Let's say I have an IC (ESP8266) with an output impedance of 39+j6 ohms, and I want to connect it to a 50-ohm antenna. The IC has a real (resistive) part of 39 ohms and an imaginary (reactive) part of 6 ohms. On the other hand, the antenna has only a real part of 50 ohms. According to the document of this link and the online designer of this link, a matching network can be developed.

If the load is a real 50 ohms, then how are the real parts matched without the presence of resistance in the matching network?

  • \$\begingroup\$ Resistance would dissipate power, not help couple it to the load. \$\endgroup\$
    – hobbs
    Sep 1 at 18:46
  • \$\begingroup\$ Don't think of it as resistance like a resistor - think of it as voltage-to-current ratio. Capacitors and inductors can "create" or "destroy" voltage or current since it's not really created or destroyed, just absorbed during one part of the wave and released during the other part. \$\endgroup\$
    – user253751
    Sep 1 at 19:47
  • \$\begingroup\$ "without the absence" <-- do you mean without the presence? \$\endgroup\$
    – Andy aka
    Sep 1 at 19:56
  • \$\begingroup\$ Yes I mean without the presence. Sorry \$\endgroup\$
    – Sadat Rafi
    Sep 2 at 13:26

2 Answers 2


If the load is a real 50 ohms, then how the real parts are matched without the absence of resistance in the matching network?

Quite simply but, it's all in the math and not obvious when looking at an L-pad for in stance: -

enter image description here

The picture above comes from my basic website and, it offers a calculator that allows you to pick values for L and C that match a 50 Ω source to a 300 Ω load. Of course, because there is a calculator you can choose to match whatever resistance value you want to which ever load resistance you want. That's up to you.

But, I also provide a derivation of the values for L and C: -

enter image description here

So, if you are interested in how the input impedance of a network formed by L, C and a load resistance looks like a resistor then take a look. For a simple L-pad it's this: -

enter image description here

How does a PI or L network matches impedance even though it does not have a real component?

For a Pi-network, break it into two back-to-back L-pads as shown on my basic website: -

enter image description here

The IC has a real (resistive) part of 39 ohms and an imaginary (reactive) part of 6 ohms.

That's easily solved; if the reactance is capacitive then use a series inductor of the same reactance to cancel it out. Then you are left with a resistive output impedance at the operating frequency you are concerned with. You can do the same with loads that are reactive and this is what people do.


...then how the real parts are matched with[out] the absence of resistance in the matching network?

If the matching network were to have any resistive component, then some power flowing through it wouldn't reach the load (it would end up as heat in that resistive component).
The matching network can/should only contain reactive components that dissipate no power. Any/all power accepted at its input must end up delivered to its output.

The assumption here is that power impinging on the matching network is in a very narrow bandwidth - we generally design these assuming a sinusoidal wave at one frequency.
The (lumped) matching network for 39ohm -> 50ohm must contain one capacitor and one inductor. The reactive part of the 39 ohm source is absorbed by the networks' series arm:


simulate this circuit – Schematic created using CircuitLab

These two examples ensure that sources V1 and V2 see no net reactance. Each delivers maximum available power to a pure 50 ohm load; at one frequency (2400 MHz).
Since the source/load impedances are close, components inside the L-network don't have to work too hard (Q is low). For the PI-network, Q is higher, which means that its 3 matching network reactances must work harder (pass more peak current). Reactive components always contain some small undesired resistance; the PI-network will dissipate more power than the L-networks shown, delivering somewhat less power to the load.

So why use a PI-network if its losses are greater?
If you must attenuate spurious emissions, the higher-Q PI-network is better, since its higher Q attenuates frequencies other than the design frequency. You have freedom to set the Q of a PI-network to anything you wish, but component tolerances become tight for a high-Q design.


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