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This is not based on any particular hardware. I am trying to understand the theory behind it by doing tutorial questions from my uni, I have completed the question but don't understand the approach taken by the worked solutions.

schematic

simulate this circuit – Schematic created using CircuitLab

Above schematic is my representation of the circuit, bascially \$i_1(t) = 0.1\cos(2\pi f)\$, f= 10 Mhz and the Amp has a dB gain of 25 (50Ω matched) and the load is 50Ω.

we are asked to calculate the output power, so i simply worked out the RMS current 0.071 and times it by R1 (50Ω) to work out the input power (0.25 W) and using the gain (G=Pout/Pinput) \$P_{out} = P_{input} \times G\$ (316.2 ratio form) making Pout = 79.7 W. However in the answer sheet they do the same steps except for \$P_{input}\$ they use \$P_{input} = (I_{rms} \times R_1)/4\$.

I don't know where this four has come from and why \$P_{input}\$ needs to be divided by four???

Nb: \$P_{out}\$= power out; \$P_{input}\$ = power in

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  • \$\begingroup\$ Maximum power transfer theorem?? \$\endgroup\$ – nidhin Jun 18 '14 at 5:44
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Here, power gain given is for the amplifier, not for the entire circuit. So you have to find the power at the amplifier input first. Then multiply by G.

The current \$i_1\$ is split equally across \$R_1\$ and \$R_2\$. Therefore current to the input of amplifier (flowing through \$R_2\$) is \$i_1/2\$. Therefore power transferred from source to amplifier input is \$(i_1/2)^2\times R_2 = i_1^2R_2/4 = i_1^2R_1/4\$.


OR

Max power transfer theorem also says that when the loads are matched, the power transferred is \$I_{rms}^2R_L/4\$. Here source is \$i_1\$ (with \$R_1\$ in parallel) and load is amplifier (input). Hence the answer.

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  • \$\begingroup\$ Thanks! Makes so much more sense now, power gain is for the amp! So if they were not match i could just use the current divider equation to find the current at the amp input? I need to work on my definition of gain. Thanks a lot! \$\endgroup\$ – user3474943 Jun 18 '14 at 13:16
  • \$\begingroup\$ @user3474943 Yes. If they don't match, you have to use the current division. \$\endgroup\$ – nidhin Jun 18 '14 at 13:18

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