Use efficiency to compare real input watts and conversion heat; use total power factor to compare input VA and RMS current. For a steady single-phase load, RMS current equals DC output watts divided by RMS input voltage, efficiency and total PF. Compare candidates at the same input, load, temperature and cooling conditions.
Table of Contents
Separate efficiency from power factor
Efficiency is output power divided by real input power. If a supply delivers 240 W at 90% efficiency, it needs approximately 266.7 W at its AC input. The difference, about 26.7 W, is the conversion loss within the defined measurement boundary.
Power factor, or PF, is real input power divided by apparent input power. Apparent power is RMS voltage multiplied by RMS current, expressed in VA. The referenced technical article on watts and VA explains these definitions and why separate voltage and current readings do not by themselves establish real AC watts.
Keep three columns in the procurement comparison: output watts, input real watts, and input VA. A statement such as “95% efficient with PFC” is incomplete until the test conditions and actual PF are known. Neither number can be substituted for the other.
Calculate single-phase input current, VA, and losses
For a steady single-phase operating point, use RMS input voltage and decimal values for efficiency η and total PF (90% efficiency means η = 0.90):
Real input power · W
Pin = Pout ÷ η
Apparent input power · VA
Sin = Pin ÷ PF = Pout ÷ (η × PF)
RMS input current · A
Iin,rms = Pout ÷ (Vin,rms × η × PF)
Conversion loss · W
Ploss = Pin − Pout
These relationships follow the power definitions. The engineering estimate becomes approximate when η and PF are assumed, rounded, or taken from typical data. the external power-supply terminology guide likewise identifies output power, input voltage, PF, and efficiency as inputs to AC current.
Use the total PF for the actual waveform. A switching supply can draw nonlinear current, so a displacement-angle value alone may be insufficient. The referenced power-supply measurement note describes input waveform distortion and treats RMS current, true power, apparent power, harmonics, and PF as separate measurements.
Compare 240 W at PF 0.60 and PF 0.95
All three cases below are hypothetical, with 230 V RMS input and 240 W DC output. Cases A and B hold efficiency at 90% and change only total PF. Case C keeps Case B’s PF and raises assumed efficiency to 92% to separate the two effects.
| Parameter | Case A (lower PF) | Case B (higher PF) | Case C (higher efficiency) |
|---|---|---|---|
| DC output power | 240.0 W | 240.0 W | 240.0 W |
| Efficiency, assumed | 90% | 90% | 92% |
| Input voltage | 230 V RMS | 230 V RMS | 230 V RMS |
| Total power factor, assumed | 0.60 | 0.95 | 0.95 |
| Real input power | 266.7 W | 266.7 W | 260.9 W |
| Apparent input power | 444.4 VA | 280.7 VA | 274.6 VA |
| Estimated RMS input current | 1.93 A | 1.22 A | 1.19 A |
| Calculated conversion loss | 26.7 W | 26.7 W | 20.9 W |
Case A current is 240 ÷ (230 × 0.90 × 0.60) = 1.932 A. Case B current is 240 ÷ (230 × 0.90 × 0.95) = 1.220 A. At this operating point, the higher PF reduces the calculated RMS current by about 36.8%; the real input watts and conversion loss remain equal because efficiency was held equal.
Use the calculated conversion loss for the supply’s enclosure heat contribution: 26.7 W in Cases A and B. Review the power supply derating guide alongside model-specific data when evaluating cabinet temperature and permitted output. The difference between VA and W does not represent extra heat inside the unit.
Case C isolates the efficiency change. Real input is 240 ÷ 0.92 = 260.9 W and conversion loss is 20.9 W, about 5.8 W below Case B. Output and PF stay constant, so this difference comes from the assumed efficiency improvement.
Compare supplier data at the same operating points
Before choosing a candidate, give each supplier the same operating points. At a minimum, request data for the actual output voltage and normal load, the lowest continuous AC input, and representative light-load operation. Add the project’s relevant temperature and cooling conditions. Do not compare one model’s peak efficiency at light load with another model’s full-load figure.
Request a table with one row per operating point and these fields:
- Exact model, output-voltage variant, document revision, and test date.
- AC voltage and frequency, DC output voltage and current, ambient temperature, mounting, and cooling.
- Real input W, input VA, RMS current, efficiency, and total PF at each requested point.
- Whether each value is typical, guaranteed, calculated, or measured, with stated measurement uncertainty.
- Inrush conditions and applicable harmonic-current evidence recorded separately from steady-state results.
The external efficiency-measurement paper explains why meter capability, measurement boundaries, temperature, and operating time affect the result. A small claimed difference is less useful when the measurement uncertainty is not supplied.
For a replacement project, carry the AC input and thermal requirements into the industrial power supply replacement checklist, alongside the output and control interfaces.
Check startup and upstream equipment separately
The example is a first-pass operating budget. It does not calculate turn-on inrush, fuse behavior, breaker coordination, conductor sizing, or UPS compatibility. Those decisions require the relevant equipment documentation and installation requirements. A steady 1.22 A estimate cannot establish that a particular protective device will tolerate startup; evaluate turn-on charging pulses separately using the multi-supply inrush current guide.
For several supplies, list each unit’s real power and operating state. Summing nameplate DC watts misses conversion losses; multiplying that total by one assumed PF can also hide mixed behavior. Where upstream capacity matters, confirm the aggregate RMS current, waveform, and operating sequence rather than assuming currents combine exactly like independent DC loads.
Keep harmonic-current evidence alongside the operating-point table. A PF value alone does not establish compliance for the exact product or finished equipment.
On the DC side, review 24 V branch protection and available fault current against the source response and the voltage limits of loads that must keep operating.
Turn the calculation into a supplier request
Download the AC input-budget record for the three worked cases and supplier comparisons. It is a static CSV; use the formulas above to recalculate after changing inputs. Keep operating conditions and data status beside each row.
Establish the required DC output with the 24 V DIN rail power supply sizing guide before requesting the exact candidate’s operating-point data. A family label or a maximum efficiency headline cannot establish the numbers needed for the final input budget.
If energy cost is part of the business case, compare real input energy across the actual duty cycle. The higher-PF case alone does not demonstrate lower kWh consumption. Any billing benefit depends on the site’s tariff and measured system behavior; upstream distribution losses also require their own evaluation.
Efficiency and power-factor FAQ
Can I use the full-load PF at light load?
Only if the exact product’s data supports it. Request the operating point that represents the machine’s standby or partial-load state.
Can these formulas be used unchanged for three-phase input?
No. The calculation here is for single-phase input. Use the appropriate three-phase relationship and the actual connection and balance conditions.
