I'm trying to design an audio amplifier output stage to drive a speaker load. I'd like to start with the design of this stage in order to calculate the input impedance to the output stage so that I can proceed to designing a buffer stage with an appropriate output impedance and a gain stage before that.

My speaker impedence is 8 Ω and I'd like to deliver a minimum of 1 W to the load and keep total harmonic distortion below 1%. One possible class AB output stage configuration is given below. I looked at a few different configurations, but this one is nice in that it doesn't require a constant current source for the biasing which makes it simpler:

Class AB with input buffer

I first began by calculating the peak voltage over the load:

\$Vp = \sqrt{2R_LP_l} = \sqrt{2(8)(1)}=4\ \textrm{V}\$

From this I calculated the maximum output current through the load to be 2 mA using Ohm's law, assuming R3 and R4 to be 0 for the time being (I can add these values back if I notice any thermal runaway effects later).

Since the voltage gain of this output stage is approximately unity, the output current which I calculated as 2 mA is related to the input current through the current gain given by:

$$A_i = \frac{i_o}{i_i}=\frac{(1+\beta)R}{2R_L}$$

The textbook I'm using has made the assumption that all of the transistors, NPN and PNP, are perfectly matched when they derive the current gain. This is obviously not true in reality. The 2N3904 NPN transistors and the 2N3906 transistors I'd like to use have very different current gains.

How can I solve for my bias resistors R1 and R2 such that I keep the total harmonic distortion low, while still supplying the output current that satisfies the power requirement? What is a more practical/real-world design approach for designing this output stage? I find that the theory in the textbook is often quite useless when it comes to designing real applications.

  • \$\begingroup\$ What do R1 and R2 do? They provide bias current for Q1 and Q2. What happens if for example the bias current of Q1 is too low? Then Q1 will be "out of current" under certain conditions. What are the conditions that Q1 + R1 need to source/sink the lowest / highest current? Well, what Base current will Q3 need? Determine the maximum collector current of Q3, divide that by the lowest \$\beta\$ you expect, take margin on that (like a factor 2), that will be the current that Q1 and R1 must be able to push/pull into Q3. \$\endgroup\$ Commented May 28, 2020 at 14:19
  • \$\begingroup\$ This circuit arrangement is tempting. A problem you haven't addressed is temperature tracking...if you can keep all transistors at the same temperature it can work very nicely. But Q3,Q4 tend to run hot with large signals while Q1,Q2 consume more-constant power. Thermal design gets a bit complicated. \$\endgroup\$
    – glen_geek
    Commented May 28, 2020 at 14:44
  • 2
    \$\begingroup\$ Here you find what you want uweb.engr.arizona.edu/~brew/ece304spr07/Pdf/… \$\endgroup\$
    – G36
    Commented May 28, 2020 at 16:03
  • \$\begingroup\$ Are you sure about that value for the current? \$\endgroup\$ Commented May 28, 2020 at 18:17

1 Answer 1


Thumbs up - discretes-based audio amps used to be my favourite toy... Size doesn't matter, the fun comes from trying to understand how these things work. Looking back, this is where I cut my teeth on feedback control :-) Class D (PWM) takes all the analog fun out of the journey (but still leaves you with some mildly annoying analog side effects).

Speaking of class AB, there are hardly any simple solutions / low hanging fruits / shortcuts. If you use plain resistors to drive transistor bases in the output stage (common emitter followers), the power transistors will run out of base drive current just where it's needed most (close to the power rails). Constant current sources work better - they're stil not perfect though, because they do not pull all the way rail-to-rail...

If you'd like to see a more complex and proven class AB output stage (still not quite rail to rail), google DPA220. While now historical, in its day it was a fairly decent design all around. Namely, the adjustable biasing stage just before the output power totem contains two transistors that were thermocoupled to the main heatsink (if memory serves) to achieve negative dependency of quiescent current on temperature...

  • \$\begingroup\$ Nice. You targeted exactly the right point -- using plain resistors for drive! I do continue to mention this, time to time, when I see it in a question. And I don't see others targeting it so quickly or directly, so it's nice to meet you here. +1 \$\endgroup\$
    – jonk
    Commented May 29, 2020 at 3:50

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