Supercap application question

Don M · 2013-03-13 08:27

I have a project that I want to use a supercap on the 3.3VDC line with a Prop and SD card so when the Prop detects a power fault it allows enough power / time for the Prop to close the SD card file. I have proven the circuit just using a large value electrolytic so the logic and method behind all that works fine.

My question is do I want the supercap connected directly to the 3.3 VDC line just as a normal capacitor

or do I want to isolate it via a low Vf schottky diode and then charge it via a resistor?

Back feeding the 3.3VDC regulator is not a problem so that isn't what I'm looking to isolate.

Just curious as I've never messed with these type caps before.

Thanks.

Hal Albach · 2013-03-13 09:34

Possibly a better solution would be to isolate the SD card, prop, and supercap 3.3 volt bus so that the supercap will only provide power to the prop and SD card during a power fault. Supercaps have very high capacitance, some up in the tens of Farads, but they also have a substantial internal resistance which would cause a significant voltage drop if current is drawn off too quickly.

Hal

LoopyByteloose · 2013-03-13 10:02

There can be significant issues of in-rush current when first turning on the device that includes a super cap. It may require a resistor to restrict in-rush current on the input side. Too much load and the voltage regulator will just shut down. You may have it attempting to provide power in fits and starts.

Also, many voltage regulators require a protective diode when the capcitance on the output side is so large. Otherwise, a disconnect from supply and possible short circuit of the inputs will cause a huge backflow through the regulator. At least, the LM7805 will not survive such damage. Others might be better protected.

Don M · 2013-03-13 10:22

This device only has a prop, eeprom, sd card so nothing else being powered by the 3.3 v.. I've ordered in a few .33MF and 1F 3.3v caps. I'll play and see...

RDL2004 · 2013-03-13 12:44

Seriously, unless the manufacturer of the voltage regulator clearly states that no bypass diode is needed you had better add one or you're almost guaranteed to blow the regulator with that much capacitance on the output.

Bobb Fwed · 2013-03-13 14:24

Or, with a 5V input, you could put the supercap on the 5V rail and power the Propeller through the regulator (5.5V supercaps are really common).

Two things you should keep in mind with supercaps:
The first (which was already mentioned) is that the series resistance of these super capacitors are very high. The ones I worked with were 30-ohm. No matter how much capacitance it has, a 30-ohm cap, charged to 3.3 volts will flatly not work with a draw of 20mA or more (due to the Propeller's 2.7V dropout). With a 1F cap, if you drop the draw by 2mA it will function for over 3 seconds. It is all about keeping the current draw low and consistent, so the power getting to the Propeller stays above the 2.7V rating.
If you do the exact same scenario, but charge the cap on the 5V rail, and regulate it down with a LDO regulator (one that looses 0.3V or less), you could operate all the way up to 50mA and the Propeller will still function for 10 or so seconds (somewhere around 66mA is the hard cut off for this scenario).
Adding a diode anywhere in the circuit and things just get worse for all scenarios.

The second is that for long term projects, if the capacitor is continually charged, there is a significant drop off in usable capacitance after a couple years of use. We ran into this potential problem with a project I was working on, and decided to go to a 9V battery for backup power.

Hal Albach · 2013-03-13 15:51

Just hypothetical meandering, but, if instead of a single 1 F supercap, suppose you wire up three or four 0.33 F supercaps in parallel. Wouldn't that effectively cut the total series resistance while still achieving 1 or more Farads and also be able to increase the current draw? Granted the harder you draw on the charge the less time it is available.

Hal

Mike Green · 2013-03-13 16:23

Sure you could wire several smaller supercaps in parallel to get more total capacitance with lower series resistance. It also takes more room. "There ain't no such thing as a free lunch" (TANSTAAFL)

Don M · 2013-03-13 18:20

RDL2004 wrote: »

Seriously, unless the manufacturer of the voltage regulator clearly states that no bypass diode is needed you had better add one or you're almost guaranteed to blow the regulator with that much capacitance on the output.

Do you mean place a diode from the output (anode) of the regulator to the input (cathode) of the regulator?

Don M · 2013-03-13 18:28

Bobb Fwed wrote: »

Or, with a 5V input, you could put the supercap on the 5V rail and power the Propeller through the regulator (5.5V supercaps are really common).

Two things you should keep in mind with supercaps:
The first (which was already mentioned) is that the series resistance of these super capacitors are very high. The ones I worked with were 30-ohm. No matter how much capacitance it has, a 30-ohm cap, charged to 3.3 volts will flatly not work with a draw of 20mA or more (due to the Propeller's 2.7V dropout). With a 1F cap, if you drop the draw by 2mA it will function for over 3 seconds. It is all about keeping the current draw low and consistent, so the power getting to the Propeller stays above the 2.7V rating.
If you do the exact same scenario, but charge the cap on the 5V rail, and regulate it down with a LDO regulator (one that looses 0.3V or less), you could operate all the way up to 50mA and the Propeller will still function for 10 or so seconds (somewhere around 66mA is the hard cut off for this scenario).
Adding a diode anywhere in the circuit and things just get worse for all scenarios.

The second is that for long term projects, if the capacitor is continually charged, there is a significant drop off in usable capacitance after a couple years of use. We ran into this potential problem with a project I was working on, and decided to go to a 9V battery for backup power.

So all this talk about series resistance with the caps. Something I've never taken into account before. I'm assuming Ohm's law is applicable here? Give me a math example maybe using your 20 mA example and what happens if maybe the load is 100 mA. At 3.3V.

Mark_T · 2013-03-13 18:35

Supercaps have very high capacitance, some up in the tens of Farads, but they also have a substantial internal resistance which would cause a significant voltage drop if current is drawn off too quickly.

Actually they have a pretty low internal resistance (compared to batteries that is), so unless you plan to short them out it is not an issue surely?

Or, with a 5V input, you could put the supercap on the 5V rail and power the Propeller through the regulator (5.5V supercaps are really common).

Supercaps (double-layer electrolytic) don't normally go beyond 3V or so, 5.5V part is likely to be two in series I believe (although they keep coming out
with newer technologies). Putting the energy storage at the higher voltage is usually the best place - your circuitry may assume that the 3V3 rail isn't
active when the 5V rail is powered down.

jmg · 2013-03-13 23:40

Don M wrote: »

This device only has a prop, eeprom, sd card so nothing else being powered by the 3.3 v.. I've ordered in a few .33MF and 1F 3.3v caps. I'll play and see...

You will need some sort of power-falling warning too.
An alternative is to use a smarter sort of regulator, like
http://www.microchip.com/wwwproducts/Devices.aspx?dDocName=en010593
this one is a Charge pump regulator, so keeps 3v3 on Vo as Vin falls from 5.5V to 2.1V
that range allows a smaller/cheaper storage cap.

Don M · 2013-03-14 05:37

How about this solution?

Bobb Fwed · 2013-03-14 11:30

Don M wrote: »

So all this talk about series resistance with the caps. Something I've never taken into account before. I'm assuming Ohm's law is applicable here? Give me a math example maybe using your 20 mA example and what happens if maybe the load is 100 mA. At 3.3V.

It's all about Ohm's law. It's pretty simple:
V = I * R -- So 0.02A (20mA) * 30-ohms == 0.6V drop
A charge at 3.3V with a drop of 0.6V gets us to 2.7V (the dropout for the Propeller -- thus no function).

At 100mA:
0.1A * 30-ohms == 3.0V drop -- and not much can run at 0.3V.

In the opposite direction: 10mA:
0.01A * 30-ohm == 0.3V drop -- resulting in a 3.0V output (it will function).
So now we do the math to see how long it will last:
( V(in) - V(drop) - V(dropout) ) = functional voltage
and V(functional) / I * C(Farads) = active time in seconds

So (3.3V - 0.3V - 2.7V) == 0.3V is our functional voltage
0.3V / 0.01A * 1F == 30 seconds
This of course assumes a linear current draw as the voltage drops.

That's why increasing the charge voltage makes such a big difference. If the input goes from 3.3V to 5V, the functional voltage jumps to 2V, and now you could run the same system for over 3 minutes! Of course that doesn't account for the drop through a regulator, so it would be less than that, but you get the point.

EDIT: heck, if you charged the cap at 5V and put the buck-boost regulator on there that jmg mentioned, you could run for over 4 minutes!
Also, here is a whole list of ~1F 5.5V super caps, some of them even have 20-ohm resistance, so all those numbers above go up! http://www.digikey.com/scripts/dksearch/dksus.dll?FV=fff40002%2Cfff8000c&vendor=0&mnonly=0&newproducts=0&ptm=0&fid=0&quantity=0&PV13=1368&PV13=1315&PV13=1263&PV13=1277&PV13=331&PV13=339&PV13=1426&PV13=1085&PV14=730&PV14=30&stock=1&rohs=1

Bobb Fwed · 2013-03-14 11:49

I built this spread sheet a few years ago. A lot of it may not be usable or applicable to many on here, but the "Amps / Watts / Ω / Volts" and "Capacitor Dropout Calculator" could be very useful for many here.

Don M · 2013-03-14 13:32

So with this discussion I have a few observations and more questions...

RDL2004 wrote: »

Seriously, unless the manufacturer of the voltage regulator clearly states that no bypass diode is needed you had better add one or you're almost guaranteed to blow the regulator with that much capacitance on the output.

If this is true then how does Martin get away with this on his DNA board? Here is his schematic of the power supply. When powered by USB the +5V is fed through mosfet Q1 onto the +5V line essentially back feeding the 5 volt regulator. There is nothing mentioned on the data sheet for the regulators he is using. So what would be different about this back feeding the regulator with the USB versus a capacitor?

@Bobb Fwed-

Thanks for your lesson here. I received the supercaps today from Digikey. The .33F have an ESR of 200 ohms. (ouch) The 1F have an ESR of 20 ohms. So using one of Martins DNA boards I set up a program that logs to the SD card to measure the current draw @ 5V. It measured about 220mA. So using your calculations with the supercap on the 3.3V line means that even with the 1F cap @ 20 ohms I'd have a voltage drop over 3.3V (4.4 V) so in essence it shouldn't even run at all after disconnecting the power?

Bobb Fwed · 2013-03-14 13:40

Don M wrote: »

...so in essence it shouldn't even run at all after disconnecting the power?

Correct.
What will happen is as soon as power is disconnected the voltage will drop too low, and everything will shut down. But then the current draw is low enough that the power will come back up to an operable voltage, but as things boot up the current will increase too much and you'll be stuck in a reboot loop until enough power has been lost from the cap.

EDIT: You would have to parallel 8 of them together to get the resistance low enough to operate. It would be down to 2.5 ESR, but then you would get 1.82 seconds of operation.
As I said in my first post, we ran into similar engineering conundrums, and ended up just popping a 9V battery on our device.
If you find a way to majorly reduce the power consumption of your project you could potentially work with supercaps, but as it is right now, it seems impractical.

Don M · 2013-03-14 14:17

I would have never thought it would have been this difficult. I thought I could add a big cap and all is well...

So what I did on another board that worked was this-

24 - 40 VDC in >> switching regulator (LM2674-12V) that had 2 x 220uF >> 5 volt regulator >> 3.3 volt regulator. I monitored the 24 - 40 volt input through a voltage divider with zener and after it got down to less than 20 volts the voltage divider would toggle a pin on the prop telling it to finish and close files. I monitored (with a logic analyzer) the time it took to finish and write to the SD card and with the 2 x 220uF caps I had more than enough time. It was running with Martins DNA board and the caps on the 3.3 and 5 volt regulators are minimal. 3 x 100uf total.

So now I'm going to need to rethink this all a bit....

Even using the IC that jmg mentioned wouldn't work as the prop, eeprom, sd card, clock chip is what draws the 220mA and is over it's capability. Looks like my solution may just be large value caps on the input to the power supply and in between the 5 and 3.3 volt regulators....

RDL2004 · 2013-03-14 14:28

I pretty much try to follow the manufacturer's recommendations as much as possible. The protection diodes are not always necessary, just a "good idea". In most cases the amount of capacitance on the output is pretty small, ranging from almost nothing up to a few hundred micro-farads maybe, you were talking about using a million times that much.

Here is a clip from the TI data sheet for the LM317.
.
.
attachment.php?attachmentid=99937&d=1363296271

attachment.php?attachmentid=99937&d=1363296271

Don M · 2013-03-14 14:55

Bobb Fwed wrote: »

As I said in my first post, we ran into similar engineering conundrums, and ended up just popping a 9V battery on our device.

So I understand using something simple like a battery for backup. But I'm wondering how you "disconnected" the battery from the circuit after the processor was done doing it's thing? Otherwise whats not to keep the battery from just continuing to power the circuit?

Bobb Fwed · 2013-03-14 15:00

Don M wrote: »

So I understand using something simple like a battery for backup. But I'm wondering how you "disconnected" the battery from the circuit after the processor was done doing it's thing? Otherwise whats not to keep the battery from just continuing to power the circuit?

Well, in our specific project we have a 12-14V input, so we just have a diode to prevent back flow. But you could setup a fairly simple power detection circuit that turns on the battery when the power is disconnected.

Bobb Fwed · 2013-03-14 15:25

Maybe something along these lines would work. There may be much better battery switching circuits on the web (I didn't even bother to look, may be have saved me some work). But this was a first crack at it.
You'd just have to keep in mind that Q2 would have to handle the load of the entire circuit. Also, the 9V would continually have a small load through R2, but R2 could be a pretty high value resistor. Again, look around for a better circuit (I'm sure there are some that don't put any load on the battery).

Don M · 2013-03-14 16:07

Taking your idea a bit further I wonder if something like this might work. When the Prop detects a power failure it turns on Q3 to enable the battery. Then it goes about finishing saving files, closing etc. When it's done it turns off Q3 and shuts itself off. The power failure detection would be ahead of the 5 volt regulator and should switch on fast enough to not miss a beat. This would save the 9 volt battery from draining down.

Don M · 2013-03-14 18:41

Well I'm back to my original design. This gives me a little over 1 second of time with 3.3V on power loss @ 220mA. The input caps to the 5 volt regulator make the biggest difference. The 5 volt regulator is one of those Murata modules.

So much for the Supercap idea....

I may play with the battery / mosfet switch idea.

Thanks everyone for your ideas, thoughts and insights.

Don

jmg · 2013-03-14 18:53

Don M wrote: »

The input caps to the 5 volt regulator make the biggest difference. The 5 volt regulator is one of those Murata modules.

That makes sense, as the Murata Module I'm guessing is a switching module, and there you gain from power conversion, and so you have serious dV available, and current will be lower than the 5V drain. (so you get close to 1/2*C*V^2 energy usage )
( 1 second is a lot of time to close a SD card)

That means you could reduce the secondary caps, and still be OK.

A little bit or surge protection would help on the IP side ?

LoopyByteloose · 2013-03-15 02:09

jmg wrote: »

That makes sense, as the Murata Module I'm guessing is a switching module, and there you gain from power conversion, and so you have serious dV available, and current will be lower than the 5V drain. (so you get close to 1/2*C*V^2 energy usage )
( 1 second is a lot of time to close a SD card)

That means you could reduce the secondary caps, and still be OK.

A little bit or surge protection would help on the IP side ?

Well, anything over about 10 mfd on the output side may cause a reverse current damage to the regulators if you don't include protective diodes.... though I am less sure you would have a problem with switching regulators. You really have to apply your real choice of regulators to super caps for any realistic comment on protection problems.

There are alternative solutions....

I discovered that while two linear regulators in parallel trying to achieve the same voltage are a known disaster; having one regulator provide 4.9 volts from a standby battery power and a wall wart regulator set to 5.0 volt will work to provide continuous uninterrupted power. This can even be done by adding one voltage regulator to an existing Propeller ProtoBoard. The trick is to separate the regulation enough to avoid having both trying to power at the same time AND to keep within the safe operating limits of the regulated devices.

Since most 5.0volt devices will tolerate up to 5.5 volts and much lower than 5.0 volts; a second power source at 4.9 volts is acceptable.

In such a scheme, the battery can even be a big Lead Acid cell with continuous trickle charge from the A/C mains. It just doesn't turn on the output until the other regulator drops below 5.0 volts (which is an off condition of the 5.0 regulator).

When the 4.9v regulator senses the presence of 5.0v on its output side, it remains dormant. The battery just sits in standby.

~~~~~~
While I found super-caps do provide a bit of power for a longer period than conventional caps, they really didn't perform long enough for me to be a good source for an 'uninterrupted power supply'.

And, they do add a lot of complications to conventional regulation goals.

Supercap application question

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