GL865 – generated heat

5 thoughts on “GL865 – generated heat

  1. GL865 Hardware User Guide mentions that heat generated during class 10 GPRS is max 2W.

    Transmission: 2 slots => Vdd*Ipeak-P_radiated=3.8V*2A-2W=5.6W
    Rest: 6 slots idle with almost no power consumption
    Average:  (2*5.6W+6*0W)=1,4W

    Why hardware user guide mentions 2W?

     

    Thanks. 

    1. 2W is the RF power….it is the power transmitted at the antenna that correspond to +33dBm.

       

      Let’s suppose that the GSM absorbs 2A during 2 TX slot (2x577us) and
      0A during the remaining 6 slot (6×577).
      Let’s suppose that the supply voltage is constant: 3.8V
       
      The average power absorbed will be: 3.8x2x2/8 = 1.9W (average)
       
      The RMS power absorbed will be: 3.8x2x(square root of 2/8) = 3.8W (rms)
       
      The peek power absorbed will be: 3.8×2 = 7.6W
       
      The heating power will be: 3.8W – 2WRFx(square root of 2/8) = 2.8W
       
      Summarizing:
       
      The power supply should be designed to withstand the 2A current peak, that means a peak power of 3.8×2 = 7.6W, a RMS power of 3.8W, an average power of 1.9W.

      1. Thanks for information.

         

        My question was about power dissipation of GL865 module during class 10 GPRS operation. Hardware user guide mentions that 2W has to be dissipated from module through PCB to ambient.

         

        Let’s suppose that no energy is radiated through antenna.

        In this case all incoming power is converted to heat that has to be dissipated from module. Incoming power is 3.8*2=7,6W during 2TX slots, 0W during remaining 6 slots.

        In this case average power during 8 slots is 1,9W.

        In my opinion it is an average power that has to be dissipated not RMS power.

        P_avg=V_dc*I_dc=V_rms*I_rms

         

        Let’s suppose that 2W is radiated through antenna.

        Incoming power is the same as for previous case but part of it (2W) is radiated through antenna to ambient and remaining power is converted to heat.

        In this case 7,6W-2W=5,6W is converted to heat and has to be dissipated from module.

        Average power during 8 slots is  1,4W.

         

        Hardware user guide mentions that 2W has to be dissipated from module through PCB to ambient. Is it correct value?

        1. Your calculations are correct, but on the real world things change a bit,
          depending
          on the antenna mismatch the peak power consumption varies [ modem
          consumptions are specified with a matched 50 ohm load/antenna]  and this
          impacts the power that needs to be dissipated.
          We provided a
          reasonable worst case  (2W to be dissipated) that shall guarantee that
          no thermal issues will be encountered if respected.