Author Topic: Protecting 2.7V supercapacitors in series, powered from USB 5V  (Read 1723 times)

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Offline JDPTopic starter

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Protecting 2.7V supercapacitors in series, powered from USB 5V
« on: December 19, 2019, 08:10:52 am »
Hi all,

I'm very much a beginner and have no engineering or electrical background. So please excuse any apparent incompetency (but do of course let me know  :)).

I'm currently working on a simple safe-shutdown implementation for a raspberry pi zero using two 10F 2.7V supercaps in series to provide reserve power while the pi shuts down.

Since I'm charging the caps in series from a 5V 1.5A USB power supply, I've gathered that i really should implement some level of protection. To do this I use this guide by Andreas Spies: http://www.sensorsiot.org/simple-and-cheap-way-to-protect-your-super-caps-video-139/

The relevant circuit diagram:

In the diagram two TL431X are used to shunt the caps in case of overvoltage (>2.7V), based on a reference voltage.

My main question is if it is correct that he uses two voltage dividers to generate separate reference voltages (R1-R2 and R4-R5) in stead of using just one voltage divider between 5V and ground?

And a side question. Wouldn't it be advisable to limit the current from the USB power supply with a current limiting resistor, or can one assume that the USB supply itself will limit the current?

I of course plan to add a diode to get rid of any backcurrent woes.

Oh, and in case you're curious. I plan on using a HT7044A-1 voltage detector with a pullup resistor to the pi's 3V rail connectet to its Vout go generate an active low signal for the pi to react to.

cheers
-JDP
 

Offline David Hess

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Re: Protecting 2.7V supercapacitors in series, powered from USB 5V
« Reply #1 on: December 19, 2019, 04:57:33 pm »
My main question is if it is correct that he uses two voltage dividers to generate separate reference voltages (R1-R2 and R4-R5) in stead of using just one voltage divider between 5V and ground?

I think the circuit as shown is overkill but TL431s are cheap.  I would have just used a pair of resistors to make a voltage divider which slowly balances the capacitors.  A single operational amplifier could have been used also to drive their common point at higher current with lower quiescent current.

Quote
And a side question. Wouldn't it be advisable to limit the current from the USB power supply with a current limiting resistor, or can one assume that the USB supply itself will limit the current?

Yes, as shown the circuit violates USB specifications for capacitance and current draw.  Some sort of current limiting should be used to isolate the capacitors from the USB supply.
 

Offline schmitt trigger

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Re: Protecting 2.7V supercapacitors in series, powered from USB 5V
« Reply #2 on: December 19, 2019, 05:19:26 pm »
Your balancing circuit should be fine.

You will have to calculate R5 and R6 to ensure that under worst case conditions, no more that 100 mA are shunted by U1 and U2.
 

Offline atmfjstc

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Re: Protecting 2.7V supercapacitors in series, powered from USB 5V
« Reply #3 on: December 19, 2019, 06:40:42 pm »
An interesting project. Will you be using some regulator downstream to boost the caps' voltage to 5V?
 

Offline Zero999

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Re: Protecting 2.7V supercapacitors in series, powered from USB 5V
« Reply #4 on: December 19, 2019, 06:59:07 pm »
The TLE2426 would be a better solution: pin 2 "in" to +V, pin 2 (out) to both capacitors and pin 3 (common) to 0V.
http://www.ti.com/lit/ds/symlink/tle2426.pdf
 

Offline JDPTopic starter

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Re: Protecting 2.7V supercapacitors in series, powered from USB 5V
« Reply #5 on: December 19, 2019, 10:07:08 pm »
I won't need to boost the voltage. The pi runs mostly on 3.3V, and it's regulator can handle down to 3.5V. That nets me about 20 seconds from when the voltage detector is triggered at 4.4V to when the voltage drops below 3.5. More than enough time for the pi to safely shut down.

I might however run into problems with startup when the voltage ramps to 5V over a couple of seconds while the caps charge. i certainly do not want the regulator to "flicker", in essence causing repeated startup and brownout cycles. That might harm the pi even more than the occasional unsafe shutdown. now that I know the circuit is sound enough (with current limiting) I'll be doing some breadboard testing before finalizing anything.
 

Online langwadt

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Re: Protecting 2.7V supercapacitors in series, powered from USB 5V
« Reply #6 on: December 19, 2019, 10:21:43 pm »
I won't need to boost the voltage. The pi runs mostly on 3.3V, and it's regulator can handle down to 3.5V. That nets me about 20 seconds from when the voltage detector is triggered at 4.4V to when the voltage drops below 3.5. More than enough time for the pi to safely shut down.

I might however run into problems with startup when the voltage ramps to 5V over a couple of seconds while the caps charge. i certainly do not want the regulator to "flicker", in essence causing repeated startup and brownout cycles. That might harm the pi even more than the occasional unsafe shutdown. now that I know the circuit is sound enough (with current limiting) I'll be doing some breadboard testing before finalizing anything.

it can quickly get a bit more complicated

you should keep the PI powered off until the caps are fully charged, so you don't boot before you have charge for a shutdown

once a shutdown has finished you should do a reset if the power has returned and caps recharged, else the PI won't reboot when power returns unless the power has been gone long enough for the caps to fully discharge

 

Offline viperidae

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Re: Protecting 2.7V supercapacitors in series, powered from USB 5V
« Reply #7 on: December 20, 2019, 10:18:10 pm »
You can pull down the global_en pin to stop the pi from booting when the voltage is too low.
It's one of the pins in the little J2 header
 


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