# Audio amplifier for school project

For a school project, we have to explain this design of a audio amplifier. We have only been able to understand to parts marked with colored rectangles.

The design comes from the book audio power amplifier design 6th edition by Douglas Self and can be seen on figure 7.16 on page 181. It is a Class A amplifier.

We know that the diodes are something called “Clamp diodes”, but we aren’t totally sure what their specific function is in the circuit.

If you could explain even just a part of it, it would be great help.

• D1,D2, R9, R11, Q8 are to create a current source. Commented Jun 14 at 12:52

D1 and D2 and R9 form a reference voltage, basically 'clamping' to use your term the bases of Q5, Q8 and Q10 to two diode drops (around 1.2 V to 1.4 V) below the positive supply voltage.

With the voltage on the bases held relatively constant by the diodes the voltage at the emitters will be this voltage minus the BE junction voltage, so one diode drop or roughly 0.6 V to 0.7 V below the positive rail. This voltage across the emitter resistors will cause the emitter current to be held constant, for instance for Q8 and Q10 it might be $$I_E = \frac{0.6~V}{100~\Omega} = 6~mA$$ while for Q5 it would be around 4 mA.

So Q5 is the constant current source for the differential stage, Q8 for the VAS stage and Q10 for the output stage.

D1 and D2 are not "clamp diodes." Together with R9, they form a voltage reference approximately 1.4 Volts below the positive supply rail.

This reference voltage is used to create three separate constant current sources formed by Q5+R6, Q8+R11, and Q10+R12. This is a standard circuit described in Wikipedia's article on constant current sources, in particular figure 4 (reproduced below).

You should have been able to quickly spot this:

Note that there are very strong hints there. There are three current sources already identified for you. And they are all tied into the same diodes + resistor structure. So it's practically given away.

One diode accounts for the base-emitter diode drop of each of the connected PNPs. The remaining diode accounts for the drop across each of the emitter resistors, leading to a current source value. The author obviously is taking it as granted that the diode drop for the 1N4148, given the current yielded by $$\R_{25}\$$, will be about $$\600\:\text{mV}\$$. (I think this is a little low, given that about $$\1\:\text{mA}\$$ will be present in the two diodes.) So it's really easy to calculate $$\\frac{600\:\text{mV}}{R_{23}}=4\:\text{mA}\$$, $$\\frac{600\:\text{mV}}{R_{29}}=6\:\text{mA}\$$, $$\\frac{600\:\text{mV}}{R_{31}}=6\:\text{mA}\$$.

This tells you the entire function and design details of the entire upper part of the schematic. And it's not hard.

Also, since the only connection from this upper section and the lower section are via collectors, the lower section is free to float with respect to this upper section. This makes it relatively independent of the supply rail values with the only real change being some slight variation in source current values. So the lower section can be treated separately for analysis, with only the current source values showing.

With only $$\6\:\text{mA}\$$ sourced by $$\Q_{22}\$$ this circuit can't drive much of a load. It must be substantially more than $$\2\:\text{k}\Omega\$$ to have any hope of the stated $$\+20\:\text{dBu}\$$. So this is just a test circuit to evaluate different ideas. Not one likely to be used for most headphones or speakers.

This is also a class-A amplifier. You've only to look at $$\Q_{23}\$$'s arrangement to be certain about that.

As seen above from the global NFB circuit there, the gain should be $$\A_v\approx 23\$$. It'll be a little less due to the finite open loop gain of the system. But it should be close.

In your example in your question, however, the gain will be less with $$\A_v\approx 11\$$ (again, perhaps a little less than that due to the finite open loop gain, but very close I expect.)