Forums

capcitors for energy reserve

Started by Richard October 2, 2014
On 10/2/2014 5:47 PM, Richard wrote:
> On Thursday, October 2, 2014 2:12:22 PM UTC-7, Tim Wescott wrote: >> On Thu, 02 Oct 2014 15:44:19 -0500, Tim Wescott wrote: >> >> >> >>> On Thu, 02 Oct 2014 13:21:05 -0700, Richard wrote: >> >>> >> >>>> I have a circuit that draws 30mA except that there is a 150ms long draw >> >>>> at 120mA. This occurs every 2s. >> >>>> >> >>>> I I am limited to 40mA current draw. I would like to provide reserve >> >>>> energy from a large capacitor. I have built a circuit that limits the >> >>>> input current to 40mA. This runs the circuit and charges the cap. >> >>>> >> >>>> I have run caps up to 3300uF, the cap voltage drops to far and in too >> >>>> short a time. Is there any methodology for calculating required >> >>>> capacitance? Any ideas on implementation beside my current limiter >> >>>> plus cap? >> >>> >> >>> Hark back to your sophomore electronics engineering courses: >> >>> >> >>> dV/dt = i / C >> >>> >> >>> For a constant current draw, you can change that to >> >>> >> >>> deltaV / deltaT = i / C >> >>> >> >>> Do some algebra, and you get >> >>> >> >>> C = i * deltaT / deltaV >> >>> >> >>> You've got 40mA in, 120mA out, which means you need 80mA out of the cap. >> >>> You know what voltage drop you can stand -- plug that figure into the >> >>> above equation along with the rest of your knowns, and you should get a >> >>> minimum capacitance. >> >>> >> >>> I must say, though, that if 3300uF isn't enough, then at those current >> >>> levels there's something else going on. If your calculations show that >> >>> 3300uF is way more than plenty, then your problem may be that you're >> >>> using aluminum electrolytics where you need to use something with >> >>> significantly lower ESR. >> >>> >> >>> I'm doing something similar to you right now, except that my current >> >>> draw is for a shorter period of time. I have my filter cap in my "raw" >> >>> power line, ahead of my local regulator. With 5V on the "raw" side and >> >>> 3.3V on the regulated side, I can accept a 1V drop during the 'on' time, >> >>> as long as the cap charges back for the next cycle. >> >> >> >> Note that I messed up my in-the-head math: 3300uF is, indeed, pretty small >> >> for this unless you charge the cap with a boost circuit, regulate after >> >> the cap, and accept large voltage swings. >> >> >> >> -- >> >> >> >> Tim Wescott >> >> Wescott Design Services >> >> http://www.wescottdesign.com > > Thanks guys for the ideas. I used 3300 because that is what I had. I am now using ~7000uF and I see better results, the voltage drops from 9V down to ~5V so I am too close to the 5V minimum. Charge time is ~1.7s. The input current is limited at 40mA. > > I think this will work with a larger cap but sure seem inelegant.
A suggestion is to use a newsreader rather than Google Groups. They double space all the quoted lines and really mess up replies. Thunderbird with eternal-september.org works well. About your problem. If you can work down to 5 volts, what does that do for your current? Many types of loads vary the current proportionally to voltage. Does yours? That will greatly reduce your total power draw and make this work better. So you could use the current limiter with a cap but then add a regulator after the cap which if it reduces the current drawn during the higher current time will make it easier to meet your requirements with a smaller cap. The regulator doesn't need to be a switcher, only a series linear regulator. -- Rick
On Friday, October 3, 2014 5:46:51 AM UTC+2, Tim Wescott wrote:
> On Thu, 02 Oct 2014 14:53:44 -0700, Klaus Kragelund wrote: > > > > > On Thursday, October 2, 2014 11:47:08 PM UTC+2, Richard wrote: > > >> On Thursday, October 2, 2014 2:12:22 PM UTC-7, Tim Wescott wrote: > > >> > > >> > On Thu, 02 Oct 2014 15:44:19 -0500, Tim Wescott wrote: > > >> > > >> > > >> > > > >> > > >> > > > >> > > >> > > > >> > > On Thu, 02 Oct 2014 13:21:05 -0700, Richard wrote: > > >> > > >> > > >> > > > >> > > >> > > > > >> > > >> > > > >> > >> I have a circuit that draws 30mA except that there is a 150ms long > > >> > >> draw > > >> > > >> > > >> > > > >> > >> at 120mA. This occurs every 2s. > > >> > > >> > > >> > > > >> > > >> > >> > > >> > > >> > > > >> > >> I I am limited to 40mA current draw. I would like to provide > > >> > >> reserve > > >> > > >> > > >> > > > >> > >> energy from a large capacitor. I have built a circuit that limits > > >> > >> the > > >> > > >> > > >> > > > >> > >> input current to 40mA. This runs the circuit and charges the cap. > > >> > > >> > > >> > > > >> > > >> > >> > > >> > > >> > > > >> > >> I have run caps up to 3300uF, the cap voltage drops to far and in > > >> > >> too > > >> > > >> > > >> > > > >> > >> short a time. Is there any methodology for calculating required > > >> > > >> > > >> > > > >> > >> capacitance? Any ideas on implementation beside my current > > >> > >> limiter > > >> > > >> > > >> > > > >> > >> plus cap? > > >> > > >> > > >> > > > >> > > >> > > > > >> > > >> > > > >> > > Hark back to your sophomore electronics engineering courses: > > >> > > >> > > >> > > > >> > > >> > > > > >> > > >> > > > >> > > dV/dt = i / C > > >> > > >> > > >> > > > >> > > >> > > > > >> > > >> > > > >> > > For a constant current draw, you can change that to > > >> > > >> > > >> > > > >> > > >> > > > > >> > > >> > > > >> > > deltaV / deltaT = i / C > > >> > > >> > > >> > > > >> > > >> > > > > >> > > >> > > > >> > > Do some algebra, and you get > > >> > > >> > > >> > > > >> > > >> > > > > >> > > >> > > > >> > > C = i * deltaT / deltaV > > >> > > >> > > >> > > > >> > > >> > > > > >> > > >> > > > >> > > You've got 40mA in, 120mA out, which means you need 80mA out of the > > >> > > cap. > > >> > > >> > > >> > > > >> > > You know what voltage drop you can stand -- plug that figure into > > >> > > the > > >> > > >> > > >> > > > >> > > above equation along with the rest of your knowns, and you should > > >> > > get a > > >> > > >> > > >> > > > >> > > minimum capacitance. > > >> > > >> > > >> > > > >> > > >> > > > > >> > > >> > > > >> > > I must say, though, that if 3300uF isn't enough, then at those > > >> > > current > > >> > > >> > > >> > > > >> > > levels there's something else going on. If your calculations show > > >> > > that > > >> > > >> > > >> > > > >> > > 3300uF is way more than plenty, then your problem may be that > > >> > > you're > > >> > > >> > > >> > > > >> > > using aluminum electrolytics where you need to use something with > > >> > > >> > > >> > > > >> > > significantly lower ESR. > > >> > > >> > > >> > > > >> > > >> > > > > >> > > >> > > > >> > > I'm doing something similar to you right now, except that my > > >> > > current > > >> > > >> > > >> > > > >> > > draw is for a shorter period of time. I have my filter cap in my > > >> > > "raw" > > >> > > >> > > >> > > > >> > > power line, ahead of my local regulator. With 5V on the "raw" side > > >> > > and > > >> > > >> > > >> > > > >> > > 3.3V on the regulated side, I can accept a 1V drop during the 'on' > > >> > > time, > > >> > > >> > > >> > > > >> > > as long as the cap charges back for the next cycle. > > >> > > >> > > >> > > > >> > > >> > > > >> > > >> > > > >> > Note that I messed up my in-the-head math: 3300uF is, indeed, pretty > > >> > small > > >> > > >> > > >> > > > >> > for this unless you charge the cap with a boost circuit, regulate > > >> > after > > >> > > >> > > >> > > > >> > the cap, and accept large voltage swings. > > >> > > >> > > >> > > > >> > > >> > > > >> > > >> > > > >> > -- > > >> > > >> > > >> > > > >> > > >> > > > >> > > >> > > > >> > Tim Wescott > > >> > > >> > > >> > > > >> > Wescott Design Services > > >> > > >> > > >> > > > >> > http://www.wescottdesign.com > > >> > > >> > > >> > > >> Thanks guys for the ideas. I used 3300 because that is what I had. I > > >> am now using ~7000uF and I see better results, the voltage drops from > > >> 9V down to ~5V so I am too close to the 5V minimum. Charge time is > > >> ~1.7s. The input current is limited at 40mA. > > >> > > >> > > >> > > >> I think this will work with a larger cap but sure seem inelegant. > > >> > > > You have 10mA surplus during normal op, right? > > > > > > Use that to charge a lower value capacitor with a boost converter > > > > > > When you need the 180mA, draw that from a buck converter, so you can > > > draw all the energy out og the capacitor > > > > > > Cheers > > > > > > Klaus > > > > Assuming 100% efficient supplies: > > > > (120mA)(150ms)(5V) = 90mJ > > > > (30mA)(150ms)(9V) = 40.5mJ > > > > Leaving you with a 49.5mJ deficit. > > > > Energy in a capacitor: W = C * V^2 / 2. Let V = 9V (it could be anything > > if we're switching in and switching out), then > > > > C = 2 * (49.5mJ) / (9V)^2 = 1222uF > > > > That's the ABSOLUTE MINIMUM capacitance you can use if you only go up to > > 9V. > >
Yes, and 1200uF is a lot less than 7000uF Cheers Klaus
On 2014-10-02, Richard <rsoennichsen@gmail.com> wrote:
> I have a circuit that draws 30mA except that there is a 150ms long draw at 120mA. This occurs every 2s. > > I I am limited to 40mA current draw. I would like to provide reserve energy from a large capacitor. I have built a circuit that limits the input current to 40mA. This runs the circuit and charges the cap. > > I have run caps up to 3300uF, the cap voltage drops to far and in too short a time. > Is there any methodology for calculating required capacitance?
The capacitor formula: Q=EC or -- umop apisdn --- news://freenews.netfront.net/ - complaints: news@netfront.net ---
On Thursday, October 2, 2014 9:21:10 PM UTC+1, Richard wrote:

> I have a circuit that draws 30mA except that there is a 150ms long draw at 120mA. This occurs every 2s. > I I am limited to 40mA current draw. I would like to provide reserve energy from a large capacitor. I have built a circuit that limits the input current to 40mA. This runs the circuit and charges the cap. > I have run caps up to 3300uF, the cap voltage drops to far and in too short a time. Is there any methodology for calculating required capacitance? Any ideas on implementation beside my current limiter plus cap?
Q=CV (or delta v in your case) Q=It Don't forget to allow for capacitance tolerance, capacitance decline over time, changing load current with voltage, load time tolerance and slight inefficiencies NT
On Thursday, October 2, 2014 11:26:51 PM UTC-4, Bill Sloman wrote:

>=20 >=20 > I'm not the original poster, who was a bit economical in describing what =
he was doing. Alkaline batteries aren't designed to be recharged, and don't= seem to be guaranteed when used that way.
>
Even though you have a Phd, you do not know everything. The following is f= rom Wiki. A rechargeable alkaline battery (also known as alkaline rechargeable or rec= hargeable alkaline manganese (RAM)) is a type of alkaline battery that is c= apable of recharging for repeated use. The first-generation rechargeable al= kaline technology was developed by Battery Technologies Inc in Canada and l= icensed to Pure Energy, EnviroCell, Rayovac, and Grandcell. Subsequent pate= nt and advancements in technology have been introduced. The formats include= AAA, AA, C, D, and snap-on 9-volt batteries. Rechargeable alkaline batteri= es are manufactured fully charged and have the ability to hold their charge= for years, longer than NiCd and NiMH batteries, which self-discharge.[1] R= echargeable alkaline batteries can have a high recharging efficiency and ha= ve less environmental impact than disposable cells.=20
>=20 >=20 > And you've snipped most of what I posted, without marking the snip. >=20 >
Most people do not need posts with every thing perfect. I try to cut posts= down to a minimum , so people can read them quickly. Dan =20
>=20 > --=20 >=20 > Bill Sloman, Sydney
On 10/02/2014 11:46 PM, Tim Wescott wrote:
> On Thu, 02 Oct 2014 14:53:44 -0700, Klaus Kragelund wrote: > >> On Thursday, October 2, 2014 11:47:08 PM UTC+2, Richard wrote: >>> On Thursday, October 2, 2014 2:12:22 PM UTC-7, Tim Wescott wrote: >>> >>>> On Thu, 02 Oct 2014 15:44:19 -0500, Tim Wescott wrote: >>> >>> >>>> >>> >>>> >>> >>>> >>>>> On Thu, 02 Oct 2014 13:21:05 -0700, Richard wrote: >>> >>> >>>> >>> >>>>> >>> >>>> >>>>>> I have a circuit that draws 30mA except that there is a 150ms long >>>>>> draw >>> >>> >>>> >>>>>> at 120mA. This occurs every 2s. >>> >>> >>>> >>> >>>>>> >>> >>>> >>>>>> I I am limited to 40mA current draw. I would like to provide >>>>>> reserve >>> >>> >>>> >>>>>> energy from a large capacitor. I have built a circuit that limits >>>>>> the >>> >>> >>>> >>>>>> input current to 40mA. This runs the circuit and charges the cap. >>> >>> >>>> >>> >>>>>> >>> >>>> >>>>>> I have run caps up to 3300uF, the cap voltage drops to far and in >>>>>> too >>> >>> >>>> >>>>>> short a time. Is there any methodology for calculating required >>> >>> >>>> >>>>>> capacitance? Any ideas on implementation beside my current >>>>>> limiter >>> >>> >>>> >>>>>> plus cap? >>> >>> >>>> >>> >>>>> >>> >>>> >>>>> Hark back to your sophomore electronics engineering courses: >>> >>> >>>> >>> >>>>> >>> >>>> >>>>> dV/dt = i / C >>> >>> >>>> >>> >>>>> >>> >>>> >>>>> For a constant current draw, you can change that to >>> >>> >>>> >>> >>>>> >>> >>>> >>>>> deltaV / deltaT = i / C >>> >>> >>>> >>> >>>>> >>> >>>> >>>>> Do some algebra, and you get >>> >>> >>>> >>> >>>>> >>> >>>> >>>>> C = i * deltaT / deltaV >>> >>> >>>> >>> >>>>> >>> >>>> >>>>> You've got 40mA in, 120mA out, which means you need 80mA out of the >>>>> cap. >>> >>> >>>> >>>>> You know what voltage drop you can stand -- plug that figure into >>>>> the >>> >>> >>>> >>>>> above equation along with the rest of your knowns, and you should >>>>> get a >>> >>> >>>> >>>>> minimum capacitance. >>> >>> >>>> >>> >>>>> >>> >>>> >>>>> I must say, though, that if 3300uF isn't enough, then at those >>>>> current >>> >>> >>>> >>>>> levels there's something else going on. If your calculations show >>>>> that >>> >>> >>>> >>>>> 3300uF is way more than plenty, then your problem may be that >>>>> you're >>> >>> >>>> >>>>> using aluminum electrolytics where you need to use something with >>> >>> >>>> >>>>> significantly lower ESR. >>> >>> >>>> >>> >>>>> >>> >>>> >>>>> I'm doing something similar to you right now, except that my >>>>> current >>> >>> >>>> >>>>> draw is for a shorter period of time. I have my filter cap in my >>>>> "raw" >>> >>> >>>> >>>>> power line, ahead of my local regulator. With 5V on the "raw" side >>>>> and >>> >>> >>>> >>>>> 3.3V on the regulated side, I can accept a 1V drop during the 'on' >>>>> time, >>> >>> >>>> >>>>> as long as the cap charges back for the next cycle. >>> >>> >>>> >>> >>>> >>> >>>> >>>> Note that I messed up my in-the-head math: 3300uF is, indeed, pretty >>>> small >>> >>> >>>> >>>> for this unless you charge the cap with a boost circuit, regulate >>>> after >>> >>> >>>> >>>> the cap, and accept large voltage swings. >>> >>> >>>> >>> >>>> >>> >>>> >>>> -- >>> >>> >>>> >>> >>>> >>> >>>> >>>> Tim Wescott >>> >>> >>>> >>>> Wescott Design Services >>> >>> >>>> >>>> http://www.wescottdesign.com >>> >>> >>> >>> Thanks guys for the ideas. I used 3300 because that is what I had. I >>> am now using ~7000uF and I see better results, the voltage drops from >>> 9V down to ~5V so I am too close to the 5V minimum. Charge time is >>> ~1.7s. The input current is limited at 40mA. >>> >>> >>> >>> I think this will work with a larger cap but sure seem inelegant. >>> >> You have 10mA surplus during normal op, right? >> >> Use that to charge a lower value capacitor with a boost converter >> >> When you need the 180mA, draw that from a buck converter, so you can >> draw all the energy out og the capacitor >> >> Cheers >> >> Klaus > > Assuming 100% efficient supplies: > > (120mA)(150ms)(5V) = 90mJ > > (30mA)(150ms)(9V) = 40.5mJ > > Leaving you with a 49.5mJ deficit. > > Energy in a capacitor: W = C * V^2 / 2. Let V = 9V (it could be anything > if we're switching in and switching out), then > > C = 2 * (49.5mJ) / (9V)^2 = 1222uF > > That's the ABSOLUTE MINIMUM capacitance you can use if you only go up to > 9V. > > If you start surveying capacitors of any given dielectric, you'll soon > find out that over a fairly wide range of voltage ratings, a capacitor > that stores a given energy is about the same size as all the other > capacitors that store the same amount of energy. Tantalums are smallish, > aluminums are bigger, ceramics are bigger yet, but the volume/energy ratio > is about constant. > > If you take into account the supply inefficiencies, and the fact that > it'll be hard to suck the last drop of energy out of a cap, you probably > can't go much smaller than a 2200uF cap, and that's taking heroic measures > with the switching supplies and all. >
I think it's the CV product that tends to be nearly constant, so since energy is CV**2/2, you win by going to higher voltage. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
On Friday, 3 October 2014 22:41:37 UTC+10, dca...@krl.org  wrote:
> On Thursday, October 2, 2014 11:26:51 PM UTC-4, Bill Sloman wrote: > =20 > > I'm not the original poster, who was a bit economical in describing wha=
t he was doing. Alkaline batteries aren't designed to be recharged, and don= 't seem to be guaranteed when used that way.=20
>=20 > Even though you have a Phd, you do not know everything. The following is=
from Wiki. =20
>=20 > A rechargeable alkaline battery (also known as alkaline rechargeable or r=
echargeable alkaline manganese (RAM)) is a type of alkaline battery that is= capable of recharging for repeated use. The first-generation rechargeable = alkaline technology was developed by Battery Technologies Inc in Canada and= licensed to Pure Energy, EnviroCell, Rayovac, and Grandcell. Subsequent pa= tent and advancements in technology have been introduced. The formats inclu= de AAA, AA, C, D, and snap-on 9-volt batteries. Rechargeable alkaline batte= ries are manufactured fully charged and have the ability to hold their char= ge for years, longer than NiCd and NiMH batteries, which self-discharge.[1]= Rechargeable alkaline batteries can have a high recharging efficiency and = have less environmental impact than disposable cells.=20 Do any of the broad-line distributors stock them? Could you post a link to an item in on-line catalogue? Too many "advances in technology" turn out to be remarkably hard to get you= r hands on. I tend not to take them seriously until a broad-line distributo= r keeps them in stock. Note that the Ph.D. is a remarkably specialised qualification - pretty much= at the sharp end of knowing more and more about less and less until you kn= ow almost everything about about practically nothing.
> > And you've snipped most of what I posted, without marking the snip. >=20 > Most people do not need posts with every thing perfect. I try to cut pos=
ts down to a minimum , so people can read them quickly. Sure. But it's polite to mark your edits. And this time, you managed to inc= lude my signature line, which did waste bandwidth. --=20 Bill Sloman, Sydney
On Fri, 03 Oct 2014 10:40:06 -0400, Phil Hobbs wrote:

> On 10/02/2014 11:46 PM, Tim Wescott wrote: >> On Thu, 02 Oct 2014 14:53:44 -0700, Klaus Kragelund wrote: >> >>> On Thursday, October 2, 2014 11:47:08 PM UTC+2, Richard wrote: >>>> On Thursday, October 2, 2014 2:12:22 PM UTC-7, Tim Wescott wrote: >>>> >>>>> On Thu, 02 Oct 2014 15:44:19 -0500, Tim Wescott wrote: >>>> >>>> >>>> >>>>> >>>> >>>>> >>>> >>>>>> On Thu, 02 Oct 2014 13:21:05 -0700, Richard wrote: >>>> >>>> >>>> >>>>> >>>> >>>>>> >>>> >>>>>>> I have a circuit that draws 30mA except that there is a 150ms long >>>>>>> draw >>>> >>>> >>>> >>>>>>> at 120mA. This occurs every 2s. >>>> >>>> >>>> >>>>> >>>> >>>>>>> >>>> >>>>>>> I I am limited to 40mA current draw. I would like to provide >>>>>>> reserve >>>> >>>> >>>> >>>>>>> energy from a large capacitor. I have built a circuit that limits >>>>>>> the >>>> >>>> >>>> >>>>>>> input current to 40mA. This runs the circuit and charges the cap. >>>> >>>> >>>> >>>>> >>>> >>>>>>> >>>> >>>>>>> I have run caps up to 3300uF, the cap voltage drops to far and in >>>>>>> too >>>> >>>> >>>> >>>>>>> short a time. Is there any methodology for calculating required >>>> >>>> >>>> >>>>>>> capacitance? Any ideas on implementation beside my current >>>>>>> limiter >>>> >>>> >>>> >>>>>>> plus cap? >>>> >>>> >>>> >>>>> >>>> >>>>>> >>>> >>>>>> Hark back to your sophomore electronics engineering courses: >>>> >>>> >>>> >>>>> >>>> >>>>>> >>>> >>>>>> dV/dt = i / C >>>> >>>> >>>> >>>>> >>>> >>>>>> >>>> >>>>>> For a constant current draw, you can change that to >>>> >>>> >>>> >>>>> >>>> >>>>>> >>>> >>>>>> deltaV / deltaT = i / C >>>> >>>> >>>> >>>>> >>>> >>>>>> >>>> >>>>>> Do some algebra, and you get >>>> >>>> >>>> >>>>> >>>> >>>>>> >>>> >>>>>> C = i * deltaT / deltaV >>>> >>>> >>>> >>>>> >>>> >>>>>> >>>> >>>>>> You've got 40mA in, 120mA out, which means you need 80mA out of the >>>>>> cap. >>>> >>>> >>>> >>>>>> You know what voltage drop you can stand -- plug that figure into >>>>>> the >>>> >>>> >>>> >>>>>> above equation along with the rest of your knowns, and you should >>>>>> get a >>>> >>>> >>>> >>>>>> minimum capacitance. >>>> >>>> >>>> >>>>> >>>> >>>>>> >>>> >>>>>> I must say, though, that if 3300uF isn't enough, then at those >>>>>> current >>>> >>>> >>>> >>>>>> levels there's something else going on. If your calculations show >>>>>> that >>>> >>>> >>>> >>>>>> 3300uF is way more than plenty, then your problem may be that >>>>>> you're >>>> >>>> >>>> >>>>>> using aluminum electrolytics where you need to use something with >>>> >>>> >>>> >>>>>> significantly lower ESR. >>>> >>>> >>>> >>>>> >>>> >>>>>> >>>> >>>>>> I'm doing something similar to you right now, except that my >>>>>> current >>>> >>>> >>>> >>>>>> draw is for a shorter period of time. I have my filter cap in my >>>>>> "raw" >>>> >>>> >>>> >>>>>> power line, ahead of my local regulator. With 5V on the "raw" side >>>>>> and >>>> >>>> >>>> >>>>>> 3.3V on the regulated side, I can accept a 1V drop during the 'on' >>>>>> time, >>>> >>>> >>>> >>>>>> as long as the cap charges back for the next cycle. >>>> >>>> >>>> >>>>> >>>> >>>>> >>>> >>>>> Note that I messed up my in-the-head math: 3300uF is, indeed, pretty >>>>> small >>>> >>>> >>>> >>>>> for this unless you charge the cap with a boost circuit, regulate >>>>> after >>>> >>>> >>>> >>>>> the cap, and accept large voltage swings. >>>> >>>> >>>> >>>>> >>>> >>>>> >>>> >>>>> -- >>>> >>>> >>>> >>>>> >>>> >>>>> >>>> >>>>> Tim Wescott >>>> >>>> >>>> >>>>> Wescott Design Services >>>> >>>> >>>> >>>>> http://www.wescottdesign.com >>>> >>>> >>>> >>>> Thanks guys for the ideas. I used 3300 because that is what I had. >>>> I am now using ~7000uF and I see better results, the voltage drops >>>> from 9V down to ~5V so I am too close to the 5V minimum. Charge time >>>> is ~1.7s. The input current is limited at 40mA. >>>> >>>> >>>> >>>> I think this will work with a larger cap but sure seem inelegant. >>>> >>> You have 10mA surplus during normal op, right? >>> >>> Use that to charge a lower value capacitor with a boost converter >>> >>> When you need the 180mA, draw that from a buck converter, so you can >>> draw all the energy out og the capacitor >>> >>> Cheers >>> >>> Klaus >> >> Assuming 100% efficient supplies: >> >> (120mA)(150ms)(5V) = 90mJ >> >> (30mA)(150ms)(9V) = 40.5mJ >> >> Leaving you with a 49.5mJ deficit. >> >> Energy in a capacitor: W = C * V^2 / 2. Let V = 9V (it could be >> anything if we're switching in and switching out), then >> >> C = 2 * (49.5mJ) / (9V)^2 = 1222uF >> >> That's the ABSOLUTE MINIMUM capacitance you can use if you only go up >> to 9V. >> >> If you start surveying capacitors of any given dielectric, you'll soon >> find out that over a fairly wide range of voltage ratings, a capacitor >> that stores a given energy is about the same size as all the other >> capacitors that store the same amount of energy. Tantalums are >> smallish, aluminums are bigger, ceramics are bigger yet, but the >> volume/energy ratio is about constant. >> >> If you take into account the supply inefficiencies, and the fact that >> it'll be hard to suck the last drop of energy out of a cap, you >> probably can't go much smaller than a 2200uF cap, and that's taking >> heroic measures with the switching supplies and all. >> >> > I think it's the CV product that tends to be nearly constant, so since > energy is CV**2/2, you win by going to higher voltage. > > Cheers > > Phil Hobbs
The last time I looked (which, granted, was a while ago) it was C * V^2. It wasn't by any authority -- it was primary research. I just pulled out a DigiKey or Mouser catalog, chose a package size, and looked at the highest C rating available in any particular V rating, and plotted a line. For that matter I think I only actually checked electrolytic caps -- so I could be wrong about ceramics. -- www.wescottdesign.com
On Thu, 02 Oct 2014 17:32:39 -0700, mike <ham789@netzero.net> wrote:
> On 10/2/2014 1:21 PM, Richard wrote: >> I have a circuit that draws 30mA except that there is a 150ms long >> draw at 120mA. This occurs every 2s. >> >> I I am limited to 40mA current draw. I would like to provide reserve >> energy from a large capacitor. I have built a circuit that limits the >> input current to 40mA. This runs the circuit and charges the cap. >> >> I have run caps up to 3300uF, the cap voltage drops to far and in too >> short a time. Is there any methodology for calculating required >> capacitance? Any ideas on implementation beside my current limiter >> plus cap?
> Often, the best place to fix something is "elsewhere". > You're going to a lot of trouble to limit the current then store it. > Think about fixing the thing that's limiting your peak current and put > the storage there.
Or... replace or modify the offending circuit so the peak draw never exceeds 30mA. Jes' a thought... Frank McKenney -- Where all are guilty, no one is; confessions of collective guilt are the best possible safeguard against the discovery of culprits, and the very magnitude of the crime the best excuse for doing nothing. -- Hannah Arendt -- Frank McKenney, McKenney Associates Richmond, Virginia / (804) 320-4887 Munged E-mail: frank uscore mckenney aatt mindspring ddoott com
On Fri, 03 Oct 2014 11:10:35 -0500, tim <tim@seemywebsite.com> wrote:

>On Fri, 03 Oct 2014 10:40:06 -0400, Phil Hobbs wrote: > >> On 10/02/2014 11:46 PM, Tim Wescott wrote: >>> On Thu, 02 Oct 2014 14:53:44 -0700, Klaus Kragelund wrote: >>> >>>> On Thursday, October 2, 2014 11:47:08 PM UTC+2, Richard wrote: >>>>> On Thursday, October 2, 2014 2:12:22 PM UTC-7, Tim Wescott wrote: >>>>> >>>>>> On Thu, 02 Oct 2014 15:44:19 -0500, Tim Wescott wrote: >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>> >>>>> >>>>>>> On Thu, 02 Oct 2014 13:21:05 -0700, Richard wrote: >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>>> >>>>> >>>>>>>> I have a circuit that draws 30mA except that there is a 150ms long >>>>>>>> draw >>>>> >>>>> >>>>> >>>>>>>> at 120mA. This occurs every 2s. >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>>>> >>>>> >>>>>>>> I I am limited to 40mA current draw. I would like to provide >>>>>>>> reserve >>>>> >>>>> >>>>> >>>>>>>> energy from a large capacitor. I have built a circuit that limits >>>>>>>> the >>>>> >>>>> >>>>> >>>>>>>> input current to 40mA. This runs the circuit and charges the cap. >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>>>> >>>>> >>>>>>>> I have run caps up to 3300uF, the cap voltage drops to far and in >>>>>>>> too >>>>> >>>>> >>>>> >>>>>>>> short a time. Is there any methodology for calculating required >>>>> >>>>> >>>>> >>>>>>>> capacitance? Any ideas on implementation beside my current >>>>>>>> limiter >>>>> >>>>> >>>>> >>>>>>>> plus cap? >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>>> >>>>> >>>>>>> Hark back to your sophomore electronics engineering courses: >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>>> >>>>> >>>>>>> dV/dt = i / C >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>>> >>>>> >>>>>>> For a constant current draw, you can change that to >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>>> >>>>> >>>>>>> deltaV / deltaT = i / C >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>>> >>>>> >>>>>>> Do some algebra, and you get >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>>> >>>>> >>>>>>> C = i * deltaT / deltaV >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>>> >>>>> >>>>>>> You've got 40mA in, 120mA out, which means you need 80mA out of the >>>>>>> cap. >>>>> >>>>> >>>>> >>>>>>> You know what voltage drop you can stand -- plug that figure into >>>>>>> the >>>>> >>>>> >>>>> >>>>>>> above equation along with the rest of your knowns, and you should >>>>>>> get a >>>>> >>>>> >>>>> >>>>>>> minimum capacitance. >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>>> >>>>> >>>>>>> I must say, though, that if 3300uF isn't enough, then at those >>>>>>> current >>>>> >>>>> >>>>> >>>>>>> levels there's something else going on. If your calculations show >>>>>>> that >>>>> >>>>> >>>>> >>>>>>> 3300uF is way more than plenty, then your problem may be that >>>>>>> you're >>>>> >>>>> >>>>> >>>>>>> using aluminum electrolytics where you need to use something with >>>>> >>>>> >>>>> >>>>>>> significantly lower ESR. >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>>> >>>>> >>>>>>> I'm doing something similar to you right now, except that my >>>>>>> current >>>>> >>>>> >>>>> >>>>>>> draw is for a shorter period of time. I have my filter cap in my >>>>>>> "raw" >>>>> >>>>> >>>>> >>>>>>> power line, ahead of my local regulator. With 5V on the "raw" side >>>>>>> and >>>>> >>>>> >>>>> >>>>>>> 3.3V on the regulated side, I can accept a 1V drop during the 'on' >>>>>>> time, >>>>> >>>>> >>>>> >>>>>>> as long as the cap charges back for the next cycle. >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>> >>>>> >>>>>> Note that I messed up my in-the-head math: 3300uF is, indeed, pretty >>>>>> small >>>>> >>>>> >>>>> >>>>>> for this unless you charge the cap with a boost circuit, regulate >>>>>> after >>>>> >>>>> >>>>> >>>>>> the cap, and accept large voltage swings. >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>> >>>>> >>>>>> -- >>>>> >>>>> >>>>> >>>>>> >>>>> >>>>>> >>>>> >>>>>> Tim Wescott >>>>> >>>>> >>>>> >>>>>> Wescott Design Services >>>>> >>>>> >>>>> >>>>>> http://www.wescottdesign.com >>>>> >>>>> >>>>> >>>>> Thanks guys for the ideas. I used 3300 because that is what I had. >>>>> I am now using ~7000uF and I see better results, the voltage drops >>>>> from 9V down to ~5V so I am too close to the 5V minimum. Charge time >>>>> is ~1.7s. The input current is limited at 40mA. >>>>> >>>>> >>>>> >>>>> I think this will work with a larger cap but sure seem inelegant. >>>>> >>>> You have 10mA surplus during normal op, right? >>>> >>>> Use that to charge a lower value capacitor with a boost converter >>>> >>>> When you need the 180mA, draw that from a buck converter, so you can >>>> draw all the energy out og the capacitor >>>> >>>> Cheers >>>> >>>> Klaus >>> >>> Assuming 100% efficient supplies: >>> >>> (120mA)(150ms)(5V) = 90mJ >>> >>> (30mA)(150ms)(9V) = 40.5mJ >>> >>> Leaving you with a 49.5mJ deficit. >>> >>> Energy in a capacitor: W = C * V^2 / 2. Let V = 9V (it could be >>> anything if we're switching in and switching out), then >>> >>> C = 2 * (49.5mJ) / (9V)^2 = 1222uF >>> >>> That's the ABSOLUTE MINIMUM capacitance you can use if you only go up >>> to 9V. >>> >>> If you start surveying capacitors of any given dielectric, you'll soon >>> find out that over a fairly wide range of voltage ratings, a capacitor >>> that stores a given energy is about the same size as all the other >>> capacitors that store the same amount of energy. Tantalums are >>> smallish, aluminums are bigger, ceramics are bigger yet, but the >>> volume/energy ratio is about constant. >>> >>> If you take into account the supply inefficiencies, and the fact that >>> it'll be hard to suck the last drop of energy out of a cap, you >>> probably can't go much smaller than a 2200uF cap, and that's taking >>> heroic measures with the switching supplies and all. >>> >>> >> I think it's the CV product that tends to be nearly constant, so since >> energy is CV**2/2, you win by going to higher voltage. >> >> Cheers >> >> Phil Hobbs > >The last time I looked (which, granted, was a while ago) it was C * V^2.
E = CV**2/2, just as (mechanical) E = MV**2/2
>It wasn't by any authority -- it was primary research. I just pulled out >a DigiKey or Mouser catalog, chose a package size, and looked at the >highest C rating available in any particular V rating, and plotted a line.
Did you plot cost, too? Volume?
>For that matter I think I only actually checked electrolytic caps -- so I >could be wrong about ceramics.
HV ceramics get expensive in the larger sizes. 50V 1210s can get pricey.