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Confusion on using Duty Cycle Factor on FCC Part 15.35 Part 15.209 and Part 15.247

Other Parts Discussed in Thread: CC2530, CC2591

I would like to ask if the final average limit is deduce by equation a) or equation b)?? . Our device was set to 100% duty cycle during testing.

a) Final average value = Measured Peak value + 20log(duty cycle)

b) Final average value = Measured Average value + 20log(duty cycle)

Which one is correct?

I have check a test report from Ember, they use equation b).

http://ember.com/faq/files/EH0533-2_EM260_RCM_Report.pdf

http://ember.com/faq/fcc-certification

However, our local test lab (SGS) said the calculation should base on equation a). The lab tested the device with both peak detector and average detector.

Now, i have some problems with CC2530 that we over the limit for 2dB in the average value at 3rd harmonics. I calculate our worst duty cycle is about 17ms/100ms = -15dB.

If DCF is applied for equation b), i have no problem in releasing the product at all. If not, gotta to redesign the product.  :((

Please see if anyone can give me some advise ......

  • Hi Leo

    My understanding is similar to the interpretation by your test lab. Measure spurious emission in continuous transmit (modulated random data) and compensate (subtract) for maximum theoretical duty cycle in any 100mS interval. (20log(DSmax).
    I have sometimes seen that un-modulated carrier have been used in this measurement resulted in increased values.  This is not normal operation and modulated signals should be used.

    Sometimes I have encountered test houses that have forgot to use band stop filter on fundamental when measuring harmonics. In these tests harmonics originating from test setup LNA significantly influenced final measurement result. However this is usually mainly a problem when design consists of CC2530 and CC2591 front-end.

    In Ember document I suspect Havg relates to VBW in measurement while peak is measured with higher VBW to determine absolute max peak. Remember no peak emission is allowed above +20dB to average requirement under any circumstances.  

    If your device have been tested according to this, you only have one alternative to redesign. Reducing output power usually significantly reduces harmonics. i.e a 2dB power reduction generally result  >>2dB drop on harmonic level. Using current design at reduced output power might hence be a alternative to redesign.

    Tor-Inge

     

  • Hi Tor-Inge,

    thx for your fast reply.


    The TI document made me confuse is that
    http://www.ti.com/lit/an/swra396/swra396.pdf

    Qoute
    "3.2 Duty Cycle
    "Duty cycle is used to reduce the value measured with an average detector. ZigBee devices typically have low duty-cycle TX operations. This makes average radiated emission lower and consequently enables the device to use higher output power with respect to regulations""


    First of all, i am not really familiar with EMC measurement and not a professional on this. And my native speaking language is not English...


    The contain of 15.35b is mainly focused on measurement of any spurious emission >1GHz should be done by an average detector. My interpretation is that measurement done by an average detector should be compared with the average limit. For 2nd and 3rd harmonics of 2.45GHz, the known limit of average field strength @ 3m is 54dBuv/m. In addition, the peak limit must also be comply with "average limit+20dB", which is 74dBuv/m for 2nd and 3rd harmonics.


    In 15.35c, it said "when the radiated emission limits are expressed in terms of the average value of the emission..." This is why I assume the value taken by an average detector is the average value. Thus, previously, I suppose I can use DCF defined in 15.35c and add it into the average value (the value compared to the 54dBvu/m limit....) to make it become the final adjusted average limit.

    Anyway, it is a hard time to redesign the layout....

  • Hi all,

    The right calculation of average factor is as follows :

    1) for pulse train shorter than 100 ms: 20*log[ (PulseDuration/PulsePeriod) * (BurstDuration/TrainDuration) * NumberOfBurstsWithinPulseTrain]

    2) for pulse train longer than 100 ms: 20*log[ (PulseDuration/PulsePeriod) * (BurstDuration/100ms) * NumberOfBurstsWithin100ms]

  • and the final result is:

    Average calculated= Peak measured + average factor

  • Also see this

    Regardless of whether or not you choose 15.247 or 15.249 for compliance, if your device is NOT operating at
    100 % duty cycle, you take the peak measurement and apply a duty cycle correction factor to derive the
    average value

  • Actually, the correct equation is:
    Final average value = Measured Average value + 20log(duty cycle) for linear voltage averaging mode.

    However, since your device was transmitting continuously (100% duty cycle), then the 20log(1/duty cycle) is 0, and the equation simplifies to:
    Final average value = Measured Average value

    See FCC KDB 558074 V04 dated 4/5/2017

    apps.fcc.gov/.../GetAttachment.html
  • Final average value = Measured Average value + 20log(1/duty cycle) for linear voltage averaging mode.

    However, since your device was transmitting continuously (100% duty cycle), then the 20log(1/duty cycle) is 0, and the equation simplifies to:
    Final average value = Measured Average value