Large bandwidths and Wilkinsons are somewhat challenging. For that you'd have to employ an intermediate impedance and basically do a multi-stage Wilkinson with stepped impedances.
Hi sir, I realised a simple Wilkinson divider with 4 outputs, but I have a problem, the fouth port is not matched S55 greater than -10 dB. Please I need help, I want to know why I have this problem. Thanks
4 kW? That would pose extreme challenges if it's supposed to be a PCB design. Even neglecting any components on the PCB, just the dielectric strength of the PCB itself would have be extremely large.
Question: I ran an ADS model for a 2way Wilkinson combiner with 0dBm 0deg on each leg and output was +3dBm. Can you explain why when I either open or short one of the combiner inputs my output power goes to -3dBm? Also, is there a similar equation for an N way combiner as there is for an N-way splitter, that is, like dB=10log(1/N) ? thanks
Your model is correct: if you remove one input from a hybrid combiner such as this, you've obviously removed half the input power, so there's -3dB. Now, the power from the remaining input is split 50:50 between the output port and the Wilkinson resistor, so the resistor sees -3dBm and the output power is -3dBm. It's a consequence of preserving the input impedance when another input is not Zo. You can look at this hybrid in another way: each C-L-C leg is a narrow-band approximation to a quarter-wave transmission line, and for a two-way combiner this line should have a Zo of 70.7 ohms (i.e square root 50 x 100) . Quarter-wave line impedance Z transforms a resistance R1 at one end to R2 at the other according to the equation R1.R2 = Z(squared) . For our combiner, we're paralleling two lines at the output, so each needs to present 100 ohms to the common port. So, 50 ohms in, 100 ohms out, square root 50x100 = 70.7.
![Building a Wilkinson Splitter/Combiner [RF Circuit Design]](http://i.ytimg.com/vi/3IAVJ4gVunE/mqdefault.jpg)
![Resistive Splitters and Combiners [RF Circuit Design]](http://i.ytimg.com/vi/QkTryokCEZo/mqdefault.jpg)
Nice. I'm looking forward to the video on the etched Wilkinson that you tweeted pictures of!
It'll come, hang in there ;)
instablaster.
bester mann
Nice Video of Wilkinson theory! If I am making a combiner for 88-108MHz, how do I make the combiner wide enough to pass the Full 88-108 RF spectrum?
Large bandwidths and Wilkinsons are somewhat challenging. For that you'd have to employ an intermediate impedance and basically do a multi-stage Wilkinson with stepped impedances.
How do you design an N-way unbalanced-to-balanced power divider? Can you make a video about that?
Hi sir, I realised a simple Wilkinson divider with 4 outputs, but I have a problem, the fouth port is not matched S55 greater than -10 dB. Please I need help, I want to know why I have this problem. Thanks
May you show the math for power-handling-capability?
Do you have theory for building a PCB Wilkinsion N Way combiner 50 ohm in, 50 ohm out, w-4KW RF Output???
4 kW? That would pose extreme challenges if it's supposed to be a PCB design. Even neglecting any components on the PCB, just the dielectric strength of the PCB itself would have be extremely large.
KF5OBS How to draw the layout for Wilkinson microstrip line in ADS...?
Question: I ran an ADS model for a 2way Wilkinson combiner with 0dBm 0deg on each leg and output was +3dBm. Can you explain why when I either open or short one of the combiner inputs my output power goes to -3dBm? Also, is there a similar equation for an N way combiner as there is for an N-way splitter, that is, like dB=10log(1/N) ? thanks
Your model is correct: if you remove one input from a hybrid combiner such as this, you've obviously removed half the input power, so there's -3dB. Now, the power from the remaining input is split 50:50 between the output port and the Wilkinson resistor, so the resistor sees -3dBm and the output power is -3dBm. It's a consequence of preserving the input impedance when another input is not Zo. You can look at this hybrid in another way: each C-L-C leg is a narrow-band approximation to a quarter-wave transmission line, and for a two-way combiner this line should have a Zo of 70.7 ohms (i.e square root 50 x 100) . Quarter-wave line impedance Z transforms a resistance R1 at one end to R2 at the other according to the equation
R1.R2 = Z(squared) . For our combiner, we're paralleling two lines at the output, so each needs to present 100 ohms to the common port. So, 50 ohms in, 100 ohms out, square root 50x100 = 70.7.
wow