{"id":43544,"date":"2024-11-07T11:06:39","date_gmt":"2024-11-07T16:06:39","guid":{"rendered":"https:\/\/www.tedpublications.com\/fr\/?p=43544"},"modified":"2024-11-07T11:06:39","modified_gmt":"2024-11-07T16:06:39","slug":"class-a-amplifiers","status":"publish","type":"post","link":"https:\/\/www.tedpublications.com\/fr\/en\/class-a-amplifiers\/","title":{"rendered":"Class A Amplifiers"},"content":{"rendered":"<p><strong><span style=\"color: #999999;\">Sketches :<\/span> <span style=\"color: #800000;\"><em>Normand Daigle<\/em><\/span><\/strong><\/p>\n<p style=\"text-align: left;\"><strong><span style=\"color: #800000;\">The first article in this series clarified some technical points to aid in understanding the amplification circuits we will analyze, and the second article addressed the power supply section, which is crucial for quality performance. Today, we finally dive fully into the study of different classes of power amplifier operation, beginning with the Class\u00a0<em>A<\/em>. But before moving forward, we will examine two other concepts that will be useful for understanding these circuits: the voltage divider and current<br \/>\n<\/span><\/strong><\/p>\n<p>A <em>voltage divider<\/em> is essentially a chain of two or more resistors that distribute voltage, as the name implies. In the following circuit, if the values of resistors <em>R1<\/em> and <em>R2<\/em> are identical, the referenced voltage at the common point would be exactly half of the supply voltage. If, however, the value of <em>R2<\/em> is reduced, the resulting voltage decreases accordingly, and vice versa. As we will see later, this feature is fundamental in setting the bias of transistors for specific applications.<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-large wp-image-43548 aligncenter\" src=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_1-263x550.jpg\" alt=\"\" width=\"263\" height=\"550\" srcset=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_1-263x550.jpg 263w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_1.jpg 294w\" sizes=\"auto, (max-width: 263px) 100vw, 263px\" \/><br \/>\n<strong>Voltage Divider<\/strong><\/p>\n<p>The concept of a current source, which aims to provide a constant current, is a bit harder to grasp as it can seem counterintuitive. In its simplest form, as shown in the following circuit, it consists only of a resistor (<em>R1<\/em>) placed in series with the power supply and the load (<em>R2<\/em>, where the diagonal arrow indicates that its resistive value is variable). In such a case, <em>R1<\/em> must have a much higher value than <em>R2<\/em> to ensure that changes in <em>R2<\/em> have minimal impact on the overall current, as the latter is determined by the equation <em>I<\/em>\u00a0= <em>V\u00a0\/\u00a0R<\/em>, where <em>R<\/em> is the sum of <em>R1<\/em> and <em>R2<\/em>. This configuration is very simple but far from ideal, especially when dealing with a low impedance load like a speaker. The solution is then an <em>active<\/em> current source, which will provide a stable and very precise current, regardless of the load, through the use of semiconductors, whether in an integrated unit or with discrete components. Any true Class\u00a0<em>A<\/em> amplifier will necessarily use a current source in its output stage. I emphasize <em>true<\/em> here because many amplifiers advertised as Class\u00a0<em>A<\/em> are actually advanced Class\u00a0<em>AB<\/em>; that is, their idle current will be very high, but we will leave that point for later<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 50%;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-large wp-image-43549 aligncenter\" src=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_2-317x550.jpg\" alt=\"\" width=\"317\" height=\"550\" srcset=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_2-317x550.jpg 317w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_2.jpg 350w\" sizes=\"auto, (max-width: 317px) 100vw, 317px\" \/><\/td>\n<td style=\"width: 50%;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-large wp-image-43550\" src=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_3-496x550.jpg\" alt=\"\" width=\"496\" height=\"550\" srcset=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_3-496x550.jpg 496w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_3.jpg 538w\" sizes=\"auto, (max-width: 496px) 100vw, 496px\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\"><strong>Active vs. Passive Current Sources<\/strong><\/p>\n<p>As mentioned previously, an amplification circuit must be designed to reproduce the entire audio waveform, meaning both the negative and positive halves. In its simplest form, a Class\u00a0<em>A<\/em> amplifier would look like the following schematic. Here, a single <em>NPN<\/em> transistor in a common-emitter configuration is used with a single 20\u00a0VDC power supply. The resistor values <em>R1<\/em> and <em>R2<\/em> form a voltage divider that biases the transistor, chosen to achieve a midpoint of 10\u00a0VDC at the collector, thus allowing for maximum amplified waveform amplitude with symmetrical clipping. The ratio of resistors <em>R3<\/em> and <em>R4<\/em> determines the circuit&rsquo;s voltage gain (<em>Av<\/em>); that is, the factor by which the input signal is multiplied. The input impedance of the circuit will be essentially equivalent to <em>R2<\/em>, and the output impedance to <em>R3<\/em>, which acts as the current source. The coupling capacitors <em>C1<\/em> and <em>C2<\/em> separate the audio signal from the bias voltages, ensuring that only the audio passes from one stage to the next or to the output, which in this case is the speaker. Assuming that the selected resistor values provide a gain of\u00a010, an input signal of 2\u00a0Vpp (<em>Vpp<\/em> meaning <em>peak-to-peak<\/em>) will yield a maximum of 20\u00a0Vpp (or 7.07\u00a0Vrms, i.e., 0.707\u00a0x 20\u00a0V\u00a0\/\u00a02) at the output before clipping, imposed by the supply voltage. Notably, in this circuit, the output signal will be phase-inverted relative to the input, as illustrated.<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-large wp-image-43551\" src=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_4-732x550.jpg\" alt=\"\" width=\"732\" height=\"550\" srcset=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_4-732x550.jpg 732w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_4-647x486.jpg 647w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_4-768x577.jpg 768w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_4.jpg 1080w\" sizes=\"auto, (max-width: 732px) 100vw, 732px\" \/><\/p>\n<p style=\"text-align: center;\"><strong>Simple Class <em>A<\/em> Amplification Circuit<\/strong><\/p>\n<p style=\"text-align: left;\">A relatively simple way to improve the performance of this circuit is to replace <em>R3<\/em> with an active current source. Advantages of this type of component include a very high internal impedance and good immunity to power supply fluctuations, which significantly improves circuit linearity. In the modified circuit shown below, the transistor can be considered a variable resistor controlled by the input signal. The signal appearing at the collector will vary with this resistance fluctuation, like a voltage divider. However, since the transistor\u2019s internal resistance decreases with an increase in base signal, the amplified result at the collector will be phase-inverted from the original.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"size-large wp-image-43552 aligncenter\" src=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_5-515x550.jpg\" alt=\"\" width=\"515\" height=\"550\" srcset=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_5-515x550.jpg 515w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_5-606x647.jpg 606w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_5.jpg 763w\" sizes=\"auto, (max-width: 515px) 100vw, 515px\" \/><\/p>\n<p style=\"text-align: center;\"><strong>Integration of an Active Current Source<\/strong><\/p>\n<p>So far, things are relatively simple, and this configuration and its many variations are found in almost all linear amplification circuits with a relatively high output impedance, including the input and intermediate stages of a power amplifier. However, with power amplifiers, things become more complicated as we introduce the concept of significant currents and power in the output stage, as these devices are designed to drive speakers with impedances that can fall below 4\u00a0\u03a9. The currents involved become very high, resulting in substantial thermal dissipation, which is characteristic of Class\u00a0<em>A<\/em> amps and quickly drives up their manufacturing cost. Their theoretical efficiency is only 25\u00a0%, so 3\u00a0watts of heat must be dissipated for every watt of audio signal, impressive figures when compared to 75\u00a0% efficiency for Class\u00a0<em>AB<\/em> and 95\u00a0% for Class\u00a0<em>D<\/em> configurations.<\/p>\n<p>The simplified circuit below shows a power amplifier with its output stage in Class\u00a0<em>A<\/em>, with values allowing us to verify what has just been mentioned. To simplify, the emitter resistor has been omitted here. Performing calculations that disregard the internal losses of the different components, we deduce that the 20\u00a0VDC power supply allows a maximum signal of 7.07\u00a0Vrms, as previously discussed. The resulting current will be 0.88\u00a0Arms (7.07\u00a0Vrms\u00a0\/ 8\u00a0\u03a9), for a maximum available power of 6.22\u00a0Wrms (7.07\u00a0Vrms\u00a0x 0.88\u00a0Arms). To achieve this power, the current source must always provide 0.88\u00a0Arms. At rest (static state, without signal), this current will flow from the power supply to ground through the transistor, creating a dissipation of 8.8\u00a0WDC (10\u00a0VDC\u00a0x 0.88\u00a0Arms) both in the source and the transistor, or 17.6\u00a0WDC in total. Comparing the maximum audio power to the continuous dissipation of the components gives us an efficiency of about 35\u00a0% (100\u00a0x 6.22\u00a0Wrms\u00a0\/ 17.6\u00a0WDC). However, since neither the transistor nor the current source are ideal components, they introduce losses that reduce efficiency to 25\u00a0%. These numbers may seem insignificant at first glance, but modern speakers are generally inefficient and require much more than 6.22\u00a0Wrms to achieve reasonable acoustic level and dynamics. A more realistic power level would be around 50\u00a0Wrms, resulting in 150\u00a0WDC of continuous dissipation per channel, which is considerable. Another significant factor is that Class\u00a0<em>A<\/em> amps are often optimized to almost double their power on a 4\u00a0\u03a9 load, as is the case with a Class\u00a0<em>AB<\/em> (provided its power supply allows), meaning doubling the static current and reducing efficiency to 12.5\u00a0%, or 300\u00a0WDC dissipation for 50\u00a0Wrms of audio power. For these reasons, it&rsquo;s common to measure temperatures around 60\u00b0\u00a0C on the heat sinks of these amps.<\/p>\n<p><strong>NOTE<\/strong>: The suffixes <em>DC<\/em> (<em>direct current<\/em>) and <em>rms<\/em> (<em>root mean square<\/em>) used for voltage, current, and power refer respectively to static and audio values.<\/p>\n<p style=\"text-align: center;\"><strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-large wp-image-43553\" src=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_6-890x525.jpg\" alt=\"\" width=\"890\" height=\"525\" srcset=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_6-890x525.jpg 890w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_6-647x381.jpg 647w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_6-768x453.jpg 768w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_6.jpg 1069w\" sizes=\"auto, (max-width: 890px) 100vw, 890px\" \/><br \/>\nSimple Power Amplifier<\/strong><\/p>\n<p>Although functional, the simplified circuit we have just seen has a disadvantage; as half of the supply voltage (1\/2\u00a0VDC) is present at its output, a capacitor is required to decouple the speaker, allowing only the audio to pass, and it must be very large, on the order of several thousand <em>uF<\/em> if we do not want the bass response to be overly affected. For a 4\u00a0\u03a9 load and a &#8211;\u00a03\u00a0dB cutoff at 10\u00a0Hz with a 6\u00a0dB\u00a0\/ octave slope (decibels per octave, so \u2013\u00a09\u00a0dB at 5\u00a0Hz), a value of approximately 4,000\u00a0uF will be required, implying a costly electrolytic capacitor that is not ideal for optimal performance. The solution is then to adopt, as we saw in <em>Part\u00a02<\/em> of this series, a symmetrical power supply configuration, as represented in the image below. The output capacitor can then be eliminated as the common point between the transistor and the current source will be at 0\u00a0VDC. Despite the added power supply complexity, performance will be improved through direct coupling and the better <em>signal-to-noise<\/em> performance of the symmetrical power supply.<\/p>\n<p style=\"text-align: center;\"><strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-large wp-image-43554\" src=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_7-800x550.jpg\" alt=\"\" width=\"800\" height=\"550\" srcset=\"https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_7-800x550.jpg 800w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_7-647x445.jpg 647w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_7-768x528.jpg 768w, https:\/\/www.tedpublications.com\/fr\/wp-content\/uploads\/2024\/11\/FIGURE_3_7.jpg 921w\" sizes=\"auto, (max-width: 800px) 100vw, 800px\" \/><br \/>\nClass\u00a0A Amplifier with Symmetrical Power Supply<br \/>\n<\/strong><\/p>\n<p>As we have seen, the current demand of a Class\u00a0<em>A<\/em> amplifier is very high, posing significant power supply challenges. To reduce <em>hum<\/em> to an acceptable level, the filtration level must be very high, often on the order of hundreds of thousands of\u00a0<em>uF<\/em>, with a substantial associated cost. Since this demand is constant and does not vary with audio power demand, all components of the power supply section are heavily stressed, which affects their lifespan. As a result, transformers reach very high operating temperatures, and even filter capacitors can overheat due to <em>ripple current<\/em> (the current flowing through the capacitor as it charges and discharges at each half-cycle of the 60\u00a0Hz mains power).<\/p>\n<p>An interesting solution for powering Class\u00a0<em>A<\/em> amplifiers is to use a regulated version, as mentioned in the second section. The active regulation circuit acts as a capacitance multiplier, which reduces the value of the filter capacitors needed, and allows for additional ripple reduction, which helps to minimize hum. However, there is a cost to everything; not only do these regulators increase the circuit&rsquo;s complexity, but they also generate heat themselves. Their operation requires a voltage drop, which, when multiplied by the current flowing through them, represents a significant power dissipation that adds to that of the amplifier.<\/p>\n<p>Despite the numerous challenges inherent in designing and using Class\u00a0<em>A<\/em> in the output stages of power amplifiers, they remain highly valued by audiophiles and music enthusiasts. Their high manufacturing costs\u00a0\u2014 due to oversized power supplies and heat sinks\u00a0\u2014 are balanced by excellent audio reproduction and a distinctive sound quality. Among the strengths of this configuration is their constant current demand on the power supply, regulated or not, which has the advantage of not causing fluctuations tied to audio power demands as in other types of amplifiers. Such fluctuations tend to negatively affect the reproduction of dynamic ranges and can impact the amplifier&rsquo;s linearity and stability. Another major advantage of Class\u00a0<em>A<\/em> is the improved inherent linearity of the circuit, which allows for simpler design without relying on high levels of negative feedback, which can introduce other issues.<\/p>\n<p>That being said, using Class\u00a0<em>A<\/em> in the output stage of a power amplifier inevitably comes with a high price tag that is not accessible to everyone. Therefore, it makes sense that a different yet high-performing and more cost-effective approach is desirable to meet market demands. This is where the other classes of operation come into play, which we will explore in our future discussions.<\/p>\n<p><em>For reference<span style=\"color: #993300;\"> ;<\/span><\/em> <span style=\"font-size: 24px;\"><em><span style=\"font-family: Abril Fatface;\"><span style=\"color: #800000;\"><span style=\"color: #993300;\"><a style=\"color: #993300;\" href=\"https:\/\/www.tedpublications.com\/fr\/en\/parlons-technologie-partie-2-2\/\" target=\"_blank\" rel=\"noopener\">Let&rsquo;s get technical &#8211; Part 2<\/a><\/span><br \/>\n<\/span><\/span><\/em><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Sketches : Normand Daigle The first article in this series clarified some technical points to aid in understanding the amplification circuits we will analyze, and the second article addressed the power supply section, which is crucial for quality performance. Today, we finally dive fully into the study of different classes of power amplifier operation, beginning [&hellip;]<\/p>\n","protected":false},"author":154564,"featured_media":43555,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2723,2950,2725],"tags":[],"class_list":["post-43544","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-hifi","category-technology-en","category-whats-new","entry","has-media"],"_links":{"self":[{"href":"https:\/\/www.tedpublications.com\/fr\/wp-json\/wp\/v2\/posts\/43544","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.tedpublications.com\/fr\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.tedpublications.com\/fr\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.tedpublications.com\/fr\/wp-json\/wp\/v2\/users\/154564"}],"replies":[{"embeddable":true,"href":"https:\/\/www.tedpublications.com\/fr\/wp-json\/wp\/v2\/comments?post=43544"}],"version-history":[{"count":7,"href":"https:\/\/www.tedpublications.com\/fr\/wp-json\/wp\/v2\/posts\/43544\/revisions"}],"predecessor-version":[{"id":43561,"href":"https:\/\/www.tedpublications.com\/fr\/wp-json\/wp\/v2\/posts\/43544\/revisions\/43561"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.tedpublications.com\/fr\/wp-json\/wp\/v2\/media\/43555"}],"wp:attachment":[{"href":"https:\/\/www.tedpublications.com\/fr\/wp-json\/wp\/v2\/media?parent=43544"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.tedpublications.com\/fr\/wp-json\/wp\/v2\/categories?post=43544"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.tedpublications.com\/fr\/wp-json\/wp\/v2\/tags?post=43544"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}