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I am looking at using a linear regulator on a load that will be turned on and off once every 10 seconds(3153600 cycles per year). For context I am currently looking at L7812CV. Will this damage the linear regulator? Will a linear regulator be reliable at this level of on-off cycles? Is there a different component you would recommend?

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    \$\begingroup\$ Probably the regulator itself will be OK. Design details may still matter. For example, the output capacitors could partially determine how much stress is applied to the regulator each time it starts up. Probably a good idea to do a rapid cycle test for 10 million cycles or failure, whichever comes first. Don't cycle at 10 second intervals. Cycle 2x per second or something in order to get the data faster. You may need to add additional circuitry to force discharge any output capacitors (to make the simulation more valid). \$\endgroup\$
    – mkeith
    Apr 8 '21 at 21:23
  • \$\begingroup\$ @mkeith, Okay awesome! I didn't think about just testing it myself; that's a good idea! I will start with the recommended capacitor sizes on the spec sheet and go from there. Thanks for your help! \$\endgroup\$
    – ericnutsch
    Apr 8 '21 at 21:36
  • \$\begingroup\$ Diode in reverse between input and output nessessary for IC protection \$\endgroup\$
    – user263983
    Apr 8 '21 at 22:00
  • \$\begingroup\$ Why not put a P-FET as a power switch and you have no worries. \$\endgroup\$
    – Gil
    Apr 8 '21 at 23:37
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In your other question you talk about powering a solenoid over a variable length wire ('1000+ft away') using the regulator to drop 24 V down to 12 V at the solenoid. Your solenoid draws 6.5 W at 12 V, which corresponds to ~0.542 A.

At this current a 7812 needs a minimum of about 2 V of 'headroom' to maintain regulation, which will be the voltage drop at maximum wire length. At 0.542 A this corresponds to a power dissipation of ~1.1 W. Shorter wires will make the regulator drop more voltage and dissipate more power, eg. at 4 V it would dissipate ~2.2 W. Maximum power dissipation (with appropriate heat sink) could be as high as 25 W and is internally limited, so you might think it isn't a problem. But...

The bare TO220 package has a thermal resistance of 50 °C/W, so at an ambient temperature of 25 °C the die temperature could reach 135 °C - close to the thermal limit of ~150 °C. Without a heat sink it may work, but put a lot of thermal stress on the die each time the power is turned on and off. Putting a good sized heat sink on it will help, but the thermal stress issue remains.

The cycle time of 10 seconds is bad because it gives the die time to cool down when turned off and heat up again when turned back on. Different parts of the regulator expand at different rates as the temperature changes, which could result in early failure due to fractures between the die and bonding wires. Accelerated testing may not pick this up.

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    \$\begingroup\$ Just thought I would chime in to say I agree that rapid cycling would probably not thermally stress the device in the same way as the real cycling would. Maybe a heatsink is a good idea for that reason even if it might otherwise not be needed. Of course OP can test this, too by doing 10 second cycles to see how much the IC heats up and cools down and what the peak temp rise is after cycling for a long time. \$\endgroup\$
    – mkeith
    Apr 9 '21 at 2:08
  • \$\begingroup\$ @Bruce Abbott, Thanks for looking up my other question. I didn't link it since everyone seems to hate it lol. The social finesse of asking an SE question that people like is something that escapes me. Thermal cycling is a good point. I didn't think about that. I guess a DC-DC converter may be a better solution in light of the thermal considerations. At some point I guess I should just put in one DC-DC converter and another bank of SSRs. Thanks for your help! \$\endgroup\$
    – ericnutsch
    Apr 9 '21 at 5:15

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