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How to design a solid state power amplifier in a few (dozen) easy steps
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<blockquote data-quote="ingenieus" data-source="post: 359646" data-attributes="member: 13716"><p>Now for some math. If we assume that the voltage on the base of Q1 is zero (don't want any nasty DC offset) we can apply Kirchhoff's voltage law (KVL):</p><p></p><p>VCC - IE.RE -VBE1 = 0;</p><p></p><p>As an educated guess, we use 2mA as IE. This current will be split 50/50 between Q1 and Q2 giving 1mA flowing through each. For the base-emitter voltage, we look at the data sheet</p><p></p><p><a href="http://s1300.photobucket.com/user/ingenieus2/media/MPSA_voltages_zps3fbd7b54.jpg.html" target="_blank"><img src="http://i1300.photobucket.com/albums/ag93/ingenieus2/MPSA_voltages_zps3fbd7b54.jpg" alt="" class="fr-fic fr-dii fr-draggable " style="" /></a></p><p></p><p>At 1mA, VBE is about 0.6V. Therefore</p><p></p><p>40 - 0.002*RE - 0.6 = 0</p><p>RE = 39.4/0.002 = 19 700 ohm</p><p></p><p>We use the closest standard 5% resistor value, which is 18k.</p></blockquote><p></p>
[QUOTE="ingenieus, post: 359646, member: 13716"] Now for some math. If we assume that the voltage on the base of Q1 is zero (don't want any nasty DC offset) we can apply Kirchhoff's voltage law (KVL): VCC - IE.RE -VBE1 = 0; As an educated guess, we use 2mA as IE. This current will be split 50/50 between Q1 and Q2 giving 1mA flowing through each. For the base-emitter voltage, we look at the data sheet [URL=http://s1300.photobucket.com/user/ingenieus2/media/MPSA_voltages_zps3fbd7b54.jpg.html][IMG]http://i1300.photobucket.com/albums/ag93/ingenieus2/MPSA_voltages_zps3fbd7b54.jpg[/IMG][/URL] At 1mA, VBE is about 0.6V. Therefore 40 - 0.002*RE - 0.6 = 0 RE = 39.4/0.002 = 19 700 ohm We use the closest standard 5% resistor value, which is 18k. [/QUOTE]
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DIY & Tutorials
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How to design a solid state power amplifier in a few (dozen) easy steps
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