Transformer coupled class a power amplifier

TRANSFORMER-COUPLED CLASS A POWER AMPLIFIER

This is also sometimes referred to as single ended power amplifier. The term “single ended” (denoting only one transistor) is used to distinguish it from the push-pull amplifier using two transistors.

In case of a direct-coupled class A power amplifier shown, the quiescent current flows through the collector resistive load and causes large wastage of dc power in it. This dc power dissipated in the load resistor does not contribute to the useful ac output power.

Furthermore, it is generally inadvisable to pass the dc through the output device such as in a voice coil of a loudspeaker. For these reasons an arrangement using a suitable transformer for coupling the load to the amplifier is usually employed, as shown. This arrangement also permits impedance matching.In a power amplifier circuit shown Rt and R₂ provide potential divider bias ing and emitter resistor R_E is meant for bias stabilization. The emitter bypass capacitor C_E is meant for R_E to prevent ac voltage. The input capacitor C_in couples ac signal voltage to the base of the transistor but blocks any dc from the previous stage. A step-down transformer of suitable turn ratio is provided to couple the high impedance collector circuit to low impedance load.

Impedance Matching. The power transferred from the power amplifier to the load (say a loudspeaker) will be maximum only if the amplifier output imped­ance equals the load impedance R_L. This is in accord­ance with the maximum power transfer theorem. If we were not able to achieve the above condition, lesser power will be transferred to the load R_L, though the amplifier is capable of delivering more power, and rest of power developed would be lost in the active device. Hence for transfer of maximum power from amplifier to the output device matching of amplifier output impedance with the impedance of output device is necessary. This is accomplished by using a step-down transformer of suitable turn-ratio. The transformer impedance matching circuit is shown separatel. where R_L‘ is the resistance looking into the primary of the transformer and is given as

R_L‘/R_L = (V1/I1) ÷ (V2/I2) = V1 I2/V2 I1 = (N1/N2)² where V1/V2 = N1/N2 and I2/I1 = N2/N1

Thus the ratio of the transformer input and output resistances varies directly as the square of the trans­former turn ratio :

R_L‘/ R_L = (N1/N2)² = a² or R_L‘ = a² R_L

where a is ratio of primary to sec­ondary turns of step-down trans­former, R_L is the resistance of load connected across the transformer secondary and R’_L is effective resist­ance looking into the transformer primary.

Circuit Operation. In this circuit dc (winding) resistance deter­mines the dc load line. Typically, this resistance is quite small (assumed to be zero) providing dc load to be a vertical line rising from V_cc, as shown. When an ac signal is applied to the base of the transistor the collector current will vary around the operating point Q.

In order to have maximum ac power output, the peak value of collector current due to input ac signal alone should be equal to the zero-signal collector current. To achieve this, the operating point Q is located at the centre of the ac load line. This is achieved by adjusting the biasing circuit (R_1, R₂ and R_E). When ac signal is applied, collector current fluctuates from maximum to minimum (zero), and operating point Q moves up and down the load line. At the peak of the positive half cycle of the input signal, the total collector current I_{c max} = 2 I_c and collector-emitter – voltage V_{cg min} = 0 while at the peak of the negative half cycle of the input signal, the collector current I_{c min} = 0 and collector-emitter voltage V_{ce max} = 2 V_cc. Thus collector-emitter voltage varies in opposite phase to the collector current. The variation of collec­tor voltage appears across primary of the transformer. Now ac voltage is induced in the transformer secondary which in turn develops ac power and supplies to the load.

Circuit Analysis. In an ideal transformer, there is no voltage drop in primary so V_CEQ = V_CC and power input to the transistor becomes equal to the dc power drawn from the collector supply V_cc as power loss in the primary is negligibly small. Thus power input to the transistor, P_tr = Power drawn from collector supply, P_{in (dc)} = V_CC I_{CQ  and overall efficiency becomes equal to collector efficiency and = Pout (ac)/ VCC ICQ}

Under condition of development of maximum ac power, voltage swings from V_{ce max} to zero and collector current from I_{c max} to zero. So

V_rms = 1/√2 { [V_{ce max} – V_{ce min}]/ 2 } = V_{ce max}/2√2 = 2V_CC/2√2 = V_CC/√2

And

I_rms = 1/√2 { [I_{c max} – I_{c min}]/ 2 } = I_{c max}/2√2 = 2I_CQ/2√2 = I_CQ/√2

AC power developed across the load,

P_{out (ac)} = V_rms I_rms = (V_CC I_CQ)/2

Collector efficiency = P_{out (ac)}/ (V_CC I_CQ)/2 ÷ V_CC I_CQ = 0.5 or 50%.

Thus for a transformer-coupled class A power amplifier the maximum theoretical efficiency is 50%. In practice, the efficiency of such an amplifier is somewhat less than 50%. It is about 30%.

The efficiency of a transformer-coupled class A power amplifier can be given as

Efficiency = 50 *{ [V_{ce max} – V_{ce min}]/ [V_{ce max +} V_{ce min}]} %

The larger the value of V_{ce max} and smaller the value of V_{ce min} the closer the efficiency approaches the theoretical limit of 50%. Well-designed circuits can approach the limit of 50%. The larger the amount of power handled by the amplifier, the more critical the efficiency becomes!