How it works
An LDO dissipates the voltage difference between input and output as heat. Its ground (quiescent) current Iq also flows from the input, so it adds to the loss.
PD = (Vin − Vout) × Iout + Vin × IqTj = Ta + PD × θJAη = (Vout × Iout) / (Vin × (Iout + Iq))Dropout headroom = Vin − Vout
Worked example
With Vin 5 V, Vout 3.3 V and a 500 mA load, the 1.7 V difference dissipates 0.85 W. The loss from Iq of 50 µA is 0.25 mW, which is negligible. With θJA of 60 °C/W, the temperature rise is about 51 °C, so with Ta 25 °C the junction is at about 76 °C. The efficiency is about 66 %.
Design tips
- The loss rises linearly with the input-output difference. If the input is high and the output low, consider a buck converter ahead of the LDO to reduce the voltage difference.
- θJA depends strongly on the PCB copper area, thermal vias and package type. Check that the datasheet layout conditions match your board.
- At light loads with long idle periods, Iq dominates the efficiency. Choose a low-Iq part, or use a structure that turns the LDO off during standby.
- Confirm that Tj at the maximum ambient temperature is well below the datasheet maximum. A margin of 10 to 20 °C is common.
Limitations
- Only the steady-state average loss is calculated. With short load pulses, the instantaneous junction temperature can be higher because of thermal time constants.
- θJA is a standard test-board value, so it can differ on a real board.
- Vin must be higher than Vout. Near dropout the output falls, and the efficiency calculation may not match reality.
Frequently asked questions
Where do I get θJA for the junction temperature calculation?
Find the JEDEC board-condition θJA in the thermal characteristics table of the datasheet. If your board has different copper area or thermal vias, the value changes, so it is best to calibrate it by measurement where possible.
When should I use a switching regulator instead of an LDO?
When the input-output difference is large and the current is high, the LDO loss becomes heat, so a switching converter is better. For noise-sensitive analog or RF supplies, an LDO can be preferable. Compare the efficiency of both approaches for your conditions.
How much does the ground current Iq affect efficiency?
With load currents in mA, the effect of a µA-level Iq is negligible. But when the load drops to µA in standby, Iq dominates the efficiency. Consider a low-Iq part or a structure that cuts power in standby.