# Battery Hot Swap Circuit with Capacitors

#### guiale

Joined Dec 30, 2020
12
Hi everyone,

I'm trying to design a circuit whose battery can be hot swapped. I have a 3.6V battery connected to a boost converter, outputting 9V. I'm powering a small computer that requires a 9-20V input, and at idle consumes ~10W. I figured that, so that the computer stays on for 10s (with a disconnected battery) the capacitor needs to output 100W. Using 1/2*C*(V^2) I calculated that a 3F supercap should work. I didn't have a 3F cap, but did have 1.5F caps that had a 5.5V rated voltage. Thus I made this circuit:

However I soon as I disconnect my battery, my computer turns off. Perhaps I'm doing something wrong?

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#### ericgibbs

Joined Jan 29, 2010
17,430
hi g,
Welcome to AAC.
Look at this LTSpice discharge curve.
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#### djsfantasi

Joined Apr 11, 2010
8,903
How long did you have the battery attached in the first place? And what is the minimum voltage required for the “computer” to operate?

To perform a “reasonability” test, assume the wiring and internal battery resistance is 1Ω. Thus, it takes 60 seconds to charge the capacitors to near 9V. 60 seconds = 5 x 1Ω x 12F.

To discharge the capacities below 6V will take around 12 seconds. 6V ~= 9V x 63.8%. Time equals 1 x 1Ω x 12F.

Within these parameters, what is the cutoff voltage/current of your computer?

UPDATE: I see @ericgibbs simulated this In LTSpice. The numbers are close!

#### guiale

Joined Dec 30, 2020
12
How long did you have the battery attached in the first place? And what is the minimum voltage required for the “computer” to operate?

To perform a “reasonability” test, assume the wiring and internal battery resistance is 1Ω. Thus, it takes 60 seconds to charge the capacitors to near 9V. 60 seconds = 5 x 1Ω x 12F.

To discharge the capacities below 6V will take around 12 seconds. 6V ~= 9V x 63.8%. Time equals 1 x 1Ω x 12F.

Within these parameters, what is the cutoff voltage/current of your computer?

UPDATE: I see @ericgibbs simulated this In LTSpice. The numbers are close!
The computer in question is a Xavier NX, and it's minimum voltage is 8.8V. Perhaps that's the issue - the input voltage from the boost converter is just too close to the minimum voltage. I let the computer run for ~60 seconds, perhaps more before disconnecting the battery.

#### djsfantasi

Joined Apr 11, 2010
8,903
Using values read from ericgibbs simulation, the capacities will drop below 8.8V in about 0.6 seconds. Not too much time!

#### guiale

Joined Dec 30, 2020
12
Agreed. Time to break out a 12V boost converter.

@ericgibbs @djsfantasi Thanks for the help!! I'll update once I change my boost converter. I will also need more capacitors as mine can only handle 5.5V each.

#### guiale

Joined Dec 30, 2020
12
Hi everyone, just wished to give an update. I've used LTspice to simulat the same circuit, this time at 12V. I also found a "deep sleep" mode on the computer that makes it use only ~2W, so I can use a small capacitor (least in theory). I put 2x 1F caps in series both rated at 5.5V. 12V is over the limit of 2x 5.5V caps but not dangerously above the limit.

You can look at my screenshot and see that the computer should stay on for 8 seconds before shuting down (the computer cannot take a voltage lower than 8.8V).

I setup the circuit and let the battery charge the caps for 10 minutes (the battery was hooked up to a 12V boost converter). Unfortunatly... it did not work . I'm wondering if the caps simply cannot output enough current (at 12V, 2W I the computer would need a little over 0.15A). For reference, here is their datasheet.

I've seen some UPS online for a raspberry pi and these use 10F caps, so perhaps mine are simply too small

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#### BobaMosfet

Joined Jul 1, 2009
2,073
Hi everyone, just wished to give an update. I've used LTspice to simulat the same circuit, this time at 12V. I also found a "deep sleep" mode on the computer that makes it use only ~2W, so I can use a small capacitor (least in theory). I put 2x 1F caps in series both rated at 5.5V. 12V is over the limit of 2x 5.5V caps but not dangerously above the limit.

You can look at my screenshot and see that the computer should stay on for 8 seconds before shuting down (the computer cannot take a voltage lower than 8.8V).

I setup the circuit and let the battery charge the caps for 10 minutes (the battery was hooked up to a 12V boost converter). Unfortunatly... it did not work . I'm wondering if the caps simply cannot output enough current (at 12V, 2W I the computer would need a little over 0.15A). For reference, here is their datasheet.

I've seen some UPS online for a raspberry pi and these use 10F caps, so perhaps mine are simply too small
You're oversimplifying things without understanding the relationship between current, voltage, and resistance (Ohm's Law). Put the proper values in, based on reality, and use this equation to determine time. Your 9V cap charge won't support an average of 14.5V (per your original statement), at the estimated 700mA load.

t = - log(V/E)(RC)

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#### guiale

Joined Dec 30, 2020
12
@BobaMosfet, thanks for the advice. Using the equation you provided, I would require an 80 Ω resistor. Unfortunatly this would create a significant voltage drop and the voltage at the computer would therefore be too low.

#### DanSohan

Joined Jan 6, 2021
18
Maybe be barking up the wrong tree here, but would it not be easier to integrate an internal backup battery that replenishes its charge when the external battery is connected?

After playing with super caps myself, i found they are not quiet at the energy density yet required to replace most battery applications. Putting them in series increases the operating voltage, but reduces the total capacitance to at least half with 2 in series. (I believe)

#### guiale

Joined Dec 30, 2020
12
@DanSohan That is a potential solution, and I am looking into it. I'm thinking of using rechargable lithium button/coin cells. But this does add complexity and a circuit is required to keep it charged, and to prevent a deep discharge.

#### DanSohan

Joined Jan 6, 2021
18
Is it a Raspberry Pi you need to power? an 18650 should provide a good amount of power, but is on the large side for small applications.

Maybe an older mobile phone battery would work as they are usually very low profile. Yes there is the safety elements to consider, but a Battery management PCB for 1 cell should cover all of that. you may also need some additional buck / power converters to get the desired voltages.

I pulled some very small lithiums from some disposable e-cigarette devices. measured abound 5mm diameter and 30mm long. Also bought a tiny 50mA hour lithium cell for a micro RC car project, measured about 10mm by 6mm by 3mm thick. Might give enough juice to run while a battery swap occurs, just need to consider the discharge C ratings are within spec.

If safety is big concern, maybe the LIFEPO4 cells would be better, bit heavier, but most safe. (still needs overcharge / discharge protection), but they also come in small form factors like AAA size.

#### guiale

Joined Dec 30, 2020
12
Nowadays you can find Lithium cells of all shapes and sizes.

I'm actually powering an Nvidia Xavier NX. At no load it consumes 7-10W of power, and in a deep sleep mode it consumes 2W. It can take 9-20V as an input and 5A max. I'm acutally powering it via an 18650 from samsung, and boosting the voltage of the battery up to 9V.

I'm trying to make a circuit that would alllow me to change the 18650 without the NX shutting down.

#### Sensacell

Joined Jun 19, 2012
3,180

1) A DC-to DC converter is a constant-power load, it needs to draw more current as the input voltage drops to maintain the constant power output.

2) If you just connect super caps in series as you show, the charge voltages will not be equal, the smaller cap will attain a higher voltage.
This can lead to over-voltage very easily, you need to have a charge balance circuit to make sure this doesn't happen.

Super caps ain't so super.

#### BobaMosfet

Joined Jul 1, 2009
2,073
@BobaMosfet, thanks for the advice. Using the equation you provided, I would require an 80 Ω resistor. Unfortunatly this would create a significant voltage drop and the voltage at the computer would therefore be too low.
Um, not the answer I got based on factors you mentioned. Instead, why don't you look at this and try it another way- this should help you calculate R (resistance):

http://labman.phys.utk.edu/phys222core/modules/m3/RC circuits.html

#### guiale

Joined Dec 30, 2020
12
I think something stinks with the computer (xavier NX) that I'm using. I tested my circuit but replaced the NX with a small DC motor. In this instance the motor was able to run for ~10s and with a peak current of 0.25.

#### ericgibbs

Joined Jan 29, 2010
17,430
Hi,
What is lowest specified input voltage that the 3.3V to 9V converter will accept in order to power the computer is Sleep mode.?
E

#### guiale

Joined Dec 30, 2020
12
Hi,
What is lowest specified input voltage that the 3.3V to 9V converter will accept in order to power the computer is Sleep mode.?
E
This shouldn't matter as my supercaps are after the the boost converter. However, to answer your question, the minimum voltage required is 3V.

#### ericgibbs

Joined Jan 29, 2010
17,430
I know where your super caps are.

Have you considered tackling the problem at the 3.3V input side of the project.?

E

#### guiale

Joined Dec 30, 2020
12
I know where your super caps are.

Have you considered tackling the problem at the 3.3V input side of the project.?

E
I have, ish. When running an LTspice simulation, however, the voltage drop curve I get is rather underwhelming, hence why I havn't tried this option yet. Attached is the simulation. As I've changed the power draw of my device from 10W to 2W, I've also simplified the circuit