Supercapacitors vs batteries. Capacity.

Buk___

10 kW
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Supercapacitors are sized in farads: batteries are sized in watt.hours. Can they be compared?

Farad is defined as Coulombs/Volts; and has equivalences as Amp.seconds/Volt and Watt.Seconds/Volt^2.

Watt.hours is Amp.hours*Volts. Amp.seconds is 3600*Ah. Watt.seconds is Ah*3600*volts.

A 3.2V 3300mAh Cell has a (nominal) capacity of 3.2V * 3.3Ah * 3600 = 38016 Watt.seconds.
Divide by 3.2V^2 = 3712.5 Farads???

If so, that's also 3.3Ah * 3600s /3.2V = 3712.5 Farads. Does that make any sense?
 
Well, that's about right. I'd compare the energy stored in each one but you get the same result. 38,016 Joules is what the battery holds, and to get the same energy from a capacitor at the same voltage, the capacitance needed is huge, as you calculated. Also keep in mind the voltage on the capacitor needs to go from full to zero in order to use up all the energy, while the battery maintains a useful voltage over the discharge cycle.

Since the energy in a capacitor is a function of V squared, using a higher voltage is generally beneficial.

It's why we don't see capacitor powered vehicles. Yet.
 
fechter said:
Well, that's about right. I'd compare the energy stored in each one but you get the same result. 38,016 Joules is what the battery holds, and to get the same energy from a capacitor at the same voltage, the capacitance needed is huge, as you calculated. Also keep in mind the voltage on the capacitor needs to go from full to zero in order to use up all the energy, while the battery maintains a useful voltage over the discharge cycle.

Since the energy in a capacitor is a function of V squared, using a higher voltage is generally beneficial.

It's why we don't see capacitor powered vehicles. Yet.

I want to model a battery using a capacitor because the simulator gets flaky when you send recharge current to a voltage source. It's only for simulation purposes, so whether the resulting capacitor is physically possible doesn't matter.

That said, whilst looking for the W.h to Farads conversion formula I did just come across these Maxell 3400F 2.85VDC Supercaps labelled as 3.4Wh

I calculated that to replace my £200 2kg 20cm x 13cm x 4cm 36V 10.4Ah pack with those,

I'd need 13s8p (37V 10.77Ah). 104 * $60 = $6,200 and 78cm v 48cm x 14cm & 54kg . :shock:

But boy would it charge fast (if I could find a source of enough amps :) ).
 
fechter said:
lso keep in mind the voltage on the capacitor needs to go from full to zero in order to use up all the energy, while the battery maintains a useful voltage over the discharge cycle.
This is the biggest issue with using capacitors in place of or in addition to battery.

Let's say your useful controller/motor voltage range (HVC-LVC) is from 52v HVC down to 44v LVC. That's a total of 8v range, out of the 52v. All of the watt-seconds the battery has are contained in that range.

But in a capacitor, only 8/52 (I think; might have forgotten to apply a curve?) of the watt-seconds are in that range, meaning most (44/52) of it's useful energy can never be accessed. However, you still have to input that energy into it before you can use the other 8/52 of it, so it's less efficient to use than a battery, as you can only get that 8/52 back out.

(unless you use some form of DC-DC converter, which adds bulk and wastes power, though probably not as much as just leaving all that other energy on the table).
'
Also remember that supercapacitors are generally lower voltage units, so you'd have to put several in series for a useful voltage range.

When you do this, the capacitances add inversely (not sure taht's the right word), rather than directly, so you have to then parallel more of them to get the same capacitance. Meaning, if you have to use 10 in series to get the voltage you're after, you then only have 1/10 the capacitance, so you now have to parallel 10 series strings of 10 capacitors to get the same capacitance you would have had with just one capacitor at the lower voltage.
 
amberwolf said:
Also remember ... you have to use 10 in series to get the voltage you're after, you then only have 1/10 the capacitance, so you now have to parallel 10 series strings of 10 capacitors to get the same capacitance you would have had with just one capacitor at the lower voltage.

I forgot. As if $6000/54kg wasn't enough penalty :)(
 
amberwolf said:
But in a capacitor, only 8/52 (I think; might have forgotten to apply a curve?) of the watt-seconds are in that range, meaning most (44/52) of it's useful energy can never be accessed.
For voltage, you are correct. But for energy, it's 1/2CV^2. So if you are going from 52 to 44 volts, you are using only 15% of the voltage range, but 30% of the energy range.
 
Buk___ said:
I want to model a battery using a capacitor because the simulator gets flaky when you send recharge current to a voltage source. It's only for simulation purposes, so whether the resulting capacitor is physically possible doesn't matter.

Could you simulate the pack with a zener diode? Might be closer to what actually happens.
 
fechter said:
Buk___ said:
I want to model a battery using a capacitor because the simulator gets flaky when you send recharge current to a voltage source. It's only for simulation purposes, so whether the resulting capacitor is physically possible doesn't matter.

Could you simulate the pack with a zener diode? Might be closer to what actually happens.

The simple answer to that is I don't know. My electronics fu isn't strong enough to begin to see how that would work.

I need a voltage source that reduces slowly over time as it supplies current and will accept current if connected to a voltage that is higher than its current voltage. That says capacitor to me. I'm not sure where a Zener fits it?
 
The capacitor will probably be better then. A zener diode will not source any current but will will act as a load when the applied voltage exceeds the zener voltage.
 
A zener has a threshold voltage below which it acts essentially open-circuit. Above that, it begins to conduct, "shorting" the voltage to hold it at that point, limiting current thru it via a resistor in series with it (across which voltage appears based on the current).

I think the simulation idea would be to have the zener in parallel with the voltage source that represents the battery, and voltage specified as just a hair above the voltage source, so it only conducts ("accepts" regen) when the system is creating a voltage higher than the voltage source.

It still wouldn't likely work the same way a battery does, with a range of voltages for the source.

I suspect you'd have to program your own simulation "part" for the battery to cause it to simulate the way it would really work, and I have no idea how to do that. :(
 
amberwolf said:
I suspect you'd have to program your own simulation "part" for the battery to cause it to simulate the way it would really work, and I have no idea how to do that. :(

The simulator I'm using, because it is simple enough that I didn't have to look anything up to construct my model, is also too simple to allow me to program bespoke components :(

I have installed several other more sophisticated simulators, a few of which do have built-in scripting capabilities that allow the construction of bespoke components, but they are -- for now at least-- beyond my ability to use for a variety of reasons.

One -- LTSpice -- has what (IMO) is probably the worst UI I ever tried to use: every action requires switching to a different 'mode'; and nothing is remembered, so you end up doing the same things over and over; laboriously.

Tina-TI seems quite powerful, but is 'free version limited' in very subtle ways that make it a pain to use.

GeckoCircuits: Incredibly powerful, fully programmable; and so sophisticated I think I'd need a degree course to work out how to get it to do the most simple of things :(


I found an SLA equivalent circuit which looks like it might be close enough for my purpose; and simple enough that I can construct it easily. I'll give that a go.
 
Buk___ said:
I want to model a battery using a capacitor because the simulator gets flaky when you send recharge current to a voltage source. It's only for simulation purposes, so whether the resulting capacitor is physically possible doesn't matter.
That works pretty well. Be sure to include a series resistor to simulate the ESR of the battery.
 
billvon said:
Buk___ said:
I want to model a battery using a capacitor because the simulator gets flaky when you send recharge current to a voltage source. It's only for simulation purposes, so whether the resulting capacitor is physically possible doesn't matter.
That works pretty well. Be sure to include a series resistor to simulate the ESR of the battery.

Thanks. I discovered what happens if you omit it! A huge GA surge when I connected a voltahe source to initially charge the battery. Makes for messy graphs :)

I'm currently using a 1uOhm because it make the initial charging of the capacitor almost instant (at simulation speed) without generating stupidly high current flows.

Do you have a notion of what value ESR would be realistic?
 
Buk___ said:
Do you have a notion of what value ESR would be realistic?
Depends completely on the battery. To get a very rough estimate, you can assume your battery will droop about 10% under full load, then calculate from there (via R=V/I.) For example, a 42V fully charged battery might droop to 38V under full load. If full load is 30 amps, then R=(4/30)=.2 ohms.

To get a better estimate, actually put a load on a battery and measure the current and voltage drop.
 
According to the Cycle Analyst v3.1, my EIG NMC pack (14s2p 40Ah "52v") is about 38 milliohms.

Some of the similar voltage 18650 packs people have reported 200-400 milliohms or higher.

I don't remember what ranges RC LiPo / etc tend to be.
 
amberwolf said:
I don't remember what ranges RC LiPo / etc tend to be.

Typically under 10 mOhm for the High C-rate stuff, but it also depends on what the SOC is. It's higher at 4.2 V resting than it is at 3.7 V resting. But it goes up again at low states of charge.
 
amberwolf said:
According to the Cycle Analyst v3.1, my EIG NMC pack (14s2p 40Ah "52v") is about 38 milliohms.

Some of the similar voltage 18650 packs people have reported 200-400 milliohms or higher.

I don't remember what ranges RC LiPo / etc tend to be.

I'm specifically interested in the 12V SLA I'm trying to model.

I found the formula ( Vnoload / Vloaded - 1 ) / Rload and then tested my SLA with 5W, 21W and 55W bulbs.

I'm calculating ESRs between 3mΩ and 30mΩ, which seems a bit too big a range, and then noticed the battery low symbol on my MM.

It's down to <6V (from 9V) so all the measurements are suspect. I'll go out later today and buy a new battery (or 3) and try again.
 
Sorry; I didn't mean to start an OT resistance thing. :oops:

I'll leave it at the below...to keep the thread back where it belongs. ;)
jonescg said:
Typically under 10 mOhm for the High C-rate stuff, but it also depends on what the SOC is.
From teh voltages you gave, I'd assume that's per-cell? Then for a 14s pack that would be around 140milliohm?
 
Yes that sounds about right. I find this calculator especially helpful, particularly for sizing precharge resistors:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html

I think supercapacitors are appealing only because they use fewer exotic materials. However, somewhat ironically, those that are able to deliver more charge in a useful voltage range tend to have ... exotic materials :)

There is a new kind of home storage product out now using supercaps, and it appears to take up 6x the space of a Li-ion battery. It's pretty expensive but not prohibitive, and most of the bulk is the DCDC converter to provide a useful voltage.

Makes LiFePO4 look pretty good...
 
jonescg said:
There is a new kind of home storage product out now using supercaps, and it appears to take up 6x the space of a Li-ion battery. It's pretty expensive but not prohibitive, and most of the bulk is the DCDC converter to provide a useful voltage.

According to this
European car maker PSA Peugeot Citroen has been using them in its cars since 2010 - supercapacitor maker Maxwell Technologies says more than a million vehicles now incorporate its products.
.

Typically lightweight reporting, but it would be interesting to read how Peugeot Citroen are using them in conjunction with Li-ion. I get that they are great for absorbing regen, and giving it back on demand for acceleration, but presumably that requires bi-directional DC-DC down/up conversion at a pretty high rate?

Anyone know if the split-pi DC-DC circuit is unique in that ability? If not, what else are they using?
 
Hillhater said:
the PSA supercap system.. http://articles.sae.org/8412/
And Mazda use a super capacitor system in their "iEloop" regen braking system.
http://articles.sae.org/11845/

That's really intriguing!

The text describes:
The alternator, although about the same weight as its 12-V predecessor, is a varying-voltage design that operates in a 12- to 25-V range. So the circuit also requires a dc-dc converter (which weighs 1.8 kg/4.0 lb) to provide 12-V power for electrical accessories. As soon as the driver lifts his foot off the accelerator, the regenerative mode begins, and the alternator uses the kinetic energy of deceleration to produce electricity at maximum possible voltage for efficiency. The “free” electricity goes through the dc-dc converter, and if there is any available electricity beyond the car’s electrical load, it goes to charge the 12-V battery.

Which implies that the SuperCap can accept 25V and they buck convert it down to 12V for the car systems and battery.

But AW pointed out above that if you stack (Super)caps serially to get higher voltage, you drop the capacity by the 1/(1/c1+1/C2+...) formula; and most SuperCaps are 2.somethingV to 3V; so how do they get/accept 25V?

Or ... can you charge 3V SCs from 25V SuperCap without harming them?

(And generally, charge low voltage parallel SC arrays from high(er) voltage supplies without harm???)

If so, that would make using them a whole bunch easier.
 
Buk___ said:
But AW pointed out above that if you stack (Super)caps serially to get higher voltage, you drop the capacity by the 1/(1/c1+1/C2+...) formula; and most SuperCaps are 2.somethingV to 3V; so how do they get/accept 25V?
?? By the method you just listed - you put them in series.
 
billvon said:
Buk___ said:
But AW pointed out above that if you stack (Super)caps serially to get higher voltage, you drop the capacity by the 1/(1/c1+1/C2+...) formula; and most SuperCaps are 2.somethingV to 3V; so how do they get/accept 25V?
?? By the method you just listed - you put them in series.

Okay. Let's do the math with the Maxell 3.2V 3.3Ah SCs from above.

To get 25V you need 8 in series. 3.3Ah translates to 3712F.

1/ ( 1/3712 * 8 ) = 464.0625F which translates back to 0.4125Ah. Is that enough to be useful?

0.4125Ah / 320A (typical small car starter current draw) = 4.64 seconds of power?
 
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