I am trying to Design a center feed PCB Dipole antenna for an 868.6Mhz frequency.
I have since come up with this discrete circuit that will be positions at the end of a 50 Ohm microstrip. The values shown is default and still has to be calculated but before I can do that I need to complete other considerations first. After following THIS white paper on dipole design I came up with the following results. Baluns for this frequency are to expensive. Ground clearance on 2 layer board is 20.00mm away from dipole legs top and bottom Parameters
FR-4 dk= 4.2
Substrane thickness 1.0mm
Freq 868.6Mhz
13dBm Tx power
Full wave length 345.1444mm
half wave lenght 172.5722mm
Using the quick cheat and times half the wave length with 0.47 I see that one leg of the dipole will be 81.1089mm. To confirm this I tried working it out manually but not sure the results are accurate as they seems very far out of per portion when using a trace width of 0.6mm. I get the following result
Impedance 90.73 ( I assume this does not apply same as microstrip since its an antenna but if 50 ohm is required the trace will have to be 2.0mm wide) Erff = 2.959 the square of that is 1.7201.
Deviding this value into the actual speed of light I get 1,74*10 power of 8 then divide in my frequency into this value I get= 20.065 x 0.47 = 90.43mm
This result compared to the 81.1089 seems far off. why is this ? when I tweak the values ( mainly trace width ) the trace width needs to be almost 10mm wide to get close, is this normal? what is the other approach. See attachment for antenna layout.
Help required.
1) Does the math make sense?
2) Will this discrete network solution work ?
I have since come up with this discrete circuit that will be positions at the end of a 50 Ohm microstrip. The values shown is default and still has to be calculated but before I can do that I need to complete other considerations first. After following THIS white paper on dipole design I came up with the following results. Baluns for this frequency are to expensive. Ground clearance on 2 layer board is 20.00mm away from dipole legs top and bottom Parameters
FR-4 dk= 4.2
Substrane thickness 1.0mm
Freq 868.6Mhz
13dBm Tx power
Full wave length 345.1444mm
half wave lenght 172.5722mm
Using the quick cheat and times half the wave length with 0.47 I see that one leg of the dipole will be 81.1089mm. To confirm this I tried working it out manually but not sure the results are accurate as they seems very far out of per portion when using a trace width of 0.6mm. I get the following result
Impedance 90.73 ( I assume this does not apply same as microstrip since its an antenna but if 50 ohm is required the trace will have to be 2.0mm wide) Erff = 2.959 the square of that is 1.7201.
Deviding this value into the actual speed of light I get 1,74*10 power of 8 then divide in my frequency into this value I get= 20.065 x 0.47 = 90.43mm
This result compared to the 81.1089 seems far off. why is this ? when I tweak the values ( mainly trace width ) the trace width needs to be almost 10mm wide to get close, is this normal? what is the other approach. See attachment for antenna layout.
Help required.
1) Does the math make sense?
2) Will this discrete network solution work ?
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