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One ACM 7135 provides 350mA. Nine can fully power Cree XM-L2 U3. But that's 10W.

ACM7135 only have Vin, Vout and GND pins. People are using them with ATtiny (example), apparently without any other electronic there. It's hard to believe they are pushing 10W of power thorough micro controller. So how do they do it? How should it be done?

Transistors would be inconvenient, because at around 5V, voltage drop on transistor might mean a significant amount of power to dissipate. Or am I wrong here? Also, it seems there are no transistors on flashlight drivers anyway.

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  • \$\begingroup\$ People are most definitely not routing that kind of power through an ATtiny, so it seems you likely misunderstand what is going on. There does not appear to be any mention of an ATtiny on the page you link. \$\endgroup\$ – Chris Stratton Mar 10 '16 at 19:07
  • \$\begingroup\$ @ChrisStratton It's ATtiny13a on the photo. And here ATtiny 85 is used - written explicitly, sadly not in English. \$\endgroup\$ – Mołot Mar 10 '16 at 19:11
  • \$\begingroup\$ @ChrisStratton I'm not trying to understand specific circuit, I'm trying to find out how can ATtiny be used to control ACM's. Examples was posted only to show it actually was done. (this was an answer to comment now deleted, I'll leave it because it explains my intent). \$\endgroup\$ – Mołot Mar 10 '16 at 19:14
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The fundamental problem with your question is that you are just guessing at what is going on, without providing a schematic. If you had read the data sheet for the ACM7135, you would see in the suggested application circuit that the "Vdd" input is only for powering the internal control circuitry. The device functions as a low-side switch, so the load current flows between the "OUT" and "GND" terminals without involving the "Vdd". The LED anode gets its own connection to the supply voltage.

In fact, the maximum supply current on the "Vdd" pin is given at 200 uA, around two orders of magnitude below what the ATtiny is capable of.

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  • \$\begingroup\$ I have read datasheet for ACM7135. I just failed to understand it. Thanks for pointing out important part. \$\endgroup\$ – Mołot Mar 10 '16 at 19:19

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