Electronusa Mechanical System [Research Center for Electronic and Mechanical]

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The Impedance Matching in The Audio Signal Processing

Umar Sidik.BEng.MSc*Director of EngineeringElectronusa Mechanical System (CTRONICS)

*umar.sidik@engineer.com

1. IntroductionCommonly, impedance is obstruction to transfer energy in the electronic circuit. Therefore, theimpedance matching is required to achieve the maximum power transfer. Furthermore, theimpedance matching equalizes the source impedance and load impedance. In other hand, theemitter-follower (common-collector) provides the impedance matching delivered from the base(input) to the emitter (output). The emitter-follower has high input resistance and low outputresistance. In the emitter-follower, the input resistance depends on the load resistance, while theoutput resistance depends on the source resistance. In addition, this study implements the radialelectrolytic capacitor 10 100⁄ .

2. Analytical WorkIn this study, and form the Thevenin voltage, while and deliver ac signal as and

(figure 1).

(a) (b)Figure 1. (a). The concept of circuit analyzed in the study

(b). The equivalent circuit

2.1 Analysis of dcFirst step, we have to calculate the Thevenin’s voltage in figure 1:

= + ×For this circuit, is 5 , then:

= 24 Ω10 Ω + 24 Ω × 524 Ω34 Ω × 5= (0.71) × 5= 3.55

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Actually, in this circuit = , so = 3.55 .

The second step, we have to calculate : = −= 3.55 − 0.7= 2.85The third step, we have to calculate :

== 2.85150Ω= 19

2.2 Analysis of acIn the analysis of ac, we involve the capacitor to pass the ac signal and we also involve the internalresistance of emitter known as (figure 2).

(a) (b)Figure 2. (a). The ac circuit

(b). The equivalent circuit for ac analysis

The first step, we have to calculate in the figure 2:

= 25= 2519= 1.32Ω

The second step, we have to calculate ( ):( ) = ( + 1) ( + )‖

( ) = (200 + 1) (150Ω + 8.2Ω)‖1.32Ω

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( ) = (201) (158.2Ω)‖1.32Ω( ) = (201) 1158.2Ω + 11.32Ω

( ) = (201) 1.32208.824Ω + 158.2208.824Ω( ) = (201) 159.52208.824Ω

( ) = (201)(0.764Ω)( ) = 153.564Ω

The third step is to calculate :

= ( )= 1153.564Ω= 0.0065= 6.5

The fourth step is to calculate : == (200)(0.0065 )= 1.3The last step is to calculate : == (1.3 )(0.764Ω)= 0.9932= 993.23. Simulation WorkThe simulation work can be classified into the dc analysis and the ac analysis.

3.1 Analysis of dcIn the simulation, is 3 (figure 3), while in the analytical work is 3.55 .

The different of the analytical work and the simulation work is:

(%) = ( ) − ( )( ) × 100%(%) = 3.55 − 33.55 × 100%

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(%) = 0.553.55 × 100%(%) = 18.33%

Figure 3. in the simulation

In the simulation, is 2.25 (figure 4), while in the analytical work is 2.85 . The different of theanalytical work and the simulation work is:

(%) = ( ) − ( )( ) × 100%(%) = 2.85 − 2.252.85 × 100%

(%) = 0.62.85 × 100%(%) = 21.05%

Figure 4. in the simulation

In the simulation, is 15 (figure 5), while in the analytical work is 19 . The difference is:

(%) = ( ) − ( )( ) × 100%(%) = 19 − 1519 × 100%

(%) = 419 × 100%(%) = 21.05%

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Figure 5. in the simulation

3.2 Analysis of acIn the analytical is 6.5 (0.0065 ), while in the simulation is 0.07 (figure 6). Thedifference is:

(%) = ( ) − ( )( ) × 100%(%) = 0.07 − 0.00650.07 × 100%

(%) = 0.06350.07 × 100%(%) = 90.71%

(a) (b) (c)

(d) (e)Figure 6. (a). in the simulation at 1Hz

(b). in the simulation at 10Hz(c). in the simulation at 100Hz(d). in the simulation at 1kHz(e). in the simulation at 10kHz

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In the simulation, is 14.9 (figure 7), while in the analytical is 1.3 . The difference is:

(%) = ( ) − ( )( ) × 100%(%) = 14.9 − 1.314.9 × 100%

(%) = 13.614.9 × 100%(%) = 91.275%

(a) (b) (c)

(d) (e)Figure 7. (a). in the simulation at 1Hz

(b). in the simulation at 10Hz(c). in the simulation at 100Hz(d). in the simulation at 1kHz(e). in the simulation at 10kHz

In the simulation, is 0 at 1Hz, is 0 at 10Hz, is 0.05 at 100Hz, is 0.94 at 1kHz, 9.61 at10kHz, and 15.2 at 16kHz (figure 8). The difference is:

For 1Hz, (%) = ( ) − ( )( ) × 100%(%) = 1.3 − 0.011.3 × 100%

(%) = 1.30000 − 0.000011.3 × 100%(%) = 1.299991.3 × 100%

(%) = 99.99%

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For 10Hz, (%) = ( ) − ( )( ) × 100%(%) = 1.3 − 0.421.3 × 100%

(%) = 1.3000 − 0.000421.3000 × 100%(%) = 1.299581.3000 × 100%

(%) = 99.967%For 100Hz, (%) = ( ) − ( )( ) × 100%

(%) = 1.3 − 4.361.3 × 100%(%) = 1.3000 − 0.004361.3000 × 100%

(%) = 1.295641.3000 × 100%(%) = 99.66%

For 1kHz, (%) = ( ) − ( )( ) × 100%(%) = 1.3 − 391.3 × 100%

(%) = 1.3000 − 0.0391.3000 × 100%(%) = 1.2611.3000 × 100%

(%) = 97%For 10kHz, (%) = ( ) − ( )( ) × 100%

(%) = 1.3 − 83.31.3 × 100%(%) = 1.3000 − 0.08331.3000 × 100%

(%) = 1.21671.3000 × 100%

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(%) = 93.59%For 16kHz, (%) = ( ) − ( )( ) × 100%

(%) = 1.3 − 84.21.3 × 100%(%) = 1.3000 − 0.08421.3000 × 100%

(%) = 1.21581.3000 × 100%(%) = 93.52%

(a) (b) (c)

(d) (e) (f)Figure 8. (a). in the simulation at 1Hz

(b). in the simulation at 10Hz(c). in the simulation at 100Hz(d). in the simulation at 1kHz(e). in the simulation at 10kHz(f). in the simulation at 16kHz

In the simulation, is 0 at 1Hz, is 0 at 10Hz, is 0.32 at 100Hz, is 5.36 at 1kHz, is 53.8at 10kHz, and 85.3 at 16kHz (figure 9). The difference is:

For 1Hz, (%) = ( ) − ( )( ) × 100%(%) = 993.2 − 0.1993.2 × 100%

(%) = 993.1993.2 × 100%(%) = 99.98%

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For 10Hz, (%) = ( ) − ( )( ) × 100%(%) = 993.2 − 2.37993.2 × 100%

(%) = 990.83993.2 × 100%(%) = 99.76%

For 100Hz, (%) = ( ) − ( )( ) × 100%(%) = 993.2 − 24.6993.2 × 100%

(%) = 968.6993.2 × 100%(%) = 97.52%

For 1kHz, (%) = ( ) − ( )( ) × 100%(%) = 993.2 − 218993.2 × 100%

(%) = 775.2993.2 × 100%(%) = 78.05%

For 10kHz, (%) = ( ) − ( )( ) × 100%(%) = 993.2 − 466993.2 × 100%

(%) = 527.2993.2 × 100%(%) = 53.08%

For 16kHz, (%) = ( ) − ( )( ) × 100%(%) = 993.2 − 472993.2 × 100%

(%) = 521.2993.2 × 100%

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(%) = 52.47%In this study, the simulation shows that the and became stable started at 1 kHz.

(a) (b) (c)

(d) (e) (f)Figure 9. (a). in the simulation at 1Hz

(b). in the simulation at 10Hz(c). in the simulation at 100Hz(d). in the simulation at 1kHz(e). in the simulation at 10kHz(f). in the simulation at 16kHz