# capcitors for energy reserve

Started by October 2, 2014
```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

```
```On Thursday, October 2, 2014 1:21:10 PM UTC-7, 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.

What voltage?

>
> 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?

3300uF is around 0.003C or 3mA equivalent.  It won't last very long on 120mA draw.  1F supercap is not too expensive, if you can limit it to 2.7V each.
```
```On Thursday, October 2, 2014 1:34:54 PM UTC-7, edward....@gmail.com wrote:
> On Thursday, October 2, 2014 1:21:10 PM UTC-7, 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.
>
>
>
> What voltage?
>
>
>
> >
>
> > 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?
>
>
>
> 3300uF is around 0.003C or 3mA equivalent.  It won't last very long on 120mA draw.  1F supercap is not too expensive, if you can limit it to 2.7V each.

Yeah, that is an issue, it is 9V.

Rich
```
```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.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
```
```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?
>A 1 Farad capacitor supplying a 1 Amp load will drop 1 Volt in 1 second.
So, if you want to supply a 120 mA load for 150 ms, that is
.12 A * .15 sec = .018 C.  If you want to limit the voltage drop
to, say, .1 V, then the calculation is (.018 C / 0.1 V) = C in farads,
or .18 F  That is 180,000 uF, which is a pretty big cap.  Possibly
a stack of a few supercaps would do it.

Jon
```
```On Thursday, October 2, 2014 1:39:35 PM UTC-7, Richard wrote:
> On Thursday, October 2, 2014 1:34:54 PM UTC-7, edward....@gmail.com wrote:
>
> > On Thursday, October 2, 2014 1:21:10 PM UTC-7, Richard w>
>
> > > 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.
>
> > What voltage?
>
> > > 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?
>
> > 3300uF is around 0.003C or 3mA equivalent.  It won't last very long on 120mA draw.  1F supercap is not too expensive, if you can limit it to 2.7V each.
>
> Yeah, that is an issue, it is 9V.

You can use 4 in series and shunt regulate them.  However, you power budget is very tight and can't afford too much power lost.  Any other way to reduce power usage in your circuit?

```
```On Thu, 02 Oct 2014 15:47:52 -0500, Jon Elson wrote:

> 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?
>>A 1 Farad capacitor supplying a 1 Amp load will drop 1 Volt in 1 second.
> So, if you want to supply a 120 mA load for 150 ms, that is .12 A * .15
> sec = .018 C.  If you want to limit the voltage drop to, say, .1 V, then
> the calculation is (.018 C / 0.1 V) = C in farads,
> or .18 F  That is 180,000 uF, which is a pretty big cap.  Possibly a
> stack of a few supercaps would do it.

Or learn how to accept a 6V drop.

Whatever you do, opening up the amount of voltage drop you can accept will

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
```
```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
```
```On 02/10/2014 21:21, 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
>
>
Look up CAPx expensive but can source loads for miliseconds... used in
GSM stuff
```
```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.

Rich
```