L matching network Q????

l match network

Hi there

I want to match 2.3 ohm (resistive) to 3.3 -j 3.94 ohm using a L network at 4.32 MHz, my problem comes in getting the right loaded Q. I designed the matching network and get a Q of 1.5. But i need a Q of between 3 and 4!!!. How do i solve this? can i augment the network? how do i do this?

Any help will be great.
Thanks

l matching network

maybe you can try T or PI network

high pass l-network

I specifically want a L network so as to minimize components. A T and pi network has a high amount of insertion loss as well which is undesireable in this application.

I want to know how to augment the circuit so as to increase the Q, i know it can be done but am not sure of the method.

compare pi and l matching network losses

Is this a measured or calculated result?

You are working at low impedance. Have you considered the losses? You may need thick wire for the coil and an expensive capacitor.

pi match network q

The actual implementation is not a problem at the moment, getting the desired Q is the main factor.

matching network losses

L networks have the Q determined by the ratio of load and source impedances. You have to use a three reactive element network to be able to specify the Q. If you want low loss, use the shunt capacitance, series inductance pi type.

l-matching network

Is there no way to increase the q of a L matching network? for instance by adding a capacitor in series with a inductor, the way you augment a resonator when building oscillators.

matching network q

Once you add a third component, you are back to the three component network. You have several choices for this, high pass and low pass, T or pi, to get four possibilities. You can also make a band pass filter type with a tapped capacitor.

In general, the network losses will increase with higher Q because of the higher circulating currents.

One thing you might try is to convert the complex load to parallel form mathematically. Then design a filter and absorb the parallel reactance into the end element.

I attach a picture of a network designed by someone else. This network was the same as the one i designed but by adding the capacitor the q was increased to 3 from 1.5. I want to know how this was done, as i dont have contact with the original designer.

In the picture C1 was added in series with L1 in order to increase the Q. L1 and C2 are the standard low pass L network components.

Any help?

Hi Willem,

I just want to know the Q you mentioned is for which network. As we know for L network, the Q is fixed by the input resistance and output resistance. Since you want to match 2.3 ohm to 3.3 -j 3.94, you should include the -j3.94 to your network first, so the matching is from 2.3 to 3.3, the Q is fixed if you still want to matching it by a L network (here the L network contains the -j3.94, the Q is for the entire LC network). If you regard the "L" network between 2.3 ohm and 3.3-j3.94, then the real network between 2.3 and 3.3 is not a L network, thinking about the -j3.94 it is a cap at the right side of C2. Then you say the Q of L1, C1 or C2, it is not fixed by the resistance ratio. This is a problem of how to specify the Q of T or PI network.

If you match 2.3 to 3.3-j3.94 by the network you mationed, then in fact you are not using a L network. You can reduce L1 and C1 to a component if you have a suitable valued L or C. L serials with C, it is nothing, it is still a single L or C, depended by the value of L1 and C1. unless you don't want to L1 affects the DC bias of the right side circuit, C1 is no use.

In fact, all the matching is performed between real resistance, the image part of impedance is absorbed into the matching network. Hope it helps.