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I'm trying to design an audio amplifier. The preamp stage is a voltage divider CE, while the power stage is a Darlington Class A:

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My goal is to amplify a \$48.7\ \mathrm{mV},\ 1.07\ \mathrm{mA}\$ (RMS) signal from my phone to have a power of \$1\ \mathrm{W}\$.

Therefore, I'm choosing \$A_v = 20\$ and \$A_i = 1000\$.

The design works well for the aforementioned signal. I would also want the amplifier to work when the volume is maximum (RMS voltage is almost \$500\ \mathrm{mV}\$). However, the signal gets cut off quite badly for the maximum volume signal:

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I have tried and considered the followings:

  • Reducing \$A_v\$ and move the Q-point of Q1 to a higher \$I_c\$ and lower \$V_{CE}\ \$value : I tried \$A_v=6\$ and Q-point to \$I_C = \frac{2}{3}I_{Cmax}\$ but there is still some cutoff:

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I'm not really sure about if going further for the values is a good thing. I'm already violating the "Q-point in the middle of DC loadline" rule.

  • Getting a higher DC voltage: I'm thinking \$24\ \mathrm{V}\$ would help, but I'm already having a \$15\ \mathrm{V}\$ rectifier circuit. Increasing it to \$24\ \mathrm{V}\$ would require a bulkier and more expensive transformer (my current one is sending \$21\ \mathrm{V}\$ ripple voltage into a LM7815).
  • Switching to a class AB amplifier: Each Darlington pair of the class AB amplifier would only need to work with more than half of the signal cycle, so almost double the headroom for the amplified signal. However, I don't have much experience on designing this so please explain to me in details if you think this is a good idea.

Other than the main problem mentioned, I would greatly appreciate any other comment on the design of my amplifier.

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    \$\begingroup\$ You should apply a negative feedback network to eliminate unwanted effects at the cost of a reduction in voltage gain. \$\endgroup\$
    – Franc
    Jan 14 at 6:49

3 Answers 3

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Looks, like you want at first to lift in the first amp stage the signal voltage to about 6 V peak to peak - that's about 1W power to the 4 ohm load.

And your Darlington stage is intended to be a current buffer - no voltage gain, but high enough output current to get that 1W to the 4 ohm load.

You obviously have noticed that it's not a good idea to connect the speaker directly to the emitter of Q3 because there's a permanent DC voltage and the speaker maybe wouldn't stand the dissipation it would cause - so there's R6 for the DC and the speaker gets only the AC part of the transistor's output voltage. The hefty 4700 uF capacitor would also let some bass to reach the speaker.

Unfortunately this doesn't work. Nothing pulls enough current to make also full amplitude negative voltage peaks possible to your 4 ohm speaker. We can test it:

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I run it at 1 kHz. V2 is 6Vpp as needed for 1W output. The output capacitor is reduced to make the circuit to reach the steady state soon enough. The capacitance 250 uF is still plenty for 1 kHz. You see that the negative peak is severely clipped when the steady state is reached.

The operation can be improved substantially by reducing the DC path resistor R1 from 47 ohm to 4 ohm:

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You could do the same, but there would be continuous about 7,5V DC voltage over R1. You should see that in the idle state with no signal it would dissipate continuously 14 watts. And so will do the output transistor - the total dissipation is 28 watts. That's intolerable for an amp which outputs max. 1W to the speaker. It could be reduced by having lower operating voltage.

What to do

Obviously you are not going to use some audio amp IC nor pick a ready to use design from manufacturer's application notes. In that case get some texbooks of proper audio amp design. And read also web articles of the subject. Start from the basics like how to calculate the needed currents and voltages for single transistor class A amps and for push-pull amps when the output power and the load are given.

An earlier comment tells that negative feedback should be used to reduce the unwanted effects of the non-idealities (=distortion and the obscurity of the operating point). That's true, but it works only if the basic relations of voltages and currents are at first designed right. Negative feedback is also a new source of challenges if one wants to avoid its own unwanted effects (which are caused by the slowness of practical components).

BTW. Class AB doesn't mean it's a push-pull amp, it states only there's some idle current, but less than the peak current.

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emitter-follower output stage

You have a great answer (I'll +1 it, shortly) exposing the problems with your output stage (emitter follower.) It's only active in one quadrant and therefore depends entirely upon a resistor (your \$47\:\Omega\$) for the other quadrant. As shown in that answer, if you match it up with your nameplate value for the speaker, then it can be made to perform better. But it will never be an active 2-quadrant output stage and as unawriter points out, the total dissipation reaches \$28\:\text{W}\$ to deliver \$1\:\text{W}\$.

Here's a picture of an old schematic template I put together almost a decade ago, which allows me to specify some details and then automatically generates the values to run the test (Note the title?):

enter image description here

The speaker gets \$1.0\:\text{W}\$. For this, the \$5.6\:\Omega\$ emitter resistor dissipates \$19.3\:\text{W}\$ and the BJT dissipates \$7.0\:\text{W}\$. This \$26.3\:\text{W}\$ total is surprisingly close to what unawriter wrote.

If I lower the VCEMIN parameter to \$1\:\text{V}\$ the emitter resistor goes to \$6.5\:\Omega\$ and this total circuit dissipation drops to about \$24.7\:\text{W}\$. Not much of a difference. Hopefully, you get the idea.

You can take it as gospel that this topology is not so good. Don't do it unless you are making ElectroBOOM-like YouTube videos.

one class-A 2-quadrant approach

The following is a 2-quadrant design class-A design where I've set the quiescent current to a scrape-by level:

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Here, the upper quadrant \$Q_1\$ dissipates \$4.3\:\text{W}\$ and the lower quadrant \$Q_2\$ dissipates \$3.6\:\text{W}\$ for a circuit total of \$7.9\:\text{W}\$.

Note the significant improvement by switching out a passive resistor in the 1-quadrant design, replacing it with an active component in the 2-quadrant design?

another class-A 2-quadrant approach

Another design based upon something I wrote here, recently, might look like this:

enter image description here

(I used a lower \$V_{_\text{CC}}\$ since that was all that was needed here.)

The upper quadrant \$Q_1\$ dissipates \$3.5\:\text{W}\$ and the lower quadrant \$Q_2\$ dissipates \$3.0\:\text{W}\$ for a circuit total of \$6.5\:\text{W}\$.

summary

The above makes some comparisons.

Both examples of 2-quadrant designs use global NFB to control the gain. So, unlike your approach, these provide stable gain against device variations and operating temperatures. They will also have much better THD and waste about \$3\times\$ less power.

But class-AB does much better, dissipating perhaps only a little more than the speaker itself. It introduces the possibility of cross-over distortion, which can be mitigated by running it a little 'hotter' and wasting a little more dissipation. But the global NFB covers many ills. So they are worth understanding and pursuing. (Especially when you are talking about delivering \$1\:\text{W}\$ and more into a speaker load.) At the bottom of this answer I do show an example of a class-AB. You can see the key ideas there as well as a specific implementation.

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I just noticed that I made a miscalculation for R5, so the Q-point is way off the middle of the DC loadline, potentially cutting off some of the signal in the negative cycle.

However, as user unawriter has pointed out, the speaker can only pull as much as the emitter DC current, so regardless of where the Q-point is, there will always be cutoff in the negative cycle should the signal's current surpasses the emitter DC current.

I write this answer also to say thank you to Franc, unawriter, and periblepsis for pointing me to the right direction in terms of what the problem is and what to do.

I wish I could mark all of your answers as correct but that doesn't work in StackExchange, so I hope this should do.

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