Question E7D15:
What is the purpose of a step-start circuit in a high-voltage power supply?  

Choices:  
A: To provide a dual-voltage output for reduced power applications  
B: To compensate for variations of the incoming line voltage  
C: To prevent arcing across the input power switch or relay contacts  
D: To allow the filter capacitors to charge gradually    

Correct Answer is  D: To allow the filter capacitors to charge gradually  

Purpose and operation:
- Large HV filter capacitors look like a short when first powered. Without limiting, the initial inrush current can be huge.
- A step-start (soft-start) inserts a series resistor (often in the primary) at turn-on so the capacitors charge gradually; after a short delay a relay shorts the resistor for normal operation.
- This protects rectifiers, transformers, switches, and fuses, and prevents nuisance breaker trips.

Why D is correct:
- “Allow the filter capacitors to charge gradually” is exactly what the step-start does by limiting the inrush current during the first second or so.

Why the others are incorrect:
- A: Dual-voltage output is a function of a different supply design (e.g., multiple taps or regulators), not step-start.
- B: Line-voltage compensation requires regulation (buck/boost, ferroresonant, tap switching). Step-start does not regulate; it only limits inrush at startup.
- C: While limiting inrush can reduce contact stress, the primary purpose is not preventing arcing across the input switch/relay; it’s controlling capacitor charging current.

Useful formulas and rules of thumb:
- Initial inrush current with a series resistor $R$ and uncharged capacitor:
  $$I_0 = \frac{V}{R}$$
  where $V$ is the applied voltage at that point in the circuit.
- Capacitor charging time constant:
  $$\tau = R C$$
  About $5\tau$ reaches ≈99% of final voltage during the “resistor-in” step.
- Energy stored in the filter capacitors:
  $$E = \tfrac{1}{2} C V^2$$
  Highlights why the surge can be severe at high $V$ and large $C$.
- Practical selection:
  - Choose $R$ so $I_0$ is within safe limits for the transformer/rectifiers (often <2–3× normal operating current).
  - Delay to short the resistor typically 0.5–2 s (long enough for substantial charge, short enough to limit resistor heating).
  - If placed on the primary: approximate $R \approx \dfrac{V_{\text{mains}}}{I_{\text{limit}}}$.
- Alternative: an NTC inrush limiter can provide a “soft” step-start automatically, but classic step-start uses a resistor plus a time-delayed bypass relay.