Pro Micro Power - 3.3 or 5 and What's this J1 ?

So I picked up this Pro Micro device from EBAY. Seems to be a copy of the Sparkfun Pro Micro.

I am running it from the Micro USB cable and it seems to be working predictably.

As I study the device, I see there is a place to mark 3.3v or 5v on the bottom. On my device, neither is marked.

I read further and there is a J1 solder jumper on the component side of the board. On my board, J1 is open.

This should mean that the ATmega32U4 is running at 3.3v, right?

This should also mean that I should measure 3.3v at the VCC pin, right?

In any event, I am running from my laptop USB cable and I measure about 4.8v (which seems correct) and my J1 is open.

On the genuine Arduino J1 just bypasses the inbuilt regulator which is VCC. Are you sure you are not measuring UVCC which is the USB pin

Thanks for the schematic...

I don't know where one can measure the UVCC except at the pin on the USB connector where I measure 4.9V.

At VCC I measure about 4.7 V. Just about the drop in D1 as I would suspect.

At RAW I measure measure a fraction more than VCC.

Maybe they built the device with something other than a 3.3v regulator. J1 measures open with the VOM

Its a bit of a mystery.

I believe its working at 5V.

Hey, the thing costs $2 I can just wait to see if it blows up.



What is the current voltage regulator on it?

I believe I have bought clones in both 3.3v and 5v

That's right. 16 MHz needs 5 volts. 8 MHz will run reliably on 3.3 volts.

I check the blueprint, and i think, that for 5V boards, if You use 5V power source, this regulator on board have slight voltage drop, and arduino will not have exactly this same voltage, bypass jumper j1 is to go around that regulator.

But if You will use another source, like 6V to 12V beater is disconnect that jumper to get on board 5V.

On schematic is misleding peoples including me that jumper is to chose what voltage i want operate. This is true for board with 3.3V if Your source like USB is connecting to Your board. Then this jumper should be always be opened.

But if You have board working on 5V, and You want VCC pin have more juice to drive external modules, and Your iinput voltage is already regulated to 5V, like from usb, then You have an option to bypass that regulator.

Looking at the schematics from Sparkfun, the Arduino Pro Micro can have either a 5V or 3.3V regulator on it, the jumper can just bypass this regulator as previously stated. The UVCC (USB power/VBUS) goes through a fuse and diode before it meets with the RAW pin and into the regulator. The diode protects the RAW power from being fed into the USB device it's connected to and into the Arduino if the jumper is closed. Unlike the original Arduino Micro and the UNO, there is not a second regulator for 3.3V with a main 5V regulator; it's one or the other.

Personally I like that the jumper is open because your USB cable may not be connected to a perfectly stable 5V (like a homemade power supply or battery). It also allows for the manufacturer to use the same board but with a different regulator for each variant. If the voltage drop is an issue then you can always short the jumper.

Sparkfun Pro-Micro Datasheet

I also noticed that my Pro Micro clone, even being 5V (has 16MHz resonator), has the J1 jumper open. I am reading 4.65V between the 5V and GND pins and 4.70V at RAW pin. Tried powering it from two different computers. On another Micro (not Pro) board I get 5.07V on the 5V pin.

Looking for info about that jumper I got to this thread. Anyone shortened it in his Pro Micro clone? Any improvement?

Could that drop (about 0.3V) come from the forward voltage of the diode D2?

In case someone is interested or finds this thread looking for this info, eventually I closed the J1 jumper and since then I get 4.97V both at VCC and at RAW. Much better!

I too have a chinese Pro Micro clone board, but when I connect the board via USB to my computer and measure VCC, there is no output at all. Is this common for certain boards? I have a 5V 16Mhz version.

Haven't measured RAW and have not jumped J1. So will test that out too.

There seems to be a lot of confusion on how this circuit works. I'm a newbie with C and C++ but do know my analog circuits.

Say the board is a 3.3V Pro Micro. This means the jumper, SJ1 is open. The voltage regulator can work with as little as 0.5V between input and output. This means that as long as the input is above 3.3+0.5 = 3.8V, you will get 3.3V out. At high current, the voltage drop across the diode, D2, could get up to 0.8V.

The voltage from the USB (UVCC) is nominally 5V. The fuse, F1, will drop some voltage but since this board does work, that drop is not enough to drop the voltage at the regulator's input below 3.8V.

Now, if we connect a voltage source to RAW greater than 3.8V, it will directly power the regulator. However, given a UVCC of 5V and a diode drop of 0.65V (smaller drop at lower current), a RAW voltage less than 5 - 0.65 = 4.35V would cause diode D2 to be on and some of the USB current might flow back into RAW or force RAW's voltage to rise up. With a RAW voltage greater than 5V, the diode will be off and UVCC will be isolated from RAW. Without the diode, The RAW voltage would feed into the USB and could blow up the PC's interface circuit.

Next, consider a 5V Pro Micro. SJ1 is closed. This means that the voltage coming from the USB is fed through the fuse and directly to Vcc. No need to regulate it down to 5V because we come in with the correct level.

Apply a voltage source to RAW greater than 5V and the diode will be off. So the same isolation between RAW and UVCC is achieved.

Note that with the jumper shorted, the diode does nothing (good). It is likely in there to keep manufacturing simple. By only changing out the voltage regulator and processor, we go from 3.3V to 5V boards.

There is one downside to the diode in the case of a 5V Pro Micro. Say I have other circuits drawing power from RAW. When I want to program the board, I want to just plug in the USB and not bother with supplying RAW power. But with just the USB supplying 5V, I will get a current flow through the diode and OUT the RAW wire. This can drag down the USB voltage enough to shut down the processor.

In my application using a 5V Pro Micro, I have the above problem. My plan to remove the diode. One way to do this is to use a current limited power supply. Connect the positive output to Vcc and the negative output to RAW. This will pass current only through the diode. I will raise the current until the diode gets hot and desolders itself. Then I will just lift it up to remove the device from the circuit board.


check the voltage regulator, mine was LG50 5v regulator, which is pretty pointless unless you use raw in.

if it had been a 3.3v regulator it would work as advertised.

hope the helps someone Kind regards Paul

I've just buzzed through a Chinese Pro Micro as I was curious. One of the J1 contacts goes to VCC and the other to the 5V pin on the USB connector via the fuse.

So with the jumper open you can power it with 5V through the USB port or whatever through the Raw pin. A diode prevents the Raw voltage getting onto the USB connector, and either way, the processor is supplied through the regulator. On a 5v part this is a 5v low drop-out regulator which means if powered from USB the processor gets about 0.3v less than the USB port supplies.

With the jumper closed, the 5v from the USB port bypasses the regulator so the processor gets the full whack, but the regulator is still in circuit for power via RAW. So this would be the preferred configuration for a 5v 16MHz part if you're going to supply it from the USB port, though in practice you're not likely to notice the difference.

The Atmega has a AVCC pin, (pin 20 AVCC on the 328p) which you can hook up to a 3.3v regulator output, or on a 5v voltage one, the logic levelsof the output/inputs will be at that level. but it's the question will there be a 8MHz crystal on the 3.3v board, or also a 16MHz one ? reading the value of the crystal would be an indication between boards, if this is done correctly.