# battery in parallel with capacitor capacity needed

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#### yassin.kraouch Hi,

i have a 3.3V battery with a capacity of 10Ah, and i want to connect in parallel to this battery a capacitor in order to charge the capacity, i have only 0.027 ms to charge this capacity, if C=TxI so C = 0.027 ms x 10 A = 465uF, is my analysis true ? or the capacity can absorbe more then 10 A ? can i put a capacity more then 465uF ???? please i have ambiguity and i don't know to resolve this problem,

#### jpanhalt We don't know what type of battery you have. If it is a 10 Ah lithium, in all likelihood, it will be able to provide far more than 10A for short periods. The maximum current is often stated as a multiple of the capacity (C), such as 10C, 20C, 25C, etc. The lithium polymer batteries I use, for example, can provide 30C discharge currents. Thus a 2.2 Ah battery can provide 30X2.2A = 66A for short periods (e.g., 30 seconds) without harm to the battery.

John

#### yassin.kraouch yes i use a lithium battery, but what is the current that the capacity in parallel will absorbe ?

---------- Post added at 13:00 ---------- Previous post was at 12:56 ----------

sorry i have a mistake the current is 10 A not 15 ==>C=0.027 ms x10 = 270uF.
in other word can i in 0.027 ms charge acapacity more the capacity that i calaculate (270uF) ?

#### jpanhalt Capacitors have an equivalent resistance and inductance that limit their maximum charge and discharge rates. Also, if it is going to be done repetitively, you need to consider heating, and if the capacitor has leads, remember those leads are often plated soft steel, not copper, so lead resistance can become a factor. One would have to know more about the capacitor you are planning to use to even guess at its maximum charge rate.

I thought your original question had to do with limits of what the battery could provide, as it seemed you had used the capacity figure (Ah) for its ability to provide current.

John

#### andre_luis

##### Super Moderator
Staff member Q = C . V
I.T = C . V
10 . 0.000027 = C . 3.3
C = 8.8uF

However, I´m not sure if that amount of current could damage capacitor.
Take a search with one is the proper type to perform that ( electrolictic, Polypropilen, Polyester ).
Maybe you will need to assembly a bank in a set array.

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#### yassin.kraouch my question is : can i charge a capacity (full charge ) that have a value >270uF, and connected to a battery 10Ah in 0.027 ms?

---------- Post added at 13:29 ---------- Previous post was at 13:22 ----------

However, I´m not sure if that amount of current could damage capacitor.

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current does not damage the capacitor, the voltage only if exceded will domage the capacitor

#### andre_luis

##### Super Moderator
Staff member Regarding above calculus, seems to be too large to fit at avaliable time.

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#### jpanhalt Notice the circuit. It ignores inductive effects. You can try substituting various values for R. Again, without know anything about the characteristics of your capacitor, we can only tell you what some of the things to consider are.

John

#### andre_luis

##### Super Moderator
Staff member ...current does not damage the capacitor, the voltage only if exceded will domage the capacitor...

Each capacitor type, uses different kind of electrolit substract, and each one have its proper current limitations.

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#### yassin.kraouch the question is in other words : can the battery supply more the 10 h current to the capacitor ? in order to fast charging it ?

#### umesh49

##### Full Member level 4 the question is in other words : can the battery supply more the 10 h current to the capacitor ? in order to fast charging it ?
For a very short time, it should be able to supply, but you cant predict the charging time due to;
1. Internal and terminal resistance of battery
2. You will actually will not get 3.3V when battery will be supplying so much current. It might go less than 2.8V for that short duration.

#### jpanhalt the question is in other words : can the battery supply more the 10 h current to the capacitor ? in order to fast charging it ?

That was answered in Post #2. You are confusing amp-hour capacity with current. (I think you mean 10A, not 10h.)

Batteries have an equivalent resistance too. For some batteries, that resistance is relatively high and they can only deliver low currents. Other battery types have very low internal resistance and can deliver large currents, LiPo, LiFePO4, and NiCd can usually deliver substantial currents

Edit:

Here are some additional reference. The first is Wikipedia. Search on "inductance."

The second is from a random site on coil guns. I did not run the applet, but at least it is claimed to consider inductance.

Finally, this site gives a practical discussion and shows actual tests of how parasitic inductance can affect charging rate.

John

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#### KerimF From the way you present the problem, it seems you like to charge the capacitor manually using a heavy wire to decrease its resistance (though there are already two internal series resistances; one of the battery and the other of the capacitor). If the charging will not be done manually (and heavy wires) then you will need to consider the resistance of the device that will pass the charging current.

#### yassin.kraouch From the way you present the problem, it seems you like to charge the capacitor manually using a heavy wire to decrease its resistance (though there are already two internal series resistances; one of the battery and the other of the capacitor). If the charging will not be done manually (and heavy wires) then you will need to consider the resistance of the device that will pass the charging current.

the problem is can i charge a 470uF or a value greater than 470 ( FULL CHARGE), with this battery (3.3V , 10Ah)

#### KerimF Of course you can if there is no maximum limit for the charging time... Otherwise it may be impossible... since the following formula should be satisfied
V = I * T / C
(assuming I, the charging current, is constant. Actually it will decrease exponentially with the time constant RC where R is the sum of all series resistances between the ideal voltage source of the battery 3.3V and the ideal capacitance C of the capacitor).

To get some numbers, let us assume the initial current I_max is 9.9A
R = V_bat / I_max = 3.3 / 9.9 = 1/3 Ohm (which is the sum of the internal resistances in the loop)

The time constant:
R*C = 1/3 * 480e-6 = 160us

So after 3RC = 160 * 3 = 480us , the capacitor reaches 95% of 3.3V
at T = 5*RC = 800us , the capacitor reaches 3.278V

In general:
Vc % = 100 * [1 - e^(T/RC)]

Kerim

From above if we accept that 3.278V is acceptable then if we know the time, we can calculate C
T = 5*RC ===> C = T/5/R

Only you, who can know how low is R by knowing/measuring the initial I_max
R = 3.3 / I_max

Let us remember again that R = R_bat + R_wire + R_c

R_wire can be made low using thick wires
R_c can be made low by using a high voltage capacitor (this means it expects higher initial current hence its series resistance should be low... just a guess :wink: )
R_bat is fixed for each type. If the battery has a datasheet, an estimated value of its internal fresh resistance would likely be given.

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#### yassin.kraouch Of course you can if there is no maximum limit for the charging time... Otherwise it may be impossible... since the following formula should be satisfied
V = I * T / C
(assuming I, the charging current, is constant. Actually it will decrease exponentially with the time constant RC where R is the sum of all series resistances between the ideal voltage source of the battery 3.3V and the ideal capacitance C of the capacitor).

and what is the time required to charge this capacity ? the battery can supply more than 10 A ?

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