200–300 mV dropout vs. 1–2 V for older designs.
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Calculate required heat sink thermal resistance: θHS = (T_max − T_ambient) ÷ P_dissipated − θJC − θCB, where θJC and θCB are die-to-case and case-to-board resistances (from datasheet). Standard aluminum heat sinks range from 5 to 50 °C/W.
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Input and output capacitors improve transient response and stability. Input capacitors (10–100 µF) reduce input impedance, helping during load steps. Output capacitors (1–100 µF) provide charge to the load during transients. Both reduce effective dropout voltage seen by the load.
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Quiescent current (I_q) is the regulator's own supply current when no load is connected. Modern low-power LDOs have I_q < 100 µA. High-current applications should account for this: P_q = V_in × I_q is wasted power, independent of load.
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Yes, but each stage wastes power as heat. Total voltage drop is the sum of individual dropout voltages plus load current times series resistances. For high voltage reduction, a buck converter followed by an LDO is more efficient.
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As temperature increases, regulator accuracy, ripple, and thermal margin degrade. Temperature coefficient (ppm/°C) quantifies this drift. Always derate the regulator's current rating at high temperatures (check datasheet derating curves).