動作原理
When a rectifier charges a large DC-link capacitor directly from the mains, the only things limiting the current are the line impedance and the capacitor's ESR. The resulting surge can reach hundreds of amperes, stressing the rectifier diodes, tripping breakers and welding relay contacts. A precharge (inrush limiting) resistor in series with the DC link keeps that first current pulse within a safe value.
This calculator sizes the resistor from the worst case: the capacitor is empty and the input is switched on at the voltage peak. The peak current is then roughly the peak line voltage divided by the resistance, so the minimum resistance follows directly from the allowed peak current.
When a capacitor is charged from a fixed DC voltage through a resistor, the resistor dissipates exactly the energy that ends up stored in the capacitor, ½·C·V², regardless of the resistance. With a rectified AC input the loss is somewhat lower, but ½·C·V² is the safe upper bound to design for. That energy arrives in a short pulse, so the resistor must be rated for pulse energy (or an NTC for its energy rating), not just continuous power. The temperature rise is estimated adiabatically from the resistor mass and heat capacity.
The time constant τ = R·C sets how long precharge takes. After about 3τ the capacitor reaches 95% of its final voltage, which is a common point to close the bypass relay. Cement and wirewound resistors must be bypassed after precharge; NTC thermistors heat up and drop their resistance on their own but still benefit from a bypass in high-power designs.
計算式
- Peak line voltage
- V_pk = √2 · V_line
- DC-link voltage
- V_dc ≈ 0.9 · V_pk (single-phase), 1.35 · V_LL (three-phase)
- Minimum resistance
- R ≥ V_pk / I_max
- Energy per charge
- E = ½ · C · V_dc²
- Time constant
- τ = R · C, t₉₅ ≈ 3τ, t₉₉ ≈ 5τ
- Adiabatic temperature rise
- ΔT = E / (m · c_p)
計算例
A 230 V single-phase drive with a 2200 µF DC link and a 50 A peak-current limit needs R ≥ 325 V / 50 A ≈ 6.5 Ω. Each power-up dumps about ½ · 2200 µF · (293 V)² ≈ 94 J into the resistor, and τ ≈ 14 ms, so the bypass relay can close after roughly 45 ms.
よくある質問
Why is the energy in the resistor independent of its resistance?
For an RC charge from a fixed DC voltage, the energy lost in the resistor equals the energy stored in the capacitor. A larger resistor lowers the peak current and stretches the pulse, but the total stays ½·C·V². With rectified AC the loss is a little lower, which the interactive explorer shows.
Should I use an NTC thermistor or a fixed resistor with a relay?
NTCs are simple and cheap for low and medium power, but they need time to cool down between power cycles and still dissipate some power in operation. A fixed resistor with a bypass relay handles repeated power cycling and high power better.
What pulse rating do I need?
Choose a resistor whose single-pulse energy rating comfortably exceeds ½·C·V² at the highest DC-link voltage, including any mains overvoltage. The calculator applies a safety factor and shows the required rating.
When should the bypass relay close?
Typically after three to five time constants, when the capacitor is at 95–99% of its final voltage, so the relay closes with very little current step.
計算結果は工学的な推定値です。データシート、シミュレーション、実測で設計を確認してください。