82 lines
3.2 KiB
HTML
82 lines
3.2 KiB
HTML
<!doctype html>
|
|
<html lang="en">
|
|
<head>
|
|
<meta charset="utf-8">
|
|
<meta name="viewport" content="width=device-width, initial-scale=1">
|
|
<title>Power Output vs. Impedance</title>
|
|
<script>
|
|
window.MathJax = {
|
|
tex: { inlineMath: [['\\(', '\\)']], displayMath: [['\\[', '\\]']] },
|
|
svg: { fontCache: 'global' }
|
|
};
|
|
</script>
|
|
<script defer src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-svg.js"></script>
|
|
<style>
|
|
:root { color-scheme: light dark; }
|
|
body { font: 16px/1.55 system-ui, sans-serif; margin: 0; padding: 2rem; }
|
|
main { max-width: 760px; margin: auto; }
|
|
code, pre { font-family: ui-monospace, SFMono-Regular, Menlo, monospace; }
|
|
pre { padding: 1rem; border-radius: .5rem; overflow-x: auto; background: #f3f4f6; color: #111827; }
|
|
table { width: 100%; border-collapse: collapse; margin: 1rem 0; }
|
|
th, td { padding: .6rem; border: 1px solid #9ca3af; text-align: right; }
|
|
th:first-child, td:first-child { text-align: left; }
|
|
@media (prefers-color-scheme: dark) { pre { background: #1f2937; color: #f9fafb; } }
|
|
</style>
|
|
</head>
|
|
<body>
|
|
<main>
|
|
<h1>Power Output vs. Impedance</h1>
|
|
<p>For a sine wave:</p>
|
|
\[
|
|
V_{\mathrm{RMS}} = \frac{V_{\mathrm{peak}}}{\sqrt{2}},
|
|
\qquad
|
|
I_{\mathrm{RMS}} = \frac{I_{\mathrm{peak}}}{\sqrt{2}}
|
|
\]
|
|
\[
|
|
P = V_{\mathrm{RMS}} I_{\mathrm{RMS}}
|
|
= \frac{V_{\mathrm{RMS}}^2}{R}
|
|
= I_{\mathrm{RMS}}^2 R
|
|
\]
|
|
|
|
<p>An amplifier is limited by whichever it reaches first: its maximum output voltage or its maximum output current.</p>
|
|
|
|
<h2>Example: ±16 V supply rails</h2>
|
|
<p>Ideally, ±16 V rails allow 16 V peak at the output:</p>
|
|
\[
|
|
V_{\mathrm{RMS}} = \frac{16}{\sqrt{2}} = 11.32\ \mathrm{V}
|
|
\]
|
|
<p>Into an 8 Ω speaker:</p>
|
|
\[
|
|
I_{\mathrm{RMS}} = \frac{11.32}{8} = 1.414\ \mathrm{A}
|
|
\qquad
|
|
P = \frac{11.32^2}{8} = 16\ \mathrm{W}
|
|
\]
|
|
<p>So the ideal maximum is 16 W into 8 Ω. More available current alone cannot increase that figure because the amplifier is already at its voltage limit.</p>
|
|
|
|
<h2>As impedance falls</h2>
|
|
<p>If the amplifier can maintain 11.32 V RMS and has enough current capacity:</p>
|
|
<table>
|
|
<thead><tr><th>Load</th><th>Power</th><th>RMS current</th><th>Peak current</th></tr></thead>
|
|
<tbody>
|
|
<tr><td>8 Ω</td><td>16 W</td><td>1.414 A</td><td>2 A</td></tr>
|
|
<tr><td>4 Ω</td><td>32 W</td><td>2.828 A</td><td>4 A</td></tr>
|
|
<tr><td>2 Ω</td><td>64 W</td><td>5.657 A</td><td>8 A</td></tr>
|
|
</tbody>
|
|
</table>
|
|
<p>At a fixed voltage, halving load resistance doubles current and doubles power.</p>
|
|
|
|
<h2>Current-limited example</h2>
|
|
<p>If the amplifier can provide only 1.414 A RMS into 8 Ω:</p>
|
|
\[
|
|
P = I_{\mathrm{RMS}}^2 R
|
|
= (1.414)^2 \times 8
|
|
\approx 16\ \mathrm{W}
|
|
\]
|
|
<p>That is 2 A peak. Maintaining the same voltage would require 4 A peak for 32 W into 4 Ω, and 8 A peak for 64 W into 2 Ω.</p>
|
|
|
|
<h2>Real-world caveat</h2>
|
|
<p>Real amplifiers need some voltage headroom below the supply rails and their supply voltage can sag under heavy current draw. Their actual output is therefore often below these ideal figures.</p>
|
|
</main>
|
|
</body>
|
|
</html>
|