Forums

capcitors for energy reserve

Started by Richard October 2, 2014
On Thu, 02 Oct 2014 15:44:19 -0500, Tim Wescott
<seemywebsite@myfooter.really> 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
Linear regs are inefficient, and waste valuable joules when the cap voltage is high. There must be a place to add an inductor to help things out. Whether that's practical, can't say yet. Cartainly a 9-to-5 switcher would make better use of the energy in the cap, by roughly 3:1. -- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
On Fri, 03 Oct 2014 13:17:31 -0400, krw wrote:

> 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
I'm well aware of that. And capacitor size seemed to go by <some constant> * C * V^2. You can roll the 1/2 into the constant or not, your choice.
>>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?
Volume went more or less by energy stored. Cost, except at the very high voltage end, pretty much went by volume. It wouldn't be a bad thing to do over again, if someone out there has time on there hands.
>>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.
-- www.wescottdesign.com
On 10/3/2014 1:33 PM, John Larkin wrote:
> On Thu, 02 Oct 2014 15:44:19 -0500, Tim Wescott > <seemywebsite@myfooter.really> 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 > > > Linear regs are inefficient, and waste valuable joules when the cap > voltage is high. There must be a place to add an inductor to help > things out. Whether that's practical, can't say yet.
That is a generalization that is not at all universal. Switchers have inefficiencies that are proportional to the power delivered (with a base power drain) while the efficiency of a linear is determined by the input and output voltages (typically the quiescent power waste is relatively small). I've seen use cases where a linear was *more* efficient across the current range required.
> Cartainly a 9-to-5 switcher would make better use of the energy in the > cap, by roughly 3:1.
Which may or may not be useful. The comparison would be several extra parts of a switcher vs a linear and a slightly larger cap. I bet the larger cap wins out by most metrics: size, complexity, cost. -- Rick
On 10/3/2014 1:12 PM, Frnak McKenney wrote:
> 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...
I believe that is what is being done here. Modifying the existing circuit to even out the power draw. -- Rick
On Friday, October 3, 2014 10:54:50 AM UTC-4, Bill Sloman wrote:


> > A rechargeable alkaline battery (also known as alkaline rechargeable or=
rechargeable alkaline manganese (RAM)) is a type of alkaline battery that = is capable of recharging for repeated use. The first-generation rechargeabl= e alkaline technology was developed by Battery Technologies Inc in Canada a= nd licensed to Pure Energy, EnviroCell, Rayovac, and Grandcell. Subsequent = patent and advancements in technology have been introduced. The formats inc= lude AAA, AA, C, D, and snap-on 9-volt batteries. Rechargeable alkaline bat= teries are manufactured fully charged and have the ability to hold their ch= arge for years, longer than NiCd and NiMH batteries, which self-discharge.[= 1] Rechargeable alkaline batteries can have a high recharging efficiency an= d have less environmental impact than disposable cells.=20
>=20 >=20 >=20 > Do any of the broad-line distributors stock them? >
Do not knew, but there are vendors on both Amazon and Ebay that have them f= or sale. I have not looked for any since about 1974. They were available at that ti= me. But you can recharge a regular alkaline if the cell voltage is about 1= .3 volts or higher. We used to do that at work where flash lights were use= d a lot.=20
>=20 >=20 > Could you post a link to an item in on-line catalogue? >=20 >=20 >=20 > Too many "advances in technology" turn out to be remarkably hard to get y=
our hands on. I tend not to take them seriously until a broad-line distribu= tor keeps them in stock.
>=20 >=20 >=20 > Note that the Ph.D. is a remarkably specialised qualification - pretty mu=
ch at the sharp end of knowing more and more about less and less until you = know almost everything about about practically nothing.
>=20 >=20
Prety much the reason I studied up on vacuum systems. A good vacuum is clo= ser to practically nothing than anything.
>=20 > > > And you've snipped most of what I posted, without marking the snip. >=20 > >=20 >=20 > > Most people do not need posts with every thing perfect. I try to cut p=
osts down to a minimum , so people can read them quickly.
>=20 >=20 >=20 > Sure. But it's polite to mark your edits. And this time, you managed to i=
nclude my signature line, which did waste bandwidth.
>=20 >=20 >=20 > --=20 >=20 > Bill Sloman, Sydney
On Fri, 03 Oct 2014 12:47:30 -0500, tim <tim@seemywebsite.com> wrote:

>On Fri, 03 Oct 2014 13:17:31 -0400, krw wrote: > >> 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 > >I'm well aware of that. And capacitor size seemed to go by <some >constant> * C * V^2. You can roll the 1/2 into the constant or not, your >choice.
OK, the /2 *is* a constant if you're talking about ratios. It made no sense to make a point of it unless you didn't know.
>>>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? > >Volume went more or less by energy stored. Cost, except at the very high >voltage end, pretty much went by volume.
Yep. The energy is "stored" in the dielectric. Just asking if you'd looked at it that way. I've found cost is a higher order function, at least above some point (depending on package).
>It wouldn't be a bad thing to do over again, if someone out there has >time on there hands. > >>>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.
On Fri, 03 Oct 2014 00:27:33 -0700, Klaus Kragelund wrote:

> 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
I'm often a proponent of just doing the simple solution -- there's no reason to spend a month searching out the best epoxy to use to glue something to the ground if you can just set a rock on it. But in this case, you could buy a lot of switching supply for the price difference between 1200uF and 7000, even before you factor in the real and opportunity cost of using up that much space in the system. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
I have used the trick of running the cap at higher voltage to lower the physical size of the capacitor a couple of times. The CV product is pretty constant for the same package size

Cheers

Klaus
On 10/3/2014 2:35 PM, Tim Wescott wrote:
> On Fri, 03 Oct 2014 00:27:33 -0700, Klaus Kragelund wrote: > >> 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 > > I'm often a proponent of just doing the simple solution -- there's no > reason to spend a month searching out the best epoxy to use to glue > something to the ground if you can just set a rock on it. > > But in this case, you could buy a lot of switching supply for the price > difference between 1200uF and 7000, even before you factor in the real and > opportunity cost of using up that much space in the system.
Really? How much do you suppose the "opportunity cost" is for the OP's system? -- Rick
On Fri, 03 Oct 2014 13:35:09 -0500, Tim Wescott
<seemywebsite@myfooter.really> wrote:

>On Fri, 03 Oct 2014 00:27:33 -0700, Klaus Kragelund wrote: > >> 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 > >I'm often a proponent of just doing the simple solution -- there's no >reason to spend a month searching out the best epoxy to use to glue >something to the ground if you can just set a rock on it. > >But in this case, you could buy a lot of switching supply for the price >difference between 1200uF and 7000, even before you factor in the real and >opportunity cost of using up that much space in the system.
Several people make switcher blobs that are cheap drop-in replacements for a 7805. OP probably has a 5 volt regulator already. -- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com