# Lesson 10 Flashcards — Power Converters

Click a question to reveal the answer.

<div data-flashcards data-deck="b1l10"></div>

<details>
<summary><strong>1. Why does a modern platform need power converters at all?</strong></summary>
<div class="card-answer"><p>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.</p></div>
</details>

<details>
<summary><strong>2. Name the four types of conversion.</strong></summary>
<div class="card-answer"><p><strong>AC→DC</strong> rectification, <strong>DC→DC</strong> level change, <strong>DC→AC</strong> inversion, and <strong>AC→AC</strong> (changing voltage or frequency).</p></div>
</details>

<details>
<summary><strong>3. What does a rectifier do?</strong></summary>
<div class="card-answer"><p>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 <strong>ripple</strong>.</p></div>
</details>

<details>
<summary><strong>4. What is a buck converter? A boost converter?</strong></summary>
<div class="card-answer"><p>Both are DC-DC converters that switch the input rapidly and filter the result. A <strong>buck</strong> steps the voltage down; a <strong>boost</strong> steps it up. Typical efficiencies are 85–95%.</p></div>
</details>

<details>
<summary><strong>5. What does an inverter do?</strong></summary>
<div class="card-answer"><p>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.</p></div>
</details>

<details>
<summary><strong>6. How does a converter enter an efficiency calculation?</strong></summary>
<div class="card-answer"><p>As just another stage: <span class="math notranslate nohighlight">\(P_\text{in} = P_\text{out}/\eta_\text{conv}\)</span>, and cascaded stages multiply.</p></div>
</details>

<details>
<summary><strong>7. When does power factor enter a converter chain calculation?</strong></summary>
<div class="card-answer"><p>Only on <strong>AC segments</strong>. On a DC segment <span class="math notranslate nohighlight">\(\text{pf} = \cos\phi = 1\)</span> and the term disappears.</p></div>
</details>

<details>
<summary><strong>8. Give the method for analyzing a converter chain.</strong></summary>
<div class="card-answer"><p>Find the real power at the converter's <em>output</em>, divide by <span class="math notranslate nohighlight">\(\eta\)</span> to get its input power, then turn that into a current at the same point. On an AC segment that is <strong>two steps</strong>, the ones from Lesson 7: <span class="math notranslate nohighlight">\(S = P/\text{pf}\)</span> first, then <span class="math notranslate nohighlight">\(I_\text{rms} = S/V_\text{rms}\)</span> — never a one-shot <span class="math notranslate nohighlight">\(I = P/(V\,\text{pf})\)</span>. On a DC segment <span class="math notranslate nohighlight">\(\text{pf} = 1\)</span>, so <span class="math notranslate nohighlight">\(I = P/V\)</span>. Repeat upstream to the source.</p></div>
</details>

<details>
<summary><strong>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?</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(P_\text{in} = 600/0.90 = 666.7\ \text{W}\)</span>. Then <span class="math notranslate nohighlight">\(S = P_\text{in}/\text{pf} = 666.7/0.95 = 701.8\ \text{VA}\)</span>, so <span class="math notranslate nohighlight">\(I = S/V = 701.8/115 = 6.10\ \text{A}\)</span>. The 66.7 W difference is heat inside the converter.</p></div>
</details>