This video explains how to design impedance matching circuits using the Smith Chart.
Пікірлер: 17
@siddhantjaisalАй бұрын
At 3:25, Why in the example 1, the normalised load impedance is inside the circle? The distance of normalised ZL (from Euclidean distance formula) is coming 1.414 which is greater than 1. So option 2 should be taken.
@tomkuiper7662 жыл бұрын
At 3:44, if z_L = 2-1j, then y_L = 1/z_L = 0.4+0.2j. Confusion arises because gamma = 0.4-0.2j . The labels on the Smith chart are correct for admittance.
@soman61332 жыл бұрын
yah
@m9405043 жыл бұрын
2:28 Why can’t use type 1 circuit when the load outside 1+jx ? For example, if zl=0.55+0.7j,frequency=1 GHz I can parallel the 1.33 pF capacitance first, it will let the load on the boundary of 1+jx circle, then series 4.85 pF capacitance, the load will go to center.
@hamidk47723 жыл бұрын
Thanks 👍.
@ProCipher3 жыл бұрын
nice video thank u
@MaitreBart Жыл бұрын
About part 2: What if, hypothetically, the length of the transmission line is 50 cm (i.e. less than lA = 90.9 cm)?
@arenaengineering80703 жыл бұрын
Thanks.
@ibrahimozcan4452 Жыл бұрын
Thank you!
@empossible1577 Жыл бұрын
BTW...This is simply a video explaining how to do impedance matching on a Smith chart and does not go too deeply into the theory. If you are looking for the theory, try... Topics 7 and 8 here: empossible.net/emp3302/ Topic 2 here: empossible.net/academics/emp4301_5302/
@disasterprepper2 жыл бұрын
Really nice, but at 4:09, I think he means to say "we look at the admittance here..." rather than impedance.
@empossible15772 жыл бұрын
Thank you!
@CircuitShepherd Жыл бұрын
👍
@itsverygreen5322 жыл бұрын
Nope. kzfaq.info/get/bejne/gphklKVpnpPRZas.html You are correct that it removes reflection between the transmission line and the matching network (looking into the matching network it will look like 50 ohms, or whatever impedance you are designing for) ... but you are incorrect that "reflections between the load impedance and the matching network can occur" ... looking into the matching network from the load, it will match the exactly the impedance of the load, eliminating reflections. The reflection coefficient should be very low for a well deigned network. That's reslly the whol point of the matching network ... looking into the source side, it looks exactly like the source impedance (50R or whatever) and looking into the load side, it looks exactly like the load impedance.
@gittesilberglarsen1262 Жыл бұрын
Technically emposible is correct. Reflections between load and matching can occour, Actually this is the normal case. To see this easily, imagine the simple case of load =z0+jsL. The mathing network is a series capacitor = -jsL. Now what this means is that the energy is flowing back and forth between the L and the C. So reflection do occour always with reactive loads. But you are also correct the impedance seen from the load back towards the generator is a conjugate match _IF_ the source is z0 in the case illustrated in the video. But this may not be the case when dealing with power outputs where output impedance /= load impedance. So all in all the right answer is subtle.
@juniedmohammedtazrian4122 Жыл бұрын
it would be better if you show it by drawing on paper instead of showing it here. we could realize this better
@empossible1577 Жыл бұрын
Agreed. These are notes from an in-person class where we do just that. Having these animated and visualized alongside the circuit would also be great for the online content.