hi Seabreezers,
I recently purchased a replacement battery for a Garmin GPS, Garmin no longer make the battery and suggested I try a third party.
I found a seller here in Australia that listed a replacement for the Garmin product with matching part numbers and the image on their website showed the battery marked with the same voltage as mine, 8.4V
however, the battery I received from them has the correct part number but is labelled 7.4VDC instead of my original 8.4V
Is it ok to use the replacement battery?
My GPS is hard wired to a 12v system and only uses battery when the master switch is off and I assume to run the internal clock when everything is shut down.
The battery charges while in use with the master switched on.
here's a photo of the 2 batteries,
thanks and cheers, Doug.
A typical lithium ion battery is 4.2 volts when fully charged and 3.7 volts when 20% charged. It is quite common to quote the 20% charged voltage since it is recommended practice not to use a lithium ion battery to lower than 20% charge.
The batteries in your photo will have two cells and one of the manufacturers has stated the fully charged voltage, 2 x 4.2 = 8.4 volts, while the other manufacturer has stated the the 80% discharged voltage, 2 x 3.7 = 7.4 volts, so they are the same thing.
You're good to go.
Lithium Ion only comes in increments of the cell voltages, so you can be pretty sure that "7.4v" is the same as "8.4v", it's just that one will be the maximum charged voltage and one the nominal voltage.
learn.adafruit.com/li-ion-and-lipoly-batteries/voltages
I'm not a chemistry quiz, but I thought an electro chemical cell voltage is the same regardless (in Lion's case - 3.7V). It's related to the energy of the electron, which is constant for a given element. The apparent voltage change is caused by the increasing serial resistance of the battery. This forms a voltage (potential) divider with the load resistance (ie, the GPS unit, multimeter, internal battery losses etc). Am I wrong here ?
As you said FN, 8.4V is the charge voltage. Probably the result of warranty department lobbying marketing department to change it from 7.4 to 8.4V
Select to expand quotejn1 said..
I'm not a chemistry quiz, but I thought an electro chemical cell voltage is the same regardless (in Lion's case - 3.7V). It's related to the energy of the electron, which is constant for a given element. The apparent voltage change is caused by the increasing serial resistance of the battery. This forms a voltage (potential) divider with the load resistance (ie, the GPS unit, multimeter, internal battery losses etc). Am I wrong here ?
That hasn't been my experience with Lithium batteries. If you were correct then you wouldn't be able to tell the state of charge with no load, but you can, just by measuring the voltage. The voltage doesn't drop as much between full and empty as most other chemistries, that is, when it's almost empty you normally can't tell that it's getting low just by the performance of your device. If you notice that it's not working as well as it used to, you've normally got seconds before the low voltage protection circuits will just cut out.
Also, the internal resistance of Lithium batteries is much lower than other chemistries, so the voltage sag under load is almost nothing. This also goes against your voltage divider theory.
They're also much lighter for a given energy rating, I could go on...
Lithium chemistry is awesome 
Ah, I forget about that. There is an active circuit in series with the battery. Could this circuit distort your observations ?. The battery's active circuit could be keeping source resistance (as seen by the load) constant, which would support your observations.
Select to expand quotejn1 said..
I'm not a chemistry quiz, but I thought an electro chemical cell voltage is the same regardless (in Lion's case - 3.7V). It's related to the energy of the electron, which is constant for a given element. The apparent voltage change is caused by the increasing serial resistance of the battery. This forms a voltage (potential) divider with the load resistance (ie, the GPS unit, multimeter, internal battery losses etc). Am I wrong here ?
As you said FN, 8.4V is the charge voltage. Probably the result of warranty department lobbying marketing department to change it from 7.4 to 8.4V
Battery internal resistance certainly creates a voltage drop across the battery terminals when there is a load on the battery, but that doesn't explain the difference in the measured voltage of a 100% charged and 20 % charged battery when there is no load attached. A multimeter has an internal resistance of around 10 mega-ohms, so the amount of current flowing when taking a voltage reading is infinitesimally small and would have no discernible impact on the voltage measurement (which of course is the whole point of a good quality multimeter having such a high internal resistance).
I haven't answered your question, but just saying it can't be because of internal resistance since there is no current flowing when the 8.4 and 7.4 volts are measured. I'm guessing the idea we were all taught at high school of different metals having a certain potential difference is just the ideal case of two clean strips of metal with an ideal electrolytic solution, but in practice, that value changes based on a bunch of factors such as concentrations of ions, etc. as the battery discharges.
I work with lithium batteries a lot with my RC planes and helis and the no load voltage measurement is very accurate for determining the state of charge.
A good explanation Harrow. I think I only have a basic understanding of batteries (a bit of knowledge is dangerous), but I would view internal resistance as both series and parallel resistances. However, I'm not clear how the parallel resistance effect would work. I spent half a day trying to find this answer earlier in the week (and my googling skills are not great), but to no avail. But, the L-ion battery should be able to be simplified to a Thevevin equivalent circuit to see these effects (actually, I'll try googling Thevevin equivalent models - watch this space..). To complicate matters (as Nebbian pointed out), there is an active circuit in series with the battery.
I probably should not have mentioned a multi-meter as a load, but in some other chemistries, you do get significant internal resistances to see the potential divider effect with a multimeter.
Sorry for this technical hijack of the thread
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Update: Nebbian and Harrow. You are right. I found this article:
mdpi-res.com/d_attachment/energies/energies-11-02305/article_deploy/energies-11-02305.pdf
It's not L-ion chemistry (but it will be similar to L-ion), and there are lots of omissions from their simulation. But according to their model, the voltage source does change during battery charge/discharge (See Figure 4 and Figure 6). I've always assumed this does not. Therefore, I'm wrong. Although, this article does not explain why the voltage source in their model changes over the state of charge. But, I will take it as God says.
Select to expand quotejn1 said..
Although, this article does not explain why the voltage source in their model changes over the state of charge.
I asked Chat GPT, and it told me this...
"A battery produces electricity through electrochemical reactions between its electrodes and electrolyte. As the battery discharges, these reactions consume the active materials, and the concentration of the reactants decreases. This leads to a reduction in the electrochemical potential difference between the electrodes, which is the voltage of the battery."
I know it's true because I see it myself, but I'm still stuck like you are..... if there is no current, even if there are difference concentrations, why isn't the 'no load' voltage the same until the last of the reactants is gone. I can't find a decent detailed explanation.
Maybe we need a chemist to chime in... Mr Milk, do you have anything to say on the subject?
Less negatively charged particles, or electrons at the the Negative post and more at the Positive post, as the battery discharges, results in a reducing potential difference (voltage) between the battery terminals.
The term "voltage" in a battery refers to the difference in electric potential between the positive and negative terminals of a battery. A greater difference in potential results in a greater voltage.
Electric potential means the difference in charge between two points--in this case, the two terminals of a battery. One is positively charged, and the other is negatively charged. A negative charge simply means that there is an excess of negatively charged particles, or electrons on the terminal, while a positively charged terminal has a lack of those electrons. Physical separation of the two terminals prevents the electrons from traveling from the negatively charged terminal to the positively charged one.
Select to expand quoteHarrow said..jn1 said..
Although, this article does not explain why the voltage source in their model changes over the state of charge.
I asked Chat GPT, and it told me this...
"A battery produces electricity through electrochemical reactions between its electrodes and electrolyte. As the battery discharges, these reactions consume the active materials, and the concentration of the reactants decreases. This leads to a reduction in the electrochemical potential difference between the electrodes, which is the voltage of the battery."
I know it's true because I see it myself, but I'm still stuck like you are..... if there is no current, even if there are difference concentrations, why isn't the 'no load' voltage the same until the last of the reactants is gone. I can't find a decent detailed explanation.
Maybe we need a chemist to chime in... Mr Milk, do you have anything to say on the subject?
LOL. Lucky you didn't ask Bing AI. You would have got a snotty reply 
Basically localised depletion of the electrolyte and electrodes. The voltage is the average. Local depletion explains why a lead acid battery is able to partially recharge itself when left to rest.
Select to expand quoteCrusoe said..
Less negatively charged particles, or electrons at the the Negative post and more at the Positive post, as the battery discharges, results in a reducing potential difference (voltage) between the battery terminals.
The term "voltage" in a battery refers to the difference in electric potential between the positive and negative terminals of a battery. A greater difference in potential results in a greater voltage.
Electric potential means the difference in charge between two points--in this case, the two terminals of a battery. One is positively charged, and the other is negatively charged. A negative charge simply means that there is an excess of negatively charged particles, or electrons on the terminal, while a positively charged terminal has a lack of those electrons. Physical separation of the two terminals prevents the electrons from traveling from the negatively charged terminal to the positively charged one.
Then you could ask why isn't the relationship linear as the charge depletes? Wish I could find an comprehensive explanation somewhere.
It basically is linear, except for the extremes (see graph below). The voltage is driven by how many free ions are available in the electrolytic solution (= resistance as they decrease) as well as the resistance on the terminals when all those ions gunk it up. Once that gets terminal (badoom) the voltage
drop will become more extreme and no longer linearthe voltage has to change because of the ionic loss. Every battery is fundamentally the same, whether it is a salt water electrolyte or a fancy lithium one.
learn.adafruit.com/assets/979
I acknowledge that the above is a pretty average explanation of batteries... but what i'm getting at is that voltage, current and resistance are inter-related, a battery does its best to deliver a constant voltage using a sloppy chemical reaction, but there is only a small window in each discharge cycle where it can do that.
This sorta answers one of your earlier questions about voltage with no load.
physics.stackexchange.com/questions/126971/what-maintains-constant-voltage-in-a-battery
Select to expand quoteCH3MTR4IL5 said..
This sorta answers one of your earlier questions about voltage with no load.
physics.stackexchange.com/questions/126971/what-maintains-constant-voltage-in-a-battery
The explanation in that link is the problem I was trying to solve, not the answer. Sure, for a circuit where this is a current flowing, there is a relationship between voltage, current, and resistance. That link explains why the no-load voltage is constant, but it doesn't explain why the 'constant' voltage is less for a partially discharged battery that has no load connected to it. And the answer isn't just 'less electron charge means less voltage'...that's a symptom. Why is there less charge? This isn't a capacitor slowly being discharged. The battery is a combination of elements, so for a 50% discharged battery, there is still around half of the original reactant remaining and those reactants have a known and fixed electronegativity which gives the level of no-load voltage required to achieve the point of electrochemical stability, so why don't the chemical reactions continue inside the battery until the no-load voltage is at the original level? I assume it's got something to do with concentration of ions, waste products, etc, but I've been looking for more than general arm waving, I'd like to see the actual chemical equations and mathematical calculations that give the reduced voltage level. I did find a half-reasonable explanation of the cause somewhere, but it was still a little vague. Again, when there is current flowing, I can understand that the internal resistance of a partially discharged battery is higher due to several factors and that this degrades the performance of a partially discharged battery under load, but it's the drop in the no-load voltage that I'd like to properly understand.
I guess I am not understanding your basic supposition.
Internal resistance increases as a battery discharges, and the battery voltage drops. Not proportionally like a capacitor, but in a non-linear fashion. You want to calculate this reduced voltage level? Why?
(genuinely interested, not being argumentative!)
No he doesn't need to calculate it. Just wants to know why the equilibrium no current voltage is lower. So do I. If you have a tank of water with a huge low resistance hole it will empty quicker than one with a small half clogged hole. But the equilibrium for both is empty.
Select to expand quoteCH3MTR4IL5 said..
I guess I am not understanding your basic supposition.
Internal resistance increases as a battery discharges, and the battery voltage drops. Not proportionally like a capacitor, but in a non-linear fashion. You want to calculate this reduced voltage level? Why?
(genuinely interested, not being argumentative!)
Sure, internal resistance increases as the battery discharges, which impacts battery terminal volts under load, but I'm talking about no-load conditions.
My question is, therefore, what is the exact precise reason that the no-load battery terminal voltage is reduced for a partially discharged battery? The battery materials have the same electronegativity, so until they are fully consumed, why don't they produce the same battery terminal voltage under no-load conditions? (Even though they won't deliver the same 'under-load' performance due to increased internal resistance.)
Why do I want to know this..... because.
(I have an idea what the answer might be, which I'll post later, but I can't find confirmation of it anywhere.)Select to expand quoteIan K said..
No he doesn't need to calculate it. Just wants to know why the equilibrium no current voltage is lower. So do I. If you have a tank of water with a huge low resistance hole it will empty quicker than one with a small half clogged hole. But the equilibrium for both is empty.
I've been waiting for you to chime in on this one....preferably with the answer! 
Select to expand quoteHarrow said..
Ao why don't the chemical reactions continue inside the battery until the no-load voltage is at the original level? I assume it's got something to do with concentration of ions, waste products, etc, but I've been looking for more than general arm waving, I'd like to see the actual chemical equations and mathematical calculations that give the reduced voltage level..
The chemical reaction within the battery causes a buildup of opposite charges between the terminals, creating a voltage difference, whether the battery is connected or not. When the battery is disconnected, the charge builds up until it is sufficient to stop those chemical reactions (an equilibrium state which is our no-load voltage). The battery is not dead when its disconnected, it is in an equilibrium state, and that equilibrium is impacted by the internal resistance of the battery (I would think surface resistance on the electrodes being the main player).
eg a fully charged battery at equilibrium has a higher voltage than a partially discharged battery at equilibrium. when we apply a load to the battery, other factors such as mass transport loss and activation overpotential will reduce the voltage further.
In terms of actual chemical equations and mathematics calculations, you're asking a lot of a watersports forum. If this was an easy calculation, then battery optimisation would be a much simpler affair. It is inherently nonlinear. Sciencedirect has some good articles on battery modelling, and if you have matlab then you can download some simulations from mathworks. Much easier to do empirically than to calculate it!
Select to expand quoteIan K said..
No he doesn't need to calculate it. Just wants to know why the equilibrium no current voltage is lower. So do I. If you have a tank of water with a huge low resistance hole it will empty quicker than one with a small half clogged hole. But the equilibrium for both is empty.
(My comment above drafted before the most recent posts btw so sorry if it doesn't exactly answer)
The above is not the same - thats not an equilibrium state, its a discharge state. If you have a tank of water with a hole in it, it is discharging until it is empty (or for a battery, no chemical reaction).
This would be closer to reverse osmosis - when there is a current, the current overcomes osmotic pressure - when the current is removed, the system will revert to an equilibrium state. As the membrane becomes fouled over time, the system will have a different state of equilibrium.
Select to expand quoteCH3MTR4IL5 said..
it is in an equilibrium state, and that equilibrium is impacted by the internal resistance of the battery (I would think surface resistance on the electrodes being the main player).
Yes, agreed on all that you say, and the crux of it is somewhere in that statement. This is the bit I want to fully understand. There is going to be an 'internal resistance' equivalent for the no-load case that is not the same thing as the internal resistance of the battery under load, although there will be some commonalities between them. I guess there could be some kind of internal leakage current flow, not through the battery because it is open circuit, but instead localised to the electrodes, and this increases with discharge state due to the presence of compounds produced during battery operation, and all of this will resolve at the equilibrium state to give the no-load terminal voltage. It seems I just want to understand the details of the electrochemical equilibrium properly.
I don't actually want to calculate the answer, it's just that the theoretical equations would inherently contain the answer to my questions. (As opposed to empirically derived graphs, which I agree are far more practical and just as useful for analysing real-world battery applications.)
I think the answer to how that process works might be in here somewhere, but i'm at a loss to absorb it all:
cpb.iphy.ac.cn/article/2016/1806/cpb_25_1_018210.html
Only remember a little bit of chemistry, but an equilibrium was when the forward reactions equal the reverse reaaction. In a battery at a certain open circuit voltage the concentration of ions above the surface that can accept or emit ions should spread through the whole of the electrolyte regardless of whether there is a square inch or a square foot of electrode space left. Like the vapour pressure of water in a drum depends only on the temp. Doesn't matter if the drum is half full or there's a teaspoon of water left in the bottom corner.
Did you know that only a small fraction of the battery weight is actually lithium? And all the charge between charged and flat is carried by most but not all of them? The rest is just overheads So a 2 tonne tesla with a 600 kg battery gets up and goes using less than 20 kg of lithium. I've seen 50kg ladies jump in a 2 tonne tesla to go to the shops to pick up 1 litre of milk. Not an efficient use of lithium!
Select to expand quoteCH3MTR4IL5 said..
I think the answer to how that process works might be in here somewhere, but i'm at a loss to absorb it all:
cpb.iphy.ac.cn/article/2016/1806/cpb_25_1_018210.html
Thanks for posting this Chemtrails. Altough, this is very heavy reading. Most of it went above my head. The first half of the paper discussed open circuit voltage using Fermi energy and Gibbs Free Energy principles. Fermi energy discussion I could follow, but the Gibbs Free energy discussion lost me !
As Harrow alluded to, the open circuit voltage is defined by electrochemical potentials in the battery. The open circuit voltage is defined as the difference between the (higher) Fermi energy state at the anode and the (lower) Fermi energy state at the cathode (See Eq 1 in paper). As the battery discharges, the chemicals at the electrode/[anode | cathode] will change configuration (shift around electrons - at different energy levels - causing Fermi energy to change in those chemicals). Hence, this explains the open circuit state of voltage profile people have observed above.
From a Thevenin equivalency model perspective (the way I simply the electric world), the voltage source is variable.
Learnt something ! 
