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OPA548 overheating with 1.75A load

Rajesh Kota
Prodigy 20 points
Other Parts Discussed in Thread: OPA548

We are using OPA548(TO-220,Staggerd lead type) for driving load upto 2-3A. As per data sheet, OPA548 can sink and source current upto 5A to the load .We are sourcing current upto 1.75A to the load of 12V,21W bulb. Even then IC is getting overheated quickly when heat sink is also attached.We are connecting the circuit as per datasheet fig 14, in Non-inverting mode with gain of 6,Supply voltage & current- 35V,2A,input for Vin+ is 0-2V. Please tell the solution why the IC is getting overheated and how to overcome the overheating of IC.Please find the attached document for reference.OPA548.pdf

  • 0
    Expert 1220 points

    Your schematic highlights a 48V supply. If your input is 2V with a gain of 6, that's 12V across the load. With 12V across the load you 36V across the output transistor. 36V across the output transistor at 1.75A is 63 watts of power dissipation. Of course, I am presuming a DC signal.

    Look at figure 2 on pg. 8 of the datasheet as well as figure 7. Figure 7 presumes an infinite heat sink and at 25°C ambient, the device can only dissipate 50W.

    Things will get hot because it is a power amplifier, so long term reliability is critical. What are the specifications of your heat sink? If it is the PCB, how many layers and what is the copper weight?

    -Ken

  • 0 in reply to Ken Dillinger
    Prodigy 20 points
    Thank you Ken. We are not using PCB. Just we are using dot board. For OPA548, we are giving input range 0-5V, varying step by step with each step varying at 0.1V.Supply voltage of op-amp is 35V, 2A and load is 28V, 2A.Maximum power dissipation at full load i.e., 28V is 56W and power dissipation for full supply voltage i.e., 35V is 70W.So, the power dissipated is 70W-56W= 14W.
    As per OPA548 datasheet page no.11, Calculating Junction temperature for selecting heat sink is as follows:
    TJ = TA + PD*θJA
    where, θJA = θJC + θCH + θHA

    TJ = Junction Temperature (°C)
    TA = Ambient Temperature (°C)
    PD = Power Dissipated (W)
    θJC = Junction-to-Case Thermal Resistance (°C/W)
    θCH = Case-to-Heat Sink Thermal Resistance (°C/W)
    θHA = Heat Sink-to-Ambient Thermal Resistance (°C/W)
    θJA = Junction-to-Air Thermal Resistance (°C/W)
    θHA = (TJ-TA)/PD – (θJC+ θCH)
    = (125-40)/14 – (2.5+1)
    = 2.57°C/W
    θJA = θJC + θCH + θHA
    = 2.5+1.0+2.57
    = 6.07° C/W
    TJ = TA + PD*θJA
    = 40 + (14*6.07)
    = 124.98° C
    For this junction temperature does heat sink needed, if needed which type of heat sink should be used. It is getting over heated.
  • 0 in reply to Rajesh Kota
    Expert 1220 points

    If your expected junction temperature is 124.98C, in practice that is 125C or a little more. The calculations do not include quiescent current (which will include the current through the current limit circuitry).

    Your calculations look correct from my eyeball calculator, and since the junction temperature is 150C that gives you 25C of margin with the given supply, load voltage/current, and ambient temperature. Also, this assumes there are no other heat sources in the application.
    If the conditions you calculated are your expected worse case then I personally would feel ok with 25C of margin. Only you can answer if that is acceptable in your application.

    -Ken