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Circuit Sim Waves AC response

This circuit is supposed to be a low noise method of getting significant (12dB or 4V/V) voltage gain. I noticed that when I impedance match the secondary load to the source resistance, the voltage at the primary (V_load1) is half the source voltage 50mVp-p. The differential outputs, are only 100 mVp-p indicating a gain of 2 relative to the primary voltage, but a gain of 0 relative to the source voltage.

This circuit will be going into a differential ADC eventually, so if I look at the differential output voltage (v+-v-) I would say I have a 6dB gain from the source, but thats just do to the nature of differential signaling, .

Is this 6dB loss due to impedance matching unavoidable if I want to impedance match (this is a very general question about impedance matching in general)?? Would I need to impedance match this circuit at all?

As well, is it possible to get 12dB gain out of a 1:4 transmission line balun circuit? Or would I need a 1:8 to actually get a 12dB gain with proper impedance matching?

My only idea would be to increase the secondary load resistance so that more voltage gets transferred, but then I'm no longer impedance matched so would that cause unwanted signal reflections and poor power transfer?

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2 Answers 2

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All matched Impedance circuits transfer at -6 dB

  • so with a gain of 2 differential out , or 6 dB with a matched load, you are still 12dB greater than the source or x4.

But by comparing source voltage of each leg out, the voltage is the same as your source but now at 8x the impedance.

No such thing as a free lunch from passive splitters.

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  • \$\begingroup\$ So did I really get 12dB greater than the source? It feels like my difference signal is 12dB greater than the -6dB matched impedance signal. \$\endgroup\$
    – Max W
    Commented Jul 1, 2021 at 0:24
  • \$\begingroup\$ That’s what I said \$\endgroup\$ Commented Jul 1, 2021 at 6:18
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A passive balun circuit cannot create power from nowhere. If you have a 50 Ω source driving a 50 Ω load then the voltage across that load will be 50% of the unloaded output voltage. That is not only inevitable but desirable.

So, with a 1 volt open circuit 50 Ω source, the terminal voltage when you connect a 50 Ω load is 0.5 volts. The power into the load is therefore \$0.5^2 / 50\$ = 5 mW.

If you used a transformer/balun of 1:4 step up ratio, that power is transferred to an impedance of 50 Ω x \$4^2\$ (800 Ω) at a voltage level of 4 x 0.5 volts. That's 2 volts and, of course 2 volts and an 800 Ω load is a power of \$2^2 / 800\$ = 5 mW.

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  • \$\begingroup\$ This helped clear up the conservation of energy perspective thank you. \$\endgroup\$
    – Max W
    Commented Jul 1, 2021 at 22:01

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