Q of Pie matching network
I am designing a matching network to transform a 50Ω source to a complex impedance 300-j450Ω. I want a BW of 100MHz. My center freq is 1.42GHz. I use the formula Q=f0/BW, to find the required Q for my matching network. It is calculated to be 14.2. But if i design it this way i cannot achieve the BW i want. I need to design for Q=3 to achieve the 100MHz BW. Why the formula doesn't stand? Any ideas. Am i getting something wrong?
Thanx in advance
George
Using Pi (or T) network can increase the bandwidth by giving there an extra degree of freedom to control the value of Q in addition of performing impedance transformation.
A Pi matching network can be seen as two L-type networks back to back configured to match the load and the source to a virtual resistance located at the junction between the two networks. The virtual intermediate resistance must be smaller than either Rs or RL because is connected to the series arm of each L-section.
vfone thank you for your responce. These networks can be used only if you want to achieve a Q greater than the one provided by the L networks.. If you want to achieve something more wideband you need something else. However i have achieved the BW i need. My question is why the Pie network doesn't present the theoretical BW estimated by the Q=f0/BW formula.
George
Q=fo/BW is the Q factor of one LC resonator.
Always adding an extra element to the matching network increase the bandwidth. And because Pi-type actually it's an L-type plus another L-type, it will have greater bandwidth than a single one. Adding more sections, the bandwidth increase further.
In additional to that..
is valid Unloaded Resonant Circuits only...
Just a reminder...
Sorry, Bigboss, your formula also works for loaded circuits. Its the traditional way of working out the losses of the pi output matching network for transmitters, efficiency = (Qunloaded-Qloaded) / Qunloaded. Best circuit of all has a loaded Q of 0 (transformer), typical has Q unloaded of 100 and a loaded Q of 10, has 10% loss.
I think the problem is the -j450 term, because what is actually happening is you are designing a matching section to 300 ohms and loading it with an extra +j450 to resonate out the -j450. I wonder what the Q would be if you just looked at the circuits natural resonant frequency (without the +j450) and bandwidth?
Frank