# How to Calculate the Derating Factor of a Semiconductor?

• shaiqbashir
In summary, the conversation discusses the operating conditions of a semiconductor with a maximum rated dissipation of 5W at 50 degrees Celsius case temperature. The question is whether the device can be operated at 5W without auxiliary cooling at an ambient temperature of 40 degrees Celsius. The answer is no, as the maximum permissible dissipation at 40 degrees Celsius is only 2W. It is not clear what is being asked in the third question, as the wording is unclear.
shaiqbashir
Okz my question is like this:

A certain semiconductor has a max rated dissipation of 5W at 50 degree Celsius case Temperature and must be derated above 50 degree celisus case temperature $$\theta_{CA}$$ = 5 degree C/W

1) Can the device be operated at 5W of dissipation without auxiliary cooling (without heat sink or fan) when ambient temperature is 40 degree celsisus?

2) if not, then what is the max possible dissipation, with no auxiliary cooling at 40 degree celsius?

3) What derating factor should be applied to the device in watts per ambient degree?

okz so now this is what i have solved myself:

1) we can find

$$P_{d} = \frac{T_{C}-T_{A}}{\theta_{CA}}$$

THis give me

Pd = 2 W

which clearly is less then the required 5W. So it it means that the device cannot be operated without auxiliary cooling.

2) The maximum permissible dissipation at 40 degree celsius will be 2W as calculated above.

3) no idea

Please tell me have i solved the first two parts correctly or not. If not then what is the correct way to solve them. Secondly, please tell me how to solve this third part.

I shall be thankful to u for ur precious and quick replies.

Take carez!

Good Bye!

SB--

You are fine on 1 & 2. I think 3 is poorly worded. What temperature scale are they asking for? Celsius? Kelvin? Could you post the exact wording of the question please?

Your calculations for the first two parts seem to be correct. The maximum permissible dissipation at 40 degrees Celsius would indeed be 2W, as calculated using the given formula.

As for the third part, the derating factor is the ratio of the maximum permissible dissipation (at a certain temperature) to the maximum rated dissipation. So in this case, the derating factor would be 2W/5W, or 0.4 watts per ambient degree. This means that for every 1 degree Celsius increase in ambient temperature, the maximum permissible dissipation of the device should decrease by 0.4 watts.

I hope this helps! Good luck with your studies.

## 1. What is a derating factor?

A derating factor is a value used to reduce the maximum capacity or load of a system or component in order to ensure safe and reliable operation.

## 2. What is the purpose of derating factors?

The purpose of derating factors is to prevent components or systems from operating beyond their designed capacities, which could lead to failures, malfunctions, or safety hazards.

## 3. How is the derating factor calculated?

The derating factor is calculated by dividing the maximum rated capacity by the actual operating conditions. This can include factors such as temperature, humidity, altitude, and voltage.

## 4. What is the difference between a safety factor and a derating factor?

A safety factor is a multiplier applied to the calculated load or capacity to provide a margin of safety. A derating factor is a reduction applied to the maximum rated capacity to ensure safe and reliable operation.

## 5. What are some common causes of derating factor problems?

Common causes of derating factor problems can include incorrect calculations, improper installation or usage, environmental factors, and changes in operating conditions. It is important to regularly review and update derating factors to ensure safe and efficient operation of systems and components.

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