LM338 is getting hot and stop working when we try to flow current more then 1A

Electrical Engineering Asked by Aditya Prajapati on December 5, 2020

I try to make a 30 V, 5 A power supply using LM338 (without heatsink) voltage regulator IC.

As given in datasheet

In my case the value of R1 is 100 Ω and R2 is 2.2 kΩ. then it gives 29.6 V as output.

when we connect a 10 Ω 10 W resistor, the current is exceeded 1 A, the IC gets hot within a few seconds output current and voltage become zero. (I think the IC thermally shuts down).

After cooling the IC when we reconnect power then we got no output voltage and no current, IC is also not getting hot.

we can’t conclude why it’s happening. Is Ic got damaged?

After that, we did one more experiment.

Constant Current Circuit as given in the datasheet.

Constant Curreent

In my case R1 = 10 Ω, 10 W and load is again 10 Ω 10 W.

Here again within few seconds no output voltage and current.

One Answer

Here's your problem:

"We are not using any heatsink."

You are drawing a lot of current from your regulator. Linear regulators drop voltage by acting like a resistor in series with the source. The waste power as heat.

$$P = (V_{in}- V_{out}) times I$$

You don't say what your input voltage is, but the LM338 allows a maximum of 40 V difference between input and output. At 1 ampere that's 40 watts. At 5 amperes that's 200 watts.

That amount of power is wasted as heat. You must use a heat sink when drawing high current from a linear regulator.

Answered by JRE on December 5, 2020

Add your own answers!

Ask a Question

Get help from others!

© 2024 All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP