high input voltage into an L7805 - would like to add a resistor to drop Vin, but not sure how to calculate

GGibb.1 · ‎2021-05-18

In the datasheet, it gives a formula, R = (Vi(min)-Vxx -Vdrop(max)) / (IO(max)+Id(max))

My IO is 0.1A, My output voltage is 5V. My input is currently 24V. I saw some discussion that I think I should consider the Vdrop to be 7V. Is Vxx the voltage I want after my resistor? Min input should be 7.5, but I was thinking 10-12V. I don't know what Id is.

See Pae 27 of link

https://www.st.com/content/ccc/resource/technical/document/datasheet/41/4f/b3/b0/12/d4/47/88/CD00000444.pdf/files/CD00000444.pdf/jcr:content/translations/en.CD00000444.pdf

Peter BENSCH · ‎2021-05-19

To give you an idea Ive put the data into a screenshot of that picture, assuming the minimum input voltage is 24V-10% and taking the other value from the tables of the electrical characteristics. Therefore for distributing some power you could use a 120ohms resistor with a power rating of at least 1.5W.

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Peter BENSCH · ‎2021-05-19

Vxx is the typical output voltage of the L78 in question, in your case 5V.

A linear regulator like the L78 generally "burns" the voltage difference between input and output, which is shown in more or less strong heating by the dissipate power (P = Vdrop * Iout).

For example, if you imagine the extreme case that your L7805 gets an input voltage of 35V and you would draw the max current of 1.5A from it, then it "burns up" 35V-5V = 30V, which would result in a power dissipation of 45W. With the 5K/W of the TO-220 package this would rise the temperature by 45W*5K/W = 225°C.

Therefore, the schematic in Fig. 23 of the datasheet shows one possibility of reducing the power dissipation within the L78 by distributing the power between itself and an external resistor.

In your case you had given the parameters Vin=24V, Vout=5V, Iout=0.1A, which resultes in a power dissipation of (24V-5V)*0.1A = 1.9W. All you need is a heat sink on the L7805, the resistor discussed above is not necessary.

Please keep the thermal resistance of that heat sink in mind, as it adds to the one of the package of your L7805. Suppose you are using the TO-220 with 5K/W and a heat sink with e.g. 2K/W then this will result in a temperature rise of (5K/W+2K/W)*1.9W = +13.3K for that power dissipation.

Good luck!

If the problem is resolved, please mark this topic as answered by selecting Select as best. This will help other users find that answer faster.

/Peter

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GGibb.1 · ‎2021-05-19

Thank you! Unfortunately I am not actually using the TO220 package any more. A previous circuit iteration used it, and the heat was not noticed to be a problem. Then the board was redesigned to be surface mount. Now we are having issues.

I found this circuit in the old datasheet and thought it should still be useful.

I have also realized that 12V is now accessible on the board for the circuit as well, and was planning to switch to that for my Vin, but I thought that possibly I could also use a resistor to drop the dissipation in the device even more.

But I am still not sure what some of the variables are in that formula in order to calculate an appropriate resistor value.

Can yo clarify the formula? A heatsink is not an option at the moment with the surface mount component.

Peter BENSCH · ‎2021-05-19

To give you an idea Ive put the data into a screenshot of that picture, assuming the minimum input voltage is 24V-10% and taking the other value from the tables of the electrical characteristics. Therefore for distributing some power you could use a 120ohms resistor with a power rating of at least 1.5W.

In order to give better visibility on the answered topics, please click on Accept as Solution on the reply which solved your issue or answered your question.

Tesla DeLorean · ‎2021-05-19

Doesn't sound like a very green/eco-friendly design. Try to design more efficiently.

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GGibb.1 · ‎2021-05-19

I am trying. That is why I am switching to 12V. There were bigger issues to solve in the layout initially, and the voltages were overlooked as something to improve, because the initial circuit worked for what was needed, and 24V was the only voltage available at the time.

But thank you for your input.

GGibb.1 · ‎2021-05-19

Thank you. So for a 12V supply, possibly something more in the 27 ohm range with a 0.5W rating might work.

Piranha · ‎2021-05-19

If your requirements allow it, I would suggest just taking L7983PU50R, L6984 or a similar buck converter in QFN/SON package. At 24V input and 5V 0,1A output the L7983PU50R will have 85% efficiency, consume 88mW and will heat up only 4,4C above ambient temperature. And even better efficiency at 12V input: 90%, 56mW and 2,8C.