relationship between battery voltage and discharge rate

[guywithaquestion]

i guess my question roots from me not understanding what a load represents to the battery.

is a 4V battery able to discharge at a higher rate than a 2V battery?

or does it completely depend on the load which will always extract current at the same rate regardless of the battery voltage.

in my current understanding i feel as if a higher voltage battery will allow for faster discharge rates if needed.

its difficult for me to test this with a potentiostat because in my understanding it can always force negative currents on discharge (GCPL mode lets say) which can be higher than the maximum discharge rate a battery can deliver to a load.

so to clarify my question. if the load is huge (a fridge). which battery will discharge faster if connected to the fridge circuit, given they contain the same energy and same internal resistance, a 4V battery or a 2V battery.

could someone please clarify this for me. i am very confused.

 

[IRstuff]

is a 4V battery able to discharge at a higher rate than a 2V battery?

> Not necessarily, but I = V/R, with all things being equal, there are constraints imposed by parasitic resistances, and overall capacity, since many batteries' ampacity ratings are constrained by the discharge rate

or does it completely depend on the load which will always extract current at the same rate regardless of the battery voltage.

> I = V/R

in my current understanding i feel as if a higher voltage battery will allow for faster discharge rates if needed.

> I = V/R

its difficult for me to test this with a potentiostat because in my understanding it can always force negative currents on discharge (GCPL mode lets say) which can be higher than the maximum discharge rate a battery can deliver to a load.

> GCPL mode??? negative current means that you are charging the battery, not discharging

so to clarify my question. if the load is huge (a fridge). which battery will discharge faster if connected to the fridge circuit, given they contain the same energy and same internal resistance, a 4V battery or a 2V battery.

> If the "fridge" is 12V or 24V input, then neither is appropriate and will discharge without doing much of anything related to keeping things cold. Your battery MUST meet BOTH voltage and current requirements; this is not a suggestion, it's a requirement.

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[guywithaquestion]

thank you for your response.

as a chemist, i think i still fail to understand what could be common knowledge for en engineer so please explain. what generates current flow when a battery is connected to an external circuit (this would be the load right?)? Lets say a battery connected to a light bulb. By itself, the light bulb is off (wires are cool). now i connect a battery to it. the light bulb lights up. why?

My understanding of battery chemistry tells me that the cathode discharges spontaneously once the circuit is closed based on Free Gibbs energy of the reaction which would occur between anode and cathode. Chemically speaking, there is reduction occurring at the cathode and oxidation occurring at the anode with all electrons moving through the external circuit through the current collector instead of transferred directly between the redox couple. The fact that the circuit is closed through the light bulb causes these electrons to heat up the coils inside the bulb and due to that heat and the gas inside the bulb, light is emitted.

What i fail to understand is the rate at which this discharge occurs. Does the rate of current flow depend on the light bulb or on the battery? This is assuming there are no additional regulators of any kind. My understanding is that the rate depends on the battery chemistry. So based on the active materials chosen, a nominal voltage is established and based on the resistance through the external and internal circuits current will flow at a certain rate.

If I carry this logic through it would always mean that a higher voltage battery (discounting differences in resistance) will always be able to spontaneously push higher currents, so higher rates of discharge. Carried a step forward it would mean that Energy Densities based on higher voltages (E=V x Q) are more desirable (from a rate point of view at least) than Energy Densities based on higher capacities.

Am I right with all this?

 

[IRstuff]

It's both.

Again, I refer you to Ohm's Law I = V/R; the current is equal to the electric potential divided by the total resistance. Therefore, for a fixed resistance, which is the load, the more potential the higher the current. But, there are constraints on how much potential or current any given system can tolerate. The total resistance would be a combination of the actual load plus parasitic resistances like the series resistance of the battery itself. There is a general trade for a given battery volume that a high voltage output will mean a higher series resistance and the higher current draw will drain the battery sooner; there is no free lunch anywhere to be found here.

The energy density basically tells you that power*time/volume = V * I * t / volume says that for a fixed volume and energy density, if you double the voltage, you can get double the current, but the battery will drain 4 times faster. Moreover, in many circuits, higher currents incur higher parasitics as well, which result in increased parasitic losses that detract from the available energy. Again, there is no free lunch; there is no general rule that says 4V is better than 2V, everything depends on the overall system requirements and constraints.

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[guywithaquestion]

thank you for the response.

could you please explain P x t/vol = V x I x t/vol ? This is equivalent to energy, E = V x I x t, which is simply the formula for Energy = Voltage x Capacity, (Capacity, Q = I x t, units of Ah, etc).

So if I take E = V x I x t and I double the voltage, then E = 2V x I x t, which means that i double the Energy, not the current. As a matter of fact to obtain double the energy, the current (I) has to remain the same (if i double the voltage), and the time (t) has to remains the same as well.

I thought this was actually an advantage because from my limited understanding of how circuits work, voltage (V) does not incur a penalty in series, but current (I) does due to higher resistance the current experiences when it travels through a longer resistor (more wiring). So in my current understanding it always advantageous to increase the energy density by increasing the voltage, NOT the capacity (Q=I x t) so the losses due to resistances observed when transmitting a higher amount of current are not incurred. Do I understand this properly?

What this basically means is that moving amount X1 of current at a high speed Y1 through a resistor (in this example a higher voltage will provide more pressure for current to move through a resistor so if the resistor is fixed, higher voltage will mean current moves faster through it) incurs fewer losses than moving a larger amount of current X2 at a lower speed Y2, where X1 x Y1 = X2 x Y2, such that the Energy delivered is the same at the end in either case. Why is that? Wouldn't losses be incurred through a resistor, if current moves faster, just as when more of it moves through the resistor, albeit at a slower pace?

In addition, i don't understand why a higher voltage battery would nominally experience higher internal resistances than a lower voltage battery of the same energy density if operating at the same C rate, aside from specific examples, where specific high voltage battery chemistries have higher internal resistances than specific lower voltage battery chemistries. Based on Ohm's Law, V = I x R so if V increases, that could be caused by an increase in R, but not necessarily.

 

[IRstuff]

There are lots of resources on the web. I suggest that you take advantage of them

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[guywithaquestion]

so i ask questions on a board i find on the web and i get sent back on the web to look for more resources. thx for trying to help i guess.

 

[IRstuff]

I refer you to the subtitle on this website:

INTELLIGENT WORK FORUMS FOR ENGINEERING PROFESSIONALS

This site is intended for for work-related questions from engineers. You signed in ostensibly as a chemical engineer or professional, yet your questions are at the high school level.

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[guywithaquestion]

cmon guy... take it easy. if they are high school level, answer them.

im a chemist by trade.