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If I have a LiPo battery with the maximum voltage of 12V and I want to use it to power a 5v(L7805) voltage regulator. Assuming the maximum current passing through the regulator is 500mA. From datasheet, the thermal resistance is 50°C/W.

Does that mean that it would generate a power loss of 3.5W and it would raise the voltage regulator's temperature by 175°C? The maximum operating temperature of the voltage regulator is 150°C. Would that means I need to drop the input voltage of the regulator?

Additional info from OP comments:

I am not able to get a switching regulator and the great power loss is not a concern. Would the thing work if I stack multiple voltage regulator together?

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    \$\begingroup\$ That's exactly what it means. Alternatively, you could use a heatsink. But keep in mind that all that power is just being wasted--you should look into a switching regulator, not a linear one, for this level of voltage difference and current. \$\endgroup\$
    – Hearth
    Commented May 8, 2017 at 14:56
  • \$\begingroup\$ If I am not able to get a switching regulator and the great power loss is not a concerned. Would the thing work if I stack multiple voltage regulator together? I did some rough calculation. If I am constructing the circuit without using resistor, using four voltage regulator would raise each temperature by 90°C. \$\endgroup\$
    – Daozui
    Commented May 8, 2017 at 15:07
  • \$\begingroup\$ Remember that heatsinks are an option as well. With a heatsink, you can reduce the case-to-ambient thermal resistance. \$\endgroup\$
    – Hearth
    Commented May 8, 2017 at 15:10
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    \$\begingroup\$ Have you calculated the maximum value of a series resistor between 12V and the regulator? And what would the worst case regulator and resistor power be? \$\endgroup\$
    – scorpdaddy
    Commented May 8, 2017 at 15:16
  • \$\begingroup\$ By the way, if I use heatsink instead of resistor to handle to power loss. It would give me a larger allowance on the battery voltage drop, right? \$\endgroup\$
    – Daozui
    Commented May 8, 2017 at 15:21

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Your calculations are absolutely right.

Given your restrictions and goals: YES you could use several cascaded linear regulators (with different output voltages, of course, and respecting the required dropout voltages) and/or series resistors, in order to distribute the heat dissipation across several components.

However, a heatsink looks like a far simpler and better solution. Let's say you want to derate the maximum junction temperature to 80% while being able to operate the regulator up to 50 C ambient temperature.

That would mean a maximum junction temperature rise of 70 C. At 3.5 W dissipation, you'll need a 20 C/W junction-to-ambient thermal resistance. The TO-220 junction-to-case thermal resistance is just 5 C/W, so you would need a heatsink providing a 15 C/W case-to-ambient thermal resistance. That can be perfectly achieved with a passive heatsink (which could even by realised by bolting the regulator to your case/chassis if available).

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