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nashelsky_15_4

# nashelsky_15_4 - For the ideal transformer the voltage...

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Unformatted text preview: For the ideal transformer, the voltage delivered to the load can be calculated using Eq. (15.9): The power across the load can then be expressed as Vﬂrms) RL PL: and equals the power calculated using Eq. (15.5c). Using Eq. (15.10) to calculate the load current yields I — I — N11 L 2 N2 C with the output ac power then calculated using PL = 1i(nns)RL EXAMPLE 15.4 Calculate the ac power delivered to the 8-!) speaker for the circuit of Fig. 15.10. The circuit component values result in a dc base current of 6 mA, and the input signal (Vi) results in a peak base current swing of 4 mA. VCC=IOV 18 = 6 mA Due to V,: I _ 4 Ficure 15.10 Transformer— b — mA ‘5 . . peak coupled class A amplifier for Example 15.4. Solution The dc load line is drawn vertically (see Fig. 15.11) from the voltage point: VCEQ Z Vcc = 10 V For 13 = 6 mA, the operating point on Fig. 15.11 is VCEQ = 10 V and ICQ = 140 mA The effective ac resistance seen at the primary is 1 N1 2 2 N2 The ac load line can then be drawn of slope —1/72 going through the indicated operating point. To help draw the load line, consider the following procedure. For a current swing of Chapter 15 Power Amplifiers A [C (mA) A [C (mA) dc load line 400 14 mA 400 14 mA 350 350 300 300 250 1C ax: 255 mA 250 200 NC 200 Operating point / 150 ICQ 100 50 . > [Cmin = 25 mA 0 5 10 15 2o 25 ch (V) VCE (V) (a) (b) Figure 15.11 Transformer—coupled class A transistor characteristic for Examples 15.4 and 15.5: (a) device characteristic; (b) dc and ac load lines. We 10V C R’L 729 139 mark a point (A): [C130 + IC = 140 mA + 139 mA = 279 mA along the y-axis Connect point A through the Q-point to obtain the ac load line. For the given base current swing of 4 mA peak, the maximum and minimum collector current and collector—emitter voltage obtained from Fig. 15.11 are VCE = 1.7V 1C. =25mA min min VCE : 18.3 V [C = 255 mA mu min The ac power delivered to the load can then be calculated using Eq. (15.13): (VCEm. — VCEminXICm _ 1cm) 8 (18.3 v — 1.7 V)(255 mA — 25 mA) — — 0.477 W 8 P0(ac) = Efficiency So far we have considered calculating the ac power delivered to the load. We next consider the input power from the battery, power losses in the ampliﬁer, and the over- all power efﬁciency of the transformer-coupled class A ampliﬁer. The input (dc) power obtained from the supply is calculated from the supply dc voltage and the average power drawn from the supply: P,-(de) = VCCICQ (15.14) For the transformer-coupled ampliﬁer, the power dissipated by the transformer is small (due to the small dc resistance of a coil) and will be ignored in the present calcula- 15.3 Transformer-Coupled Class A Ampliﬁer 759 ...
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nashelsky_15_4 - For the ideal transformer the voltage...

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