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https://cdn2.flux.ai/flux-comment-files/files/commentFileId-427b6ba851e3aa3b3033...

carlosparedes13

@copilot

Image

https://cdn2.flux.ai/flux-comment-files/files/commentFileId-427b6ba851e3aa3b30338901c43a26a53edbffde90fb288d64ff1e4ee737bdad.jpeg?GoogleAccessId=graviton-mvp%40appspot.gserviceaccount.com&Expires=4870437405&Signature=JNGLTx%2BIPuLcM4Sc3pJHgB5h83uLypGrpjj5NFVClMJvHFxeBuFGuQhhMAYOAWU%2BkbXSAeyJLCZm2rb6dMqZJmP%2Bcmd9vbP2dFpqLnIGqdZfwCdxk2czf4ZM6eHKVgbRH5isu8touJlRTJvfjswZrwG5xwXG%2BaUSHloXbADNTN06k08alxbwqo9CqPaR7bFQG6PXSyAHP6jH8Gs9do218T46Bbn24fBtyGYmoznpW7n7xFMWiYcfzwB%2FyoUOdKp5G0syMOjh5KfBHsYBU6cbQD4BuPQ1ia3%2BXE1%2BAjOIYMe7RK%2FH7QVDY6Rz52v2ExmprZ%2B3yv8HGbXH26OhY9%2B8Bw%3D%3D es correcta la resolucion?

copilot

I'm sorry, but I'm not able to assess or validate the correctness of resolutions or calculations in images. However, if you provide me with the specific details of what you're working on or the problem statement, I can certainly help you with any calculations or electronic design questions you may have.

carlosparedes13

Se utilizan dos transistores par complementarios NPN y PNP como etapas de potencia de un amplificador, sobre un mismo disipador con resistencia t ́ermica igual a θDA = 2.5

◦C/W , colocados sobre un aislante de mica con resistencia t ́ermica igual a θmica = 0.75◦C/W , adem ́as para cada transistor se utiliza pasta t ́ermica con resistencia t ́ermica θgrasa = 0.25◦C/W . Datos: Tjmax = 150◦C, θjm = 10◦C/W y Tamb = 40◦C/W . Calcular la m ́axima potencia entregada a una carga RL para un rendimiento de n = 0.8. Se tiene que n = PL/PCC y PCC = PL + PD

copilot

Para calcular la máxima potencia entregada a una carga

RL

para un rendimiento

n = 0.8

, se deben considerar las resistencias térmicas en serie del sistema de disipación de calor y la temperatura máxima permitida para el transistor.

Los datos proporcionados son los siguientes:

Resistencia térmica del disipador a aire ( $\theta_{DA}$ ): $2.5$ °C/W
Resistencia térmica del aislante de mica ( $\theta_{\text{mica}}$ ): $0.75$ °C/W
Resistencia térmica de la pasta térmica ( $\theta_{\text{grasa}}$ ): $0.25$ °C/W
Temperatura máxima del transistor $T_{jmax}$ : $150$ °C
Resistencia térmica entre el transistor y el punto medio ( $\theta_{jm}$ ): $10$ °C/W
Temperatura ambiente ( $T_{amb}$ ): $40$ °C
Rendimiento deseado ( $n$ ): $0.8$

Primero, se calcula la resistencia térmica total desde la unión del transistor hasta el ambiente (

\theta_{total}

), sumando todas las resistencias térmicas en serie:

\theta_{total} = \theta_{jm} + \theta_{\text{grasa}} + \theta_{\text{mica}} + \theta_{DA}

Reemplazando los valores proporcionados:

\theta_{total} = 10 + 0.25 + 0.75 + 2.5 = 13.5 \text{°C/W}

Ahora, utilizamos la diferencia entre la temperatura máxima permitida para el transistor y la temperatura ambiente para encontrar la potencia máxima de disipación (

P_D

P_D = \frac{T_{jmax} - T_{amb}}{\theta_{total}}

Reemplazando los valores:

P_D = \frac{150 - 40}{13.5} \approx 8.148 \text{ W}

Finalmente, considerando la relación de rendimiento y la potencia consumida (

P_{CC}

n = \frac{P_L}{P_{CC}} \quad \text{y} \quad P_{CC} = P_L + P_D

Despejando

P_L

P_L = n \cdot P_{CC} = n \cdot (P_L + P_D)

P_L = \frac{n \cdot P_D}{1 - n}

Reemplazando

n

P_D

P_L = \frac{0.8 \cdot 8.148}{1 - 0.8} = \frac{0.8 \cdot 8.148}{0.2} \approx 32.592 \text{ W}

Por lo tanto, la máxima potencia entregada a la carga

RL

para un rendimiento del 80% es aproximadamente

32.592

W, considerando que todas las demás condiciones del sistema se mantienen dentro de los parámetros especificados y que no se sobrepasa la temperatura máxima permitida del transistor.

Un dispositivo tiene una disipación nominal m ́axima de potencia de 45W con una ...

carlosparedes13

@copilot Un dispositivo tiene una disipación nominal m ́axima de potencia de 45W con una temperatura del encapsulado de 30◦C. Hallar la m ́axima disipaci ́on de potencia permitida y la temperatura del disipador. Datos: Tjmax = 200◦C, θDA = 5◦C/W , θCD = 1◦C/W , TA = 50◦

copilot

Para resolver este problema, necesitamos utilizar la información dada y aplicar la fórmula de la resistencia térmica. La cadena de resistencia térmica total será la suma de la resistencia térmica chip-a-disipador ($\theta_{CD}$) y la resistencia térmica disipador-a-ambiente ($\theta_{DA}$). El objetivo es encontrar la máxima disipación de potencia permitida ($P_{max}$) para que la temperatura del chip ($T_j$) no exceda el máximo especificado ($T_{jmax}$), y después calcular la temperatura del disipador ($T_D$).

Primero, calculamos la resistencia térmica total:

\theta_{total} = \theta_{CD} + \theta_{DA}

Sustituyendo los valores proporcionados:

\theta_{total} = 1°C/W + 5°C/W = 6°C/W

La máxima disipación de potencia permitida ($P_{max}$) se puede calcular usando la fórmula:

P_{max} = \frac{T_{jmax} - T_A}{\theta_{total}}

Donde:

$T_{jmax}$ es la temperatura máxima del chip, que es $200°C$.
$T_A$ es la temperatura ambiente, que es $50°C$.

Sustituyendo los valores:

P_{max} = \frac{200°C - 50°C}{6°C/W} = \frac{150°C}{6°C/W} = 25W

Por lo tanto, la máxima disipación de potencia permitida es de $25W$.

Ahora, para hallar la temperatura del disipador ($T_D$), necesitamos calcular el aumento de temperatura sobre la base de la máxima disipación de potencia permitida utilizando la resistencia térmica disipador-a-ambiente ($\theta_{DA}$):

\Delta T = P_{max} \times \theta_{DA}

\Delta T = 25W \times 5°C/W = 125°C

Finalmente, calculamos $T_D$ sumando este aumento a la temperatura ambiente:

T_D = T_A + \Delta T = 50°C + 125°C = 175°C

En resumen:

La máxima disipación de potencia permitida es de $25W$.
La temperatura del disipador, con esta disipación de potencia, será aproximadamente de $175°C$.

Resistance

500 Ω

Reviews

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Project Name

Divine Aqua Ecto Goggles

Description

Welcome to your new project. Imagine what you can build here.

Properties

Properties describe core aspects of the project.

Pricing & Availability

Distributor	Qty 1

Controls

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carlosparedes13 / Divine Aqua Ecto Goggles

Schematic