I've been reading J.White's High Frequency Techniques, and in one passage is says:

If the two port is a passive network, one might expect that the insertion loss could range from 0 dB (no loss) to some finite loss (a positive decibel value). But this is not necessarily so. For example, if the load is 25 Ohm and the generator impedance is 50 Ohm, installing a two port that is a low-loss transformer might increase the power delivered to the load, resulting in a negative decibel loss value, or power gain, and this would be obtained with a passive two-port network.

I do not quite understand how exactly can a transformer or other not externally powered element provide gain. Does it mean that the power is just transferred more efficiently? But then again, I'm not sure if this would be considered power gain.

  • \$\begingroup\$ "the generator impedance is 50 W" - this is gobbledygook. \$\endgroup\$ – Andy aka Aug 21 '17 at 10:24
  • \$\begingroup\$ Sorry, it didn't copy the Ohm symbol correctly, edited. \$\endgroup\$ – Anthropomorphous Dodecahedron Aug 21 '17 at 10:27
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    \$\begingroup\$ @AnthropomorphousDodecahedron You can type Ω for it if you haven't got an Ω to copy. And it should be "ohm", not "Ohm" if you go that route. \$\endgroup\$ – Andrew Morton Aug 21 '17 at 11:06
  • \$\begingroup\$ @AnthropomorphousDodecahedron better yet, use mathjax, \$50\Omega\$, spelt slash dollar 50 slash Omega slash dollar \$\endgroup\$ – Neil_UK Aug 21 '17 at 11:22
  • \$\begingroup\$ @Neil_UK But how to get a space between the magnitude and the unit? \$\endgroup\$ – Andrew Morton Aug 21 '17 at 13:23

You can match a 50 ohm source to a 25 ohm load using a resistive pi network however this will naturally attenuate the signal by no less than about 7.67 dB (see this calculator here and try it out your self).

So for a 50 ohm input and a 25 ohm output the resistors are: -

enter image description here

  • Shunt in = 30.015 kohms
  • Series = 35.44 ohms
  • Shunt out = 35.36 ohms
  • Attenuation = 7.67 dB

Or you could match using a pi network of reactive components and this will give less attenuation but makes it frequency dependent.

Or you can use a step down impedance matching transformer. This will have a step down ratio that is the square root of the impedance ratio i.e. 1.4142 for a 50 ohm source and 25 ohm load. This will give zero dB loss and so, with respect to matching with a resistive pi network, it represents an insertion loss of -7.67 dB or a gain of 7.67 dB.

  • \$\begingroup\$ Thanks. From what you said it seems like using matching transformer has significant advantages over over the pi network. As far as I understand a transformer is a more expensive and maybe space demanding solution, any other reason why one would use a pi network instead of a transformer? \$\endgroup\$ – Anthropomorphous Dodecahedron Aug 21 '17 at 11:01
  • \$\begingroup\$ The pi network is a general solution and, as you can see, it becomes what is known as an L network when you need the lowest loss i.e. the input shunt resistance tends to become infinite. So, if you are dealing with low powers and signal to noise ratios are not compromised, an L or pi network is convenient due to space and cost. \$\endgroup\$ – Andy aka Aug 21 '17 at 11:09
  • \$\begingroup\$ If you need decent power transfer then use a transformer although, as frequency rises to 100 MHz the transformer losses will increase and it becomes more effective to use a reactive L or pi network. This uses inductors and capacitors to match impedance and only "loses" maybe a couple of dB BUT it becomes usable only at a single frequency nominally. \$\endgroup\$ – Andy aka Aug 21 '17 at 11:13

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