Understanding of two stage DC coupled complementary amplifier

Question

I am struggling with an complementary amplifier and unable to compute the value and equations. My target is to identify the working of the circuit, its total gain and equations. After studying many circuits I learned that:

Gain of 1st stage A1 = 4700 / 200
Gain of 2nd stgge A2 = 1600 / 200  (I dont know if this is correct)
total gain A = A1 * A2  (I dont know if this is correct)

Q point for active region for 1st tage is 
Icq1 = (V1 / (R6 + R1)) / 2

Q point for active region for 2st tage is 
Icq2 = (V2 / (R2 + R3)) / 2  (I dont know if this is correct)

V6 = V4 + 0.7

I0 = I6 + I4

I6 = Icq1

I4 = Ib1 + Ie2 => Ib1 + Icq2

Ic1 = B * Ib1  With this equation I can calculate the required Ib for beta and Ic1 = Icq1

This is how far I go but I still don't know how this works. I want equations for the total gain and for the resister value selection because this is reference circuit and I don't have actual transistor Q7 and Q8 with reference beta values. Now if I use any other good transistor then the beta will change so we need to calculate new resister values to set the circuit in active region and total gain of > 32 dB. Also the bandwidth is > 3 Mhz. Input type is video signal.

Answer 1

First I must say that it is easier to design such a circuit then to find the DC operating point. But let us try

^{simulate this circuit – Schematic created using CircuitLab}

You can use any network analysis technique you are familiar with.

I will use KVL.

First KVL loop looks like this:

So the first equation will look like this:

$$(I_{C1} - I_{B2})R_{C1} =(I_{E2} + I_{B1})R_{E2} + I_{E2}R_{E3} + V_{BE2} \:\:\: (1) $$

The second KVL loop will look like this:

And the KVL equation around this loop: $$V_{CC} = (I_{E2} + I_{B1})R_{E2}+I_{B1}R_F+V_{BE1}+I_{E1}R_{E1} \:\:\: (2) $$

And from "transistor action" we know that:

\$I_E = I_B(\beta +1)\$ ; \$I_B =\frac{I_C}{\beta} \$ and \$I_B =\frac{I_E}{\beta+1} \$

Thus

$$(I_{C1} - \frac{I_{E2}}{\beta_2 +1})R_{C1} = V_{BE2}+ I_{E2}R_{E3} + (I_{E2} + \frac{I_{C1}}{\beta_1})R_{E2} \:\:\: (1) $$

$$V_{CC} = (I_{E2} +\frac{I_{C1}}{\beta_1} )R_{E2}+V_{BE1}+\frac{I_{C1}}{\beta_1 }\left(R_F+(\beta_1+1)R_{E1}\right) \:\:\: (2) $$

And for \$\beta_1 = 67 ;\: \beta_2=130\$ and \$V_{BE1} = V_{BE2} = 0.6V\$

I've got:

\$I_{C1} = 1.343mA\$

\$I_{E2} = 2.53mA\$

$$I_{C1} \approx \frac{(R_{C1} + \beta (R_{E2} + R_{E3}))V_{CC} - (R_{C1} + \beta R_{RE3}) V_{BE}}{\frac{R_{C1} (R_{E2} + \beta(R_{E1} + \beta R_{E2}) + R_F)}{\beta} + (R_{E2} + R_{E3})R_F + \beta R_{E1}(R_{E2} + R_{E3}) + R_{E2}R_{E3}}$$

And the voltage at \$Q_1 \$ collector is:

$$V_{C1} = V_{CC} - I_{C1}R_{C1} = 12V - 1.343mA \times 4.7k\Omega = 12V - 6.3V = 5.7V$$

and the voltage at \$Q_2\$ collector is:

$$V_{C2} = I_{C2}R_{C2} = I_{E2}\frac{\beta_2}{\beta_2+1} \times R_{C2} = 2.51mA \times 1.6k\Omega = 4V $$

For the AC (small-signal) analysis the equivalent circuit will look like this:

^{simulate this circuit}

And the mid-frequency voltage gain is equal to:

$$\frac{V_{OUT}}{V_{IN}} \approx \frac{R_{C1}||(\beta +1)(R_{E3} + r_{e2})}{R_{E1} + r_{e1}} \times \frac{R_{C2}||R_L}{R_{E3} + r_{e2}}$$

If \$C_1\$ and \$C_2\$ have a higher capacity so that in the "mid-frequency" we can treat them as short circuit the gain will increase to:

$$\frac{V_{OUT}}{V_{IN}} \approx \frac{R_{C1}||(\beta +1)r_{e2}}{r_{e1}} \times \frac{R_{C2}||R_L}{r_{e2}}$$

Where

\$r_{e1} = \frac{V_T}{I_{E1}} \approx \frac{26mV}{I_{C1}}\$ \$r_{e2} = \frac{V_T}{I_{E2}} \approx \frac{26mV}{I_{C2}}\$

The End