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For Christmas my company handed me a small LED "work light" that wasn't bright enough to light up any real work, so I figured I'd tear it down see what was inside hoping the the lesson would be worth more than the light.

I expected a tiny circuit with a button and resistor. I got two of those things, but how did the resistance work in this circuit? I didn't take a before photo, but the only thing different is a mound of dried, opaque white paste where there is only a paste circle now.

I tried Googling resistive paste, and there is such a thing, but I don't think this is it. Did I chip off a tiny resistor in there that I didn't see?

Board up to light Board

Edit: After some comments pointing out that this was likely COB, I've come to the conclusion that the momentary switch couldn't, on it's own, maintain the state of the light. Therefore there would need to be some logic in an IC. The original question wasn't so much focused on what COB is (although I learned that), but instead why a COB would be utilized instead of a much simpler circuit.

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closed as off-topic by Voltage Spike, ThreePhaseEel, DoxyLover, Tom Carpenter, Harry Svensson Jan 4 '18 at 15:35

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  • "Questions on the repair of consumer electronics, appliances, or other devices must involve specific troubleshooting steps and demonstrate a good understanding of the underlying design of the device being repaired. See also: Is asking on how to fix a faulty circuit on topic?" – Voltage Spike, ThreePhaseEel, DoxyLover, Harry Svensson
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    \$\begingroup\$ That is not a resistor, it is an integrated circuit bare die wirebonded to the board and covered with a blob of epoxy. \$\endgroup\$ – τεκ Jan 4 '18 at 1:23
  • \$\begingroup\$ IC I understand, however "bare die wirebonded" is a term I've never heard before, can you elaborate? And also speculate on how this simple LED driver circuit operated? \$\endgroup\$ – Joey Gennari Jan 4 '18 at 1:28
  • \$\begingroup\$ And also what advantage this type of construction would have? I only imagine it could've been cost for this cheap light. \$\endgroup\$ – Joey Gennari Jan 4 '18 at 1:32
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    \$\begingroup\$ Wires are bonded between the integrated circuit die and the lead frame of typical IC parts that have packages. In the case of no IC package the wires are bonded between the die and the PCB the die is glued too. The the die is covered with a blog of epoxy for protection. Why would they do this? Cost. The epoxy blob is cheaper than an putting the die into an injected plastic package with lead frame. \$\endgroup\$ – TimB Jan 4 '18 at 1:54
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    \$\begingroup\$ It's called "Chip-On-Board" technology COB for short.. see this learn.sparkfun.com/tutorials/how-chip-on-boards-are-made \$\endgroup\$ – Trevor_G Jan 4 '18 at 2:47
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Was there a resistor under this paste?

No.

The "paste" was an epoxy coating that was protecting a "chip on board" IC.

Most of us are familiar with ICs in rectangular black plastic packages with metal legs. There's an unseen tiny sliver of silicon embedded in the centre of those. However, for very high-volume manufacturing, it is sometimes cheaper to dispense with the plastic and metal leg packaging and instead attach the tiny silicon IC directly to the PCB.

I expected a tiny circuit with a button and resistor.

What you got was more sophisticated than that.

how did the resistance work in this circuit?

Either the IC was regulating the current or (more likely) the circuit design relied on the internal resistance of the battery. For example, a 3V lithium coin cell has about the right resistance to make a small-output LED work safely - see LED throwies

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    \$\begingroup\$ Perfect, that explains it. Now that I think about it, that switch was a simple momentary switch. Therefore the IC must have been maintaining the state of the light and switching it on and off. It's hard to imagine how that's cheaper than an on/off switch, but I'm guessing at the right volume anything can make sense. \$\endgroup\$ – Joey Gennari Jan 4 '18 at 16:06

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