Question

A Schottky diode carrying current (If) of 5A with forward voltage drop (Vf) of 0.5V is used in a power converter. If the diode has junction to ambient thermal resistance Rth_ja = 40deg.C/W and operating at ambient temperature Ta = 40deg.C. Calculate the junction temperature (Tj) accounting only for conduction losses.​

Answer 1

Answer:

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Answer 2

Final Answer:

The junction temperature ($T_j$) accounting only for conduction losses is $140\textdegree C$.​

Given:

A Schottky diode carrying current (If) of 5A with forward voltage drop (Vf) of 0.5V is used in a power converter.

The diode has junction to ambient thermal resistance $R_{th}_{ja}$=40deg.C/W and operating at ambient temperature $T_a$ = 40deg.C.

To Find:

The junction temperature ($T_j$) accounting only for conduction losses.​

Explanation:

The following points are noteworthy for a  Schottky diode.

• The conduction losses appears as the power dissipated in the diode.
• The power dissipated in the diode is the product of the square of the current and the diode resistance.
• The power dissipated in the diode is the difference between the temperature at its junction and the ambient temperature divided by the thermal resistance.

Step 1 of 4

The diode resistance of the Schottky diode carrying current (If) of 5A with forward voltage drop (Vf) of 0.5V is

$=R_f\\=\frac{V_f}{I_f} \\=\frac{0.5}{5} \\=0.1\;A$

Step 2 of 4

So, the power dissipated is

$=5^2\times 0.1\\=2.5\;W$

Step 3 of 4

Again, the power dissipated is

$=\frac{T_j-T_a} {R_{th}_{ja}}\\=\frac{T_j-40}{40}$

Step 4 of 4

Solve for the following equation for the power dissipated in the diode.

$\frac{T_j-40}{40} =2.5\\T_j-40=40\times 2.5\\T_j=140\textdegree C$

Therefore, accounting only for conduction losses in the diode, the required junction temperature ($T_j$) of the diode is $140\textdegree C$.​

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