Optimizing Class A Amplifier Design - Page 1
Forum section: electronics / beginners

author: DatKnight23
author_rank: Newbie
timestamp: May 30, 2021, 11:44:10 pm
content: A few weeks ago I tried to design a CW transmitter using Experimental Methods in RF Design, but made a few mistakes with the output amplifier design.https://www.eevblog.com/forum/beginners/why-is-my-ce-amp-not-working/Starting again I used the same oscillator filtered through a Crystal bandpass filter buffered into a Class A power amplifier. I realise that designing a Class B or C might have been more efficient but my reasoning was that a higher drive level would damage the crystals used.Using EMFRD as well as other resources including an MIT opencourseware lecture I designed the amplifier for 200mW from a 9V rail voltage.https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-976-high-speed-communication-circuits-and-systems-spring-2003/lecture-notes/guest3.pdfTweaking the values to match the resistors I had available, I ended up with the circuit shown in my diagram, with an approximate output power of 277mW with a rail voltage of 9V through a 292 Ohm load. As I only have 220 ohm resistors, I tested the output power using that.Initially, I measured about 1.75Vpp across the load with my scope and according to EMFRD that indicates an output power of 1.7mW, degenerating the resistor to increase the voltage gain for more swing distorts the signal but only provides 4Vpp which should correspond to 9mW.At this point I'm not entirely sure where I'm going wrong, I am aware of the inefficiency problems of Class A compared to Class B or C but surely my amplifier can't be 3% efficient even when I push the gain so much to distort the signal in the second case?The only other problems I can think of might be that I'm not driving the amplifier enough from the previous stage, or that I'm missing some considerations from the design process.Perhaps I'm simply barking up the wrong tree with Class A amps? I'd like to at least try to make this design work before starting over again.

author: bob91343
author_rank: Super Contributor
timestamp: May 31, 2021, 03:46:21 am
content: You would do well to scrap your analysis method and do it the old fashioned way. Draw an equivalent circuit and do the math to make the swing what you want and get the power you seek.I see so many cases of simulations gone wrong. The software is supposed to help, not get in your way. But if you don't know what you are doing to begin with, you will get crazy results.Class A means the current flows continuously. If you adjust the operating point somewhere near the middle of the range so that you get maximum swing, calculate the power.

author: gcewing
author_rank: Regular Contributor
timestamp: May 31, 2021, 11:53:34 am
content: Your power calculation assumes the output voltage can swing by +/-9VRMS, but it can't get anywhere near that, for a number of reasons:1. With a single-ended 9V supply you can't get more than +/-4.5V at most.2. You won't even get that, due to the quiescent voltage across the emitter resistor.3. That's a limit on the *peak* voltage, not RMS.The way the transistor is biased, the quiescent voltage at the collector will be about 3.4V below the 9V rail. So the output can't swing up or down by more than that. With a sine wave and a 220 ohm load, that gives a maximum power of about 25mW.To get the 200mW into 200 ohms that you want, you'd need a supply of at least 20V or so to allow for a +/-9V peak swing with some room to spare:Vrms**2 / 200 ohms = 200 mWVrms = sqrt(200 * 0.2) = 6.32 VVpeak = sqrt(2) * Vrms = 8.94 V

author: Zero999
author_rank: Super Contributor
timestamp: May 31, 2021, 01:18:00 pm
content: The output stage doesn't have any real voltage gain, 1.1 at most. It has an output impedance of 220R and an unloaded gain of 2.2, which is reduced to half, when loaded with 110R. You're better off putting the voltage amplification stage, before the emitter follower, which just amplifies the current.