Thermal Consideration for SMB Diode - Page 1
Forum section: electronics / projects

author: tarun172
author_rank: Regular Contributor
timestamp: November 07, 2013, 03:22:46 am
content: Hello,I am trying to figure out a problem with a design I am working on. I have a SMB diode (Vf = 1.1 V @ 1A) used as a clipper to limit voltage down to 3.3 V from 120 V AC input. Circuit looks something like this: Three of these diodes are connected in shunt to give me Vf of total = 3.3 V. The input to this shunt is bridge rectifier circuit (120 VAC, 60 Hz). When I look at the diode datasheet; the thermal calculations do not look very good. Meaning that at a forward voltage of 1.1 V @ 1A, the power dissipation would be very high possibly exceeding junction temperature of 150C!!.I do not have whole lot of PCB space but I can work on having a copper pour for better heat dissipation. Besides making changes to the design, can copper pour be efficiently utilized for reduce temperature rise?- tarun172

author: nickm
author_rank: Regular Contributor
timestamp: November 07, 2013, 06:00:30 am
content: If you look at the datasheet they will spec the thermal resistance from the diode junction to the ambient air. They might give you two numbers, one with the minimum amount of copper, and another with about 1 square inch of copper. You are dissipating 1.1W per diode so multiply the deg C/W by 1.1 to find out what the temperature rise of your diode junction will be.You could get a pcb with thicker copper, or use vias and utilize more layers for heatsinking. Posting more information about you circuit and project might get you better results.

author: SeanB
author_rank: Super Contributor
timestamp: November 07, 2013, 07:55:30 pm
content: What current limiting resistor are you using? At 1A the resistor will dissipate 120W of heat, and will be a really big unit. If you are just protecting an input the current should be well under 1A, likely in the under 5mA region where the diode will barely get above ambient.

author: tszaboo
author_rank: Super Contributor
timestamp: November 07, 2013, 09:40:06 pm
content: You need about 7x7cm copper pour to dissipate that heat, said the rule of thumb. But why don't you build a proper power supply?

author: tarun172
author_rank: Regular Contributor
timestamp: November 07, 2013, 11:27:25 pm
content: All,The PCB is relatively small, i.e few mm. The diode datasheet lists typical thermal resistance as 30 C/W. Since, I do not want diode to reach maximum junction temperature during temp tests (-40 to 80C). So, I need heat sink which can dissipate the generated heat. Going by these numbers, here is what I have:Worst case Power dissipation = 1.1 WTemperature rise worst case 150 -25 = 125So, heat sink thermal resistance required = 125/1.1 - 30 = 83.6 C/WAm I being too conservative or just going nuts? I also want to move to 2 oz of copper instead of 1 oz.

author: Niklas
author_rank: Frequent Contributor
timestamp: November 08, 2013, 08:35:18 am
content: The worst temp rise condition would actually be at 80'C ambient temperature. If you are going to let one diode dissipate that 1.1 Watts of heat, then the maximum allowed thermal resistance of junction-ambient is (125-80)/1.1 = 41 K/W without any kind of safety margin. Then you must also take into account those other two diodes as well as the space is constrained. Just as an approximation divide 41 K/W by 3 and then add some margin to that and the heat sink you are looking for is closer to 5 K/W or less. Also remember that the diode packages are small and that means bottlenecks for the dissipated heat to transport through. Even if you can manage to find a small heatsink with a fan you might not be able to transport the heat out of the small hot spots.Some questions: Are you going to regulate the current through the diodes in any way? If so, how? The voltage-current graph is not very linear and will change fast with a change in either voltage or temperature. That 120 Vac will also vary depending on the load and might be higher than 120 Vac...

author: poorchava
author_rank: Super Contributor
timestamp: November 12, 2013, 11:56:08 am
content: You can use larger number of diodes at lower current and lower forward voltages. This would reduce power dissipation, while still giving 3.3V.What exactly are you trying to accomplish? Some sort of uber-cheap voltage supply? Mains power detection?

author: w2aew
author_rank: Super Contributor
timestamp: November 12, 2013, 03:24:39 pm
content: Quote from: SeanB on November 07, 2013, 07:55:30 pmDitto this! If you've got 100+ volts across the limiting resistor ahead of this shunt regulator arrangement, the dissipation in the this resistor is the real problem!

author: Jon86
author_rank: Frequent Contributor
timestamp: November 12, 2013, 08:19:14 pm
content: Definitely second what's been said up there ^Build a real supply, trying to drop all that voltage through something like a resistor is never going to go well.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: November 12, 2013, 08:24:46 pm
content: Hint- when working with AC, a reactance doesn't dissipate any power. Choose a low loss non-polar capacitor with sufficient voltage rating such that the reactance at line frequency is equal to the resistor you were going to use. There are lots of other considerations, safety, overvoltages and such, but it's one way to get the job done.

author: tarun172
author_rank: Regular Contributor
timestamp: December 20, 2013, 04:01:21 am
content: Thank guys for all your help.. Yes, using diodes of lower current and voltage would help. Thermal resistance of diodes would also matter in this case. Lower thermal resistance the better.I did some testing and played around the circuit for last few weeks. The first thing I noticed was the temperature reading across the current limiting resistor was slightly higher than the rest of the circuit. I do not plan to use more than 200 mA across this indicator circuit. Using ground planes definitely helped. When I compared temperature reading with and without copper pour PCBs; the temperature difference was more than 25C!Of course the two are used for different purposes but lower temperature reading really amazed me.

author: tarun172
author_rank: Regular Contributor
timestamp: December 20, 2013, 04:03:02 am
content: @w2aew. I agree. That is exactly what I found during testing. @poorchava, @NANDBlog. This is just an indicator circuit not used for voltage regulation.

author: free_electron
author_rank: Super Contributor
timestamp: December 20, 2013, 05:51:55 am
content: indicator circuit ; do not use full wave rectification but half wave. this brings your dissipated power to 1/4 !couple it in capacitively and you can drop it even further

author: djococaud
author_rank: Regular Contributor
timestamp: December 20, 2013, 09:30:28 am
content: Hi everyone...Just if anyone is interested, here's a simple schematic for a 3.3 Vdc PSU.This is not isolated (so,carefullwith it) and low power (10-20mA) but it's simple and low heat dissipated (thanks to the reactance of C1).regards.P.S: Yes, V1 should be SINE(0 325 50) instead of SINE(0 230 50) in my simulation...(just a small mistake...). And there is two R1...(I was too speedy renaming the comps !)

author: tarun172
author_rank: Regular Contributor
timestamp: December 21, 2013, 03:19:43 pm
content: @free_electron. Very true. Great, simple idea:clap:I guess I could use half wave circuit. Thanks.