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I am trying to design a Class A Audio Amplifier with the following specs:

  • 9V DC Power Supply, Maximum withdrawn current 15mA
  • Used with a Low Impedance (8 Ohm) Speaker
  • Provide a minimum variable gain of 10
  • Can work with frequencies from 20 Hz - 5 kHz
  • Input signal is 0.2 V(peak to peak)

I tried several designs and I faced the following problems:

  1. Is it possible to actually drive that low impedance load with only 15mA, given this input signal? All the designs I did drew about 500mA from the DC Supply
  2. The signal always had some sort of clipping.
  3. I had some problems choosing coupling capacitors for the design.
  4. While choosing a Q point for each transistor, I had some problems figuring out which Ic should I assume to start working with

The design is mainly built on a common emitter stage then a common collector stage to act as a buffer. I will attach two design concepts I tried (none of which actually worked).

I would like to know if one can clarify the correct procedure of calculating the correct Q point, and then walk me quickly through the steps of making the required calculations of designing such an amplifier.

DC Coupled Circuit

AC Coupled Circuit

EDIT So I have been working for some time on this, and I reached this final design, but we neglected that 15mA current limitation. enter image description here

I am not sure how this performs on real life, I have just been testing that on simulators.

MarkU
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Amr Mohamed
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  • Welcome to EE.SE. You have made several mistakes mate. Delete C2 and R3 and let R5 be your coupling bias. As it is right now R5 will saturate Q2 fully ON. Also you should have some feedback to self-stabilize the circuit. –  Apr 29 '18 at 06:31
  • @Sparky256 If I remove C2 and R3 the circuit just has zero output on the 8 Ohm load. – Amr Mohamed Apr 29 '18 at 07:07
  • That is because Q2 is saturating, thus not working. Connect R2 from Q1 base to Q2 emitter to get negative feedback, so both Q1 and Q2 should be 50% saturated. Note that the value of R2 may have to be increased for this to work. An ideal loop is when Q2 has 1/2 Vcc on its emitter. (4.5 volts) –  Apr 29 '18 at 08:08
  • Let's see. Class-A means 25% efficient or worse. This means a max of \$33.75\:\text{mW}\$ to the speaker and that implies \$V_{\text{OUT}_\text{PK}}=735\:\text{mV}\$. Given the input spec is \$V_{\text{IN}_\text{PK}}=100\:\text{mV}\$, I think you will not achieve a voltage gain greater than 10. You will have to change the power supply specs, your input specs, or your gain specs. Now, if this were class-AB, then you just might barely skin by on this. But you said class-A. – jonk Apr 29 '18 at 08:42
  • @Sparky256 I don't quite get it, so can you verify that this circuit has the edits you mentioned ? https://imgur.com/a/SwjN65F – Amr Mohamed Apr 29 '18 at 08:55
  • @jonk I am not that experienced in circuits, so can you please tell me how you got that 33.75 mW ? Regarding the VoutPK = 735 mV, I believe this whole circuit needs re-designing. So If I'm to start with just a common emitter stage, and then design a common collector stage to cascade the two, what should be the collector current (Ic) to choose so that I can proceed with the calculations ? – Amr Mohamed Apr 29 '18 at 08:58
  • @AmrMohamed Power: \$9\:\text{V}\cdot 15\:\text{mA}=135\:\text{mW}\$. Class-A best efficiency, 25%. So power to the speaker is at best \$\frac{135\:\text{mW}}{4}=33.75\:\text{mW}\$ due to class-A. So then \$V_{\text{OUT}_\text{PK}}=\sqrt{2\cdot8\:\Omega\cdot 33.75\:\text{mW}}=735\:\text{mV}\$. Were this class-AB, and if you kept your biasing currents really low and used 2 BJTs for each output quadrant, then you just might get a voltage gain of 10. But it's cutting it fine. – jonk Apr 29 '18 at 09:13
  • @AmrMohamed Also, if you are using a \$9\:\text{V}\$ battery (explaining the low power), then you have to take into account the near \$2\:\Omega\$ output impedance for the battery, too, as well as its droopy voltage behavior as it gets used up. Probably need a capacitor directly across it. – jonk Apr 29 '18 at 09:18
  • @AmrMohamed Actually, I failed to notice another important detail. You are running off a voltage that you cannot even hope to fully use, so you'll be wasting more power than the approximations I used before suggest. That fact plus your tight current requirement from the supply suggests you could make good use of an efficient buck converter, too. – jonk Apr 29 '18 at 09:41
  • @jonk I got that. I will move to a DC power supply that can give me up to a 1A, yet I will still be restricted to the 9V. – Amr Mohamed Apr 29 '18 at 11:57
  • @AmrMohamed Are you still stuck with a class-A design? Since you can move the goal posts with the power supply, can you also move those, too? Or can it now be class-AB? – jonk Apr 29 '18 at 19:30
  • @AmrMohamed I think you could reasonably expect, assuming \$9\:\text{V}\$ as the supply and \$8\:\Omega\$ as the speaker impedance, that you could do a reasonable job with an average of \$100\:\text{mA}\$ from the power supply to provide about \$400\:\text{mW}\$ into the speaker. The voltage gain would then be \$A_\text{V}=25\$, or so, assuming your \$100\:\text{mV}\$ input peak. But this would be class-AB. An actual \$9\:\text{V}\$ alkaline battery could perhaps supply that current at maximum volume for an hour (maybe two.) – jonk Apr 30 '18 at 00:57
  • @jonk I have been making my way into calculations and edited the circuit to be like the last one I added in my question. Can you tell me if this is a better circuit? Would it perform as expected ? The simulator shows a gain of 17, but a withdrawn current of abut 300mA. Is that realistic ? – Amr Mohamed Apr 30 '18 at 06:11
  • @AmrMohamed No, it's not realistic. For one, it's just brute force for an emitter follower to operate in this mode. Plus, \$R_6\$ isn't adequate to driving half a quadrant of your load. And the emitter follower's DC operating point is highly dependent on beta (which is NOT dependable.) It's just not good. Please read through [class-A amplifier design overview](https://electronics.stackexchange.com/questions/368614/why-cant-class-a-amp-drive-8-ohm-speaker-with-just-one-bjt/368660#368660). – jonk Apr 30 '18 at 06:18
  • @jonk I don't get exactly what the circuit configuration is from the link, as I tols you I'm just still a beginner. – Amr Mohamed Apr 30 '18 at 09:21
  • @jonk by brute force with common Collector, you mean that it is a waste of power or just won't work at all, because the simulator shows it is working. Also, I noticed something, the simulator shows the circuit giving a gain, but the waves don't seem semetric, do you know why this could be? – Amr Mohamed Apr 30 '18 at 09:22
  • @AmrMohamed It sounds as though you need to read a book, then. And I still don't know if you must use class-A or can use class-AB. The link I have provided both options. The waves won't be symmetric because your quadrants aren't symmetric. It sounds as though you really need many chapters of writing. Especially if you cannot follow the link I gave you. – jonk Apr 30 '18 at 09:27
  • @AmrMohamed Do you see why \$15\:\Omega\$ cannot pull down well on an \$8\:\Omega\$ speaker as \$Q_2\$'s base drives downward and thereby pinches its \$V_\text{BE}\$ and is no longer actively participating? – jonk Apr 30 '18 at 11:23
  • @jonk I understand that biasing problem. My problem was that whenever I used a higher value the gain would always drop when I plug in the 8 Ohm. Is it a normal issue to be solved in a certain way, or is it because of a totally wrong configuration of the whole circuit? – Amr Mohamed Apr 30 '18 at 17:14
  • @AmrMohamed The topology you are using is pretty much wrong from the start. An emitter follower can be coerced into making "passable" output. But the price is just way too high and it's never going to be decent, anyway. Plus your \$R_6\$ needs to be a large, high dissipation resistor. None of this is worth the trouble except perhaps to prove a point to someone, even if you could get it going. – jonk Apr 30 '18 at 17:35

1 Answers1

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Just based on your specs you're going to have current draw issues.

You are driving the circuit with a 0.2 V signal and want a maximum gain of 10, meaning a peak output voltage of 2 V.

$$ \frac{V}{R} = I,\quad \frac{2}{8} = 250\ \text{mA} $$

This would also assume a 100% efficient system which you won't get anywhere close to in a class A amp.

I would recommend starting with a minimum working example, i.e. instead of starting with a multistage amp try and get a common-emitter amplifier working to your specs then see if you can improve it by extending it to a second stage.

loudnoises
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  • I had my suspicions about that current draw spec. Okay, now while designing the common emitter phase, what correct value of Ic should I work with, given I'm using the BC337 NPN ? I know the concept of the load line, but I can't seem to figure it out correctly. – Amr Mohamed Apr 29 '18 at 07:09