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I am attempting to deliver 9V @ 1.1A to a PCB circuit design using an L7809CV voltage regulator which is rated for 1.5A max.

Using a power supply, I can deliver the required current and voltage to my PCB (9V/1.1A) and all is well.

Once I turn up the voltage to ~12-14V (emulating an automotive application) and feed it into the L7809, the regulator will output 9V as expected with NO LOAD on the device. Once the load is applied, the device will only deliver 6.9V/0.1A, well below what is expected.

The device does get hot to the touched during extended use, but I would expect the correct current/voltage to (9V/1.1A) to be delivered when the device is cold. Is this not true?

Datasheet for voltage regulator: http://www.mouser.com/ds/2/389/CD00000444-249828.pdf

Circuit Requiring 9V @ 1.1A

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  • \$\begingroup\$ What is the impedance of your 12-14V supply? What does it measure before the regulator when its output drops down like that? On a semi related note, where are your input and output caps? \$\endgroup\$
    – PlasmaHH
    Commented May 2, 2016 at 14:42
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    \$\begingroup\$ The die inside heats up very quickly and shuts down, long before you feel the heat on the outside. \$\endgroup\$
    – Dave Tweed
    Commented May 2, 2016 at 14:46
  • \$\begingroup\$ And is the regulator mounted to a heatsink? This regulator has built-in thermal protection, which could be kicking in almost instantly as Dave alludes to. \$\endgroup\$
    – rdtsc
    Commented May 2, 2016 at 14:49
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    \$\begingroup\$ Measure the input voltage of the 7809 with the load in place. That voltage needs to be at least 9 V + 2 V = 11 V but in practice I would want it to be 12 V. Also 1.1 A through one 7809 is pushing it. Why not use 2 of them each feeding half of the LEDs. Apply heatsinks to those 7809s. Apply supply decoupling caps. I would opt for an LM2596 based switched mode regulator module, they can be bought cheap on ebay and are much more efficient. \$\endgroup\$ Commented May 2, 2016 at 14:49
  • \$\begingroup\$ When I apply a 12.8V input voltage, them multimeter reflects that. The voltage regulator puts out 7.7V. \$\endgroup\$ Commented May 2, 2016 at 15:09

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The chip die will get hot almost instantly and may go into limit before the heat conducts out to the heatsink tab.

Power dissipated in the device is \$ (V_{IN}-V_{OUT})I = (14-9)\cdot 1.1 = 5.5~W\$.

I don't see any decoupling capacitors on the regulator input and output. You may have a high frequency oscillation going on.

Even with a heat sink, can this regulator handle the the stress we're asking from it?

It can if we can keep the inner chip temperature low enough.

enter image description here

Figure 1. Thermal information from TI datasheet section 6.4.

I've highlighted the two relevant figures here. With no heatsink the chip temperature rise will be 22.9°C/W so you could expect a 125°C temperature rise which puts it just about on the 150°C max.

enter image description here

Figure 2. TO220 heatsink.

This one has a thermal resistance of 13.4°C/W. If we get good thermal conductivity between the tab and the heatsink (use thermal paste) the total resistance will be the series sum of all the thermal resistances = 13.4 + 1 = 14.4°C/W. At 5.5 W the temperature rise will now be 80°C. This is a drop of 45°C and will be a big improvement. Bigger heatsinks will reduce this further.

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  • \$\begingroup\$ Even with a heat sink, can this regulator handle the the stress we're asking from it? From what I understand, the input/output voltage difference in combination with the required current seems to generate too much heat to make this device a candidate for our application. \$\endgroup\$ Commented May 2, 2016 at 16:29
  • \$\begingroup\$ See update regarding heatsinking. You should now be beginning to understand the advantages of switched mode power supplies. \$\endgroup\$
    – Transistor
    Commented May 2, 2016 at 17:08
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With reference to the datasheet, the thermal resistance from silicon junction to ambient is 50 °C/W. At 20°C ambient and 14 V from power supply, the silicon junction would rise to (20 + 50*(14-9)*1.1) = 295 °C. The datasheet limit for junction temperature is only 125 °C. The temperature does not need to rise to the full theoretical value of 295 °C before the thermal limit engages. AT 14 V input, the thermal limit would engage much sooner.

enter image description here

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  • \$\begingroup\$ Can a heat sink dissipate this kind of heat, or are we plain and simple asking too much from this voltage regulator for this application? \$\endgroup\$ Commented May 2, 2016 at 16:26
  • \$\begingroup\$ @scorpdaddy: Where did you read 62.5°C/W? See my answer for the info on the TI datasheet. Have I misread? \$\endgroup\$
    – Transistor
    Commented May 2, 2016 at 17:09
  • \$\begingroup\$ I stand corrected. 50 °C/W for this package. \$\endgroup\$
    – scorpdaddy
    Commented May 3, 2016 at 13:57
  • \$\begingroup\$ I'm still curious. The TI sheet that I used (linked in the answer) shows 22.9°C/W for a TO220 (which I assumed). Did you pick another package? \$\endgroup\$
    – Transistor
    Commented May 3, 2016 at 14:15
  • \$\begingroup\$ Inserted into the answer a scrape of page 7 of the Mouser datasheet link above. \$\endgroup\$
    – scorpdaddy
    Commented May 3, 2016 at 15:20

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