Can somebody please analyse this circuit
Contrary to what other commentators have suggested, there's nothing wrong with the circuit topology as shown. In fact, it rather cleverly sets the Q point of Q3's collecter at almost exactly 6V, half of the supply voltage.







simulate this circuit &ndash; Schematic created using CircuitLab

For the DC analysis, start by calculating the Thevenin equivalent of the bias network: 8.0 V and 16.67 k&Omega;. This sets the emitter of Q1 at about 7.3 V.

Now, consider R6. Since the voltage across it is limited to 0.7 V, it is carrying at most about 0.15 mA. If we assume for the moment that the contribution of Q3's base is negligible (we'll verify this later), then that same current is flowing through R12, which gives it a voltage drop of 1.5 V, setting the collector voltage of Q3 at about 5.8 V.

This means that R10 is carrying a total current of about 1.23 mA, which means that the remaining current (1.08mA) is flowing through Q3. If Q3 has a gain of 100, then its base current is about 10.8 &micro;A, which is less than 10% of the R6 current, as surmised.

You could iterate through this analysis a few more times to get more exact figures, but that's what circuit simulators are for. I've redrawn the circuit in CircuitLab so that people can play with it.

As far as the AC analysis goes, Q1 by itself has a gain that is set by R6 and R8 to about 21, but since Q3 has no emitter resistor, its voltage gain is very large. Therefore, the overall gain of the circuit is almost entirely controlled by the negative feedback (R12 and R8), which makes it about 46&times;.