Lesson 10 Flashcards — Power Converters#
Click a question to reveal the answer.
1. Why does a modern platform need power converters at all?
Because it runs both kinds of power at once — avionics and electronics are DC, actuation and high-power motors are often AC, generation and distribution are AC, and batteries are DC.
2. Name the four types of conversion.
AC→DC rectification, DC→DC level change, DC→AC inversion, and AC→AC (changing voltage or frequency).
3. What does a rectifier do?
Uses diodes to pass current in one direction only, folding or blocking the negative half of the sinusoid, then a filter capacitor smooths the result. The residual wiggle is called ripple.
4. What is a buck converter? A boost converter?
Both are DC-DC converters that switch the input rapidly and filter the result. A buck steps the voltage down; a boost steps it up. Typical efficiencies are 85–95%.
5. What does an inverter do?
Switches DC back and forth to synthesize an AC waveform. Output quality ranges from a crude square wave to a clean sinusoid — solar arrays, UPS units, and aircraft emergency AC buses all use them.
6. How does a converter enter an efficiency calculation?
As just another stage: \(P_\text{in} = P_\text{out}/\eta_\text{conv}\), and cascaded stages multiply.
7. When does power factor enter a converter chain calculation?
Only on AC segments. On a DC segment \(\text{pf} = \cos\phi = 1\) and the term disappears.
8. Give the method for analyzing a converter chain.
Find the real power at the converter's output, divide by \(\eta\) to get its input power, then turn that into a current at the same point. On an AC segment that is two steps, the ones from Lesson 7: \(S = P/\text{pf}\) first, then \(I_\text{rms} = S/V_\text{rms}\) — never a one-shot \(I = P/(V\,\text{pf})\). On a DC segment \(\text{pf} = 1\), so \(I = P/V\). Repeat upstream to the source.
9. A 600 W DC load is fed by a 90% efficient converter from a 115 V AC bus at pf 0.95. What bus current flows?
\(P_\text{in} = 600/0.90 = 666.7\ \text{W}\). Then \(S = P_\text{in}/\text{pf} = 666.7/0.95 = 701.8\ \text{VA}\), so \(I = S/V = 701.8/115 = 6.10\ \text{A}\). The 66.7 W difference is heat inside the converter.