How reduce battery current to zero when external power connected - Page 1
Forum section: electronics / projects

author: sebmadgwick
author_rank: Regular Contributor
timestamp: September 14, 2015, 12:31:34 pm
content: I'm designing a carrier board for a GPS module (GNS 902 MT-3333) which will need a coin cell back up battery. The module has a pin for VDD and a pin for VBACKUP. I have measured the current drawn by VBACKUP as 7-17 uA when VDD is not present, and 129 uA when VDD is present. This is a problem, ideally VBACKUP current would not increase when the device is powered.I have considered using a rechargeable coin cell but I would prefer the capacity/lifetime of a CR2032. I would therefore like to create a simple circuit to 'disconnect' the battery when VDD is present, with VDD powering VBACKUP instead.I built the attached schematic. It works but the diode reverse-voltage current seems to be ~200 uA; my circuit assumes an ideal diode in this respect. I am also concerned that the reverse-voltage current of the diode in the FET could cause the non-rechargeable CR2032 to sink current when VDD is present.Your thoughts?

author: max666
author_rank: Frequent Contributor
timestamp: September 14, 2015, 06:10:27 pm
content: Wait, you're telling me that VBACKUP is drawing more current when VDD is present?You sure the current direction didn't switch, when VDD is present, so that there are 129 uA flowing out of VBACKUP and back into the battery?EDIT: Did you use a schottky diode for D1? Schottky diodes have higher reverse current, significantly higher.

author: free_electron
author_rank: Super Contributor
timestamp: September 14, 2015, 07:44:05 pm
content: use a BAS116 for this kind of application.Those things leakage is in the order of 3 pico ampere.

author: sebmadgwick
author_rank: Regular Contributor
timestamp: September 14, 2015, 08:15:56 pm
content: Quote from: max666 on September 14, 2015, 06:10:27 pmYes, peaking at ~180 uA with a mean of 129 uA.Quote from: max666 on September 14, 2015, 06:10:27 pmI did use a schottky diode.Quote from: free_electron on September 14, 2015, 07:44:05 pmAh, I see, they are even marketed as "Low-leakage diode".Thanks for the tip. I'll test the circuit again with a BAS116.The BAS116 forward voltage of 900 mV at 1 mA might be an issue. The GPS datasheet states that VBACKUP "should be kept between 2.8V~4.3V, Typical 3.3V." (page 16)... maybe a 3V CR2032 isn't the best choice. I'm not sure what kind of battery they had in mind with this voltage range.

author: onlooker
author_rank: Frequent Contributor
timestamp: September 15, 2015, 12:34:23 am
content: Were you measuring the current at the battery terminal or at the VBACKUP pin? VBACKUP may draw more current when in normal operation, but, this current should be supplied from the 3.3V source, instead of, the battery.Cheap diode such as 1N4148 has a reverse leakage about 25nA@20V; Cheap Schottky such as 1N60P is about <1uA@15V. That is, reverse leakage of diode is unlikely to be the cause of the problem you saw.

author: mos6502
author_rank: Frequent Contributor
timestamp: September 15, 2015, 01:06:25 am
content: Are you sure it's not a software problem? Some kind of register you have to set? Because it seems to me that the module should switch between Vdd and Vbackup by itself.

author: max666
author_rank: Frequent Contributor
timestamp: September 15, 2015, 04:15:46 am
content: Quote from: mos6502 on September 15, 2015, 01:06:25 amI only took a quick look at the datasheet, but they explicitly state that during normal operation both Vdd and Vbackup have to be connected. So I guess they kept two completely separate supply rails for the RTC and the rest. Otherwise why would Vbackup drain more current, when the unit is turned on with a Vdd > Vbackup.I don't know what battery they had in mind, but the only other alternative I can think of would be3.6V Lithium Thionyl Chloride 1/2 AA

author: mos6502
author_rank: Frequent Contributor
timestamp: September 15, 2015, 04:38:55 am
content: Another thing, the OP didn't mention what voltages he was using for Vdd and Vbackup. The datasheet says typical value for Vdd is 3.3V and Vbackup 3.0V. So it looks like Vdd needs to be higher than Vbackup. I would disconnect the circuit and experiment with different voltages at Vdd and Vbackup.If that fails, I would replace the diode with the B-E junction of a 2N3904. That has a leakage of less than 5pA. I would also replace the FET with a 2N3906, which has a maximum leakage of 50nA.

author: Marco
author_rank: Super Contributor
timestamp: September 15, 2015, 06:34:21 am
content: Can't you use a CMOS SPDT switch? Use Vbackup as it's supply voltage, Vdd as it's control voltage and connect Vbackup to either the battery or Vdd. Add say 10pF of capacitance to Vbackup to hold it up during transition (these switches are pretty fast).

author: sebmadgwick
author_rank: Regular Contributor
timestamp: September 16, 2015, 12:08:52 pm
content: Quote from: onlooker on September 15, 2015, 12:34:23 amI was measuring the current from the battery terminal using Dave's uCurrent GOLD.Quote from: onlooker on September 15, 2015, 12:34:23 amI don't have the circuit to hand right now. I assume the leakage was the diode because connecting/disconnecting the diode within the above circuit (with no VDD) would cause the battery current to switch between 17uA and ~200uA. I therefore assumed that the current was leaking through the diode and into (a) the VDD pin, and (b) the 1M pull-down resistor.Quote from: mos6502 on September 15, 2015, 04:38:55 amI using 3.3V for VDD and a fresh CR2032 for VBACKUP (~3V).The datasheet recommends a typical voltage of 3.3V and "should be kept between 2.8V~4.3V") for both VDD and VBACKUP (page 16). I would like to take advantage of the form-factor and availability of a CR2032 oppose to something like a 3.6V Lithium Thionyl Chloride 1/2 AA. I am considering using something like the MCP1259 charge-pump 3.3V regulator. This would be a slightly more complicated circuit that I'd like, I'm yet to calculate actual battery life given efficiency, and I'd have to make sure that VBACKUP doesn't brownout when switching between power sources.This is all a bit of headache really; why didn't they just design their RTC to work with common RTC batteries

author: onlooker
author_rank: Frequent Contributor
timestamp: September 16, 2015, 12:23:47 pm
content: Actually on page 14, the datasheet says 2.0V~4.3V for VBACKUP twice. If the 2.0V is correct, Q1 can be replaced with a Schottky. And make sure to test the diode or Schottky for leakage before using them. You may test to see if 2.0V will still retain data for warm starts.

author: sebmadgwick
author_rank: Regular Contributor
timestamp: September 16, 2015, 10:31:30 pm
content: Page 16 states that VBACKUP "should be kept between 2.8V~4.3V", page 14 contradicts this so the only safe decision is design for the highest minimum voltage. I could test if the RTC continues to operate down to 2V but I cannot base a design on this; the design must be based on the manufacturer's specification.

author: sebmadgwick
author_rank: Regular Contributor
timestamp: September 19, 2015, 03:24:03 pm
content: I tested the minimum VBACKUP voltage by leaving VDD disconnected for 30 mins with VBACKUP at 2V. The module settings were retained. The suggests that the VBACKUP minimum is 2V as stated on page 14. I have contacted the manufacture to ask if I should design for 2V (page 14) or 2.8V (page 16).I also test the above schematic using a BAS116. The VBACKUP current has reduced from 200 uA to 17 uA when VDD is not present. However:Disconnecting the diode from the schematic reduces the current further to 7 uA which suggests there is still leakage involving the diode. Note: The above mentioned 17 uA seems to slowly reduce over time, e.g. after 5 minutes it is now 10 uA; the current when the diode is disconnected does not change.VBACKUP current is still high when VDD is present if (VDD - BAS116 Vf) > (VBACKUP - FET Vf). BAS116 Vf = 900 mV. I have found that VBACKUP cannot exceed 3V with my current breadboard circuit. However, variations in Vf may result in allot of battery energy being wasted in other circuits.How would you account for the observed current leakage? Can you recommend an alternative low-leakage diode with a Vf of 300 mV or less?

author: max666
author_rank: Frequent Contributor
timestamp: September 19, 2015, 03:49:25 pm
content: Just out of interest I would measure the leakage current of the BAS116 by itself.

author: sebmadgwick
author_rank: Regular Contributor
timestamp: September 19, 2015, 04:21:21 pm
content: Quote from: max666 on September 19, 2015, 03:49:25 pmGood idea. I measured it to be ~40 uA and slowly reducing over time. This is familiar to me! ...so I washed the BAS116 breakout board to make sure any residue water soluble flux was removed and then measured the leakage again to be 54 nA.The above circuit now achieves a VBACKUP current of 7 uA when VDD is not present and -3 nA (i.e zero) when VDD is present.The last issue to resolve is that of the Vf tolerances. VBACKUP will draw a mean up to 129 uA if (VBACKUP - FET Vf) > (VDD - BAS116 Vf). TheBAS116 datasheetstates a maximum Vf or 900 mV. TheIRLML6401 FET datasheetstates a maximum Vf or 1.2 V. I do not know what the variation is between parts but I see VBACKUP current increase from zero when VBACKUP exceeds ~2.92 V.This circuit is a significant improvement compared to connecting a CR2032 directly to VBACKUP but it would be nice to achieve a VBACKUP current of zero when VDD is present for all expected CR2032 voltages.

author: sebmadgwick
author_rank: Regular Contributor
timestamp: September 19, 2015, 04:58:17 pm
content: The solution is kinda obvious: use another FET in place of the diode. See attached schematic.The FETs are identical so we can assume that Vf is the same for both (max 1.2 V). Therefore the VBACKUP current will be zero when VDD is present provided that the VBACKUP voltage is less than VDD. This will be true for a CR2032 and VDD of 3.3V.This circuit assumes that VBACKUP minimum if 2V as stated on page 14 oppose to 2.8V as stated on page 16. I am yet to receive clarification from the manufacturer.A nice bonus of this design is that it only adds two parts to the BOM, the resistor and 2 identical FETs; even better if I use a resistor value already on the BOM. I'm quite pleased with this design

author: max666
author_rank: Frequent Contributor
timestamp: September 19, 2015, 05:05:08 pm
content: So what's the leakage current of the schottky diode in Q2?

author: sebmadgwick
author_rank: Regular Contributor
timestamp: September 19, 2015, 05:36:15 pm
content: Quote from: max666 on September 19, 2015, 05:05:08 pmI measure a mean of 8 nA. For this measurement, the IRLML6401 Source and Drain are connected directly to a power supply (3V with 10 mA current limit) in series with a uCurrent, the Gate is shorted to the Source as shown in the schematic. The uCurrent output is measured using a Fluke 115.

author: max666
author_rank: Frequent Contributor
timestamp: September 19, 2015, 06:09:21 pm
content: Then I would assume a VBACKUP current of up to 7 + 8 uA = 15 uA now when VDD is not present. That could halve your battery life?

author: sebmadgwick
author_rank: Regular Contributor
timestamp: September 19, 2015, 06:36:15 pm
content: Quote from: max666 on September 19, 2015, 06:09:21 pmYou misread the units. The FET leakage is 8 nano Amps. So the current increases from 8 uA to 8.008 uA.

author: technix
author_rank: Super Contributor
timestamp: September 19, 2015, 06:50:29 pm
content: I would use a LIR2032 (3.7V Li-ion rechargeable in the form factor of CR2032) battery here with proper battery protection circuitry and a charger like TP4057 that sources current from VDD should it be present, no diode required. This will cause the battery charger circuit to supply both the current VBACKUP draws and charge the battery when VDD is on, and since the battery is charged the VBACKUP draw when VDD is not on is no longer an issue.

author: max666
author_rank: Frequent Contributor
timestamp: September 19, 2015, 06:56:18 pm
content: Quote from: sebmadgwick on September 19, 2015, 06:36:15 pmD'oh! My bad

author: sebmadgwick
author_rank: Regular Contributor
timestamp: September 20, 2015, 11:54:33 am
content: Quote from: technix on September 19, 2015, 06:50:29 pmI decided against a rechargeable battery because I expect the GPS module may be left un-powered for long periods of time. The35 mAh of a LIR2032is so much less than the240 mAh of a CR2032. A LIR2032 may last something like 50 days whereas as a CR2032 is around a year.I may change my mind... or change GPS module. It is far from ideal that the GPS module also requires the battery to maintain device settings!

author: technix
author_rank: Super Contributor
timestamp: September 21, 2015, 03:28:11 am
content: Quote from: sebmadgwick on September 20, 2015, 11:54:33 amYou can use some other shapes of Li-po or Li-ion cells like a 300mAh 502035-size Li-po or a 2600mAh 18650.

author: sebmadgwick
author_rank: Regular Contributor
timestamp: October 02, 2015, 06:08:03 pm
content: Here's a link to the repo containing the carrier board CAD files, schematic, etc.https://github.com/xioTechnologies/GNS-902-Carrier-Board/tree/master

author: onlooker
author_rank: Frequent Contributor
timestamp: October 03, 2015, 03:20:40 am
content: I guess your Q1 had drain and source connected backward. Should IRLML6401 be a negative drain device? Though, I can see by a strange way Q1 as is will still work: when no 3.3V supply, the source of Q1 is initially pulled up through the diode in it and that give the needed negative Vgs. Also, a small cap between VBACKUP and GND may ease the supply transition and help to turn on Q1.As l mentioned you might even replace Q1 with a schottky, or otherwise R1=1k is way too small. It drians the 3.3V supply unnecessarily. Make it 1M should be fine.Q2 is acting not more than the diode in it. And worse yet, the diode has 1.2V drop(max). A schottky is better. Did you measure the voltages drops over Q2, or did I miss something?Edited:

author: onlooker
author_rank: Frequent Contributor
timestamp: October 03, 2015, 04:31:47 pm
content: Ok I found this quote to explain the working of Q1,QuoteSwitching drain and source should work too. But it looks like either way may have potential problems depends on other factors. With Q1 as is, there is the need to make sure to prevent both sources to compete to supply VBACKUP (Q1 through the body diode); while with Q1 flipped, one run the risk of the 3.3V supply charging the backup battery when the backup battery is low, unless some rechargeable battery scheme can be considered.

author: NiHaoMike
author_rank: Super Contributor
timestamp: October 04, 2015, 12:14:24 am
content: Quote from: sebmadgwick on September 20, 2015, 11:54:33 amCould you consider using a small microcontroller to reinitialize the GPS on every boot?