Balun tranmission Line, Please help

I'm assuming it is this ref design CC2530 Reference Designs  (ZIP  408 KB) from the CC2530 page and you are replacing the SMA connector with the ref design for the F antenna as shown in DN007 . Please provide the correct link if this is not it.

LZH said:

PCB is FR4, 1.6mm 2 layer PCB

 

LZH said:

so is my transmission also be 50 ohm?

A1.) The CC2530 is on .8mm(32mil) FR4 and you plan to use 1.6mm(63mil) FR4. You should not have a problem provided you (A2) increase the transmission line width to maintain its w/h. It is currently 1.4mm/.8mm which is 1.76 which is very close to 50 ohms.  On 1.6mm FR4 the line will be 1.76 x 1.6mm = 2.8mm

A3.)

LZH said:

"MSUB1" default parameters

 The default microstrip substrate parameters may be for any material, you need to set these for FR4. Default FR4 parameters should be sufficient given the simplicity of the design.  Yes. Microstrip Line (MLIN) is appropriate for the transmission line. The extent to which you can use the tool to model the circuit is limited since s-parameters are not available for the CC2530 and the antenna is not microstrip since there is no ground plane.

A4.)

LZH said:

What else shall I take note?

Stay close to the reference design for the RF portion and bypass cap placement and it will work. Be sure to scale the copper pour to transmission line gap for the thicker substrate.  Note the reference design does not have a RC / pull up on the /reset line. This is OK when connected to the programmer but when disconnected from the programmer /RESET must be high for the device to run. /RESET is toggled to enter debug / flash so R pull up can not be too small or C too large. If you are using the Smart RF04 a pull up will interfere with the board detecting the device.

 

Hi, Stewart,

Thank you for your reply. But I still have something not clear. According to TI DN007. "Since the impedance of the Inverted F Antenna is matched directly to 50 ohm no externalmatching components are needed" (at page 3), Is this mean the impedance of the inverted F antenna is 50 ohm already? If yes, according to the Johanson balun, at the datasheet page 7, its unbalanced impedance is already 50 ohm. I use Agilent ADS to calculate the impedance of the transmission line between the balun and antenna (ref CC2530 Reference Designs ), which is only about 20 ohm. So I'm quite confuse about this part of the impedance matching. Could you please give me some idea about this impedance match? Thank you very much.

Hi, Stewart,

Thank you very much. You helped me quite a lot. I need to design a 2 layer PCB using Johanson balun, and meanwhile design a 4 layer PCB using cc2591 in 2 weeks from now on. Hope I can finish that. May I ask one more question, why balun output 50ohm, transmission line 50ohm, and antenna 50ohm? I mean why not balun + transmission line totally 50 ohm? Thank you.

LZH said:

why balun output 50ohm, transmission line 50ohm, and antenna 50ohm? I mean why not balun + transmission line totally 50 ohm?

It is hard to give a short answer without over simplifying the problem but I'll try. The antenna, T-line and balun are not like resistors where you can simply add the numbers or measure the impedance with an ohm meter. As an example the T-line is close to zero ohms from one end to the other and infinite ohms from the line to ground on a regular ohm meter.

The 50 ohm impedance refers to the characteristic impedance of the item. That is the value of a real load (a resistor)  that results in all of the power being absorbed and none of it is reflected due to mismatch.

 It I feed a line that has a 50 ohm impedance with a signal generator that has a 50 ohm impedance and place a directional coupler on the other end of the line with it's output open or shorted to ground I will measure all of the power being reflected back towards the generator. With a 25 or 100 ohm resistor I will see about 1/10 of the power reflected back toward the generator and with a 50 ohm resistor I will see 0% of the power reflected toward the generator. The key is reflected power does not make it to the load. It is loss.

                                                                         Power meter = 0mw
                                                                                     |
             50 ohms                     50 ohms                  |         <<<      50 ohms
Generator- /\/\/\/\//\/\/\/\/\\--LINELINELINELINE-COUPLER-/\/\/\/\/\/\\/\/\/\GND

So a balun output 50 ohm, transmission line 50 ohm, and antenna 50 ohm results is 100% power transfer and no power lost due to reflections. In reality parts are designed and tuned to be close to 50 ohms but typically have some mismatch loss, as an example see the balun data sheet

Questions?

 

Thank you very much, Stewart.

Best regards.

Hi, Stewart,

 

First of all, thank you very much for helping me quite a lot. But new problem occurs.As you 

 

know, I was asked to design the cc2530 with balun in 1.6mm FR4 PCB. I used the ADS to calculate the TL width = 120mil. However, my boss ask me direct use 120mil polygon to fill from the unbalanced pin to the anttena input.As I know, the TL should be gradually increase and decrease, but I cannot say something wrong with my boss. If use my way, I dont know how to calculate the TL impedance, if use my boss way, what are the pros and cons for that? Thank you very much. Please help.

 

Best regards

Having a step from a narrow line to a wider line results in a little extra copper (The end corners of the wider line). Electrically these look like tiny capacitors between the line and ground (shunt). The amount of capacitance is a function of the width of the extra material and its length as a percentage of a wavelength. Also it depends on what type of circuit it is and if it is sensitive to adding a small shunt cap. An example of a sensitive circuit would be a narrow high Q filter or resonator in an oscillator.

In your case it is a very small amount of material, a very small percentage of a wavelength in length, and in a low Q fairly non sensitive part of the circuit. It should not cause a significant change in circuit performance. In fact the extra shunt capacitance may help compensate for the fact the line from the balun is narrower than ideal and therefore inductive. So keep the boss happy and stay with the step.

Using the HP design tool build a small model of a MLIN the width / length of the line from the balun followed by a MLIN of the length and wide of the transmission line, followed by a 50 ohm termination resistor and analyze it for insertion loss. Now add an open stub to the model where the two lines connect. You will find the stub has to get fairly long before you see its effect. After all at 2.4GHz a wavelength is 12.5cm (in air), a few mm of extra material is a very small percentage of a wavelength.

 

1) The balun can be scaled to keep the same impedance in the lines connecting the components.    These steps are shown in AN068.    The balun purpose is to take the two our of phase signals and add them in phase to get 2X the output signal.   The problem is the discrete components steps are not fine enough in resolution so the traces in between the two different sides create phase delay so the phases equal zero at the summing point. 

2) If I understand you are talking about a discrete transformer balun.  These come in varioius imedances 50:50, 50:100, or 50:200.  The high side is usually matched to the radio output but the impedance must be scaled so to match at the radio freqeuncy.  This is done by an external inductor and the traces between the radio and the inductor before it feeds the balun's input.  The output is 50 ohms so can be connected to a 50 ohm line.   We generally suggest a matching network pads be placed between the line and antenna's input in case there is a small difference.   If not just add a zero ohm resistor.

3)  We use ADS also,  the schematic simulator is close but not as accurate as making a layout changing to a simulation file then you get the effects of the boards magnetics.  We seen accuracy with this method of 1%.   With the method you use its more like 9-12% error.

4)  Once you are done connect to the output of the 50 ohm line and measure your output power from the radio.   You can set the radio to continuous single frequency using SmartRF Studio to 0dBm.  If you did a good job you should measure output power with less than .25 to .5 dB less than the 0dBm. 

You can also check your antenna's resonant frequency with ADS using FEM.   If you change the boards thickness the antenna will detune and its efficiency numbers will drop. The Inverted F antenna typically shows efficiencies of 93-97% for Omni-directional pattern.  I would suggest you check your antenna.  Send me your email contact and I'll provide setup information for simulating the antenna.

Rgds

RRS

Lower Power Wireless Support

 

 

Hi,

Please help me to get answer if I direct connect the balun unbalanced pin to the antenna pin? That means to ignore the transmission line, solder the 2 pad together. My boss idea is quite crazy, and I totally no idea about what will happen if I do such a design. I really cannot point what's wrong with this idea, but feel this world no one do this. Please help. Thank you very much.

 

Regards.

Hi, RRS,

Thank you for your help. I have got Microwave Engineering  2nd ed by David M. Pozer and RF circuit design : theory and applications 3rd ed. Hope it can help me to solve the environment I trapped. I just feel that the responsibility is quite large because the time is rush. However, I need to give 100 samples followed by 4000 products by NOV. I will try my best. Thank you again for your support. 

Best regards.

Using one of the reference EVM you can scale the design for new layer stack up.  This is discussed in AN068 application note.   If the traces become to large or small then you must look into other changes.   Usually for stack change of 2 to 1 you can just change the trace widths in the balun network. 

RRS

Hi, 

Sorry to interrupt you again. I asked a few manufacturers about the dielectric constant of FR4 1.6mm 2 layer PCB. The answer from they were almost same that dielectric constant is between 4.2 and 4.7, and they cannot control exactly. Is this the truth? May I know where TI find the manufacturer? Thank you.

Best regards.

First with a short 50 ohm line at 2.4GHz sensitivity to Er  in FR4 is minimal and assuming Er=4.2 will give good results. For higher Q (e.g edge coupled filters, broadband Lang couplers, etc.) or higher frequency components consider a low loss, low dispersion material such as those made by Rogers or other similar material designed with RF applications in mind.

FR4 is made from sheets of Prepreg, which is itself constructed from fiber glass matting which has been impregnated with the epoxy resin. A standard 1.6mm circuit board is made up of around 8 layers of Prepreg and the outer copper layers. As the weave of the fabric, the mixture of the epoxy, and the pressure they use to press it all into a nice flat board changes so does the dielectric constant. This is just one board. In a 4 layer "board" there are 3 boards pressed together and another opportunity to impact the dielectric constant a bit due to the heat and pressure. With these variables a Er of 4.1 to 4.7 is not bad and is true.  Another variable is the frequency they measure Er at (typically 1 MHz but not always) and how the specific material varies with frequency. See http://www.polarinstruments.com/support/cits/AP125.html for a short but interesting artical.

A good PCB house will take samples of the material they get and process, and then measure the dielectric constant over a wide frequency range. Using this information they will adjust the width of lines that are specifically called out to meet a certain impedance in your design. All for a price. If you get to know a shop well they will sometimes share the data so you can design to it.

Another variable often overlooked is the board stack up to get the 4 layers.  This can be (1) a thick core with a thin board (called prepreg) attached to the top and back, (2) two thick boards with a thin board in the middle, or (3) a thick board with 2 thin boards stacked on top.

The figure is taken from the Altium Stack Manager and shows case 1. This is likely the most common stack up and for low cost and quick turns we tend to use what they have. The overall board might be 1.6mm but with a thick core the top and bottom board could be very thin. (In fact from some off-shore very low cost suppliers the top and bottom can be a non-conductive spray-on with no tolerance at all and should not be referred to as FR4, but is) For line impedance all we care about is Top to Layer 2 (GND) thickness. Case 3 above might be the best; use the thick core for Top and Layer 2 where the lines are wider and less sensitive to width variations, the material tougher so changes less with processing, etc. Then use two thin board for layer 3 and 4. I most cases this would be a special order and expensive.

The take away is understand your PCB house's stack up for your boards and correctly set up the stacks in the CAD program.

LZH,

This does vary from board vendor to vendor.    What we learned is certain board house process is consistant enough to maintian the number will little variaiton.   We primarily use AdvancedCircuits   ( www.4pcb.com) for small quantities of boards and Sunstone (www.sunstone.com )  for larger volume tend to be cheaper.

We found they are consistant in their values.   This is probably more important than what the value they quote.   Both these board vendors at 2.45GHz have a permittiivity of 4.11 with tan functions of 0.0155. .   The Chip Con facilatity in Oslo uses a board vendor that quotes 4.22 at 2.45GHz.      This number increases as the freqeuncy drops.   So for 1Ghz the number is 4.54. 

We do not pay to have test coupons to deteremine contorled impedances but instead design the boards for minimum variation.  Your biggest concern should be how well the board vendor controls the permitivity.  They will not give this spec in terms of this number but as the tolorenace of the trace widths.    Using a line calcutar like the free one from AWR you can determine this variation by modeling each line as an impedance and calculating your reflections. 

Selecting  traces and components for values so with the vendors tolerances are minimum for the system is the goal of our EVM.     We target a low VSWR <1.5 then setting a acceptable level for production of VSWR <= 2 assures the wireless design works for Low Power Wireless Networks.  You can select tighter values but will need to go through the numbers to make sure this is achievable.    Of course this is easily done with the 3-D simulaiton packages. 

Regards,

RRS

Thank you for the professional answers above. H Stewart and RRS helped me quite a lot. Thank you very much.

Best regards

Hi, RRS,

Sorry for a few more question about the RF design for CC2530. First, the dimension of inverted F antenna in DN007, is slightly different from the CC2530-CC2591EM Reference Design. If we don’t care about that and use DN007, is it possible to insert a 120mil microstrip line since I’m using a 1.6mm 2layer FR4? I think it’s hard because 120 mil microstrip need at least 240 mil keepout region without the ground plane fill. However, as shown below, there is not enough space for the keepout even for the microstrip in the gap. Is that mean It’s not possible to use the inverted F antenna with a 1.6mm 2 layer PCB? Do you have any good suggestion about this? I’m looking forward for your helps. Thank you very much.

 

LZH,

DN007 is for 0.8mm board thickness.  I would advise you to use the dimension on the CC2430_CC2591 board layout because this is a 1.6mm (62mils)  and the antenna layout is tuned with an efficiency 94%.      Load the gerber files from  swra214a.zip file and use a gerber viewer tool to measure the various lengths. 

If you copy this one you will probably see a 2 to 3 dB drop in performance.  For this board thickness increasing the thickness changes the radiator impedance and length.  Then you are faced with changing the ground stuub and feedline impedance.  So the easy is to copy one already done for 1.6mm board.   We don't have a RF layout for the balun/filter network for a two layer thickness of 1.6mm (62mils)  so depending on the chip you are using will determine the target trace lines and components for this section.

IF you have read AN068 it takes a 31 mil board changes it to 62 mil and gives the steps to do this.  

IF your system is for a combo board then you have more work to do in terms of a 2 layer.   The CC2591 power amp has high gain and requires careful design to ensure phase margin.   This is an open loop amp at 2.45GHz so a very small amount of capacitance can create spuruors or un-stable design. 

Please advise exactly what configuration of radio devices you plane to use on this board. 

Cheers

RRS