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I am a bit confused about Impedance Matching for maximum gain and for maximum power transfer.

I knew that the conjugate matching between, for instance, an amplifier's output port and a load Zl, is given by the condition Zout = Zl*. Under this assumption, reflection at the interface between them will be zero.

Does this operation correspond to maximum gain condition, or maximum power condition, or both? Is there a difference between these two conditions?

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2 Answers 2

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There’s a difference.

  • conjugate matching reduces the VAR LOSS to real power transfer.
  • maximum power when matched results and 50% efficiency from a voltage source.

  • Higher efficiency and gain means less output power but lower input signal when load is say higher impedance than source. At extremes Gain with no load is twice the voltage but no current so no power.

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  • \$\begingroup\$ Thank you for the answer. Precisely, what do you mean with VAR LOSS? \$\endgroup\$
    – Kinka-Byo
    Jun 30, 2019 at 20:16
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    \$\begingroup\$ And, another question, so when you say "At extremes Gain with no load is twice the voltage but no current so no power.", do you mean "voltage gain", correct? \$\endgroup\$
    – Kinka-Byo
    Jun 30, 2019 at 20:19
  • \$\begingroup\$ Go search VAR... \$\endgroup\$ Jul 1, 2019 at 0:15
  • \$\begingroup\$ But if with gain you mean "power gain", i.e. the ratio between power delivered to the load and input power, maximum gain = maximum power transfer? \$\endgroup\$
    – Kinka-Byo
    Jul 1, 2019 at 1:28
  • \$\begingroup\$ Yes there is a difference between current,impedance reduction ( emitter followers) and Voltage gain with for some Zin, Zout and with same impedance in/out when designs for this produces max when matched to source. \$\endgroup\$ Jul 1, 2019 at 2:09
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Depending on narrow-band versus broad-band design, your SNR becomes a key design variable because broad-band interfaces that have 50ohm Rsource and 50ohm Rload will cost you 6dB signal level.

Yet you have the magic-incantation of "but boss, I matched it".

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