# Class A Audio Amplifier

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• 9V DC Power Supply, Maximum withdrawn current 15mA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.  EDIT So I have been working for some time on this, and I reached this final design, but we neglected that 15mA current limitation. I am not sure how this performs on real life, I have just been testing that on simulators.

• 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.
– user105652
Apr 29, 2018 at 6:31
• @Sparky256 If I remove C2 and R3 the circuit just has zero output on the 8 Ohm load. Apr 29, 2018 at 7:07
• 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)
– user105652
Apr 29, 2018 at 8:08
• 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, 2018 at 8:42
• @Sparky256 I don't quite get it, so can you verify that this circuit has the edits you mentioned ? imgur.com/a/SwjN65F Apr 29, 2018 at 8:55

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