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

# capcitors for energy reserve

Started by ●October 2, 2014

Reply by ●October 2, 20142014-10-02

Reply by ●October 2, 20142014-10-02

On Thursday, October 2, 2014 2:53:44 PM UTC-7, Klaus Kragelund wrote:> On Thursday, October 2, 2014 11:47:08 PM UTC+2, Richard wrote: >=20 > > On Thursday, October 2, 2014 2:12:22 PM UTC-7, Tim Wescott wrote: >=20 > >=20 >=20 > > > On Thu, 02 Oct 2014 15:44:19 -0500, Tim Wescott wrote: >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > On Thu, 02 Oct 2014 13:21:05 -0700, Richard wrote: >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >> I have a circuit that draws 30mA except that there is a 150ms long=draw>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >> at 120mA. This occurs every 2s. >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >>=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >> I I am limited to 40mA current draw. I would like to provide rese=rve>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >> energy from a large capacitor. I have built a circuit that limits=the>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >> input current to 40mA. This runs the circuit and charges the cap. >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >>=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >> I have run caps up to 3300uF, the cap voltage drops to far and in =too>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >> short a time. Is there any methodology for calculating required >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >> capacitance? Any ideas on implementation beside my current limite=r>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >> plus cap? >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > Hark back to your sophomore electronics engineering courses: >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > dV/dt =3D i / C >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > For a constant current draw, you can change that to >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > deltaV / deltaT =3D i / C >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > Do some algebra, and you get >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > C =3D i * deltaT / deltaV >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > You've got 40mA in, 120mA out, which means you need 80mA out of the=cap.>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > You know what voltage drop you can stand -- plug that figure into t=he>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > above equation along with the rest of your knowns, and you should g=et a>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > minimum capacitance. >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > I must say, though, that if 3300uF isn't enough, then at those curr=ent>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > levels there's something else going on. If your calculations show =that>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > 3300uF is way more than plenty, then your problem may be that you'r=e>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > using aluminum electrolytics where you need to use something with >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > significantly lower ESR. >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > I'm doing something similar to you right now, except that my curren=t>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > draw is for a shorter period of time. I have my filter cap in my "=raw">=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > power line, ahead of my local regulator. With 5V on the "raw" side=and>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > 3.3V on the regulated side, I can accept a 1V drop during the 'on' =time,>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > as long as the cap charges back for the next cycle. >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > Note that I messed up my in-the-head math: 3300uF is, indeed, pretty =small=20>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > for this unless you charge the cap with a boost circuit, regulate aft=er=20>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > the cap, and accept large voltage swings. >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > --=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > Tim Wescott >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > Wescott Design Services >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > http://www.wescottdesign.com >=20 > >=20 >=20 > >=20 >=20 > >=20 >=20 > > 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 do= wn to ~5V so I am too close to the 5V minimum. Charge time is ~1.7s. The = input current is limited at 40mA.>=20 > >=20 >=20 > >=20 >=20 > >=20 >=20 > > I think this will work with a larger cap but sure seem inelegant. >=20 > >=20 >=20 > You have 10mA surplus during normal op, right? >=20 >=20 >=20 > Use that to charge a lower value capacitor with a boost converter >=20 >=20 >=20 > When you need the 180mA, draw that from a buck converter, so you can draw=all the energy out og the capacitor>=20 >=20 >=20 > Cheers >=20 >=20 >=20 > KlausInteresting idea Klaus. I could see the enable pin of the reg being trigge= red by the higher current draw... More parts though ($). Rich

Reply by ●October 2, 20142014-10-02

On Friday, October 3, 2014 12:01:18 AM UTC+2, Richard wrote:> On Thursday, October 2, 2014 2:53:44 PM UTC-7, Klaus Kragelund wrote: >=20 > > On Thursday, October 2, 2014 11:47:08 PM UTC+2, Richard wrote: >=20 > >=20 >=20 > > > On Thursday, October 2, 2014 2:12:22 PM UTC-7, Tim Wescott wrote: >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > On Thu, 02 Oct 2014 15:44:19 -0500, Tim Wescott wrote: >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > On Thu, 02 Oct 2014 13:21:05 -0700, Richard wrote: >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >> I have a circuit that draws 30mA except that there is a 150ms lo=ng draw>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >> at 120mA. This occurs every 2s. >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >>=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >> I I am limited to 40mA current draw. I would like to provide re=serve>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >> energy from a large capacitor. I have built a circuit that limi=ts the>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >> input current to 40mA. This runs the circuit and charges the ca=p.>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >>=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >> I have run caps up to 3300uF, the cap voltage drops to far and i=n too>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >> short a time. Is there any methodology for calculating required >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >> capacitance? Any ideas on implementation beside my current limi=ter>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >> plus cap? >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > Hark back to your sophomore electronics engineering courses: >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > dV/dt =3D i / C >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > For a constant current draw, you can change that to >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > deltaV / deltaT =3D i / C >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > Do some algebra, and you get >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > C =3D i * deltaT / deltaV >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > You've got 40mA in, 120mA out, which means you need 80mA out of t=he cap.>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > You know what voltage drop you can stand -- plug that figure into=the>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > above equation along with the rest of your knowns, and you should=get a>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > minimum capacitance. >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > I must say, though, that if 3300uF isn't enough, then at those cu=rrent>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > levels there's something else going on. If your calculations sho=w that>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > 3300uF is way more than plenty, then your problem may be that you='re>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > using aluminum electrolytics where you need to use something with >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > significantly lower ESR. >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > I'm doing something similar to you right now, except that my curr=ent>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > draw is for a shorter period of time. I have my filter cap in my="raw">=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > power line, ahead of my local regulator. With 5V on the "raw" si=de and>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > 3.3V on the regulated side, I can accept a 1V drop during the 'on=' time,>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > > as long as the cap charges back for the next cycle. >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > Note that I messed up my in-the-head math: 3300uF is, indeed, prett=y small=20>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > for this unless you charge the cap with a boost circuit, regulate a=fter=20>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > the cap, and accept large voltage swings. >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > --=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > Tim Wescott >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > Wescott Design Services >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > > http://www.wescottdesign.com >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > 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. Th= e input current is limited at 40mA.>=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > > I think this will work with a larger cap but sure seem inelegant. >=20 > >=20 >=20 > > >=20 >=20 > >=20 >=20 > > You have 10mA surplus during normal op, right? >=20 > >=20 >=20 > >=20 >=20 > >=20 >=20 > > Use that to charge a lower value capacitor with a boost converter >=20 > >=20 >=20 > >=20 >=20 > >=20 >=20 > > When you need the 180mA, draw that from a buck converter, so you can dr=aw all the energy out og the capacitor>=20 > >=20 >=20 > >=20 >=20 > >=20 >=20 > > Cheers >=20 > >=20 >=20 > >=20 >=20 > >=20 >=20 > > Klaus >=20 >=20 >=20 > Interesting idea Klaus. I could see the enable pin of the reg being trig=gered by the higher current draw... More parts though ($).>=20Well, you dont actually need to do anything Run the boost continously, draw all the current it can Have a zener clamp the cap And let the buck convert down to the needed voltage. The circuit will adjus= t to the current needed Cheers Klaus

Reply by ●October 2, 20142014-10-02

More optimal: Use a fet to feed the cap and boost. When you need to draw high current, turn off the fet and let the boost drain the cap Cheers Klaus

Reply by ●October 2, 20142014-10-02

On Friday, 3 October 2014 06:21:10 UTC+10, 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?Why not look at a NiMH (nickel metal hydride) battery? They'll deliver 120mA for 150msec without effort, and the voltage-versus-charge curve is exponential rather than linear. 9V is a very popular voltage level for battery packs, but the cells tend to be expensive. A single cell and a switching charge discharge system might work. A single 1.1V AAA cell costing a couple of dollars looks as if it might work. 850mAh is way more capacity than you need, but 150mA at 9V is 1.23A at 1.1V, and you need a tolerably big cell to sustain that current without getting warm. -- Bill Sloman, Sydney

Reply by ●October 2, 20142014-10-02

On 3/10/2014 8:28 AM, Bill Sloman wrote:> On Friday, 3 October 2014 06:21:10 UTC+10, 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 havebuilt 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 methodologyfor calculating required capacitance? Any ideas on implementation beside my current limiter plus cap?> > Why not look at a NiMH (nickel metal hydride) battery? They'll deliver 120mA for 150msec without effort, andthe voltage-versus-charge curve is exponential rather than linear.> > 9V is a very popular voltage level for battery packs, but the cells tend to be expensive. > > A single cell and a switching charge discharge system might work. > > A single 1.1V AAA cell costing a couple of dollars looks as if it might work. 850mAh is way more capacity than you need, but 150mA at 9V is 1.23A at 1.1V, and you need a tolerably big cell to sustain that current without getting warm.Farnell also stock a 3V Li-Mn coil cell http://au.element14.com/seiko-instruments/ms621fe-fl11e/coin- cell-lithium-5-5mah-3v/dp/1614634 but the internal resistance - 80R is much too high http://www.farnell.com/datasheets/321180.pdf For ten times the price you can buy something a little bigger with an acceptable internal resistance http://au.element14.com/seiko-instruments/ms621fe-fl11e/coin-cell-lithium-5-5mah-3v/dp/1614634 http://www.farnell.com/datasheets/1666646.pdf The trade journals have been talking up lithium iron phosphate batteries but I've no idea where you could buy any. -- Bill Sloman, Sydney

Reply by ●October 2, 20142014-10-02

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? > > Thanks > > Rich > >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.

Reply by ●October 2, 20142014-10-02

On Thursday, October 2, 2014 6:28:48 PM UTC-4, Bill Sloman wrote:> On Friday, 3 October 2014 06:21:10 UTC+10, Richard wrote: > > > I have a circuit that draws 30mA except that there is a 150ms long draw at 120mA. This occurs every 2s. > > > > > > Why not look at a NiMH (nickel metal hydride) battery? They'll deliver 120mA for 150msec without effort, and the voltage-versus-charge curve is exponential rather than linear. > > > > 9V is a very popular voltage level for battery packs, but the cells tend to be expensive. > > Bill Sloman, SydneyYou did not provide enough info on needed life , cost restaints, etc. But a 9 volt alkaline battery might be the answer. Alkaline batteries can be charged if the open circuit voltage is above about 1.2 volts per cell. Dan

Reply by ●October 3, 20142014-10-03

On Friday, 3 October 2014 10:43:11 UTC+10, dca...@krl.org wrote:> On Thursday, October 2, 2014 6:28:48 PM UTC-4, Bill Sloman wrote: > > On Friday, 3 October 2014 06:21:10 UTC+10, Richard wrote: > > > > > I have a circuit that draws 30mA except that there is a 150ms long draw at 120mA. This occurs every 2s. > > > Why not look at a NiMH (nickel metal hydride) battery? They'll deliver 120mA for 150msec without effort, and the voltage-versus-charge curve is exponential rather than linear. > > > > 9V is a very popular voltage level for battery packs, but the cells tend to be expensive. > > You did not provide enough info on needed life , cost restaints, etc. But a 9 volt alkaline battery might be the answer. Alkaline batteries can be charged if the open circuit voltage is above about 1.2 volts per cell.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. And you've snipped most of what I posted, without marking the snip. -- Bill Sloman, Sydney

Reply by ●October 3, 20142014-10-03

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 > > KlausAssuming 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. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com