regarding question 1.12 from Sedra and Smith MicroElectronics book (2009 international edition): A transducer characterized by a voltage of 1V rms and a resistance of \$1M\Omega\$ is available to drive a \$10 \Omega\$ load. If connected directly, what voltage and power levels result at the load? if unity-gain (i.e \$A_{v_o} =1\$) buffer amplifier with \$1M\Omega\$ input resistance and \$ 10\Omega\$ output resistance is interposed between source and load, what do the output voltage and power levels become? For the new arrangement, find the voltage gain from source to load, and the power gain (both expressed in decibles).

I do get the first two answers, but I don't get the other answers.

So wev'e got: \$ V_L = 10\mu V_{rms} , \ P_L= 10^{-11} W \$

I know that we have: \$v_o = v_i \frac{R_L}{R_L + R_o}= v_i \frac{10}{10+10} =0.5 \cdot v_i\$

Now if I am not mistaken, \$v_i\$ is the voltage of the transducer which is \$\frac{1}{\sqrt 2} \$, so unless I am mistaken here I should get \$ 0.5\cdot \frac{1}{\sqrt 2}\$, which is not \$ 0.25V \$, where did I go wrong here?

For the power I get:

\$P_o = v_o ^2/R_o + v_i ^2 / R_o = (1/8+1/2)/10 = 6.25\cdot 10^{-2} W\$ and again not as in the texbook which is \$6.25\cdot 10^{-3} W \$.

Again can someone please inform me what did I do wrong? perhaps I misused the formulas.

Thanks in advance.

  • 4
    \$\begingroup\$ The image is barely readable despite its huge size. Text in images cannot be indexed by search engines and therefore nobody with the same problem will ever find your question about it. Please take the effort of typing the questions in plain text (with reference to the book and its author). \$\endgroup\$
    – jippie
    Aug 18, 2013 at 15:29
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    \$\begingroup\$ It looks like you're mixing up RMS and amplitude. \$\endgroup\$
    – mng
    Aug 18, 2013 at 20:53
  • \$\begingroup\$ @jipple can you upvote my post, I added the question from the textbook. \$\endgroup\$ Apr 20, 2016 at 18:16

1 Answer 1


As mng mentions, you are already working with the RMS voltage, so no need to multiply by 0.707. The RMS is the equivalent of the same DC voltage, the actual amplitude of the signal would be ~1.414V.

Here is a runthrough in Maxima of the calculations:

enter image description here


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