ADL5802 overheating

Hello,
I have designed a receiver including ADL5802,


1- I followed the datasheet, set enbl value to low, connecting it to ground through a cap.
2- Setted the VSET to 3.3V taking it from voltage regulator.
3- Used recommended minicircuits baluns etc.


Aside from these the design is almost same with eval board schematics,
But at the power up, (there are no RF or LO signal) the ADL5802 starts to heat up.
I have checked the soldering, but can not see a problem.

What could be the issue?

Parents
  • +1
    •  Analog Employees 
    on Feb 9, 2021 6:40 PM

    Your schematic looks OK. Although you did not show DC-blocking caps at ADL5802's LO pins, I do see 2 series caps on your board. Please make sure LO pins are ac-coupled.

    You can start troubleshooting by measuring DC voltages at the IC pins:

    1,2,5,8,11,14,17,18,21: GND

    3,4,15,16: Output pins = VPOS = 5V

    6,13,14: VPOS = 5V

    7: ENBL = GND (active low)

    9,10: LO input pins = some internal bias voltage between 0V and VPOS

    12: VSET = 3.3V according to your schematic

    19,20,22,23: Input pins = some internal bias voltage between 0V and VPOS

  • Thank you for your answer, I will check all again. I was wondering what would be a regular temperature while operating do you have any idea about that?

  • +1
    •  Analog Employees 
    on Feb 10, 2021 11:11 PM in reply to baykalyu

    From ADL5802's thermal parameters shown in the datasheet:

    Theta JA = (Tj - TA) / Pd

    Tj = (Theta JA) * Pd + TA

    Assuming ambient temp (TA) is 20degC, and power dissipation of 1.1W (5V * 220mA):

    Tj = 26.5 * 1.1 + 20 = 49.2degC

    Theta JC = (Tj - TC) / Pd

    TC= Tj - (Theta JC) * Pd

    TC = 49.2 - (8.7 * 1.1) = 39.6degC

    Note that this calculation assumes using ADI evaluation board. TC is at the IC's exposed paddle. Case top will be a little bit hotter.

Reply
  • +1
    •  Analog Employees 
    on Feb 10, 2021 11:11 PM in reply to baykalyu

    From ADL5802's thermal parameters shown in the datasheet:

    Theta JA = (Tj - TA) / Pd

    Tj = (Theta JA) * Pd + TA

    Assuming ambient temp (TA) is 20degC, and power dissipation of 1.1W (5V * 220mA):

    Tj = 26.5 * 1.1 + 20 = 49.2degC

    Theta JC = (Tj - TC) / Pd

    TC= Tj - (Theta JC) * Pd

    TC = 49.2 - (8.7 * 1.1) = 39.6degC

    Note that this calculation assumes using ADI evaluation board. TC is at the IC's exposed paddle. Case top will be a little bit hotter.

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