RTC Battery Backup Current

The problem

The following thread appeared in the forum, pointing out that the current drawn through the VB pin (RTC battery back up pin) was observed to be a lot higher than expected. Specifically, the report was 175uA, whereas the datasheet for the LPC1768 suggests it should be around 27uA.

The forum thread in question is here :



The first experiment was to make some measurements of my own to confirm what was being seen. I connected the VB pin of my mbed to 3.3v supply via 1k resistor, as the original poster had. This gives me a drop of 1mV per uA, so at the reported 175uA I should be seeing 175mV drop across the resistor - easily enough to measure.

The results came in :

  • While powered from USB, Ivb = 10uA
  • When unplugged from USB, Ivb = 320uA

This was nearly twice the figures seen by the original poster!

The Cause

Seeing the numbers above triggered a little investigation. The original mbed design was based on the LPC2368, and very little changed on the design when we moved to the LPC1768, as it was essentially a pin for pin replacement.

On the LPC2368, the RTC was powered completely separately. If the RTC was required, Vbat had to be powered, either from the main 3.3v supply, or a battery. Not wanting to run down a battery uneccessarily, a pair of ultra low Vf diodes were put on the board. Expecting a 3.0v coin cell to be used, the arrangement would allow the RTC to be powered from the main 3.3v supply while it was present, and then switch to the 3v battery when the main supply was removed.

The part of the circuit in question is shown below.


The side effect of this is that when the main power is removed, current will leak through D3 and sink into the rest of the circuit. This isnt enough to power it, but it is enough to notice a significant increase on the expected 27uA.

However, in the newer LPC1768 design this power switching is now built into the MCU itself. If power to the rest of the device is present the RTC will select the main supply to conserve the battery. When the main supply is not present, the RTC will source from the Vbat pin.

Essentially, the diode circuit above is no longer needed.

The Fix

For someone using the mbed NXP LPC1768 and wanting back up the RTC with a coin cell, the easiest fix is to simply remove D3. This will mean there is no path to leak current from the battery back to the main supply rail.

Taking a look at the UNDERSIDE of the PCB, D3 is near the at the bottom, and is the diode closest to the "Pb" symbol, highlighted in bright white.

PCB layout view The actual PCB


On future build of mbed this diode will not be fitted at the manufacturing stage, For now, it can be easily removed if you are confident with a soldering iron.

If you intend to give this a try, here is the method I used:

  1. Use a soldering iron with a wide (3mm) tip
  2. Clean the tip, then tin excessively so there is a little bulge of moulten solder
  3. While being careful not to touch any other component or via on the pcb, cover the diode with bulge of solder, so that it is touching (and therefore melting) the solder at either end of D3
  4. Feed in a little more solder if needed, to ensure the solder is melted at both ends
  5. Pull the iron away from the PCB, the surface tension of the solder should bring the diode away from the PCB with it
  6. Wipe the solder bulge onto a damp sponge, the diode should be embedded


After Sucessfully removing D3, it is time to check the result. Using the same test circuit, a 1k resistor to give 1mV per micrcoamp, we measure the current drawn through VB both with and without the USB power plugged in.

With USB power applied the current draw is down to 20uA, and this goes up to 40uA when USB power is removed. For now this is massive saving, but some further investigation with some higher accuracy measurements is needed.

With USB power Without USB power

With USB power attached - 20uA

Without USB powerattached - 40uA

function In the mean time, it is necessary to check the function of the RTC.

I wrote a very simple test program to set the RTC, then print it's value over USB to a terminal in a loop.

#include "mbed.h"

int main() {
    while (1) {
        time_t seconds = time(NULL);
        printf("It is %d seconds since January 1, 1970\n", seconds);
        wait (1.0);

With the test program compiled and running, I took the following steps:

  • Run the program with no power attached, and ensure that the RTC was actually powered and running by observing the output on Hyperterm.
  • Apply 3.v to the Vbat pin, observe that this configuration still works
  • Unplug mbed from USB and leave for 10 seconds. In this time RTC should be retained by Vbat
  • Plug USB back in, and observe RTC still running, and with a correct value.


Further Verification

The initial results measured above were a good first approximation, but some better measurements were needed. So the tests were repeated using a 10k resistor, giving 0.1uA/mV sensitivity, which led to the following results:

With USB power attached - 1.9uA Without USB power attached 4.6uA

Encouraged by this, I repeated again with a 100k resistor, giving 0.01ua/mV

With USB power attached - 1.4uA Without USB power attached 4.3uA

So it seems that upping the sensitivity of the reading has made us converge on the figures, which are 1.4uA when the LPC1768 is powered, and 4.3uA from the battery. This is still a lot higher than the 0.8uA quoted in the NXP datasheet, but in the abscence of high precision, high sensitivity  measuring equipment, the measurements are enough to satisfy us that we can expect a long backup time from a coin cell.

The Other diode

Taking another look at the circuit diagram, you'll spot that a second diode was needed inthe original circuit. With D3 removed, D4 no longer needs to be a diode, but it is doing any harm? The assessment here is that it is not necessary to have a diode there, and in the future, we may be able to fit D4 with a zero ohm link. In the mean time, the effect of the diode is that the Vbat pin of the LPC1768 is being supplied at one diode drop below VB.

The diode used was chosen for its ultra low Vf characteristic, the data sheet can be found at :

Looking at the If vs Vf graph on page 3, at 25^C which I take as room teperature, a forward current of 30uA (3 x 10^-5) we can expect a Vf of less than 100mV, I read it as about 30mV.

So while the diode isn't necessary, it appears from the datasheet that it isnt doing any harm.


The Conclusion

The higher than expected current drain through Vb is due to leakage in a circuit that was neccessary for the LPC2368 mbed, but it no longer needed withthe LPC1768. Future mbed builds will not have this diode fitted, but for those who wish to reduce thier Vb consumption and are competant with a soldering iron, it is a fairly straight forward operation.





01 Feb 2010

I looked at the drawing and the board and was puzzled for a while before I realized I should look at the bottom of the board. You might want to mention that.

02 Feb 2010

Chris, Thanks for this nice writeup!

I did the rework you propose to my 1768 board.  I checked if RTC keeps time after diode removal using my pub_iva2k_ethsntp, and it does.

Interesting... for my MBED the VB current dropped from 290mA down to 0.5uA @ 3.00V after I removed D3. I was near a bench and used Keithley 2000 (7-digit DMM) in DCI mode instead of a series resistor. Expecting a higher value, I looked up NXP datasheet, and there it says 0.8uA typ and <0.8typ without other power (see page 41 of http://www.nxp.com/documents/data_sheet/LPC1768_66_65_64.pdf), so my current is on the lower side, but not that much off (NXP does not state min or max).

I wonder what is causing this really high reading (10, 20 and 40uA) for you?

02 Feb 2010 . Edited: 02 Feb 2010
Igor Skochinsky wrote:

I looked at the drawing and the board and was puzzled for a while before I realized I should look at the bottom of the board. You might want to mention that.

Hi Igor,

Good point. I have updated the text to make it clear. I might add a picture too!


04 Feb 2010

Hi Chris,

A good investigation and an excellent outcome but I think your maths is up the creek! I suggest your current measurements are as follows:

Resistor Scaling Power On Power Off
1K 1mV per µA 0.2µA 0.4µA
10K 10mV per µA 0.19µA 0.46µA
100K 100mV per µA 0.148µA 0.437µA

I removed D3 on my mbed and the battery current dropped to 0.2µA with power on and 0.6µA with power off.

If you have some fine point tweezers there is a much easier way to remove D3:
1. Turn mbed over so it is upside down
2. Secure it in place - I stuck some map pins in my breadboard to do this
3. Apply a fine tipped soldering iron to one of D3 pads
4. Grip body of D3 with tweezers and lift it away from mbed - the lead being unsoldered will lift off the pad
5. Apply soldering iron to other pad of D3.
6. Lift D3 off mbed.

Not all mbed users will have the means or the ability to remove D3. Are you going to do the decent thing and offer a free D3 removal service to such users?



12 Oct 2010

goodwork, but you have nottaken into account that you need to raise the "battery" voltage, up untill you have 3 Volts on the mico!

if youdrop 100mV accross the resistor, then the effective voltage on the pin is 2.9 Volts, which will probably draw more current.

this is a very good why does it not work problem :)




22 Jun 2011

Is this diode removed in mbed-005.1 version? Thanks.

22 Jun 2011

Hi Alex,

Yes, this diode is no longer fitted in production. However, depending on when you bought your mbed, and how long the stock had been on the shelf where you bough it from, you might still have one with a diode.

If you have a look on the underside of your mbed at the bottom, you should be able to verify if you have a diode or not.


22 Jun 2011
Hello Alex, I can confirm that the diode is not present on the mbed-005.1 version. Regards, David
22 Jun 2012

Where do the clock for the rtc on the mbed comes from? It seems from the schematic as the 32.768hz xtal is not connected to it.

25 Aug 2014

Hello Chris,

You said : "when the main power is removed, current will leak through D3 and sink into the rest of the circuit"

However looking on that schematic, when main power it's removed, than anode of D3 it's connected to ground , then diode isn't conducting at all . On datasheet diode reverse voltage it's 10V.

Doesn't says either that would be a kind of shottky diode or a avalanche diode.

Does not make much sense to start conducting when powered from battery and main supply it's missing...



28 Apr 2019

Hello chris

I was wondering what the battery slot on my Uni provided development board was for when I came across your post. I could verify all your measurements however i had to make some modifitcationsto your code in order to be able to get it to work an i was wondering if this was due to my development board being set up wirdely or not.

I had to include an digital in pin and set a serial out port for the NXP to communicate on aka


#include "mbed.h"
Serial pc(USBTX,USBRX); // For my setup printf on its own won't work it needs to be established on a specified port/pin

DigitalIn button_A(p29); // Needed to declare a new digital input pin, this is simply a button on the provided board that when pressed will go high

int main() {
    if (button_A.read() == 1){
    pc.printf("Setting time\n");
    set_time(1256729737);  // Set RTC time to Wed, 28 Oct 2009 11:35:37       
    } // If i didn't set the time it would always return as 0 whenever i ran, but if i included it in main as with the RTC examples

     //in the documentation it would resset every time i connected the power
    while (1) {
        time_t seconds = time(NULL);
        pc.printf("It is %d seconds since January 1, 1970\n", seconds);
        wait (1.0);
    } // This is your section of code


so yeah, i was wondering if there was something i'd done weirdly or if i could set the RTC time once on the nxp and then forgot, should I run a small program straight off with any new board to set the RTC and then keep it hooked up to a 3.3v battery, or just set it again every time?

Thanks, in advance


p.s (sorry for necroing this post)

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