Example-1:

For the cascaded differential amplifier shown in fig. 5, determine:

• The collector current and collector to emitter voltage for each transistor.
• The overall voltage gain.
• The input resistance.
• The output resistance.

Assume that for the transistors used h_FE = 100 and V_BE = 0.715V

Fig. 5

Solution:

(a). To determine the collector current and collector to emitter voltage of transistors Q₁ and Q₂, we assume that the inverting and non-inverting inputs are grounded. The collector currents (I_C ≈ I_E) in Q₁ and Q₂ are obtained as below:

That is, I_C1 = I_C2 =0.988 mA.

Now, we can calculate the voltage between collector and emitter for Q₁ and Q₂ using the collector current as follows:

V_C1 = V_CC = -R_C1 I_C1 = 10 – (2.2kΩ) (0.988 mA) = 7.83 V = V_C2

Since the voltage at the emitter of Q₁ and Q₂ is -0.715 V,

V_CE1 = V_CE2 = V_C1 -V_E1 = 7.83 + 0715 = 8.545 V

Next, we will determine the collector current in Q₃ and Q₄ by writing the Kirchhoff's voltage equation for the base emitter loop of the transistor Q₃:

V_CC – R_C2 I_C2 = V_BE3 - R'_E I_C3 - R_E2 (2 I_E3) + V_BE= 0
10 – (2.2kΩ) (0.988mA) - 0.715 - (100) (I_E3) – (30kΩ) I_E3 + 10=0
10 - 2.17 - 0.715 + 10 - (30.1kΩ) I_E3 = 0

Hence the voltage at the collector of Q₃ and Q₄ is

V_C3 = V_C4= V_CC – R_C3 I_C3 = 10 – (1.2kΩ) (0.569 mA)

        = 9.32 V

Therefore,

V_CE3 = V_VCE4 = V_C3 – V_E3 = 9.32 – 7.12 = 2.2 V

Thus, for Q₁ and Q₂:
              I_CQ = 0.988 mA
            V_CEQ = 8.545 V
and for Q₃ and Q₄:
             I_CQ = 0.569 mA
          V_CEQ = 2.2 V

[Note that the output terminal (V_C4) is at 9.32 V and not at zero volts.]

(b). First, we calculate the ac emitter resistance r'_e of each stage and then its voltage gain.

The first stage is a dual input, balanced output differential amplifier, therefore, its voltage gain is

Where

 R_i2 = input resistance of the second stage

The second stage is dual input, unbalanced output differential amplifier with swamping resistor R'_E, the voltage gain of which is

Hence the overall voltage gain is

A_d= (A_d1) (A_d2) = (80.78) (4.17) = 336.85

Thus we can obtain a higher voltage gain by cascading differential amplifier stages.

(c).The input resistance of the cascaded differential amplifier is the same as the input resistance of the first stage, that is

R_i = 2β_ac(r_e1) = (200) (25.3) = 5.06 kΩ

(d). The output resistance of the cascaded differential amplifier is the same as the output resistance of the last stage. Hence,

R_O = R_C = 1.2 kΩ

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