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I have RFID reader module. For this RFID module, they recommend 5V-9V LDO power supply and here is recommended schematic :

enter image description here

I read 7809's data sheet. It says I can use 7809's input value between 11-35 V . So my questions is ,

  1. What happen if I use 24V DC instead of 12V DC for 9V ? Will RFID module's read range change?
  2. I want to use 24 V DC because I have 24V DC adapter and it has 5A current. This RFID module needs max 120mA. Is this 5A cause something for 7809 (heat or something)? Thank you.
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  • \$\begingroup\$ You can use a LDO without using dinosaur LM7809. Check out for example LT30xx family of regulators. \$\endgroup\$
    – Lundin
    Commented Dec 15, 2021 at 14:59
  • \$\begingroup\$ @Lundin I want to ask you why you suggest me to use two separate switch regulator. Can't I use 12V to 5V switch regulator then linear regulator for 5V to 3.3V. like ncp1117st33t3g . Is it because what is used by modern electronics.Like you mentioned before. \$\endgroup\$ Commented Dec 16, 2021 at 5:40
  • \$\begingroup\$ Yes sure, that's often how it's done, if you wish a dedicated 3.3V supply for the RF parts. Going from 5V to 3.3V with a LDO is not a problem. \$\endgroup\$
    – Lundin
    Commented Dec 16, 2021 at 7:20
  • \$\begingroup\$ I will use 3.3 Volt for esp01 and W5500. I guess, I need low noise for these right? \$\endgroup\$ Commented Dec 16, 2021 at 7:51
  • \$\begingroup\$ I have no idea what those are - if they are pre-made modules they might have voltage regulators on board. \$\endgroup\$
    – Lundin
    Commented Dec 16, 2021 at 7:53

2 Answers 2

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If you use 24 V instead of 12 V, you will dissipate 5X as much power in the 7809 regulator. Using your quoted current of 120 mA, with a 12 V supply you need to drop \$(12 - 9) = 3~\mathrm{V}\$ so the regulator is dissipating 360 mW (\$P=IR\$). If you change to 24 V you need to drop \$(24-9) = 15~\mathrm{V}\$, and then you need to dissipate 1.8 W. This is not a huge amount, but you will absolutely need a heatsink for it.

Your module will only draw as much current as it needs, so the 5 A rating of your supply isn't relevant. More information in the canonical answer to this question.

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  • \$\begingroup\$ Thank you. How can I decide using heat sink ? I checked data sheet I couldn't find. (I am not good at it that's why probably.) If I learn how to use it , I don't bother you with these question. Is this border 600 mW ? \$\endgroup\$ Commented Dec 15, 2021 at 13:10
  • \$\begingroup\$ No problem, we all started somewhere :) So, you need to look at the package type - for the 7809CV, it's TO-220 (good, as you'll need the big package to dissipate the heat). You'll want to limit your heat rise to less than 20-30 C or so, so find a heatsink that's rated for 20 C/1.8 W = 12 C/W. I'm neglecting the package thermal resistance, but something like this (14 C/W as it's unlikely your RFID module draws 120 mA all the time): uk.farnell.com/aavid-thermalloy/6098bg/… - feel free to ask a new question on this topic! \$\endgroup\$
    – awjlogan
    Commented Dec 15, 2021 at 13:30
  • \$\begingroup\$ Thank you awjlogan, you helped a lot. Have a nice day. \$\endgroup\$ Commented Dec 15, 2021 at 14:37
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    \$\begingroup\$ Feels like I already answered this here: electronics.stackexchange.com/a/599335/6102. I strongly advise not to use 7809 with 24V input. Even if you are within the specs and even if you use TO220 with massive heatsink, electronics don't like to live in some permanent state of 70-something dgr C. (From what I recall, these kind of LDO go muppet at 85dgr and start to oscillate on/off when the thermal shutdown kicks in.) Also we shouldn't design as if it's the 1970s, power efficiency matters. \$\endgroup\$
    – Lundin
    Commented Dec 15, 2021 at 14:39
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    \$\begingroup\$ @Tryingtogetsome - you can use the 7809 directly, but there are more modern and efficient ways of doing it, which Lundin has pointed out. I answered your question as written, but I wouldn't do 24->9 V with a linear regulator either. For your application, 9 V @ 120 mA is 1.1 W but if you use a 7809 to do 24->9 V you're burning 1.8 W in heat as well, so 62% of the energy is just wasted. The advantage is that using a linear regulator is simpler, but even that's not really as true as it used to be with modern switching regulators. \$\endgroup\$
    – awjlogan
    Commented Dec 15, 2021 at 20:48
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For a project that I'm doing, I have a 24v power supply that I use these buck converters https://www.amazon.com/gp/product/B087RHWTJW/ref=ppx_yo_dt_b_asin_title_o09_s00?ie=UTF8&psc=1 to power a Raspberry Pi, that needs 5v3A power. I'm sure you can probably find 9v ones (or variable). One caveat: they need to have at least 10% of their rated power in use, so you'll need to find 1A ones.

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  • \$\begingroup\$ Okay, thank you. I can use it for RP2040. \$\endgroup\$ Commented Dec 16, 2021 at 4:30

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