# How do I use an adjustable LDO regulator to power devices with USB that usually use AA/AAA batteries (or should I)?

I want to use USB to power devices designed to work with one, two, or three AA/AAA batteries. Three seems to be the trickiest, because it involves dropping the voltage only ~0.5 V from 5±0.25 V to 4.5 V.

After reading around (my electronics knowledge is extremely rusty), it seems an adjustable low dropout (LDO) regulator is the way to go (but please correct me if I'm wrong). The best option locally available seems to be an NCV59302, and because I'm expecting to draw 1A max @ 4.5 V, its ~175mV dropout voltage @ 1A should mean I can get 4.5 V as long as the USB power is > 4.675 V.

However I'm not exactly sure how to get it working correctly. The data sheet is here, and my interpretation of the instructions is illustrated in the circuit diagram below. The thing I'm most unclear about is the resistors, though I'm aware I could have made other major mistakes. Someone has already asked here about the resistors for an NCP59302, a very similar component (I'm not sure what the difference between the NCV & NCP is), but after reading the answers I'm still not sure which resistors will work. Nor am I 100% sure about any of the components or wiring.

• R1 is a 500 Ω potentiometer, unspecified wattage rating (does it matter? I assume this only has to carry the 100-350 nA "Adjust Pin Bias Current")
• R2 is a 180 Ω 2 Watt wire wound resistor

The formula given in the data sheet to determine the output voltage is:

$$V_{OUT} = 1.24 V × (1 + R_1/R_2) + I_{ADJ} × R_1$$

IADJ is typically 100 nA and can be up to 350 nA, but this makes a negligible difference to VOUT. From what I can tell, with R2 as a 180 Ω resistor, this would output 1.5 V, 3 V, and 4.5 V when the R1 500 Ω potentiometer was set to ~37.7 Ω, 255.5 Ω, and 473.2 Ω respectively. Or am I missing something?

As for the other components:

• CIN is a 1 μF 50V electrolytic capacitor
• COUT is a 47 μF ±20% 6.3V X5R dielectric SMD ceramic multilayer capacitor (it would cost me the same to get a 100 μF model with the same specs if there is some advantage)

I guess the most important question is: if I wired all this up as in the circuit diagram above, should it work? Or have I made a mistake?

• Have you considered using a simple series diode to give a small voltage drop? Most diodes have a voltage drop of 0.7V, which would give about 4.3V to the device. Given that NiMH cells have a nominal voltage of 1.2V/cell (or 3.6V for 3), you'd be well within the operating voltage of the device. Obviously that wouldn't be suitable for the lower voltage requirements. Does the same circuit have to provide all 3 voltages?
– LeoR
Jun 3 '14 at 12:00
• Thanks for your comment. I want to be able to supply all three voltages from a USB plug, and while they don't have to be one circuit, I'd definitely prefer it. That and I want to keep the cost down - the price of all the essentials for building another 1 or 2 separate circuits is unfortunately more than the cost of the specialised components of this circuit. On a side note, I find NiMH batteries don't power some of my devices (though 4.3V is not 3.6V). I guess I could put a switch in and have the different voltages handled separately.
– Tim
Jun 3 '14 at 12:45
• Keep in mind that very simple devices depend on the battery's internal resistance, and have no overcurrent protection themselves. Jun 3 '14 at 14:08
• That's something I didn't think of. Is there an easy way around that, or should I just avoid very simple devices? Apparently the short circuit current of a AA battery is just over 1 A, and according to an Energizer graph, their fresh AA alkaline batteries have an internal resistance of around 0.1-0.2 Ω at room temperature. But I'm not sure where to go from there.
– Tim
Jun 4 '14 at 9:24

The circuit as shown will work. Remember though that it's a linear regulator, which means that the voltage is dropped by turning excess energy into heat. At 4.5V the drop is small, but if you're dropping to 1.5V (3.5V drop) at 500mA (you shouldn't expect to draw more from USB) then you'll have to deal with 1.75W of heat. The amount of heat that the IC you linked can dissipate depends on the PCB design (did you mean to link a surface mount component?), but in any case 1.75W would be the upper end of what you could expect a TO-220 package component to dissipate. I'd probably use a heat sink or ensure that my load wasn't drawing as much current.

Resistor selection basically doesn't matter. The two aspects that you need to normally consider when selecting resistors are the power rating (1/4W, 1/2W etc) and the tolerance (1%, 5%, 10%). The power rating isn't important in this case (see below) and because you've got a manually adjustable potentiometer the tolerance really isn't important either. Almost any resistor of approximately the right value would do for $R_2$.

Both resistors can be low wattage ones. As you've said, the $I_{ADJ}$ current is negligible and can be completely ignored. There will also be a current flow through the two resistors from the output voltage to ground, which you can calculate with Ohm's law ($V=IR$). In all cases this will be about 7mA.

The capacitors are there for filtering/ripple reduction purposes and their characteristics aren't too important. If they're approximately the same as those suggested in the datasheet (and the voltage rating is above what they'll see) then there shouldn't be any issues.

Your calculations seem to correct. This is just a basic voltage divider calculation. The regulator adjusts the output voltage until it sees 1.24V on ADJ. You can confirm that your calculations are correct with the equation given on wikipedia:

$$V_{div} = V_{in} \times \frac{R_2}{R_1+R_2}$$

Where $V_{div}$ is 1.24V, $V_{in}$ is your target voltage and you want to solve for $R_1$.

$$R1 = \frac{V_{in} \times R_2 - V_{div} \times R_2}{V_{div}}$$

for the 4.5V case therefore:

$$R1 = \frac{4.5 \times 180 - 1.24 \times 180}{1.24} = 473 \Omega.$$

There's inherent variability in electronics which mean that whatever values you calculate won't be exactly right when you wire it up. Wire it up, connect the output to a volt meter and adjust the potentiometer until you have the right voltage.

• Thanks a lot, that was a fantastic answer. I wasn't worried enough about heat. All of the USB sockets I'll be using output 1-2.1A, so there's a big potential it'll get too hot. Maybe I could just put a three way switch that has a series diode coming from one contact, and two separately calibrated Step-Down Voltage Regulator Modules (that have a 1.5 V minimum voltage drop) coming from the other two. It would be easier, more efficient (?), probably cheaper, and from the sounds of it safer and more reliable.
– Tim
Jun 3 '14 at 13:16
• It's not what the port supplies, it's what the load draws. Jun 3 '14 at 14:06
• If I was to do it with a diode and the regulator modules, what kind of diode should I use? Do I just need a standard silicon diode with a rated current & voltage at least as much as the 5V 2.1A that could be supplied? For example would this IN5408 1000V 3A Power Rectifier Diode do the trick? And is there anything else I should know before trying what I mentioned in the above comment?
– Tim
Jun 3 '14 at 14:20
• @IgnacioVazquez-Abrams I did know that, but didn't really think about the load that a device running on batteries would draw. I won't be using it with anything that would last less than 6 hours with standard alkaline batteries, so I guess nothing will be drawing more than ~433mA if an alkaline AA has a max capacity of 2800mAh. But I want to be sure it won't burn out if I did use it with something that drew up to 1A. Unless I'm just way off and nothing draws that much from a AA/AAA.
– Tim
Jun 3 '14 at 14:45
• Keep in mind that although USB is nominally 500mA, not all hosts will allow devices to draw that without some sort of negotiation. Simple USB chargers probably will, but a laptop might limit you to 100mA. Wikipedia gives a bit of information
– LeoR
Jun 3 '14 at 15:53

I think you got the math right, but with only 350 nanoamps max required for Iadj, your resistors are wasting a lot of the batteries' capacity.

The easy way to do it is to set Vadj to 1.24 volts with your desired Vout at the top of the string and at least Iadj and the current through R2 both flowing through R1.

For example, Let's say we have 100 microamps we can spare for the string and that Iadj is 100nA, per the data sheet.

Then with the remaining 99.9 microamps flowing through R1 and R2, with Vout equal to 4.5 volts, and with Vadj equal to 1.24 volts referenced to ground,

           Vout - Vadj     3.26V
R1 = -------------- = ------- ~ 32.6k ohms
It          100µA


and,

           Vadj
R2 = -------- ~ 12.4k ohms
99.9µA


If, though, Iadj happened to be 350nA and R1 and R2 remained the same values, Vadj would stay at 1.24V and IR2 would stay at 99.9 µA, but the current through R1 would increase to 100.25µA, forcing the drop across R1 to increase to 3.268 volts, and Vout to 4.51 volts, so those values of resistance would ease the load on the batteries.

Your circuit is just like the one in figure 21 of the data sheet, so it should work fine.

Ceramic capacitors?

http://www.ti.com/lit/an/snva167a/snva167a.pdf

• Thanks. It seemed strange to choose two resistors with no respect for their resistance values other than their ratio to each other. This explained a lot. CIN isn't ceramic, but the data sheet specifies a ceramic capacitor for COUT.
– Tim
Jun 4 '14 at 10:13