0
\$\begingroup\$

I am trying to use a voltage regulator to regulate a 12V, 2A DC Power supply down to ~5.2V with a load current of less than 0.5A.

My schematic is as follows: Schematic

Problem: the regulator is getting quite hot, and voltage is dropping down to 4.3V or so with my fan connected. The output voltage is correct (~5.2V) with no load connected.

This is the datasheet for the regulator. The 10 uF capacitor is a tantalum capacitor (16V), and the 22 uF capacitor is an aluminum capacitor (20V). I do not have a heatsink on the regulator.

Is there something I am misunderstanding about the application? Is there a different regulator that would perform better?

\$\endgroup\$
  • \$\begingroup\$ try adding 100nF ceramic between the input of the regulator and the ground end of the 560 ohms. \$\endgroup\$ – Jasen Mar 2 at 4:03
  • 1
    \$\begingroup\$ at 3.5 W dissipation the LM317T will need a heatsink. \$\endgroup\$ – Jasen Mar 2 at 4:09
  • 1
    \$\begingroup\$ Check datasheet for capacitor recommendations. || Your regulator dissipation is about 14 Watts - you'd been a significant heatsink to handle that. (Power = V x I = (12-5)*2A = 14W. Without a heatsink "you haven't got a show" to use technical language :-). \$\endgroup\$ – Russell McMahon Mar 2 at 4:09
  • \$\begingroup\$ The datasheet does not show the thermal resistance in C/W for the T model regulator but at say 100C rise you'd need a C/W = delta_T/Power = 100 / 14 ~= 7 C/W heatsink. As a TO3 (large metal ) can has 39 °C/W thermal resistance and a TO220 will be quite a lot higher, you must have a heatsink OR get rid of some heat. || If you add a series resistor in the input you can dissipate most of the heat in the resistor. To give the 317 2v headroom and 1.25V for regulation ... \$\endgroup\$ – Russell McMahon Mar 2 at 4:15
  • \$\begingroup\$ Vdrop_R = (12-5-3-1.25) = 2.75A. R = V/I = 2.75/2 = 1.4 Ohms. Probably even more at say R=3/2 = 1.5 Ohhms (limit is when Vout is 5V at full load). Power reg is now about P=v*I = (12-5-3) x 2 A = 8W so you STILL need some sort of heatsink \$\endgroup\$ – Russell McMahon Mar 2 at 4:17
1
\$\begingroup\$

Your regulator dissipation is about 14 Watts - you'd need a significant heatsink to handle that.
Power = V x I = (12-5)*2A = 14W.
Without a heatsink "you haven't got a show" to use technical language :-).

The datasheet table 7.5 page 6 shows the thermal resistance in C/W for the T model regulator as 23 °C/W, - say 25 ° C/W, so say 100C rise you'd need a C/W = delta_T/Power = 100 / 14 ~= 7 C/W heatsink. That's a largish heastink.

If you add a series resistor in the input you can dissipate most of the heat in the resistor. The 317 needs about 3V drop / "headroom" at 2A (IF your version will handle 2A) (see datasheet)

To give the 317 3v headroom for regulation

Vdrop_R = (12-5-3) = 4V
R = V/I = 4/2 = 2 Ohms.
Limit is when Vout is 5V at full load.

Power reg is now about P=V*I = (12-5-4) x 2 A = 6W so you STILL need some sort of heatsink. For 100C rise Rth_sink ~= 100/6 = 16 C/W

The resistor power dissipation is about V x I = 4v x 2A = 8W.
So use 10W minimal and better 20W. 2 x ceramic 10 Watt air cooled cheap would work well.


THIS SE Q&A will tell you more than you want to know about regulator overheating

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.