I'm working on an arduino project. The arduino have a 5V output up to 500mA and a 3.3V output up to 50mA. The problem is i have to power several circuits under 3.3V who consume between 8mA and 20mA each. 8 circuit maximum, so 160mA max. So I can use the builtin output to power two of them but will need to drop the voltage of the 5V output to 3.3V.

I've found 3 options :

  1. Series resistors, but with 160mA I will need some 2 Watts resistor and I don't have some actually.
  2. Diodes in series, seems ok for me, got a bunch of 1N4007 but as the load can change, the voltage drop will change too, and let's say I have only one device connected, I will have much more than the 3.3V...
  3. Voltage regulator, but I can't find some for 5->3.3, or really expensive (around 10€, 15$)

Any ideas ?

  • 1
    \$\begingroup\$ pololu.com/product/2097 \$\endgroup\$
    – krb686
    Dec 16 '13 at 5:24
  • \$\begingroup\$ An LDO should not be that much costlier.. \$\endgroup\$
    – user19579
    Dec 16 '13 at 5:26
  • \$\begingroup\$ You don't now any "one chip" solution I can easily find ? As it's to integrate on a design and solder on PCB after this one is not the best I think :'( \$\endgroup\$ Dec 16 '13 at 5:26
  • \$\begingroup\$ @user19579 : The cheapest I've found in belgium is the ROHM BP5275-33 at 7.5€ (12.5$) But if there's no cheapest solution, I'll go with it. \$\endgroup\$ Dec 16 '13 at 5:28
  • \$\begingroup\$ 48 cents...digikey.com/product-detail/en/AP7333-33SAG-7/… \$\endgroup\$
    – krb686
    Dec 16 '13 at 5:29

Let's go through your list of options.

  1. Voltage divider wouldn't work. Usually, the textbook voltage divider assumes that the output voltage is measures with an open circuit. In you case, there will be a load with a varying current. Output voltage of the divider will depend on the load and will not be regulated.
  2. Series diodes. You have already found the weakness of this approach: regulation is not very good.
  3. Voltage regulator is the right method for doing this. If you use a linear regulator, then the power dissipation will be P = (5V - 3.3V) * 0.16A = 0.272 W. This is manageable. TO-92 package, for example, can dissipate 0.5 W.
    A switch-mode step down regulator (buck) will dissipate even less.
  • \$\begingroup\$ Note that a simple TL431 as shunt should be just OK for your purpose (100mA difference between minimum current and maximum current) and they're very cheap. \$\endgroup\$
    – Nicolas D
    Dec 17 '13 at 15:52

Best one I've been able to find in <10 mins of searching.

2.67 euros

Can take 5V input

500mA max output

DIP package


  • \$\begingroup\$ Everybody. Just to be clear. Sourcing issues are deprecated on EE.SE . Don't make sourcing questions a habit here. If you need to sourcing help, got to chat. \$\endgroup\$ Dec 16 '13 at 6:04

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