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Will we get tombstoning of this resistor and capacitor?

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Flyback

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Hi,

Please advise if we will get “tombstoning” if we remove the thermal reliefs to the pads of the following components..

The component ringed in yellow is an 1812 ceramic capacitor.

The component ringed in red is a 2010 chip resistor.

We want to simply merge the left hand pads of these components completely into the surrounding thermal copper pour. (for better cooling of the components). However, as you can see, the right hand pads of these components have less copper attached to them. If we merge the left hand pads completely into the thermal copper pours, then will the right hand pads heat up more during the SMT soldering process, resulting in the components tombstoning?
 

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I am not a manufacturing engineer but: I think tombstoning is caused because the temperature of the ends of the component are different.

I do not use thermal relief on power supplies and high current parts. (hot components) No thermals is hard for hand soldering. My boards are heated in a oven and the entire board is at the same temperature. Manufacturing was unhappy at first but they started making my boards and there was no problems.

Some people believe that a "via" close to a pad (touching a pad) will suck the paste off a part. That is why you can get a via that is filled. I just keep the via off the pad. Or some time I fill the pad with many via(s) and call for more paste on the pad.

If you want to pull heat out of the parts, remove the thermals. Increase the width of the traces. Use via to get heat to the other side of the board. Most of the time there is no reason to use thin traces. Use copper as a heat sink. Use copper to spread the heat out over a larger area.
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I am no manufacturing guy, but I see it as close to impossible to tombstone a 1812 capacitor due to weight of the cap the capillary forces would have to lift, and 2010 resistor due to it being very small height, so there is not enough purchase for the solder to pull it up.
 
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