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For a school project, I need to design an 8 W audio amplifier with BJTs. The power supply is from 9 V to -9 V and the input is 0.8 Vp 1 kHz. A speaker with 8-ohm impedance and a voltage gain of 10 is wanted. I tried using Class AB design but couldn't get over 700 mA of output current, the power supply makes things really hard. Do you have any tips on how can I design a working amplifier circuit?

I added the design I'm currently working on and the max output I can get. I try to increase the VOUT by increasing R6 but it gets clipped in the negative cycle.

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    \$\begingroup\$ Well, we won't hand you the design. Show us the class AB design that you made. I think the best advice for "tips on how I can design" is just to study existing successful audio amplifier designs. There are many, many of them available online. \$\endgroup\$ Jan 10, 2022 at 16:27
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    \$\begingroup\$ We don't know what tips you already used. Post your design so we can tell what is wrong with it. Unless the problem is not the amp, but the power supply. Did you happen to use 9V batteries? \$\endgroup\$
    – Justme
    Jan 10, 2022 at 16:29
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    \$\begingroup\$ the power supply makes things really hard Why? \$\endgroup\$ Jan 10, 2022 at 16:30
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    \$\begingroup\$ TIP#1 - An 18 volt supply rail could just about sustain a 16 volt p-p output waveform. Assuming that waveform is sinusoidal means an RMS value that is 16/2.82 volts = 5.66 V RMS. Into an 8 ohm speaker, that's a max power of 4 watts. This means you can't use a conventional push-pull class AB (or otherwise) output stage. \$\endgroup\$
    – Andy aka
    Jan 10, 2022 at 16:31
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    \$\begingroup\$ Bah, what a spoiler Neil. \$\endgroup\$
    – Andy aka
    Jan 10, 2022 at 17:52

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Your first problem is your choice of a Darlington follower for your output stage. The voltage across a Darlington cannot be less than two diode drops, which would amount to about 1.5 volts. So the output can never be greater than about 7.5 volts for a 9-volt supply - and this is an absolute limit. In your case your getting about 7 volts swing, which is a very reasonable number.

But this is not your big problem. That derives from

I need to design an 8W audio amplifier with BJTs. The power supply is from 9V to -9V and the input is 0.8Vp 1 kHz. A speaker with 8-ohm impedance and a voltage gain of 10 is wanted.

I'm afraid you haven't thought this through. Generally, an 8W audio amplifier means 8W RMS with a sinewave input such as 1 KHz. What will produce this? Well, obviously 8V RMS into an 8 ohm load. However, the peak voltage of a sine wave is approximately 1.414 times the RMS, which means you need a 12-volt supply, plus another 2 volts to account for your output stage limitations. Since this requires both a + and - excursion around ground, you need about a +/- 14 volt supply. That is, as long as you are trying to do it the way you are.

Pay close attention to Neil_UKs comment.

On the other hand, this implies that your required voltage gain is not 10, it's 14 and a bit. Let's say that the 8W output is 8W peak, which only requires a gain fo 10. This you can do with 9 volt supplies. The trick is not to use a Darlington configuration. Something like

schematic

simulate this circuit – Schematic created using CircuitLab

this will do. The final output transistors can be driven to produce a Vce voltage of about 0.1 to 0.2 volts, which gives you the 8 volts you need.

Of course, you can't do this open loop, or at least not reliably. You probably can tweak the base drive currents to give you what you want, but in real life it wouldn't work due to temperature instability.

Feedback is your friend.

I don't know what your course has covered, so I don't know which approach is appropriate to your situation. But your original circuit can't be made to work as you have specified.

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  • \$\begingroup\$ ALL audio amplifiers have lots of voltage gain then use negative feedback from the output to reduce gain and reduce distortion. you should use Sziklai pairs (look it up) instead of darlingtons and use bootstrapping for more output voltage swing. \$\endgroup\$
    – Audioguru
    Jan 11, 2022 at 1:27
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    \$\begingroup\$ Although it is common terminology to use RMS for power strictly speaking it should be average. Also a convenient thing to remember when doing such designs is that the average power of a sine wave is half of the peak power. So the 8W peak that 8 volts provides when driving into 8 ohms means that the average power is 4W. \$\endgroup\$ Jan 11, 2022 at 2:02
  • \$\begingroup\$ @KevinWhite - "strictly speaking it should be average" That's news to me, especially in audio work. For instance, I went to Amazon, downloaded one of the first amps I found m.media-amazon.com/images/I/91Vpn+ashZL.pdf and it specifies RMS. I remember the Bad Old Days when amps were rated in terms of "peak power", and the efforts required to get everybody to use RMS. Average power only makes sense for quasi-DC applications (long-term averages) and is not used for audio work. \$\endgroup\$ Jan 11, 2022 at 21:17
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    \$\begingroup\$ @WhatRoughBeast - Yes, it is almost universally used for audio amplifiers but when you multiply RMS voltage by RMS current, the RMS drops out leaving average. The average power will be the average over an integral number of cycles. \$\endgroup\$ Jan 11, 2022 at 23:44
  • \$\begingroup\$ @WhatRoughBeast - see for example the Measurement section at en.wikipedia.org/wiki/Audio_power \$\endgroup\$ Jan 11, 2022 at 23:50

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