AD811AN(PDIP Package) Getting Heated for +/-12V Power Supply

Hi Jigar,

Just like what you mention, the power dissipation of the part is approximately 400mW, that's why the part is getting hot. To compute the junction temperature we can use this formula: T_J = T_A + (θ_JA)(P_D) without considering the power dissipation of the output. T_J is the junction temperature, T_A is the ambient temperature, θ_JA is the thermal resistance and P_D is the power dissipation. Since the thermal resistance of AD811AN is 90°C/ W, we can say that at 400mW the part has a temperature around 61°C at an ambient temperature of 25°C. Does the heating of the part affects its functionality and your expected output?

Kindly refer to this technical article, Data Sheet Intricacies - Absolute Maximum Ratings and Thermal Resistances. Other than using heat sink, there are some tips and recommendations in the article that you can apply to decrease the power dissipation.

Another option is to switch to other ADI parts with low supply current consumption like AD829, AD844 and AD810.

Regards,

Jay

HI Jigar,

Below, please find the equation that will allow power dissipation total to be calculated.  This equation is used to compute the total average power being dissipated in the amplifier due to sine wave signal processing.  As you can see from the equation, there are a few things to consider.    So far, you've only accounted for the quiescent power (384mW) which is the small and simple equation above the red box.  Referring now to the equation in the red box, the middle piece is collector input power, and relates the supplies, signal amplitude, and the load resistance when a sine wave is being processed by a push pull output stage.  Included in that quantity is load power which is the third term and must be subtracted from the total because it's being dissipated in the load, not in the amplifier. This equation will help you determine the power that is being dissipated by amplifier.

To answer your questions:

1) Dissipating heat in a package (PDIP) that has no exposed thermal pad is not very easy nor very effective...This is due to the poor thermal conductivity of the plastic that encapsulates the amplifier.  You haven't told me anything about the Rload (RL) in the equation below so it's difficult to know exactly how much power you are talking about. But generally,  if power is a concern, select an amplifier with an exposed thermal pad so that the heat can be conveyed to the PCB thru highly conductive thermal vias and dissipated thru the PCB copper plane and into ambient air.

2) You'll have to give us more information on the load and signal content, frequency etc before we can recommend an amplifier for your needs.  And the only amplifier that dissipates no power are the ideal kind .

3) What are the frequencies of the receive signal in your ultrasound system?  <10MHz? What are the gain ranges in question?  What are your noise requirements?  Distortion requirements?  I don't know of a detailed document on ultrasound receivers but I'll look.

Regards,

Bob