4 - P In = 2 Vin R In 1.6-l I I iout ~-r Amplifier I v out...

This preview shows pages 1–5. Sign up to view the full content.

, 2 P, = Vin In R. In Amplifier ( 1.6) ,-- - - - - --l I I I i out I ~---r--'-" + I v out I I I I I I Load I '- --1 FIGURE 1.4 An illustration of the definition and use of the decibel (dB).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
The power delivered to the load is 2 P = V out out R L ( 1.7) The power gain of the amplifier is P G · Pout ·ower am -- Pin V;ut tc; --- vfn RL (1.8 ) The power gain expressed in decibels is defined as Power Gain dB - 10 loglo (P~ut) Pin (1.9 )
G ' Vout Voltage am = - Vin ( 1.10) G ' lout Current am = -,- lin (1,11) In dB these are defined as Voltage Gain dB = 20 loglo (V~ut) v in (1.12) Current Gain dB - 20 loglo (i~ut) . 1 m (1.13) ( ) 2 , V out Power Gam = - Vin R L = tc; 0.14) In dB the power gain becomes, for Ri~ = R L , ( 1.1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
In dB the power gain becomes, for Ri~ = R L , ( 1.1 from which we obtain the defining relation for expressing the voltage gain dB as in (1.12), In summary, the ratio of two quantities in dB is given by dB = 10 log,o (;:) (power) (1.1 dB = 20 loglO C:) (voltage) ( 1.1' dB - 20 log,o C:) (current) ( 1.1:
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 10

4 - P In = 2 Vin R In 1.6-l I I iout ~-r Amplifier I v out...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online