charging a supercap

[andy257]

Hi all,

I am looking for some circuit ideas on how best to charge a super capacitor. Ive tried using large power resistors however they get slightly hot and are very bulky. I was wondering if there are any circuits that can remove the bulky resistor and go for some kind of charge control circuit.

Ideally i want to charge the supercap quickly but not at the expense of high power low value resistors.

all ideas welcome

 

[Hero999]

How much current do you need?

You could use a switch mode power supply, a power resistor is only 50% efficient.

 

[Nigel Goodwin]

You could use a switch mode power supply, a power resistor is only 50% efficient.

That's a very bizzare claim - perhaps you would care to explain your reasoning?.

 

[Hero999]

You don't seem convinced.

Imagine the efficiency when the capacitor is completely discharged and the power is connected via a resistor. The efficiency is near 0% as all of the power is dissipated in the resistor. As the capacitor continues charging the efficiency approaches 100% as does the charge. It turns out that because the same current flows through the resistor as the capacitor, the same energy amount of energy is dissipated in the resistor as is stored in the capacitor.

 

[Nigel Goodwin]

Right, I see what you mean, but as it's exponential, it's not going to be 50%.

 

[Hero999]

No, the overall efficiency is 50%.

If you know the maths you can probably work it out the energy transfer and loss using calculus.

Unfortunately my maths isn't good enough to work it out but I've simulated it in LTSpice and it it's true, the efficiency is 50%.

I used a 5R resistor and a 5F capacitor.

Over five time constants (125s) the energy supplied by the voltage source was 124.16J and the energy stored in the capacitor was 61.653J which is just under 49.656% and can be explained by the fact that I didn't allow the capacitor to fully charge which of course will never happen.

E = 0.5CV² = 0.5*5*5² = 62.5J

It doesn't matter what the value of capacitor and resistor is the maximum efficiency will be the same: 50%.

In reality the efficiency will be <50% because of the ESR and other losses in the capacity.

In order to charge the capacitor efficiently, you need a constant current supply.

 

[crutschow]

An series inductor will give you near 100% charging efficiency. But you have to open the circuit at the peak of the resonant cycle, as by adding a series diode (at which point the capacitor voltage is near twice the supply voltage).

You can calculate the energy loss from these equations without calculus: Energy stored in a capacitor is E = 1/2 CV², energy delivered from a voltage source is E = QV, and charge stored in a capacitor is Q = CV. Substituting the third into the second equation gives an energy transfer from the source to the capacitor of E = CV², which is twice the energy stored on the capacitor. Thus half of the energy is dissipated in the resistive charging circuit.

 

[andy257]

Well the system has one power supply. At switch on the power supply is not only charging the cap it will power all the remaining hardware. I have worked out i can probably allow 2.5A max initially. The power supply i am using has an overcurrent protection function and connecting the supercap across its terminals puts it into a fault condition.

Power supply output is 12v, so cap will charge to close this value. 12V / 2.5A = 4.8R @ 30W.

Thinking about this i dont think i can improve charging time without increasing max current.

 

[Hero999]

That's right, you can't increase the charging time without increasing the current.

If your power supple is overcurrent protected and it goes into constant current mode and doesn't shut down or blow as fuse when the limit is exceeded, you can safely connect the capacitor across the output but you might need a diode to prevent it from discharging back into the PSU.

 

[andy257]

The power supply has a max output of 8A. If you exceed 8.5A it shuts down and pulses the output until the fault is removed. I have calculated the hardware takes around 6A, so i only have 2.5A available to charge the capacitor. Although 2.5A is the intial charge, it will decrease as the capacitor charges.

The trouble with supercaps is i need a really low resistor to charge quickly, but charging quickly increases the power requirements of the resistor. Power resistors tend to be big and bulky.

 

[crutschow]

What is the capacitance of your supercap?

 

[andy257]

ive got a 5F cap

 

[Hero999]

I understand now, the capacitor can't be allowed to short circuit the supply because it will take everything else down.

You need a Schottky diode to bypass the current limiter when the capacitor is being used as a temporary power source.

 

[MrAl]

Hi,


I have to agree that a switching constant current power supply is probably the best. A regular constant current will probably eat up more power too although i havent gone through the calculations yet. A switcher should work nicely, and you can set the current to a level the capacitor can take without a problem.
BTW, how much current can that super cap take, like what is the maximum spec?

Yes, a resistor charging a cap from a constant voltage supply eats up 50 percent of the energy.
Another way to show this is to form the equation for i for all time and square that and multiply
by R, then integrate over all time. The result is C*V^2/2, the same as the cap.
Now we can see why there was a power dissipation problem in processors going up in clock frequency.

 

[Hero999]

A linear regulator will be no different to a resistor - 50% efficiency maximum.


It's funny how I know these things but I haven't been taught them and aren't always good enough at maths to work them out. I often don't know how I figure it out so I can't show my workings. This often used to be my downfall at college, some other people in my class seemed to be able to do the maths but didn't have the understanding, then there were the real geniuses who knew both.

Fortunately a constant current source is simpler so I can calculate it. Take a 1F capacitor, charged to 1V using a perfect 1A constant current linear regulator connected to a 1V supply:

[latex]t =\frac{CV}{I} = 1s\\
E_{IN} = VIt = 1 \times 1 \times 1 = 1J\\

E_(C) = \frac{1}{2}CV^2= \frac{1}{2}=1\times 1^2 = 0.5J[/latex]

This is assuming it's a perfect linear regulator with a dropout of 0V and a perfect capacitor, in reality it'll be worse than a resistor.

 

[The Electrician]

You say you want to charge it "quickly"; how quickly is "quickly"?

After you charge it, what will discharge it; will it be fully discharged? How often will you be doing this?'

Do you know what the ESR of your supercap is? Is there a maximum charging current spec?

 

[MrAl]

A linear regulator will be no different to a resistor - 50% efficiency maximum.


It's funny how I know these things but I haven't been taught them and aren't always good enough at maths to work them out. I often don't know how I figure it out so I can't show my workings. This often used to be my downfall at college, some other people in my class seemed to be able to do the maths but didn't have the understanding, then there were the real geniuses who knew both.

Fortunately a constant current source is simpler so I can calculate it. Take a 1F capacitor, charged to 1V using a perfect 1A constant current linear regulator connected to a 1V supply:

[latex]t =\frac{CV}{I} = 1s\\
E_{IN} = VIt = 1 \times 1 \times 1 = 1J\\

E_(C) = \frac{1}{2}CV^2= \frac{1}{2}=1\times 1^2 = 0.5J[/latex]

This is assuming it's a perfect linear regulator with a dropout of 0V and a perfect capacitor, in reality it'll be worse than a resistor.

Hi again,


That's good that you can understand these things one way or another.
I am not sure i understand what you are saying here though.
You seem to be saying that the constant current generator will provide the cap with
energy and also not waste much. The only problem though is that the constant
current generator has to get it's power from somewhere too, and that is usually
from a DC power supply of some sort with a constant voltage...hence we are back
to the constant DC voltage source problem again. That's why it has to be a
switcher too.
I may not have understood you correctly though.

 

[Hero999]

No, I'm saying a linear constant current source is no more efficient that a resistor because a 1F capacitor charged to 1V will store 0.5J, yet it will take 1J to charge from 1V source powering a linear constant current regulator.

Sorry, I probably didn't explain the formula I previously posted well enough.

Ein = Energy provided by the power supply = 1J
E(c) = Energy stored in the capacitor. = 0.5J

The size of the capacitor or current makes no difference, if you look at the graph of it charging, the area on the top side of the graph is still equal to the area on the bottom.

In practise, a linear regulator will be less efficient than a resistor because it will have a certain drop out voltage.

 

[andy257]

You say you want to charge it "quickly"; how quickly is "quickly"?

After you charge it, what will discharge it; will it be fully discharged? How often will you be doing this?'

Do you know what the ESR of your supercap is? Is there a maximum charging current spec?

Well if i can charge it to around 95% or more in 1 minute that would be great. Although the more i look at this i am prob willing to sacrifice charging time for a less bulky circuit.

The cap will run a PC. The PC only has to run for 10 seconds after power interupt. The PC can run from 12 - 9V. I have an LVD circuit to disconnect cap when below 9V. what i dont have is a circuit (other than a large resistor) to control the charging current of the capcitor at switch on.

 

[Hero999]

An series inductor will give you near 100% charging efficiency. But you have to open the circuit at the peak of the resonant cycle, as by adding a series diode (at which point the capacitor voltage is near twice the supply voltage).

I've just realised the problem with that is that you'd need a huge choke to limit the current to 2.5A.

I think a buck is the best option.