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The question is. What value for R1 is correct if we want the total power amplification to become 30 dB

My calculations. 30dB is amplification of 1000 times. Last circuit got amp of 15/1,5=10 times. 1000/10=100

We therefor need amp of first circuit to be 100 times.

10/R1=100

R1=0.1kohm

We could tick in 100 ohm but according to solutions manual the correct answer is 3.2kohm and I do not understand why.

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  • \$\begingroup\$ "30dB is amplification of 1000 times". No, it isn't. Take another look into the use of decibels for voltage gain. \$\endgroup\$
    – Finbarr
    Commented Jan 11, 2018 at 17:12
  • \$\begingroup\$ "30dB is amplification of 1000 times". Yes it is, for power amplification. But without knowing source and load impedances, the question is unanswerable in current form. Informally, solve it by assuming they don't mean power gain = 30dB, but voltage gain = 30dB. \$\endgroup\$
    – user16324
    Commented Jan 11, 2018 at 18:45

2 Answers 2

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30 dB amplification as a voltage ratio is \$10^{30/20}\$ = 31.62.

Given that the 2nd stage has a gain of 10 the first stage has a gain of 3.16 hence, R1 approximately equals 3.162 kohm.

30dB is amplification of 1000 times

30 dB is a power ratio of 1000 and a voltage ratio of \$\sqrt{1000}\$

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  • \$\begingroup\$ I forgot to mention but it was 30dB power amplification and not voltage \$\endgroup\$
    – J. Doe
    Commented Jan 11, 2018 at 17:40
  • \$\begingroup\$ 30 dB of power amplification is still an equivalent voltage amplification (assuming same value input and output resistances) of 31.62. Do the math. \$\endgroup\$
    – Andy aka
    Commented Jan 11, 2018 at 17:43
  • \$\begingroup\$ Second circuit got amp of 10 times. 20*log(10)=20dB voltage amplification. First circuit must have 10dB amp which is 3.16 times. I get the correct answer when I calculate for voltage gain but not for power \$\endgroup\$
    – J. Doe
    Commented Jan 11, 2018 at 17:43
  • \$\begingroup\$ Apart from anything else, unless you show an output load resistor there is no power amplification because the load is open circuit and therefore takes zero power. \$\endgroup\$
    – Andy aka
    Commented Jan 11, 2018 at 17:47
  • \$\begingroup\$ Okay but our textbook states that: power gain = 10^(power gain (dB)/10) and voltage gain = 10^(power gain (dB)/20) so you mean that because there is no load attached I should calculate as voltage gain despite that the text tell me that the power amplification is 30dB? \$\endgroup\$
    – J. Doe
    Commented Jan 11, 2018 at 17:49
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You mixed up power and voltage gains. As per the question Power amplification is:

$$ 10 log (P_o/P_i) = 30 dB $$ Since P is proportional to \$ V^2 \$ , we can also write: $$ 10 log (V_o^2/V_i^2) = 30 dB$$ Or $$ 20 log (V_o/V_i) = 30 dB$$ $$ i.e., log (V_o/V_i) = 1.5 dB$$ Therefore the corresponding voltage amplification is: $$ (V_o/V_i) = 10^{1.5} = 31.62$$

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    \$\begingroup\$ Decibels, decibels or dB and not DB nor Db nor db or anything with an "s" on the end to pluralize it. The d is for deci and it is used lower case. \$\endgroup\$
    – Andy aka
    Commented Jan 11, 2018 at 18:12
  • \$\begingroup\$ oh yea......correct. I was thinking something looks weird in the equations... got it :D \$\endgroup\$
    – Mitu Raj
    Commented Jan 11, 2018 at 18:29

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