Should I consider a heatsink? - Page 1
Forum section: electronics / beginners

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 20, 2023, 05:02:11 pm
content: I'm simulating parts for a buck converter (theLT1765), and I'm using this diode as my catch diode and boost diode,the V35PWM15. I'm a bit concerned about whether I need a heatsink for the diode.Looking at the datasheet, they recommend a 6mm by 6 mm square pad for the cathode in their package. But then, above the dimensions, there's Figure 7, the typical resistance junction to ambient temp vs copper pad area. Going by the aforementioned square pad, that's 0.36 cm2, which results in a thermal resistance of ~55 deg. C per watt. What does that mean? Is it a measure of how much the diode temperature rises as a result of the power running through it? If so, the recommended footprint size doesn't seem to offer a lot of thermal protection. Is there something I'm not taking into account?

author: Berni
author_rank: Super Contributor
timestamp: October 20, 2023, 05:10:00 pm
content: It is exactly what the unit says it is. The number of °C of temperature rise above ambient for every 1 W of power dissipated in the device.If you need a heatsink or not depends on how hard you drive the diode. Good simulators can reasonably accurately predict the average number of Watts dissipated in a component.And yes 55°C/W is not very impressive for power dissipation, this is because PCBs are pretty terrible heatsinks unless you really pack them with lots of thick copper

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 20, 2023, 05:43:27 pm
content: Thanks for replying. I'm planning to go with a 2-oz copper pour in my end design, so hopefully that extra ounce of copper will help a bit.The simulation is kind of annoying. With the numbers I'm running at, my diode is reaching ~30 watts at roughly ~4.8 amps. That's kind of insane thinking about it. Heatsink aside, I guess the only thing I can do this late in the game is to find a different Schottky diode, or perhaps reduce my output capacitance to hopefully reduce the current.

author: Berni
author_rank: Super Contributor
timestamp: October 20, 2023, 05:58:05 pm
content: Yes at 30W of average power dissipation that diode will not live for very long (Unless it is soldered to something like a solid copper bar).Tho with that much power dissipation in just the diode you must be making a VERY powerful DC/DC converter. Or if it is a low-ish power converter then it must have horrible efficiency. Tho it could also be an inaccurate simulation since it is easy to accidentally make a circuit component unrealistically good or unrealistically bad by modeling your circuit in a bad way.The way i design DC/DC converters is usually just taking the example schematic from the datasheet or reference design as a base and mostly just replacing components with the ones i actually want to use (using the original components as a guide for approximately what kind of part is required to do the job). Usually not simulating anything, but do tend to do some rough calculations about the kind of currents id expect to flow inside the main parts (helps with choosing the more appropriate part for my particular use case).

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 20, 2023, 06:58:16 pm
content: Apparently, more power does hurt. All jokes aside, someone once told me to not put too much faith in simulations. Considering how the input voltage pin was drawing way more than the max current it was limited to in the datasheet, accuracy is subjective.I mean, I suppose, I did have too much of a high capacitance. Hopefully, the boost diode current draw will be lessened by reducing my output capacitor values. I'm running a total of 870 uF as an output capacitor. Perhaps that much is overkill for drawing 1 amp to my test load. It's hard for me to tell how much capacitance is too much. Last DC/DC converter I designed, I was running 560 uF 4x, so if that's any indication...

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 20, 2023, 09:09:15 pm
content: What you said before, the typical thermal resistance of 55 deg. C per watt...that seems pretty high. With that sort of value, it sounds like it can't handle even low wattages like 3-5 watts, if that much would increase the device temp by over 100 degrees C. Furthermore, if you look at Figure 1 on the datasheet, as the mount temperature rises, the average current decreases...from ~5 amps. Thinking about this now, really feels like this diode is not good at all.

author: TimFox
author_rank: Super Contributor
timestamp: October 20, 2023, 09:17:25 pm
content: That's why the datasheet discussed heatsinking.5 W of power into 55 K/W gives 275 Cotemperature rise, and silicon is unhappy above 175oC junction temperatureNon-conservative heatsinking is the original sin of solid-state power design.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 20, 2023, 09:43:52 pm
content: Makes me wonder how I ever got my first design to work. In any case, I hopefully have a few options here aside from looking for a new diode to model. I'm hoping that by increasing the slow-start time to an order of milliseconds, the charge/discharge will take longer, so the diode doesn't have to pass a lot of current all at once. It's funny. I try to account for high current loads by picking a Schottky diode that can handle it, but in doing so, I totally screwed over my design. This makes me wonder...if you look at the LT1765's datasheet, they use the CMDSH-3 diode in their example as the boost diode. It's only rated for 100 mA, and it's in a smaller SOT-323 package. Why would something like that be practical in a high-power circuit?

author: Berni
author_rank: Super Contributor
timestamp: October 20, 2023, 10:12:10 pm
content: The diode could be ran at 30A when it is provided with enough heatsinking.Schottky diodes are usually the best choice for DC/DCs due to the high speed and low forward drop. If you can't get the performance you need out of a diode, then perhaps you need to upgrade your DC/DC converter to a more complicated synchronous rectifier.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 20, 2023, 10:26:02 pm
content: For the moment, I'm playing around with different diodes to simulate. I don't think it's so much the diode but more the conditions I'm trying to simulate them in. I'm trying to power a buck converter and a buck-boost converter at the same time, and because of a near-instantaneous drop in the output voltage of one of them, I suppose that's causing a big discharge in the capacitors resulting in the large diode current draw (40 amps). I can think of a few reasons why this is occurring, but I just can't seem to slow it down.

author: Konkedout
author_rank: Frequent Contributor
timestamp: October 21, 2023, 04:04:50 am
content: I have been designing power supplies for 45 years. Either I am missing something, or the power numbers I am reading here are nuts. The IC is good for 3 Amps output maximum and about 24V input maximum. With 3A in a schottky diode, you should be getting power dissipation in the diode (that is heat produced by the diode, not power output) of no more than 1.8 Watts and almost definitely less than that. I think that the diode you have chosen is bigger than optimal. Schottky and other diodes can go into a thermal runaway. This is a vicious cycle where high temperature causes reverse leakage current which dissipates power and further heats the diode. But there is no reason why you should have such problem in this design if you are running in a consumer or commercial environmental temperature.I would go with a diode more like this:https://www.vishay.com/docs/89029/ss8p3c.pdf(go with the 40V, unless your input voltage is less than maybe 15V. Best to stay well under the diode voltage rating).

author: macboy
author_rank: Super Contributor
timestamp: October 21, 2023, 01:02:29 pm
content: Quote from: LoveLaika on October 20, 2023, 05:43:27 pm30 W at 4.8 A? That's implausible. That implies over 6 V forward voltage drop, or very high reverse leakage current combined with very high reverse voltage. I just don't see it. At 4.8 A, a Schottky should dissipate just a few watts.

author: Tation
author_rank: Regular Contributor
timestamp: October 21, 2023, 03:05:52 pm
content: Quote from: LoveLaika on October 20, 2023, 05:43:27 pmHow is it possible for a (non Zener) diode conducting 4.8 A to dissipate 30 W?

author: wasedadoc
author_rank: Super Contributor
timestamp: October 21, 2023, 03:37:49 pm
content: The value of output/load capacitance has minimal effect on the average current through the diode. Think about it. No direct current flows through the capacitor. The average current through the diode must equal the average current into the load.The value of capacitancedoesaffect the peak current through the diode, but as the peak curent increases its duty cycle decreases. (The average current remains the same.) As the peak current increases the voltage drop in the diode increases too but nowhere near 1:1, so the average dissipated power increases only quite slowly as the peak current increases.

author: jwet
author_rank: Frequent Contributor
timestamp: October 21, 2023, 05:24:55 pm
content: This is a confusing post.1. The CMDSH-3 diode is just to provide power for the high side driver- in this case an NPN bipolar. No load current flows through it but it will have an average of the output divided the gain of the bipolar. It's just topping off that .18 uF cap on each cycle for the high side drive. You can see how much that cap gets whacked in a cycle but its probably only 20%. Because this converter runs at fairly high speed you want a fast diode but it doesn't dissipate anything.2. The free wheeling diode current average which will drive thermals. You don't say what your Vin/Vout are but the freewheeling diode will conduct when the output isn't. Duty Cycle for a buck : D=Vout/Vin. Diode conducts 1/D. The 1765 has a 3 amp switch, so diode peak currenty won't be any more than 3 amps, average diode current would be 1.5 amps at Vout=Vin/2 - ie 50% Duty Cycle. Vf for your diode is .5v at 5amp so at 50% of 3amps will only dissipate .75W, 42 degree C rise with little copper square. Not a big deal but could be better.3. At these currents especially at low voltage, you're getting into sync rectifier territory. What is Vin/Vout and I out??

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 23, 2023, 02:29:32 pm
content: I appreciate the suggestion, but I think I'm also misunderstanding something, both with my design and my understanding of certain values. If I may, I'd like to ask you something regarding power capabilities.Let's take your suggested diode, the SS8P3C. It has a thermal resistance of 60 deg. C per W from Junction to Ambient and 3 deg. C per W from junction to case (or lead, but is that the same thing as case)? If we assume conservative temperature values (150 deg. C operating temp and 25 deg. C ambient), using the junction to ambient temp, the power required to reach the max operating value is 2.08 watts (150-25/60). Would that be the max power that the diode can handle while still being in operation? If so, that would be the upper limit of what the catch diode can handle and gives me numbers to work with. I think I'm fundamentally misunderstanding how much power my device can handle given these thermal resistance numbers.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 23, 2023, 02:57:59 pm
content: Quote from: macboy on October 21, 2023, 01:02:29 pmIt's odd to me too. To give more context, I'm using the LT1765 to create an inverting buck-boost output with an input of 14 volts and an output of -10 volts. With the way the circuit is working, the diode seems to draw 4 amps at start-up with the current coming from both the input and output caps when the LT1765 switches at the minimum input voltage to work. But, over time, the switching voltage node peaks near Vin (minus, I believe, the minimum Boost Voltage Above the Switch voltage). With a big difference in voltage between the switching node and the output, even though the diode conducts, say around ~1.25 amps, the voltage causes the power loss to be high. What I'm concerned about is that the diode won't be able to handle this type of wattage given its thermal resistances.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 23, 2023, 05:26:25 pm
content: I apologize for not fully explaining myself. I thought I could figure out this issue by simply using a heatsink, but it's becoming more complicated now.I said in another comment after yours, but I'm doing this in an inverting buck-boost configuration with the LT1765. For the moment, input is 14 volts and output is -10 volts. Odds are, I'm not going to draw more than 1 amp from it, but I'm designing in the worst case scenario. When the regulator starts conducting, the diode has a surge of current flowing through it resulting in a initial high wattage across it. The switch voltage seems to increase until it reaches a peak of VIN minus the min. voltage between switch node and boost cap, so thats ~12 volts using these values. As the output decreases slowly via slow start, the diode reaches a peak wattage as the switch voltage rises, eventually reaching a peak wattage. Once it reaches that peak value, since the switch voltage has reached its max value, the power loss across the diode decreases significantly.

author: Berni
author_rank: Super Contributor
timestamp: October 23, 2023, 09:06:11 pm
content: You seam to be overly concerned about peak values.Semiconductors can typically tolerate pretty large peak power dispensations because the piece of silicon inside has a reasonable amount of thermal mass. For switching applications you typically get graphs in the datasheet that tell you just how far you can go for various amounts of time. It is usually called a SOA (Safe Operating Area). Some devices (like MOSFETs, SCRs..etc) can handle repetitive 1000% overloads for short peaks without any harm.Since i see you are using LT Spice. The way i estimate power dissipation there is first Alt+Click the component to bring the power dissipation into the graph. Then zoom the graph into a large area of stable operation. Then Ctrl+Click the legend on the graph to get a window with the average value of that potion of the trace. This works on all the usual types of components (power sources, loads, semiconductors, passives..etc)

author: Tation
author_rank: Regular Contributor
timestamp: October 24, 2023, 06:50:32 am
content: The datasheet of the diode states a max surge current of 260 A and a transient thermal resistance (for short pulses) below 1 ºC/W. Are these enough?

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 25, 2023, 09:03:54 pm
content: Thank you for showing me that. Indeed, I am using LTspice, but I never really used that power dissipation tool before. I know that mileage may vary, but it reassured me. I was focusing on the peak values, and I was concerned about that. Though, that's what they are: peak values. Upon closer inspection, those high wattage peak values (reaching ~8 watts) only lasted a few nanoseconds, or maybe microseconds. Doing what you suggested, getting an average value, showcased a more modest average of 0.147 watts given an 83 ms runtime. That sort of metric gives me a bit more confidence in that I won't melt down my diodes when I first turn it on.