I have 3 components on my board that have different power requirements. They are as follows:

msp430fr2433 needs 3.3V (Active mode) draws 126uA/MHz. Clock freq is 8Mhz, so the draw will be around 1mA.

LCD screen needs 5V (Minimum 4.5V) 1mA (Max of 1.5mA)

LCD backlight needs 4.1V (Max 4.3V) 60mA

I was told to use 3 separate regulators on TI forums. I feel like there has to be a better way. After searching for a while I found the following:


It's expensive though at 8$. Couldn't find anything else. It's cheaper to buy 3 separate regulators than this. Any ideas?

Which way should I go to implement this efficiently?

  • \$\begingroup\$ Surely the LCD backlight does not need a separate regulator, if it is a standard LED backlight. But there is no info about the LCD so a datasheet is needed to verify that. \$\endgroup\$
    – Justme
    Commented Oct 19, 2020 at 18:05
  • \$\begingroup\$ Here is the link to datasheet(Page 6): orientdisplay.com/wp-content/uploads/2019/08/… Its Yellow Green version and it says 60mA \$\endgroup\$
    – varun
    Commented Oct 19, 2020 at 18:12
  • \$\begingroup\$ @varun - Hi, IMHO the fact that your built circuit has a problem, is a completely different question to your original question from 6-ish weeks ago about how to avoid using 3 separate regulators - too big and too late a change to be an edit. I recommend you revert (rollback) your edit from today, upvote & accept Spehro's answer to thank him and indicate that it's valuable info for future readers, and ask a new question - link to this one in the new question for context, and that is where you add the schematic and new details. I also recommend adding a photo of your problematic PCB. Thanks. \$\endgroup\$
    – SamGibson
    Commented Nov 28, 2020 at 21:19
  • \$\begingroup\$ @varun - Hi, I have done the rollback to the original version of the question. My recommendation for the following steps still stands. Thanks. \$\endgroup\$
    – SamGibson
    Commented Nov 28, 2020 at 21:29

1 Answer 1


Start with a 5V supply. Series resistor for the LCD backlight. 3.3V regulator (eg. AMS1117-3.3) for the 3.3V rail. Total cost less than a dime.

Reminds me of the old adage about asking a barber if you need a haircut.

Make sure to follow the recommendations on capacitors for the '1117, especially the output capacitor. The series resistor is the correct way to power an LCD backlight because the voltage is not well controlled, so a constant voltage supply can cause problems.

  • \$\begingroup\$ The supply voltage could vary from 10V to 30V. So I plan on using the following 5V 100mA regulator: digikey.com/en/products/detail/texas-instruments/… Now I plan on using another regulator for 3.3V and a Zenner diode of 4V for the backlight. Unless somebody has a better idea. \$\endgroup\$
    – varun
    Commented Oct 19, 2020 at 18:15
  • \$\begingroup\$ Depends how you would use the Zener, but it sounds like a bad idea. What is wrong with a simple resistor? \$\endgroup\$
    – Justme
    Commented Oct 19, 2020 at 18:23
  • \$\begingroup\$ At 10 to 30V you could use an LM7805 but that will waste about 1.5W with 30V in, which is too much for that small package (you would need a TO-220 with a small heat sink). It would be better to use a switching regulator such as this one. I don't think you need or want a zener diode. Just a resistor between the backlight LEDs and +5V of about 15 ohms. The LEDs themselves act sort of like a zener diode. \$\endgroup\$ Commented Oct 19, 2020 at 18:35
  • \$\begingroup\$ If you use the switcher I suggest using TI's webbench suggestions as to the inductor part number if you are not very familiar with specifying inductors for SMPS designs. \$\endgroup\$ Commented Oct 19, 2020 at 18:40
  • 1
    \$\begingroup\$ Sounds like a reasonable choice. Switching regulator to +5 and then a small linear regulator +5 to 3.3 and a resistor for the LEDs. \$\endgroup\$ Commented Oct 20, 2020 at 3:28

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