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**Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits**

**Marks:** 5M

**Year:** May 2015, Dec 2015

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Derive expression for efficiency for Class A transformer coupled amplifier.

written 6.8 years ago by | modified 22 months ago by |

**Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits**

**Marks:** 5M

**Year:** May 2015, Dec 2015

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written 6.8 years ago by | • modified 6.8 years ago |

**Figure 1 : Circuit Diagram of Class A Transformer coupled amplifier**

Above Figure 1 shows the circuit diagram of class A transformer coupled power amplifier. In this circuit resistors R1 and R2 & Vcc are used to bias transistor in active region. Transformer provides impedance matching with load by adjusting turns ratio of the transformer.

Apply KVL from power supply to ground through collector-emitter,

Vcc - Icq*$R_d$ - VCEq - Icq*$R_E$ = 0

Where, $R_d$ = D.C. resistance of coil ≈ 0

Hence,

VCEq ≈ Vcc-Icq*$R_E$

Hence, the dc load line is almost parallel to y axis as shown in the Figure 2.

**Figure 2 : AC & DC load line of Class A transformer coupled amplifier**

Resistance looking into primary of transformer will be given by,

**Figure 3 : Resistance looking into Primary of transformer**

Turns Ratio is given by :

$\frac{N_1}{N_2}$ = $\frac{V_1}{V_2}$ = $\frac{I_2}{I_1}$

$\frac{N_1}{N_2}^2$ = $\frac{RL'}{RL}$

**RL' = $\frac{N_1}{N_2}^2$* RL**

When ac signal is applied at the input then Ic varies around Icq which results in variation of VCEq. From dc load line shown in Figure 2 shows that peak output voltage will be approximately Vcc therefore peak to peak voltage is 2Vcc. Hence ac load line is as shown in Figure 2 above,

Hence,

Vopp = 2Vcc & Iop = $\frac{2Vcc}{RL'}$ (1)

Poac = Vorms×Iorms

Poac = $\frac{Vop}{\sqrt{2}}$×$\frac{Iop}{\sqrt{2}}$

Poac = $\frac{Vop×Iop}{2}$

From equation (1) we can obtain Vop = Vcc & Iop = $\frac{Vcc}{R_L}$

**Poac = $\frac{Vcc^2}{2R_L'}$** (2)

Similarly, Pindc = Vcc × Icq

Since Q-point is located in middle of ac load line , Icq = Iop

Pindc = $\frac{Vcc^2}{R_L'}$ (3)

Hence efficiency is given by,

From equation (2) & (3),

η(max)% = $\frac{Poac}{Pindc}$×100

η(max)% = $\frac{Vcc^2}{2R_L'}$ × $\frac{R_L' }{Vcc^2}$ × 100

**η(max) = 50%**

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