super capicitors backup power schematic

[pleiser]

super capacitors backup power schematic

I am planning to make a simple laptop (more details here http://www.picaxeforum.co.uk/showthread.php?21068-i2c-byte-word-adress)
and was thinking I could add a super capacitor for power backup ( maybe this one http://www.sparkfun.com/products/746 ) (for when battery died, power while replacing battery) but I don't know how I would hook it up, or if it is possible at all. so my question is would it work and what would the schematic look like?
Patrick

 

[nick12ab]

There's a little problem with that capacitor - it's 2.5V. As the page suggests, you can put two in series to allow a 5V supply at the expense of only having a capacity of 5F. Anyway, if that's OK, you just then connect the assembly in parallel with the entire circuit.

 

[pleiser]

There's a little problem with that capacitor - it's 2.5V. As the page suggests, you can put two in series to allow a 5V supply at the expense of only having a capacity of 5F ...

thanks,would it be better to have 2 in series at 5 volts or 4 in series at 9 volts ?(I will be using a 9 volt battery with 7805 regulator (actually using this http://www.techsupplies.co.uk/epages/Store.sf/en_GB/?ObjectPath=/Shops/Store.TechSupplies/Products/REG007 which is similar.)ie which will drain fastest, hold most power, ect,

 

[nick12ab]

thanks,would it be better to have 2 in series at 5 volts or 4 in series at 9 volts ?(I will be using a 9 volt battery with 7805 regulator (actually using this http://www.techsupplies.co.uk/epages/Store.sf/en_GB/?ObjectPath=/Shops/Store.TechSupplies/Products/REG007 which is similar.)ie which will drain fastest, hold most power, ect,

Well using 4 in series will decimate the available capacity and using a linear regulator further decimates useful capacity of the supercapacitors. If you don't need the unregulated voltage as well as the 5V then using the supercaps to store 5V would be best but if you need the unregulated supply too then either use some supercaps in parallel which are charged by a dedicated low current/cost switching regulator and use a joule thief or a boost converter to bring the voltage back up to whatever you want.

If space is no concern then you could use some really cheap "premium brand" 'Suntan' big capacitors:



What a shame Rapid discontinued them.

 

[pleiser]

thanks nick12ab, I will use 2 capacitors at 5 volts i guess, as it is simpler, even though I might use the unregulated voltage for something (I don't know what) I don't think I will need it constantly so I will use the simple solution @ 5v.
Patrick

 

[AllyCat]

Hi Patrick,

What current, running time and (regulated) voltage are you considering? I'm rather worried by the lack of units in that data sheet, I presume the "Max current" is microamps? I was very disappointed with the high internal impedance of a (rather smaller) supercap that I tried.

Bear in mind that the energy in a capacitor is proportional to the square of the voltage, so you need to run it between as high as possible and as low as possible to get the maximum energy from it. Thus you really should use a switching regulator.

What is wrong with a rechargeable battery? There are many possible technologies and geometries, they will give you virtually a constant voltage and I think a much higher energy density. You might even use a PICaxe to "manage" its charging.

Cheers, Alan.

 

[nick12ab]

What is wrong with a rechargeable battery? There are many possible technologies and geometries, they will give you virtually a constant voltage and I think a much higher energy density. You might even use a PICaxe to "manage" it's charging.

I wonder if someone will eventually post a i2c smart battery PICAXE interface?

I tried but failed since mine required multi-master i2c. My batteries had five pins - maybe the ones with more pins use separate i2c lines for master and slave operation.

 

[pleiser]

one last thing, is this proof of concept curcuit correct (made from snap curcuits)(it does not work) I have batteries hooked up to capacitor (on + side) connected to a led (with resistor) connected to negative terminal of batteries, it works without capacitor. I also tested capacitor, it works. What is worng with this curcuit?
patrick

 

[pleiser]

Hi Patrick,

...
What is wrong with a rechargeable battery? There are many possible technologies and geometries, they will give you virtually a constant voltage and I think a much higher energy density. You might even use a PICaxe to "manage" its charging.

Cheers, Alan.

just saw your post, would rechargeable batteries be as simple to use, if so i might use them instead.

 

[g6ejd]

Out of interest:

Q = CV

I*t = C*V

Therefore t = C * V / I (secs)

At 5V, 2.5F and 10uA, t = 647 Hrs or 15 days

You'd have to put the PICAXE to sleep mode for a low current consumption in standby.

 

[nick12ab]

it works without capacitor. I also tested capacitor, it works. What is worng with this curcuit?
patrick

Your circuit is essentially an AC pass filter. You need to wire the capacitor and LED in parallel.

 

[AllyCat]

Hi Patrick,

What do you expect the circuit to do? As far as I can see the capacitor will just charge to a volt or two, but the capacitor is so small that I doubt if the LED would blink visibly. Even large capacitors really don't hold much energy. You need to do the "maths" (or get someone to help).

I certainly wasn't considering anything as complex as a smart battery. A Lithium Ion battery as used in mobile phones, etc. (and also available in AA cell form) really only needs to charged until it reaches a fairly accurate 4.2 volts. At 10uA you'd probably get at least 20,000 hours (200 mAhr) or over a year if my maths is correct.

Cheers, Alan.

 

[pleiser]

Your circuit is essentially an AC pass filter. You need to wire the capacitor and LED in series.

I may be wrong but I think they are in series as they are all connected in the same path

Components connected in series are connected along a single path, so the same current flows through all of the components. Components connected in parallel are connected so the same voltage is applied to each component.

do you mean parallel?

 

[nick12ab]

I may be wrong but I think they are in series as they are all connected in the same path do you mean parallel?

Yes I do! Your current circuit appears to be in series and your new circuit needs to be parallel. Maybe I should stop posting late at night before I post some advice that blows something up.

I certainly wasn't considering anything as complex as a smart battery.

I know - I was just bouncing the idea around as another possibility.

A Lithium Ion battery as used in mobile phones, etc. (and also available in AA cell form) really only needs to charged until it reaches a fairly accurate 4.2 volts. At 10uA you'd probably get at least 20,000 hours (200 mAhr) or over a year if my maths is correct.

It will be less once self discharge is factored in.

 

[pleiser]

ally cat, the curcuit in the photo was just a proof of concept, the parts haven't even arrived yet (except the oled screen arrive 10 minuets ago, the first part) .How complicated would it be for a simple rechargeable battery? (eg. could I be using it and just plug in the charger to my laptop and it would charge, without damaging the picaxe ect.),I dont need it to be on all the time, just so I can use it without being plugged in (I dont want to have internal, non rechargeble batteries I have to change)or at least lasting long enough to change external batteries (or swap between wall outlet and batteries)that Is why I thought of capacitors, if I can just as easily use rechargeable battries that would be better.
Patrick

 

[AllyCat]

so the same current flows through all of the components.

Hi Patrck,

But a capacitor does not pass dc current so the LED cannot light.

What you need to do is first connect the capacitor to the battery (via a current limiting resistor) to charge it up and THEN apply the capacitor to the LED (again via a resistor). But the energy stored in that capacitor is so small that I doubt if you will see even a flash.

You MUST tell us what current you need and for how long.

Cheers, Alan.

 

[AllyCat]

It will be less once self discharge is factored in.

Hi,

Yes indeed, but my hasty maths actually allows over 50% internal leakage over a year (8760 hours according to my calculator) and I suspect the need is for much shorter times (after all the capacitor only gave 15 days). Similarly, the capacitor calculation assumes that you manage to drag all the energy out of the capacitor, right down to zero volts.

Patrick, A lithium battery (or any rechargeable technology for that matter) needs to be handled with some caution (because the energy density is so much higher than a capacitor), but IMHO a battery circuit is far easier than trying to use a capacitor. If you MUST have 5 volts then obviously there are some design decisions to make on what battery voltage to use, but still far easier than the wide voltage range from a capacitor.

Cheers, Alan.

 

[pleiser]

Hi Patrck,

But a capacitor does not pass dc current so the LED cannot light.

What you need to do is first connect the capacitor to the battery (via a current limiting resistor) to charge it up and THEN apply the capacitor to the LED (again via a resistor). But the energy stored in that capacitor is so small that I doubt if you will see even a flash.

You MUST tell us what current you need and for how long.

Cheers, Alan.

is there any way to do that automatcally.
I do not know how much current I will need,I am planning to make a laptop (more details in post 1).
with my test curcuit (attached) it seems to work, when I turn it on the led lights and when I turn it off the led dims (rapidly with this 470 uf capacitor (super capacitors have not arrived yet))

 

[pleiser]

alan
what battery/curcuit would you reccomend for it. I do not need the best battery life ect. just something to transition between different power sources (but long battery life is a bonus).
Patrick

 

[AllyCat]

when I turn it on the led lights and when I turn it off the led dims (rapidly with this 470 uf capacitor )

Hi Partick,

Well certainly the LED will light, but that circuit isn't even close to using the capacitor effectively. A LED is basically a forward-biassed diode so it has a moderately constant voltage across it (perhaps 1.5 volts - it depends on the colour, etc.). So it is preventing the capacitor charging to much above (say) 1.5 volts. When power is removed the capacitor voltage will fall to perhaps 1.2 volts by which time the LED will virtually stop conducting. My quick calculation suggests that would use less than 5% of the energy which could be stored in the capacitor at 5 volts.

You need to use separate resistors in series with the LED and the capacitor so that the capacitor can charge to the full battery voltage. It would be a more realistic configuration if both the LED and capacitor negatives go to the battery negative, with the two resistors to battery positive (via the switch).
_____

The simplest configuration is just to "float" a battery across the "computer" supply rails. That implies using a (nominal) "3.7 volt" rail (if Li Ion), but that may well be sufficient for a PICaxe (and OLED). Certainly worth considering until you can determine what (current and voltage) the hardware actually needs.

Cheers, Alan.

 

[pleiser]

...The simplest configuration is just to "float" a battery across the "computer" supply rails. That implies using a (nominal) "3.7 volt" rail (if Li Ion), but that may well be sufficient for a PICaxe (and OLED). Certainly worth considering until you can determine what (current and voltage) the hardware actually needs.

Cheers, Alan.

why 3.7 volts? so would all I need to do be attach a rechargeable battery across the power rails and it would charge when powered and power other devices (the rest of the computer) when unpowered?
Patrick

 

[AllyCat]

Hi,

Lithium Ion batteries have a nominal voltage of 3.7 volts (just as NiMH are 1.2 volts) but the correct procedure is to charge them until they reach exactly 4.25 volts (also limiting the maximum current) and no more. A PICaxe could potentially do that (with a little manual calibration) or small ready-made chargers can be purchased quite cheaply (unfortunately mainly from Hong Kong / China via ebay). An AA packaged rechargeable Li Ion battery (often called a 14500) should have a capacity of at least 400 mAh.

So the answer is yes, if designed and constructed correctly. The current used by the computer will probably be quite small, so the power supply can be simply designed to deliver no more current (and voltage) than the battery is rated for.

Did you read any of the "comments" on that Sparkfun webpage? Several make similar suggestions to yours, but here are what appear to be some reasonably authorative comments:

“The stored energy of a supercapacitor decreases from 100 to 50 percent in 30 to 40 days. A nickel-based battery self-discharges 10 to 15 percent per month. Li-ion discharges only five percent per month.”

"A typical 2.8 amp hour 1.5 Volt AA battery is the equivalent of 10,080 Farads. I don’t think it [the supercap] would power a circuit for very many hours."

"These don’t have the energy density of a battery, so you may be disappointed with the lifetime. There are capacitor discharge calculators out there; I put in a 10F cap, with 7.2V starting and 6.5V ending voltages (the 5V regulator needs 1.5V overhead to operate), and a 10mA load (10000000nA). The lifetime is only 11.4 minutes. "

Cheers, Alan.

 

[pleiser]

... so would all I need to do be attach a rechargeable battery across the power rails and it would charge when powered and power other devices (the rest of the computer) when unpowered?
Patrick

Hi,

Lithium Ion batteries have a nominal voltage of 3.7 volts (just as NiMH are 1.2 volts) but the correct procedure is to charge them until they reach exactly 4.25 volts (also limiting the maximum current) and no more. A PICaxe could potentially do that (with a little manual calibration) or small ready-made chargers can be purchased quite cheaply (unfortunately mainly from Hong Kong / China via ebay). An AA packaged rechargeable Li Ion battery (often called a 14500) should have a capacity of at least 400 mAh.

So the answer is yes, if designed and constructed correctly. The current used by the computer will probably be quite small, so the power supply can be simply designed to deliver no more current (and voltage) than the battery is rated for.

...
"These don’t have the energy density of a battery, so you may be disappointed with the lifetime. There are capacitor discharge calculators out there; I put in a 10F cap, with 7.2V starting and 6.5V ending voltages (the 5V regulator needs 1.5V overhead to operate), and a 10mA load (10000000nA). The lifetime is only 11.4 minutes. "

Cheers, Alan.

could you put a capacitor across the power rails too (11.4 minutes of power is ok for what I need) or are there any rechargeable batteries that need no matenence and can charge at 5 volts and output 5 for same difficulty of use and assembly as a capacitor?

 

[Haku]

Most lithium (ion/polymer) cells charge up to 4.2v, any higher and it can damage them/be dangerous, there are some that require the higher 4.25 charging voltages, it's always best to check specs.

For safely charging single lithium cells from a 5v power supply you can use a MAX1551/MAX1555 IC, but they can be expensive, or the Microchip MCP73831-2 /OT (cheaper from Farnell) which allows you to set the charging rate by resistor size so you don't put too much amps into small cells.

If you use a low drop out regulator to power the Picaxe at exactly 3v you can use a neat trick to get 5v from it to power an LCD screen (and perhaps an OLED screen): see tip #10 http://ww1.microchip.com/downloads/en/devicedoc/40040b.pdf
I've tried it and successfully gotten 5v from a 3v powered Picaxe outputting a 10khz 50% duty cycle PWM signal. The PIC powered DIY scientific wristwatch uses that technique to power its 5v LCD screen from a 3v watch battery: http://www.calcwatch.com/

 

[papaof2]

There is no SIMPLE answer.

A capaxitor across the power rails will begin to discharge immediately when the mains power is removed (thus the voltage begins to drop immediately). A large capacitor bank requires some "smart" electronics to handle its charging on equipment power-up.

Any given battery system has a charge voltage and a discharge voltage. If you can place the battery system ahead of any voltage regulator devices, then the actual voltage isn't as critical - as long as the battery voltage remains above the minimum voltage needed by the regulator.

When dealing with any recharagable battery, there is a limited number of charge/discharge cycles - typically somewhere between 100 and 1000 cycles. Completely discharging certain battery types will destroy them in a few charge/discharge cycles. Any battery you use needs some type of protection circuitry to prevent over-discharge.

How long should the life of your device be (number of charge/discharge cycles)?

 

[srnet]

I would also agree with Papaof2, there is no simple answer here.

Unless you really do understand what you are doing and have a fiar bit of experience in Electronics, I would not suggest you try to use Lithium batteries as a backup supply. These batteries have many advantages over over NiMh or even lead acid batteries.

But Lithiums also have one very major dissadvantage, get the charging wrong and they will explode in a large fireball.

 

[srnet]

For safely charging single lithium cells from a 5v power supply you can use a MAX1551/MAX1555 IC, but they can be expensive, or the Microchip MCP73831-2 /OT (cheaper from Farnell) which allows you to set the charging rate by resistor size so you don't put too much amps into small cells.

Very interesting, never realised those devices existed.

I have always avoided using a built in charger for a small Lipo, due the obvious safety problem if it goes wrong. But those little devices solve that problem for single Lipo applications.

 

[MikeGyver]

Supercapacitors are a bad choice for power backup because they have very low energy densities.

 

[AllyCat]

11.4 minutes of power is ok for what I need

Hi Patrick,

Bear in mind that calculaton was based on a 7.5v 10F capacitor, yours are rated 2.5v so you would need NINE (3 parallel strings of 3 in series) to get that rating.

As others have said, power supply design is NOT simple, batteries are complex devices (but IMHO trying to use a capacitor as a power supply is even more so). And it's pointless to attempt to design one until you know what Voltage, Current and running time you need (as I said several pages ago).

For now, I strongly suggest that you try to get the "computer" working from a 3.6 - 3.7 volt rail (the PICaxe itself is very tolerant). A real advantage of this voltage is that it can come directly from a Li Ion cell (as used by virtually ALL portable devices like phones, MP3 players, cameras, etc..) or simply 3 x AAA or AA NiMH cells (1.2 volts each). If you must have "5.0 volts" then 4 x AA(A) NiMH is an obvious answer.

Cheers, Alan.

PS: Yes, if you get a design really wrong then a Li Ion battery might catch fire, but I found the "explosions" on Youtube rather disappointing. An electrolytic capacitor going "bang" can be far more impressive.

 

[srnet]

but I found the "explosions" on Youtube rather disappointing

Charging went wrong on one of my Lipos, it swelled so I slung it out of the workshop. About 2 mins later, the explosion on the grass outside was fairly impressive. If I had not noticed it swelling, and thrown it out of the workshop in time, the fireball was big enough that I would likley have had no hair or sight left.

The other one that went wrong would have burnt down my workshop, had it not been in a fireproof bag for charging. As it was there was a fair bit of soot to clear up, and the shelf the battery was on was charred, even throght the firepoof bag.

Sorry, no pictures.