Lecture 6 - Heat in Electrical Components (DOBSON)

Question 1

What is Joule heating?

Answer 1

Heat in a transistor is generated by the interaction between moving electrons and the vibrating atoms in the materials lattice structure.

Question 2

How is the power in electrical components dissipated?

Answer 2

Heat

Question 3

What is CMOS?

Answer 3

Complementary metal-oxides-semiconductor.

Question 4

What is special about CMOS?

Answer 4

Transistors based on this technology only consume significant power while they are switching between their on and off states.

Question 5

Why is thermal management becoming a critical issue with CMOS?

Answer 5

As switching speed (cpu clock speed) and transistor densities continue to increase, this increases heat per unit area.

Question 6

Can you draw the power density vs year graph?

Answer 6

Yes or no.

Question 7

Can you draw the temp vs log time to failure graph?

Answer 7

Yes or no.

Question 8

Can you draw the transistor count vs date of intro graph?

Answer 8

Yes or no.

Question 9

Can you draw the transistor density vs time line and clock speed graph?

Answer 9

Yes or no.

Question 10

Why is silicon unusual?

Answer 10

Because its electrical resistance increases with temp up to about 160 deg then it starts decreasing and drops further when the melting point is reached.

Question 11

What is a thermal runway phenomena?

Answer 11

the resistance decreases in the regions which become heated above the threshold, allowing more current to flow through the overheated regions, in turn causing yet more heating in comparison with the surrounding regions, which leads to further temp increase and resistance decrease.

Question 12

Is joule heating commonly uniform on a PCB or semi conductor chip?

Answer 12

Seldom uniform

Question 13

What is often desirable in regards to heat distribution?

Answer 13

It is often desirable to spread the heat around a system as evenly as possible, although in some instances it is possible to use components with differing thermal tolerances in different regions.

Question 14

What is a heat sink used for?

Answer 14

To increase the surface area in contact with the cooling fluid surrounding it, such as the air.

Question 15

What are some of the factors which affect the thermal performance of a heat sink?

Answer 15

Air velocity, choice of material, fin design and surface treatment.

Question 16

In electrical components what must the temp of the heat sink be?

Answer 16

Above the temp of the surroundings to transfer heat.

Question 17

Can you draw a labelled diagram of a heat sink?

Answer 17

Yes or no.

Question 18

What causes thermal contact resistance?

Answer 18

Occurs due to the voids created by surface roughness effects, defects and misalignment of the interface.

Question 19

What are voids filled with?

Answer 19

Air.

Question 20

What if the contact area is small? (e.g. rough surfaces)

Answer 20

The major contribution to the resistance is made by gaps.

Question 21

What is a TIM?

Answer 21

A thermal interface material.

Question 22

What do TIMs do?

Answer 22

Displace the air present in voids with a material with a much higher thermal conductivity. ( Air k = 0.022 TIM k > 0.3).

Question 23

Why must care be taken when choosing a TIM?

Answer 23

Most manufacturers give a value for the thermal conductivity of a material but these do not take into account the interface resistances. Therefore if a TIM has a high k, it doesn’t mean the interface resistance with be low.

Question 24

Most common heat sink materials?

Answer 24

Al alloys

Question 25

Why is Cu good/bad as a heat sink?

Answer 25

Has double k than Al but is 3x as dense and can’t be extruded.

Question 26

Is diamond a good heat sink material?

Answer 26

Yes, k=2000 and excellent diffusivity. But expensive. Can use synthetic diamond as submounts.

Question 27

Examples of composite heat sink materials?

Answer 27

Copper-Tungesten pseudo-alloy, AlSiC, Dymalloy, E-material.

Question 28

Desirable characteristics of a heat sink material?

Answer 28

As high a k as possible and the best heat spreading characteristics possible.

Question 29

Fin efficiency equation?

Answer 29

Fin efficiency = Actual heat transferred/ heat that would be transferred if entire fin area was at base temp = n_f = tanh(mLc)/mLc where mLc=sqrt(2h_f/kt_f)*L_f. Where h_f = convection coefficient of fin, k = thermal conductivity of fin material, t_f = fin thickness, L_f = fin height.

Question 30

What is spreading resistance?

Answer 30

When thermal energy is transferred from a small area to a larger area in a substance with finite thermal conductivity. In a heat sink this means that heat does not distribute uniformly through the heat sink base. Shown by how heat travels from the heat source location and causes a large temp gradient between source and the edges of the heat sink. (some fins are at a lower temp than if the heat source were uniform across base of heat sink.. this non uniformity increases the heat sinks effective thermal resistance.)

Question 31

Ways to decrease spreading resistance in the base of a heat sink?

Answer 31

1. Increase base thickness
2. Choose a material with > k
3. Use a heat pipe in the heat sink base

Question 32

When surrounding temp and heat sink temp are between 0 and 100 degrees, what can be neglected?

Answer 32

Radiation because it is generally small compared to convection.

Question 33

When will radiation be a significant factor?

Answer 33

When convection is low, e.g. flat non-pinned panel with low airflow. Here surface properties may be an important design factor. Matte-black surfaces with radiate much more effectively than shiny bare metal.