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I am trying to use this 5V RAVPower battery pack to drive some electronics and a high-current load: http://www.ravpower.com/ravpower-dynamo-on-the-go-rp-pb07-super-high-capacity-power-bank-10000mah-capacity-1a-and-2a-dual-usb-output-for-ipod-iphone-smartphones-digital-cameras-tablets-and-e-books.html

I find that when my logic turns on the large load (~800mA), the 5V voltage drops to around 3V and my LCD screen can no longer function.

It seems that I need to regulate the 5V output of this battery, correct?

I suspect that what I need is a buck-boost converter that will boost the voltage to something higher than 5V, and then step that down back to regulated 5V.

Am I on the right track? If so, could someone recommend a through-hole part? I am having trouble locating one...

EDIT:

Some of the comments below suggest that the battery pack I am using will not be able to provide the needed 800mA at 5V. This is ok - I am willing to let the current drop, as long as I can maintain the voltage at 5V. Could I use something like this for the job? http://www.digikey.ca/product-detail/en/MIC2296YD5%20TR/576-2659-1-ND/1821819

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  • \$\begingroup\$ Specs indicate 5V @ 3A . Can you test batterypack on its own for V-I characteristic? Are you sure your 800mA load doesn't have a large capacitive surge load? You might consider a metal oxide ICL \$\endgroup\$
    – user38637
    Commented Apr 21, 2014 at 4:33
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    \$\begingroup\$ If you can spare one of the battery packs, just open it up: It contains 4 of the 18650 LiPo cells, with an actual capacity of under 8000 mAh (the cheaper ones are under 4000 mAh despite advertised capacity) and a continuous current capacity of between 0.9 and 1.4 Amperes. The "5V 3A" rating is not going to happen with that battery pack, nor with any other cheap battery bank in that size factor. At best that rating will apply for the first few seconds after applying a load, merely to overcome any initial capacitive current demand surge. \$\endgroup\$
    – Anindo Ghosh
    Commented Apr 21, 2014 at 6:54
  • \$\begingroup\$ @Joe: I just tested the battery with a simple resistive load. The battery has 5.14V when open, which drops to 4.16V with a 730mA load. I don't know if my actual load is a large capacitive surge load - it is a AA NiMH battery that I am trying to charge and I am not exactly sure what it's properties are... What does "metal oxide ICL"? \$\endgroup\$
    – Val Blant
    Commented Apr 21, 2014 at 7:00
  • \$\begingroup\$ @AnindoGhosh: I understand that I probably can't have both the voltage and the current I want from this battery, and that is ok. Is there a way for me to prevent the voltage drop and allow the current to drop as needed instead? For example, what would happen if I tried to regulate the voltage from the battery with this IC: digikey.ca/product-detail/en/MIC2296YD5%20TR/576-2659-1-ND/… \$\endgroup\$
    – Val Blant
    Commented Apr 21, 2014 at 7:04
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    \$\begingroup\$ @ValBlant As your power bank will almost definitely drop to below 5 Volts on practically any load, and from your 730 mA experiment it will probably not sustain 1 ampere under load, the simplest solution I can think of would be to use a DC-DC boost regulator rated for 5 Volts 0.5 Amperes, with brown-out shutdown or overload shutdown, and run with that. The best you can hope for with any reliability would be 0.5 Amperes or so due to the limitations of your power bank. \$\endgroup\$
    – Anindo Ghosh
    Commented Apr 21, 2014 at 7:48

2 Answers 2

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Since you can't do anything about your power source, then you should try your idea of a boost regulator.

You don't need to boost the voltage above 5v and then regulate it back down as you suggested; instead you can just boost it up to 5v using a boost regulator like the ADP1613 which will take any input voltage from 1.8 to 5.5v and regulate it to an adjustable voltage up to 20v, but in your case you will set this at 5v.

You can get the part at Digi-Key for $2; plus you will several external components. Unfortunately all of the boost regulators with the current requirements you require (~1A 5v output with 3v input) are only available in surface mount packages; this one is 8-MSOP which can be hand-soldered fairly easily.

Your battery pack has plenty of capacity (10.4 Ah) so it should be able to power your circuit for several hours.

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  • \$\begingroup\$ Thank you! This is what I was looking for. With a boost regulator like this, is it ok for Vin to exceed Vout? My battery outputs 5.1V before the large load is turned on. I can't find any data in the datasheet on how the regulator will handle this condition. \$\endgroup\$
    – Val Blant
    Commented Apr 21, 2014 at 7:16
  • \$\begingroup\$ Would I need to configure the regulator as a "SEPIC Converter", as shown on page 22? \$\endgroup\$
    – Val Blant
    Commented Apr 21, 2014 at 7:41
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    \$\begingroup\$ That battery pack doesn't actually have plenty of capacity, as I've found from opening up several such power banks. The best ones in that size factor do not go much beyond 7600 mAh or beyond 1.4 Amperes current. The cheaper ones (< $30) are around 3000 mAh and 0.5 Amperes. \$\endgroup\$
    – Anindo Ghosh
    Commented Apr 21, 2014 at 7:52
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    \$\begingroup\$ @ValBlant My experience with other boost converters is that if Vin exceeds the Vout setpoint, then Vout will equal Vin (i.e. 5.1v in your case). I'm not sure about this one because of the diode D1. You could try modeling it using their simulator mentioned on page 13. It looks like the SEPIC configuration would also handle this situation and keep your output right at 5.0v, but it's not clear if you will still get the current you need with that circuit. Again, I suggest using their simulator. \$\endgroup\$
    – tcrosley
    Commented Apr 21, 2014 at 13:16
  • \$\begingroup\$ Hmm, this is strange. The datasheet for ADP1613 says that it can handle 2A of current, yet their design tool I downloaded refuses to do the calculation for me, b/c it thinks that ADP1613 can't handle more than 0.3A. Hope its a bug in the design tool. I have calculated the values for ADP1614 instead, which I am assuming will work fine for ADP1613. \$\endgroup\$
    – Val Blant
    Commented Apr 21, 2014 at 23:35
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The battery pack already regulates its internal batteries (most likely 3.7V Li-ion cells) to 5V. The reason the output voltage droops is because the impedance of the regulator in the battery pack and the impedance of the circuit together form a voltage divider, reducing the regulated 5V. What you need is not a second regulator, but a better first regulator, along with enough battery to drive it at the current you need.

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  • \$\begingroup\$ I am not sure what you are suggesting... I don't have control over the regulator inside the battery pack, and the battery is supposed to be able to provide 2A at 5V. Obviously it doesn't, but surely there must be a solution which would allow me to keep my voltage at 5V at the expense of current if need be? \$\endgroup\$
    – Val Blant
    Commented Apr 21, 2014 at 2:59
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    \$\begingroup\$ Dump the battery pack. Get a Li-ion battery and a 5V ESC instead. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Apr 21, 2014 at 3:02
  • \$\begingroup\$ The project design constraints do not allow me to select a different battery pack. Is there a reason why my intended solution will not work? \$\endgroup\$
    – Val Blant
    Commented Apr 21, 2014 at 3:10
  • \$\begingroup\$ The regulator in the battery pack is not up to the task, and a second regulator will not make it work. Either increase the impedance of the circuit or find another power solution. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Apr 21, 2014 at 3:13
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    \$\begingroup\$ If the thing is supposed to supply 5 volts at 2 amps, and does not do so when fully charged, I suspect that the internal regulator, or the battery, is faulty. It would be normal for the output voltage to fall when the battery is nearly discharged. \$\endgroup\$
    – Peter Bennett
    Commented Apr 21, 2014 at 3:14

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