5
\$\begingroup\$

I am migrating a design from through-hole parts to use as many SMT devices as possible.

There are 7- 2 Watt power resistors in the current design. These are metal-oxide resistors that have crimped and formed leads so as to raise the body of the resistor about 0.2" above the surface of the circuit board. This current design has been reliable and trouble-free for many, many years.

I'm now looking at replacing those axial-lead resistors with multiple SMD resistors in parallel. There would be 3- 1W resistors (2512 package) in parallel to give the required resistance and power rating.

However, I don't know how much PCB copper surface area that these three resistors are going to require. Each resistor will dissipate about 250mW under normal line-voltage conditions but the dissipation increases to about 900mW under worst-case high-line-voltage conditions. That's a total of about 2.7 Watts for each of the resistor clusters.

Because the PCB material is FR-4, I want to keep the maximum PCB temperature under 75C. Assuming ambient temperature of about 30C, that gives me a temperature rise of only 45C. That suggests that I'm going to want an awful lot of surface area - but I don't know how much.

The boards will be in an enclosure with only natural convection currents inside the enclosure.

Guidance is much appreciated.


[Edit]

Based on the document link provided by John Birckhead, it's not feasible to increase the resistor current as originally desired. If I stay with the total resistance near 680R, I'll need to dissipate about 1.3 Watts in each resistor cluster with worst-case high-line-voltage input. This drops to about 0.5 Watts with nominal line voltage.

The customer has asked me to increase the size of the PCB from the current size - this will allow more room for descriptive text near the terminal blocks as well in blank areas on the board surface. I should be able to eek out darned near 3 square inches per resistor cluster - solid copper fills on both top and bottom layers as well as lots of large-diameter vias (0.036" holes) scattered all over the area. The vias will fill with solder when the board is wave-soldered

All that said: I have the option of have the copper fills on the bottom of the PCB either covered with solder-mask or left as bright solder finish. I'm inclined to have it as bright solder finish so as to increase the radiating qualities. Opinions requested.

\$\endgroup\$
  • \$\begingroup\$ So on the old design the 2W resistors were shifting 2.7W each, worst case? \$\endgroup\$ – user57709 Jun 22 '16 at 21:05
  • \$\begingroup\$ Nope. I've reduced the total resistance value from 680R down to 340R as per customer request. Original design power dissipation was about 1.3 Watts under worst-case high-line-voltage conditions. I'll roll back the power dissipation if I can't get adequate heat-sinking. \$\endgroup\$ – Dwayne Reid Jun 22 '16 at 21:08
  • \$\begingroup\$ There are cute off-board SMD resistors, but I've only seen them for current sense applications. Still cute though. \$\endgroup\$ – pipe Jun 22 '16 at 21:34
  • \$\begingroup\$ Is this a multi-layer (more than 2) PCB? \$\endgroup\$ – Peter Smith Jun 23 '16 at 9:56
  • \$\begingroup\$ You would need a lot of area for that. I would keep through-hole resistors and switch other lower-power elements to smd. \$\endgroup\$ – Darko Jun 23 '16 at 16:32
6
\$\begingroup\$

TI has done a good job on this app note http://www.ti.com.cn/general/cn/docs/lit/getliterature.tsp?baseLiteratureNumber=snva419&fileType=pdf

The salient point from the above document is: "With only natural convection (i.e. no airflow), and no heat sink, a typical two sided PCB with solid copper fills on both sides, needs at least 15.29 cm squared or ≊ 2.37 in squared of area to dissipate 1 watt of power for a 40°C rise in temperature."

\$\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.