Reading through the datasheet of the OPA549, can you confirm if the thetaHA value for 6396B heatsink is 5.6 C/W or 3.3 C/W?
This thread has been locked.
If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.
When the OPA549 datasheet was written more than 20 years ago the 6396B and 6399B heatsinks where produced by Thermalloy. It looks like the company has changed hands and name a few times, but the heatsinks are still being produced under the original model numbers.
The theta ϴHA numbers you mention from the OPA549 datasheet section represent 5.6 °C/W for the 6396B, and 3.3 °C/W for the 6399B, respectively. The 6399B is physically larger than the 6396B and has more surface area, which results in lower thermal resistance.
A more complete picture of the ϴHA performance of the two hear sinks can be found in their Boyd-AAVID datasheet. The ϴHA information is on the right hand vertical axis. OHA can be determined more precisely relative to power dissipation and air flow rate.
Precision Amplifiers Applications Engineering
As Thomas pointed out, the heat sink to ambient figures and part numbers have changed.
If you provide us with power dissipated heat Pd in OPA549, Ta operating temperature (worst case) and air flow rate across the heat sink or static etc., we may be able to calculate some of heat sink requirements for you. Normally, larger the heatsink, more dissipated surface area, higher heat conduction material etc. will have lower figure in Heat-Sink-To-Ambient (C/W), which is more efficient heat sink in cooling, and it will keep OPA549 cooler for a given application.
If you need additional help, please let us know.
I do not have all the information to calculate ThetaHA. Pd power may not be the peak power. The Pd or dissipated power is a function of supplied voltage, output voltage and load current that goes through the internal die of OPA592. The peak power here may refer to the max. wattage of a load. I need to know the delta voltage from Vcc-Vout and load current that goes through the internal OPA592 for a worst case in order to calculate Pd, which is not necessarily equal to a load's peak power.
Enclosed is a captured image that may help you to calculate Pd at OPA592.
If you have additional questions, please let us know.
Yes, this is the plot for AC input. You provided 14W in peak power, I need to know what is load current and OPA549's output peak voltage. From the figures, you are able to calculate Pd = (Vcc - Vout)Iout_max or use the following equations in AC.
From the graph, max. Pd is occurred at Iout approx. 2A or approx. Pd = 28W.
14W peak power at output can come from high voltage &low output current, or medium voltage & current or low voltage & high load current etc. in (Vcc-Vout)*Iout products. Since I do not have these figures, I am unable to calculate Pd per your inquiry.
I see, thank you.
The load is an ~3.33ohm TEC
The power supply to the op-amp is 13.5V (This is generated from LMZM33604RLXR)
below is Vout vs. Iout plot
I will do the calculations and hopefully it matches the calculated heatsink with ThetaHA = 5.2C/W.
Since OPA549 is used for TEC application, the part's output is operating in DC voltage to cool or heat TEC for temperature control.
With 14W at the output, R_load = 3.33Ω, the OPA549's output voltage is regulated at V^2/R_load = 14W or Vout = 6.83Vdc and output current is 14W=(I^2)*R_load or Iout = 2.05Adc
Since Vcc = 13.5C, Vcc-Vout = 13.5V - 6.8V = 6.7Vdc. Pd = 6.83V * 2.05A = 14 W. And your ThetaHA calculation is correct.
All content and materials on this site are provided "as is". TI and its respective suppliers and providers of content make no representations about the suitability of these materials for any purpose and disclaim all warranties and conditions with regard to these materials, including but not limited to all implied warranties and conditions of merchantability, fitness for a particular purpose, title and non-infringement of any third party intellectual property right. No license, either express or implied, by estoppel or otherwise, is granted by TI. Use of the information on this site may require a license from a third party, or a license from TI.
TI is a global semiconductor design and manufacturing company. Innovate with 100,000+ analog ICs and
embedded processors, along with software, tools and the industry’s largest sales/support staff.