0
\$\begingroup\$

casual beginner electronics hobbyist here.

I read an article online describing how to impedance match a resistive source impedance of 75 ohms to a resistive load impedance of 1k ohms. In order to accomplish this, the author of the article inserts what he calls an L-Network consisting of an inductor and a capacitor between the resistors as show below in “Image A”. He then goes on to explain how the circuit can be analyzed by taking the load resistor and capacitors and combining them into one as shown in “Image B”. The result of this combination can be seen in “Image C”.

The question is: Are both circuits (Image A and Image B) really the same in the real world? Would both circuits really provide the exact same output, performance etc?

Thanks.

enter image description here

Improve this question
\$\endgroup\$
3
  • \$\begingroup\$ Did you do the math? What is the effective impedance of the first circuit and what is the effective impedance of the second circuit? \$\endgroup\$ – The Photon Aug 15 '17 at 1:01
  • \$\begingroup\$ @The Photon: Aren't both the same impedance? Is that your point? \$\endgroup\$ – T555 Aug 15 '17 at 2:19
  • \$\begingroup\$ It only works precisely at one frequency and is a decent matching network over a small range of frequencies around that one frequency. It doesn't work at extremes of frequency. \$\endgroup\$ – Andy aka Aug 15 '17 at 7:24
1
\$\begingroup\$

These 2 circuits are equivalent indeed, given the condition that the frequency is fixed. This is implied by the fact that you have a fixed impedance for the capacitor. The impedance will change with frequency, and once that happens these circuits are no longer equivalent.

I know seems counter intuitive, but just imagine for a minute that the capacitor is like a resistor that decreases value as the frequency increases, therefore, having a "resistor" in parallel with the 1k load resistor, will make an equivalent impedance smaller than impedances. For that reason, you can say that it's equivalent to having a smaller resistor, such as the 75 ohms, plus a series capacitor.

A way to understand why they are not equivalent for every frequency is to think what would happen if your frequency increases a lot. The impedance of your capacitor would become very small, and for the circuit on the left side of Image B, it would look almost like a short to ground, whereas on the right side it would still have the 75 ohms of the resistor in series. Therefore, you have to set the frequency you want the have the equivalent circuit and chose the capacitor to achieve the desired impedance for that specific value.

Try simulating the following 2 circuits. At 10 kHz these values of capacitance will give you similar ratio as given by the article values. The waveforms are the voltages at the load of each circuit, and you can see they are the same for both circuits.

enter image description here

So if you ignore practical factors that could add small disparities between the circuits in image B, such as intrinsic capacitances, resistances, inductances and mechanical factors, you can say that they are equivalent and you can actually use it in a practical design, if you know your frequency will be fixed and steady.

Improve this answer
\$\endgroup\$
4
  • \$\begingroup\$ In the article he mentions "An often-welcomed side effect is that this L-Network acts as a low-pass filter". I have almost no experience with filters so I have no idea what I am talking about but would't the circuit in Image A have such filtering effect while the circuit in image C not have the filtering effect making them non equivalent in the real world? \$\endgroup\$ – T555 Aug 15 '17 at 3:10
  • \$\begingroup\$ Ok, I need to let this sink in a little. I figured that the circuit from Image A would filter high frequency signals (noise) because the capacitor is connected directly to ground so it would basically short those high frequency signals (much higher frequencies that the specific impedance match frequency). The circuit from Image C does not have the capacitor connected directly to ground so I figure it would not have the same effect. \$\endgroup\$ – T555 Aug 15 '17 at 3:38
  • \$\begingroup\$ Just to clarify, you cannot replace one with the other and simply forget about everything else and pretend they are really the same for 100% of the cases. You can do that if you have other parameters under control; mainly the frequency. The equivalency is just for the impedance seen from the source perspective, given that you have a fixed impedance for the capacitor and inductor, which implies that your frequency is fixed. \$\endgroup\$ – Lucas Aug 15 '17 at 3:42
  • \$\begingroup\$ Yout logic is correct. Therefore, as you increase the frequency they behave differently. \$\endgroup\$ – Lucas Aug 15 '17 at 3:44
1
\$\begingroup\$

Yep. From the "black box perspective" (that is, looking only from the ports, not at the connections inside) the two have the same steady-state behaviour. The transient behaviour might be different because the capacitor can store energy, but for RF work that's usually not important.

To help allay the natural tendency to assume this cannot be so, consider two 10k resistors in parallel and two 2.5k resistors in series. Despite the very different arrangement and values, both networks will appear the same to an outside observer.

Improve this answer
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.