OPA549: Heat Sink Design

Hello P Subrahmanya,

I believe we need a bit more information to assist you. You mention Vrms = 17 V, and Irms = 3 A. That would be an equivalent resistance of 5.67 Ohms; however, you do mention a motor which can be a more complex load. Do you have any info about the motor?

If we take 17 Vrms and convert that to a peak-to-peak voltage, then that equates to 60 Vpp, or +/- 30 Vp for a sinusoidal waveform. You indicate a Vin = +/-25 V, is that the supply voltage, or the actual input voltage to the OPA549 amplifier circuit? If that represents the supply voltage there is an issue because the output swing of +/-30 Vp exceeds the +/-25 V supplies. Also, +/-30 V is the maximum specified OPA549 supply voltage so the output swing would have to be less. The output swing is limited by the peak output current level.

Do note that in OPA549 datasheet Figure 7 shows the Maximum Power Dissipation vs Ambient Temperature, and curves are provided for several different heat sinks. One in particular, the Thermalloy 6399B, provides the lowest θJA of 5.2°C/W. This is a good, practical heat sink for many high-power OPA549 applications. However to be certain it would be suitable would require a more rigorous determination.

Can you tell us anything about the end equipment in which you plan to use the OPA549?

Regards, Thomas

Precision Amplifiers Applications Engineering

Hello Thomas,

Thank you for your reply.

Vrms = 17V corresponds to Vp of 24V. So the supply of +/- 25V is sufficient. 

The power dissipation in the IC is coming to be 60W. The figure given in the datasheet does not cover the value. So I feel I need to use different heat sink. 

I am trying to use it for induction motor driver circuit.

Regards,

P S Adiga

Hello Thomas,

Thank you for the reply.

Can you please give more explanation regarding output voltage swing vs Output Current graph? Because it is written in the datasheet that we need to minimize the power dissipation by using lowest possible power supply voltage. So if we consider the graph, then the power dissipation increases to very high level.

I have used sufficiently big heat sink with cooling fans. But still the Opamp is shutting down after some 15 to 20 sec once the motor starts. I could see series of pulse at E/S pin, that is, it shuts down and again it turns on. Is there any other possible reason for this thermal shutdown?


With Regards,
P S Adiga

Hello P S,

Like any transistor, the output transistors internal to the OPA549 have limits to how much current they can deliver and voltage they can withstand. Since they are power transistors they can deliver amperes of current up to the limit dictated by their current gain and power handling capabilities. For the OPA549 the limit is in the range of 8 to 10 amperes.

The transistors have collector-to-emitter resistance that varies with current, voltage across the transistor and its junction temperature. However, if you think of it simply as a fixed resistance from the collector to the emitter as current is increased the voltage drop across it increases. The OPA549 Output Voltage Swing vs Output Current graph tells us that the higher the output current demanded by the load, the lower the maximum output swing capability will be in order to maintain a linear, non-clipped waveform. You can see from the graph that as the output current increases, |Vsupply| - |Vout| must be increased to maintain the linear output. The curves are different for the positive and negative swings.

The OPA549 thermal shutdown circuit only reacts to the condition where the die junction temperature exceeds approximately 160 C. The op amp shutsdown until the termperature dips back below about 140 C. If the temperature of the die keeps trying to exceed160 C, the output will cycle between on and off as the temperature moves between the threshold points.

Take a look at a Burr-Brown Application Note, AB-039, on the subject of Power Amplifier Stress and Power Handling Limitations. That note may provide some help in undertanding what you need to do to get the thermal situation under control.

Regards, Thomas

Precision Amplifiers Applications Engineering

Hello Thomas,
I could run the motor at peak output voltage of +/- 24V where the supply rail voltage is +/-25V. The motor is drawing Irms of 2.5 A. But according to the graph, this should not be possible. I tested it for 1 hour. So how could it be possible?? and what could be the effect if we continuously run like this??
I tried making rail voltage as +/- 28V, considering the graph. But the device had gone to thermal shutdown within 30 to 40 s. So what is the solution for this?
Regards,
P S Adiga

Hello P S,

I would not expect the OPA549 output to be able to swing within a volt of the supply rails while supplying an IRMS of 2.5 Amps. Have you captured an image of the OPA549 output voltage when it is driving the motor in this fashion? I am curious what it shows. Better yet, if both the output voltage and the output current can be measured as the motor is being driven that might be more revealing. If you don't have a current probe, then you can add a small value resistor in series with OPA549 output and measure the voltage across it. The current can be derived from Ohm's law. Even if the particular device you are evaluating is capable of swinging that close to the rails I wouldn't plan on being the case for all OPA549 devices. The transistor characteristics will vary with the individual device's lot processing, and other devices from other lots may not swing as close to the rails.

Certainly, reducing the voltage difference between the supply voltage and the peak output swing voltage to as low value as possible reduces the OPA549 internal power dissipation. That must be the case when the OPA549 isn't heating enough to activate the thermal shutdown circuit. If the OPA549 is attached to a sufficently large heatsink, and has a well executed thermally conductive connection between the package and the heatsink, the ability to dissipate the heat is improved.

Regards, Thomas

Precision Amplifiers Applications Engineering