Why all books told that Av differential of an differential input stage in power amp is very big? Ad = Rc/re, but if there are current sources in the emitter and collector circuit they will have too big impedance and division one big "Rc" to another big "re" will not give very big "Ad" value. Where is a trick?
In the differential stage with asymmetric output, the load current can "swing" from:
\$0A\$ to \$I_{EE}\$
And the gain is
$$A_d = 0.5*R_C \approx \frac{R_C}{2r_e}$$
But by adding the current mirror we can "increase" the gain (the gain is now the same as it is for a symmetrical output). Because now the load current can swing from:
\$-I_{EE}\$ to \$+I_{EE}\$
And the gain is:
$$A_d = gm*R_L \approx \frac{R_L}{r_e}$$
In your audio amplifier the dif. stage the load resistance is equal to the input impedance of a VAS stage (Q6). Hence, the DC gain is "low" because \$r_\pi\$ is low and the Miller capacitance. http://www.ecircuitcenter.com/Circuits_Audio_Amp/Miller_Integrator/Miller_Integrator.htm
Why all books told that Av differential of an differential input stage in power amp is very big?
I find "all books" a dangerous statement. It really depends on the input stage if it has a large voltage gain (Av). Also you're talking about Power amplifiers for audio, you should have mentioned that because audio is low frequency. There are also Power Amplifiers for high frequency applications. Many tricks we can do in audio cannot be done at these high frequencies.
In the schematic you show the first stage by itself actually does not have a large voltage gain! The voltage gain is roughly equal to the gm of the input pair transistors times the load at the collector.
Here that load is Q12 and Q19 which make a current mirror. It means the (differential) currents from Q1 and Q2 are added and this current ends up the collector of Q12. This collector is loaded by the base of Q6 which is a diode to ground and this is a low impedance. Note that the signal is still a current. If you would calculate the actual voltage gain up until the collector of Q12, that gain would be quite low.
The AC signal current appearing at the collector of Q12 will go into Q6 and will then be multiplied by beta, flow through the string of diodes.
At these diodes the signal is turned into a voltage again and current amplified (buffered) to the output.
Do note that this design has negative feedback using C2, R3 and R10. This feedback sets the total gain of the amplifier. For this to work properly the open-loop gain of the amplifier needs to be high enough. That's why we want high-gain stages.
As you can see, this design is not so easy, it is not actually only the first stage that determines the gain. This circuit is basically a voltage-to-current converter - current amplifier -- active load -- current buffer with overall feedback.
You have mentioned the expression Ad = Rc/re.
Please note that this expression is (1) a very rough approximation for a common emitter amplifier with heavy negative feedback (due to re) and (2) does not apply for a diff. amplifier. Neither for a simple one (two or three transistors only) nor for the one shown by you.
This is because the resistance in the common emitter path (as seen from the left transistor of the diff. pair) is bypassed by the low-resistive input of the most right transistors emitter.
For a good understanding of the various gain expressions (common mode, single diff. mode, full diff. mode) you must know the working principle of the diff. pair.
Detailed derivation (for the diff. stage Q1, Q2, Q3 only).
Assumption: Unsymm. differential operation with Vin1=Vin and Vin2=0 [Q1 can be treated as an emitter follower loaded by the input resistance of Q2 in common base configuration].
Gain: Ad=Gain1*Gain2=A1*A2 with
A1=gm1*ro/(1+gm1*ro) with ro=re||(1/gm2).
If the dynamic resistance re is very large (re>>1/gm2) we can set ro=1/gm2.
A1=(gm1/gm2)/(1+gm1/gm2).
For identical operational points for Q1 and Q2 we have: gm1=gm2=gm and the gain is
A1=1/2.
Hence, with A2=gm*Rc (Rc=collector resistance) the total gain is
Ad=0.5*gm*Rc .
