# Driving Constant Current Through 2 LEDs from 2 Batteries

I'm working on a design where space is limited. I need to drive 2 LEDs from 2 AAAA batteries and using a switching regulator requires too many components and won't fit.

EDIT: The PCB space I'm dealing with is 0.25" x 0.25" (6.35mm x 6.35mm) with an exposed pad in the center which is about 0.1" (2.54mm).

I could use a simple circuit where I just use current limiting resistors for each LED and connect it to the batteries. This is good for the space claim but I want to try to keep a steady brightness with a constant current. There are linear constant current ICs available but I'm having trouble finding one that fits my requirements. Here are the specs I'm working with:

• LED Forward Voltage: 2.75V
• Desired LED current (for each LED): 50mA
• Power Source: 2 AAAA Batteries in series (3V nominal voltage)
• EDIT: I want power the LEDs for at least 4 hours.

NXP's PSSI2021SAY is the closest part I've found but I don't think the battery voltage will be able to turn this thing on, http://www.nxp.com/documents/data_sheet/PSSI2021SAY.pdf:

EDIT: I also found this constant current circuit. I'm not sure if I'll need the negative supply though, http://www.linear.com/solutions/1562

(source: linear.com)

• second: your precision current source based on an opamp looks about as complicated as a simple switch-mode power supply... Commented Dec 6, 2016 at 14:03
• And: a non-switch mode power supply can't ever make 2.75 V out of the battery voltage as soon as that drops below 2.75 V. I encourage you to ask another question with a simple constant current switch mode supply, while I answer this. Commented Dec 6, 2016 at 14:08
• No, as said in my previous comment, and my answer. Linear is doomed for your application, because you need the regulation exactly when the voltage starts sagging, and it's already a pretty harsh requirement for an LDO to have only 0.275V voltage drop at 50mA, normally, so it's already hard for full batteries, and it gets even harder after the first minutes (see my answer!) Commented Dec 6, 2016 at 14:11
• I think it'd be actually great for Passerby to be able to answer the question that arises from the fact that you can't use a linear supply; how to, alternatively, design a switch-mode LED driver that uses very little space; and maybe you'd want to explicitely name that space (because inductors for 50mA and ~3V aren't really space-hogs, usually, in my understanding of hog) Commented Dec 6, 2016 at 14:19
• Is this 2.75V for both LEDs in series or per LED? If it's for both LEDs in series then consider putting them in parallel instead. Commented Dec 7, 2016 at 1:35

Think about a Joule Thief style boost converter running off a single AA battery.

I gather you spec AAAA batteries because you have limited space yet believe battery voltage must exceed LED voltage. That's not true, says the Joule Thief.

Customers hate oddball batteries. We like normal battery sizes like AA, which have much more capacity while being cheaper than AAA or AAAA. AAAA is rare and nobody will buy a product that uses them.

Customers also hate devices that don't work with NiCD or NiMH rechargeable batteries (which output 1.2 volts per cell).

More than one battery also means more than four contact surfaces to get dirty and corrode. You know those horrible LED flashlights that are everywhere? They take 3 AAA's in a "cartridge" that's about the size (but not quite) of an 18650. 4 surfaces per battery, 4 for the cartridge, 2 for the bottom cap, and 2 for the switch. No wonder they never work! Into the trash it goes.

It all says a boost converter off a single battery is the way to go.

• In fact, Zetex/Diodes makes an IC specifically for this job, the ZXSC380, so there's no need to fiddle with getting a Joule Thief working -- one IC, one inductor, and the LED and you're good to go. Commented Dec 7, 2016 at 2:29
• I like the size, cost and simplicity of the ZXSC380 operating from 1.2V battery with a 100uH choke but dont like the 75% efficiency. Commented Dec 7, 2016 at 7:44
• In its favor is it can draw the battery down to a deeper state of discharge. Commented Dec 7, 2016 at 16:26
• @ThreePhaseEel, very interesting part. I will need to take a deeper look and weigh the efficiency vs. battery discharge level. Thanks! Commented Dec 7, 2016 at 20:16

If you do this with a linear regulator, you have to accept the fact that as soon as the battery voltage drops below 2.75 V, you won't be able to "create" the forward voltage for the LED(s) anymore, and thus, your current will fall, inevitably.

Hence, based on non-switch mode supplies, this approach is doomed; see the typical discharge curve for an alkaline battery below:

Notice ho fast it falls below 2.75 V /2 = 1.375 V; you'd have to add another margin for the voltage drop of the linear regulator. Best LDOs I know do about ~90 mV at 50 mA, so that'd be 1.415 V as threshold voltage.

So let's walk through the design process:

# Input voltage range

Something between 2·1.55 V = 3.10 V and what the discharge curve below tells us about what happens after drawing 50 mA for 4 h. That'd be 0.2 Ah, and we need to scale that from a 2.20 Ah battery down to the typical AAAA capacity of 0.5 Ah, so we need to look at the 0.2 Ah · 4.4 = 0.88 Ah

Discharge curve says about 1.2 V. That sounds realistic, so our overall input range, including the fact that anything has a non-100% efficiency, would be at least 2.3 V to 3.1 V.

That highlights all LDO/linear based approaches are doomed, because they can't step-up Voltage.

# Design choice

As shown, we need a switch-mode power supply. Assuming we don't first want to burn energy to always work below the target voltage of 2.75V, a switch-mode supply that is able to boost and buck is necessary.

To achieve the 40mm² space restriction, we must look into highly integrated circuitry – definitely ICs that include the switch, not only the controller, but if possible even the inductor, or use switched capacitance (because our current isn't that large).

# Component Choice

We visit the websites of the "usual suspects", being

• TI
• NXP
• Maxim
• Linear
• ST
• ONSemi

## Buck/Boost converters

Let's assume TI is doing pretty well with their simple-switcher modules that integrate the inductor. Let's see if what we can find comes close to our space restrictions – if it doesn't, we can drop that approach (we probably won't do better than TI, will we?).

Also, let's look at ON, which are known for selling high-volume, highly integrated SMPS ICs – typically in packages that are chip-scale, and they also expert in things like mobile phone flash LED controllers.

## Switched capacitance

Let's check TI's portfolio for that.

## Results

1. the smallest simple-switcher module is 99mm²: http://www.ti.com/lsds/ti/power-management/buck-boost-negative-output-module-products.page Ouch. More than twice as large as allowable,
2. On has a category for LED drivers: http://www.onsemi.com/PowerSolutions/taxonomy.do?id=16200&lctn=header . The CAT3224 seems to fit our bill very well, and has a 3 mm x 3 mm = 9mm² package – leaving plenty of room for the external components necessary for the switched capacitance design. You'd need something like (very rough guess) 25mF as supercap – but I think that should work. Needs more reading of datasheet. They'll need a lot of space, though – I'd really experiment with high-capacity MLCC and see whether that works if I don't need the "flash" mode.
3. In the category of switched capacitance, an ONsemi device seems to be the most interesting for this application: the NCP1729 is an inverting driver that would fulfill your needs and is dead simple to use:

but I'm not 100% certain from skimming the datasheet it works with 50mA with as little input voltage as 2.4 V. Further charge pumps should be considered.
• Thanks for your research, I went through a similar search. The NCP1729 is pretty interesting but it looks like the demand for the 50mA actually doesn't allow for a voltage high enough to turn on the LED. Commented Dec 7, 2016 at 20:13

Plenty of options. Simplest in mind are single chip led drivers, which often need just a diode and a small resistor-shaped inductor. The chips do come in smd packages like SOT-23, or SIP packages for low board space.

Or a current limiting diode, a one piece solution. But head voltage of your batteries may be an issue.

• yep, I agree, but these are the switch-mode supplies OP wanted to avoid. Admittedly, he wanted to avoid them for space reasons (and those reasons are false), but I tried to ask about space-efficient switch mode LED drivers in a separate question :) Commented Dec 6, 2016 at 14:15
• @MarcusMüller, the PCB space I'm dealing with is 0.25" x 0.25" with an exposed pad in the center which is about 0.1". I would need to fit all of the components in that area. Plus, of course I'm trying to keep the cost low. If you have suggestions, I'm all ears. Commented Dec 6, 2016 at 14:19
• wait, need to convert that to metric (European here) Commented Dec 6, 2016 at 14:20
• so, 6.25mm x 6.25 mm; that's not that catastrophic, though it certainly is challenging. Commented Dec 6, 2016 at 14:21
• Space claim added. Commented Dec 6, 2016 at 14:24

Battery selection depends with choices of chemistry and size

Design criteria for using LEDs with a battery, is based on;

• voltage, Vmin:Vmax, Vf and If for LED(s), capacity in (milli)amp-hour [mAh] or [VAh=Wh], cost, size, and/or efficiency.

Modern designed flashlights use tiny SMPS to efficiently regulate the current and reduce conductive losses using 18650 type cells to get long life and high intensity light. Often beginners want to avoid these for the sake of simplicity but should be aware that this also sacrifices performance.

However, let's analyze a low dropout (LDO) method of driving LED's from a battery to obtain the best compromise.

design attributes

1. minimize the dropout on the current sense to 50mV ( typical std. current shunt)
2. choose a "logic level" MOSFET with RdsOn near or less than ESR of battery
3. use a comparator or Op Amp that allows sensing near 0V input

 - Vout must >= Vgs spec using single supply from Vbat for low RdsOn

4. For 1C rates initial Voltage, Vi to final, Vf
5. Capacity C or 1C=Ah ( for h=20hour) e. the battery internal (effective series) resistance $ESR=|\frac{\Delta V}{\Delta I}|$
6. % Vbat unregulated range $\frac{\Delta V}{V_i} = \frac{V_i-V_f}{V_i}*100{\%}$

• % Vbat, Battery voltage drop from 100 to 0% state of charge (SoC)
• 8% on lead acid secondary
• 10% on Lithium 3.0V primary
• 19% on LiPo 3.7V to 3.0V secondary
• 33% on Alkaline 1.5V primary

Below is a sim. for 3 Alkaline cells. A better design uses 1 LiPo cell.

my Suggested java solution needs approval as below

• Note trace above for current shows excellent stability with 4.5 to 3.5V (not 3V) for 3x AAA batteries
• Sensitivity $\frac{I_f}{V_{bat}} = \frac{49.0-48.96 [mA]}{4.5-3.5 [V]}*100 = 4\%$

excuse: Falstad did not have a low RdsOn FET , so I used two.

Broad references in LED Driver design

It may be pointless to reinvent the wheel, but it is useful to understand how it works.