Model regulator performance using practical electrical loss inputs. See efficiency, wasted power, and thermal impact. Download clean reports for design reviews and test documentation.
| Scenario | Mode | Vin (V) | Vout (V) | Iout (A) | Pout (W) | Total Loss (W) | Efficiency (%) |
|---|---|---|---|---|---|---|---|
| Control board rail | Linear | 12 | 5 | 1.5 | 7.50 | 10.56 | 41.53 |
| Industrial DC stage | Buck | 24 | 12 | 2.5 | 30.00 | 1.272 | 95.93 |
| Portable boost rail | Boost | 5 | 12 | 1.0 | 12.00 | 1.29 | 90.29 |
The calculator uses a practical loss-budget method for engineering estimates. It combines useful output power with dissipation sources to calculate efficiency.
Pout = Vout × IoutPq = Vin × Iq (Iq converted from mA to A)Pcopper = Iout² × R (R from mΩ to Ω)Pin = Pout + TotalLossEfficiency (%) = (Pout / Pin) × 100Thermal Rise = TotalLoss × θCAFor linear mode, dropout dissipation is included:
Pdrop = (Vin − Vout) × Iout
For switching modes, switching, magnetic, rectifier, gate-drive, and auxiliary losses are added as separate terms, making the estimate easier to tune with lab data.
Regulator efficiency controls heat, runtime, and supply margin in equipment. Teams often aim above 90 percent for switching rails, while linear rails are kept for low noise paths. A converter delivering 30 watts at 95 percent efficiency loses about 1.58 watts. At 88 percent, loss rises to 4.09 watts. That heat changes enclosure temperature, spacing, and airflow needs. This calculator shows the impact during design reviews.
The calculator uses a practical loss budget instead of one ideal equation. Users can enter quiescent current, copper resistance, switching loss, magnetic loss, rectifier loss, gate drive loss, and auxiliary loss. This matches bench observations and design worksheets. For example, 40 milliohm copper resistance at 2.5 amp load creates 0.25 watt loss. Adding switching and magnetic losses changes the thermal picture before layout, helping engineers tune assumptions.
Linear regulators suit quiet analog rails, but dropout dissipation rises with voltage drop and current. With 12 volt input, 5 volt output, and 1.5 amp current, output power is 7.5 watts. Dropout loss alone is 10.5 watts, so efficiency sits near 42 percent before parasitic terms. A buck regulator at similar load often exceeds 90 percent. The calculator highlights this tradeoff and supports architecture decisions across power domains.
Efficiency drives temperature, and temperature drives reliability. The calculator estimates thermal rise using total loss multiplied by thermal resistance theta CA. If loss is 1.3 watts and thermal resistance is 18 degrees C per watt, rise is about 23.4 degrees C. With 30 degrees C ambient, estimated case temperature reaches about 53.4 degrees C. That estimate supports derating checks, silicon margins, and enclosure validation before expensive prototype revisions.
Measured mode helps during prototype validation because it compares instrument data with modeled results. Enter direct input power and output power, then the tool returns total loss and efficiency. If input power is 32.1 watts and output power is 30 watts, efficiency is about 93.46 percent and loss is 2.1 watts. Gaps between estimated and measured results often reveal trace heating, diode recovery, or quiescent assumptions during testing.
For many switching designs, 90 to 95 percent is a strong target. Linear regulators may be much lower, especially with large voltage drops, but they remain useful for low-noise rails and small currents.
Measured mode is best for validation after testing. It calculates efficiency from direct input and output power. Use the detailed loss mode during design to understand where heat is generated.
Copper resistance causes I squared R loss, which increases rapidly with current. Even small milliohm values can add noticeable heating in high-current stages and reduce overall efficiency.
Yes. Select linear mode and enter Vin, Vout, and load current. The calculator includes dropout dissipation, quiescent current loss, and optional parasitic losses to estimate practical efficiency.
Start with datasheet values, simulation outputs, or lab measurements. Then refine inputs using oscilloscope and thermal data from your prototype so the calculated efficiency matches real operating conditions.
The thermal estimate gives case temperature rise based on total loss and thermal resistance. It is a screening value, not a full junction model, but it helps compare design options early.
Important Note: All the Calculators listed in this site are for educational purpose only and we do not guarentee the accuracy of results. Please do consult with other sources as well.