Power dissipation of very small size Diode

  AI Overview shows

"
Power dissipation is the loss of energy in an electrical or electronic system, typically converted into heat.

The power dissipated by a resistor can be calculated using the following formulas, derived from Ohm's Law (V=IR) and the law of heating:

• P = I^2R (Power equals current squared times resistance)
• P = VI (Power equals voltage times current)
  
• P = V^2R (Power equals voltage squared divided by resistance)

"
The second one

• P = VI (Power equals voltage times current)  seems not power dissipated but power generated.

as R is not involved, at least not shown in the equation. However, when considering the current, the flow of charge I = Q/t, the conductor like metal wire is necessary to make Charge flow. Conductor has some R (Resistance), even very low.

"

Wiki (Resistance)

All objects resist electrical current, except for superconductors, which have a resistance of zero.

1 meter of copper wire with 1 mm diameter 0.02Ω

 "

 For Instance

1)

I-----------------I I  1.5V Battery -------------I

I                                                              I

I ----------------- 0.02Ω Wire -----------------I

Current = 1.5V / 0.02Ω = 75A

Power (Dissipation) = V x I = 1.5V x 75A = 112.5W. 112.5W / sec.

75A is a very large current for a 1.5V Batter and 112.5W is also very high. The wire becoens very hot. When for 10 sec continued, it becomes 1,125W used.

2)

I----------------- I I  1.5V Battery --------------I

I                                                               I

I ------ 1Ω Resistor + 0.02Ω Wire ----------I

Current = 1.5 / 1.02 = 1.47A

Power (Dissipation) = V x I = 1.5V x 1.47A = 2.205W

3)

I----------------- I I  1.5V Battery --------------I

I                                                               I

I ----- 1KΩ Resistor + 0.02Ω Wire ---------I

Current = 1.5 / 1000.02 = 0.0015A

Power (Dissipation) = V x I = 1.5V x 0.0015A = 0.0225W

The current reduces to 0.0015A. so Power (Dissipation)re reduces too to 0.0225W.

It generate heat but much less than 1) 112.5W.

Again, Power dissipation is a loss of energy , or more precisely time rate of loss of energy. R (Resistance) must be involved.

 P = I^2R 

1)  75^2A x 0.02Ω = 112,5W

 P = V^2R

1)  1.5^2V / 0.02Ω = 112,5W

 

Resistor function

The major function of Resistor is

Wiki

"
In electronic circuits, resistors are used to reduce current flow

"

In the above examples, when a resistance increases the current reduces accordingly, therefore so the dissipation power.

P = I^2R

The power dissipation.increases by I^2 and R. The effect of I increases / decrease by a factor of I^2 while The effect of R is simple.

P = V^2R can be said as deriving from Ohm's Law (V=IR).

 

Ohm's Law (V=IR) is well known but  <law of heating> is not.

According to Joule’s law (law of heating), the amount of heat generated is proportional to the resistance of the wire, the time duration for which the current flows, and the square of the current flowing through the circuit. The mathematical equation explaining Joule’s law of electric heating can be given by: 

Q = I²  Rt

 Q is the amount of heat generated in Joules.

 

Q is the symbol of Heat ad Energy

 

This equation reminds us

Electrical power (P) = I²·R

 

where <t> is missing. Multiplied by < t>,  P x t = Electrical energy. Electrical power (P) is commonly defined as <the (time) rate of transfer of electrical energy (within a circuit)>. 

 

However. this equation involves R (Resistance) , this undicace the energy transfer from Electrical energy to "Heat energy".

 

"Heat energy" is complicated.

Joule

Wiki

The joule (symbol: J) is the unit of energy in the International System of Units (SI).^[1] In terms of SI base units, one joule corresponds to one kilogram-metre squared per second squared (1 J = 1 kg⋅m²⋅s⁻²). One joule is equal to the amount of work done when a force of one newton displaces a body through a distance of one metre in the direction of that force. It is also the energy dissipated as heat when an electric current of one ampere passes through a resistance of one ohm for one second. It is named after the English physicist James Prescott Joule (1818–1889).

Conversely 

 

Electrical energy = Electrical power (P) x time (unit in Joule)

Therefore

Q = I⋅ Rt

is Energy but in the form of heat generated. (unit in Joule), in which case the electrical power is dissipated.

Dissipation Power (W) = I²⋅ R

Dissipation Energy (J) = I²⋅ Rt

---------

 

Another important concept is <Heat Current>.

 

<Heat Current>.is analogus to Electrical Current. The unit of <Heat Current> is is <Watt>.

Wiki shows simply

 

Other names	Thermal current
Common symbols	H
SI unit	Watt
Derivations from other quantities	${\displaystyle H={\Delta Q \over \Delta t}={\Delta T \over R}}$

 

The explanation does not explain

$$\frac{\Delta T}{}$$/ R.

Wiki (Heat)

In thermodynamics, heat is defined as the form of energy crossing the boundary of a thermodynamic system by virtue of a temperature difference across the boundary..A thermodynamic system does not contain heat. Nevertheless, the term is also often used to refer to the thermal energy contained in a system as a component of its internal energy and that is reflected in the temperature of the system. For both uses of the term, heat is a form of energy.    

Heat energy or Thermal Energy

 

Wiki (Thermal Energy)

The term "thermal energy" is often used ambiguously in physics and engineering.^[1] It can denote several different physical concepts, including:

• Internal energy: The energy contained within a body of matter or radiation, excluding the potential energy of the whole system.
• Heat: Energy in transfer between a system and its surroundings by mechanisms other than thermodynamic work and transfer of matter.
• The characteristic energy k_BT, where T denotes temperature and k_B denotes the Boltzmann constant; it is twice that associated with each degree of freedom.

 

 

The following might help but confusing.

 

:" 

$$\frac{\mathrm{1)\; https://www.thoughtco.com/heat-current-2699425}}{}$$

By

Andrew Zimmerman Jones

Updated on February 09, 2019

The heat current is the rate at which heat is transferred over time. Because it is a rate of heat energy over time, the SI unit of heat current is joule per second, or watt (W).

Heat flows through material objects through the conduction, with heated particles imparting their energy to neighboring particles. Scientists studied the flow of heat through materials well before they even knew that the materials were made up atoms, and heat current is one of the concepts that was helpful in this regard. Even today, though we understand heat transfer to be related to the movement of individual atoms, in most situations it is impractical and unhelpful to try to think of the situation in that way, and stepping back to treat the object on a larger scale is the most appropriate way to study or predict the movement of heat.

Mathematics of Heat Current

Because heat current represents the flow of heat energy over time, you can think about it as representing a tiny amount of heat energy, dQ (Q is the variable commonly used to represent heat energy), transmitted over a tiny amount of time, dt. Using the variable H to represent heat current, this gives you the equation:

H = dQ / dt

If you've taken pre-calculus or calculus, you might realize that a rate of change like this is a prime example of when you would want to take a limit as the time approaches zero. Experimentally, you can do that by measuring the heat change at smaller and smaller time intervals.

Experiments conducted to determine the heat current have identified the following mathematical relationship:

H = dQ / dt = kA (T_H - T_C) / L

That may seem like an intimidating array of variables, so let's break those down (some of which have already been explained):

• H: heat current
• dQ: small amount of heat transferred over a time dt
• dt: small amount of time over which dQ was transferred
• k: thermal conductivity of the material
• A: cross-sectional area of the object
• T_H - T_C: the temperature difference between the warmest and coolest temperatures in the material
• L: the length across which the heat is being transferred 

There's one element of the equation that should be considered independently:

(T_H - T_C) / L

This is the temperature difference per unit length, known as the temperature gradient.

Thermal Resistance

In engineering, they often use the concept of thermal resistance, R, to describe how well a thermal insulator prevents heat from transferring across the material. For a slab of material of thickness L, the relationship for a given material is R = L / k, resulting in this relationship:

H = A(T_H - T_C) / R

 

$$\frac{\mathrm{---------}}{}$$

 

The following also might help but confusing.

 

2 ) https://www.doubtnut.com/qna/642648547

Text Solution

Generated By DoubtnutGPT

Step-by-Step Text Solution

Step 1: Define Heat Current

- Heat current, also known as heat flow or thermal current, is defined as the rate at which heat energy is transferred from one body to another over a specific period of time. It is typically measured in watts (W), where 1 watt is equivalent to 1 joule per second. 

(watts (W), where 1 watt is equivalent to 1 joule per second. This is the same as the Electrical Power (P) where Resistance is not involved) 

Mathematically, it can be expressed as:

$$Q$$(Heat current) $$=\frac{dQ}{}$$(Heat current)$$\frac{\mathrm{/d}t}{}$$

$$\frac{\mathrm{(This\; is\; differential\; concept.\; but\; what\; does\; this\; mean\; ?\; This\; is\; analogous\; to\; I\; =}}{}$$$$\frac{\mathrm{/d}\mathrm{t\; in\; Electric\; current\; but\; confusing\; as\; Q\; is\; used\; as}}{}$$<Heat Current> while Q of $$\frac{\mathrm{I\; =}}{}$$$$\frac{\mathrm{/d}\mathrm{t\; is\; Electrical\; Charge.)}}{}$$

where $$Q$$ is the amount of heat transferred in (joule), and $$\mathrm{<t>}$$ is the time taken for that transfer.

Step 2: Define Thermal Resistance

- Thermal resistance is a property of a material that quantifies its ability to resist the flow of heat. It is defined as the temperature difference across the material divided by the heat current flowing through it. The formula for thermal resistance $$R$$ can be expressed as:

$$R=\frac{\Delta T}{\mathrm{/Q}}$$

where $$\Delta T$$ is the temperature difference across the material and $$Q$$ (or H) is the heat current. Additionally, thermal resistance is the reciprocal of thermal conductance, which is a measure of how easily heat can flow through a material.

"

Please do not confuse the above Heat Current <Q> with Electric charge <Q>  (actually confusing).

$$R=\frac{\Delta T}{\mathrm{/Q \; -->}}$$$$=\frac{\Delta T}{\mathrm{/R}}$$

Analogy

$$=\frac{\Delta T}{\mathrm{/R}}$$

The higher Thermal Resistance (R) the lower Heat current.

I = V (Voltage difference) / R

The higher Electrical Resistance the lower Electrical current.

Heat Resistance is defined as 

 

in unit in K/W

 

that is

 

Heat Resistance (R) = Temp difference in K (or °C ) / Heat Current (Q) in unit W (Watt)

$$\frac{\mathrm{That\; is}}{}$$

                                      $$\frac{\Delta T}{}$$
$$\frac{\mathrm{Q\; (Heat\; Current)\; =}}{}$$$$\frac{\mathrm{--------}}{}$$

                                    $$\frac{\Delta \mathrm{T/Q}}{}$$

 

While the upper $$\frac{\Delta \mathrm{T\; is\; the\; actual}}{}$$Temp difference while the lower $$\frac{\Delta \mathrm{T\; is\; the\; specified\; as\; the\; property\; of\; the\; material\; -}}{}$$ability to resist the flow of heat.

$$\frac{\mathrm{We\; must\; consider\; in\; terms\; of\; time\; dimension\; - \; differentiation\; and\; integration.}}{}$$

1 joule per second

$$\frac{\mathrm{Wiki}}{}$$

$$\frac{\mathrm{Watt}}{}$$

In terms of electromagnetism, one watt is the rate at which electrical work is performed when a current of one ampere (A) flows across an electrical potential difference of one volt (V), meaning the watt is equivalent to the volt-ampere (the latter unit, however, is used for a different quantity from the real power of an electrical circuit). $${\displaystyle \mathrm {1~W=1~V{\cdot }A} .}$$

Two additional unit conversions for watt can be found using the above equation and Ohm's law. $${\displaystyle \mathrm {1~W=1~V^{2}/\Omega =1~A^{2}{\cdot }\Omega } ,}$$ where ohm (${\displaystyle \Omega }$) is the SI derived unit of electrical resistance.

$$\frac{\mathrm{---------}}{}$$

$$\frac{\mathrm{Power\; dissipation\; of\; very\; small\; size\; Diode}}{}$$

The specifications or Dara Sheet of Electronic components usually show Max Power Dissipation as a warning - not to use the part with over Max Power Dissipation (unit Watt) or apply this to the device (component).

 

Recently I ask one Taiwan Schottky Diode supplier what Max Power Dissipation is as the spec does not show this. Their reply was

 

|"

根据公式PD=(TJ-TA)/热阻=0.15W，谢谢

" 

(TJ-TA) Unit   °C
热阻 Thermal Resistance Unit  °C / W

   °C

----------    =  W

  °C / W

 This is the point.

The Chinese spec is not available.

The English spec shows 

Maximum Thermal Resistance (Note 1) RθJA 650  °C / W

NOTES : 1.Mounted on an FR4 PCB, single-sided copper, mini pad 

 

根据公式PD=(TJ-TA)/热阻=0.15W，

TJ -TA = Temp junction - Temp ambient
热阻 = Thermal Resistance 

Temp junction must be max allowable but not shown in the spec.
Temp ambient 20/25 °C but not shown in the spec.

0.15W can be come from like

(125 - 25)./ 650 = 0.1538

TJ  Temp junction 125°C s not very high. usually 150°C

Also

Maximum Thermal Resistance (Note 1) RθJA 650  °C / W

is very high. But it depends on the dissipation method. A similar part Nexperia PMEG4002EJ.The spec shows

Ptot total power dissipation Tamb ≤ 25 °C

[2] [3]   385 mW
[2] [4]   695 mW
[1] [2]   1.045 W

[1] Device mounted on a ceramic PCB, Al2O3, standard footprint.
[2] Reflow soldering is the only recommended soldering method.
[3] Device mounted on an FR4 Printed-Circuit Board (PCB), single-sided copper, tin-plated and standard footprint.
[4] Device mounted on an FR4 PCB, single-sided copper, tin-plated, mounting pad for cathode 1 cm2

while

Rth(j-a) thermal resistance from junction to ambient in free air

[1] [2] [3]  max  325 K/W
[1] [2] [4]  max  180 K/W
[1] [2] [5]  max  120 K/W

[1] For Schottky barrier diodes thermal runaway has to be considered, as in some applications the reverse power losses PR are a significant part of the total power losses.
[2] Reflow soldering is the only recommended soldering method.
[3] Device mounted on an FR4 PCB, single-sided copper, tin-plated and standard footprint.
[4] Device mounted on an FR4 PCB, single-sided copper, tin-plated, mounting pad for cathode 1 cm2.
[5] Device mounted on a ceramic PCB, Al2O3, standard footprint

-------

Ref

 

How to calculate thermal resistance of diodes?

The junction-to-ambient thermal resistance, Rth(j-a), of a small-signal diode can be calculated from its power dissipation as follows:

Equation: R_th(j-a) ＝ (T_j Max - T_a） / P Max

The thermal resistance can be calculated from its power dissipation in the technical datasheet of the diodes.
The following is the formula for calculating the thermal resistance from the junction to the ambient.

Equation: R_th(j-a) ＝ (T_j Max - T_a） / P Max

R_th(j-a): Junction-to-ambient thermal resistance
T_j Max: Maximum junction temperature
T_a: Ambient temperature (temperature condition of P Max)
P Max: Maximum power dissipation

 

ACT