CMS50E oximeter battery dead :( replacement ??

[-SWS]

I would be extremely cautious before making any assumptions about the battery. The shape and look, and the 3.7V nominal suggest a Lithium-ion polymer rather than a Lithium-ion.

What does the 3.7 nominal voltage for all the other batteries linked above suggest?

...and the two have very different charging requirements.

The ones used for MP3 players, GPS, etc. typically finish regulating charging on the battery pack's circuit board. They are designed to charge with ordinary USB-level voltage as input---either from the computer or a standard wall charger that accomodates a USB connector. The same is true for Uncle Bob's oximeter above I believe.

The former is also rather explosive, if handled improperly.

Your point is well-taken. DO NOT CRUSH. Best to not mount it outside the case now that you mention it. The other main risk of explosion lies in overcharging. Here's a typical on-board PCM description I would look for:

•Built in over/under voltage protection circuit

http://www.ebay.com/itm/Rechargeable-Li ... 415b145b05

Here is one that is probably exactly like it: http://www.alibaba.com/product-gs/40125 ... _Pack.html and this onehttp://www.radioshack.com/product/index ... Id=2453467 might work, but the point is, if you can't be sure, do you really want to take the chance?
McSleepy

Take a chance replacing his potentially explosive polymer softpack with one that might be even LESS inclined to explode? Sure. Epecially since they're only about $6 in the links above instead of $30 at Radio Shack.

[McSleepy]

What does the 3.7 nominal voltage for all the other batteries linked above suggest?

You tell me; I wasn't referring to the other batteries above but making a statement about the battery the OP is interested in.

The ones used for MP3 players, GPS, etc. typically finish regulating charging on the battery pack's circuit board. They are designed to charge with ordinary USB-level voltage as input---either from the computer or a standard wall charger that accomodates a USB connector. The same is true for Uncle Bob's oximeter above I believe.

I don't have time to explain in details the difference in charging requirements between Li-Ion and Li-Po, you can Google it. I will just mention a couple that are rather important: Li-Po have much lower internal resistance (one consequence is it can take a much higher charging current) and they have much tighter tolerances in operating voltage range. The regulating circuit is almost certainly built into the CMS and in the case of Li-Po different strategies (based on current and voltage) are used. The difference can confuse the regulator, at best, and destroy it, at worst.

•Built in over/under voltage protection circuit http://www.ebay.com/itm/Rechargeable-Li ... 415b145b05

Li-Ion are more tolerant, that kind of protection would not help - in this case it's the regulator in the CMS that needs protecting. Looking at the trip points: high is 4.275 and low is 2.30V, that is well wider than the requirements for a Li-Po, whose useful operating voltage is 4.2V to 3.75V. If the original battery is Li-Po and has the regulating circuitry in the device (almost certain), and the difference doesn't end up otherwise killing it, that Li-Ion battery will not end up being charged/used properly.

Take a chance replacing his potentially explosive polymer softpack with one that might be even LESS inclined to explode? Sure. Epecially since they're only about $6 in the links above instead of $30 at Radio Shack.

Actually, because of the higher internal resistance of Li-Ion (heat is proportional to resistance) and other reasons I cited above, they are in fact more likely to explode than Li-Po (in general). And because of the liquid content of the Li-Ion (the electrolyte is an organic solvent versus the polymer in Li-Po), the explosion of a Li-Ion is more violent. This is why bigger Li-Ion packs are always enclosed in sturdy metal jackets, which is not the case with the tiny battery discussed here.

When I say "take on chances" I mean do not try to replace the battery at all, since it could be dangerous but mostly it is quite likely that he will end up wasting his money. My point is, like most Chinese-design electronics, the design tolerances are already so small that they tend to malfunction in their original form; making any changes is very likely to lead to disappointment.

McSleepy

[Uncle_Bob]

Thanks for the feedback. I'm not sure i like the idea of soldering in the battery. I know Ni-cads are sensitive to heat but not so sure about other types. Either way looking at the battery I don't fancy my chances even if i do get my hands on the correct replacement battery, it just looks too much like "go ahead punk, make my day" kinda job which is maybe what the Contec intended.

[chunkyfrog]

I guess I'll plug mine in and top off the charge.
Just to cover my little green hiney.

[McSleepy]

I guess I'll plug mine in and top off the charge.
Just to cover my little green hiney.

While for Li-Ion the best charging strategy is to keep it to 100% at all times (with periodic discharges), the strategies for Li-Po vary and the one I tend to trust says it should always be stored at 50% charge. While with the CMS you don't know the exact percentage, the idea is - don't let it sit discharged for long and no need to keep it fully charged at all times. So, yes, if you don't use it often, give it a charge once in a while to prolong its life.
McSleepy

[rosacer]

I don't think there is a problem soldering on the battery, put some flux and try not to keep too long the solder tool on it ; use a very thin solder wire. I really don't think that will explode, they solder it too and I guess in a very poor conditions. Hopefully my oximeter uses one or two (I don't remember ) AAA battery I'm so glad of it.

[-SWS]

Actually, because of the higher internal resistance of Li-Ion (heat is proportional to resistance) and other reasons I cited above, they are in fact more likely to explode

Heat and current are both inversely porportional to resistance.

I'm not sure i like the idea of soldering in the battery.

I would clip the battery off the circuit board but leave the old leads soldered on. I would strip the ends of both old and new wires before splicing and taping. There's no need to solder. Don't chance it if you'd rather not. I would. But my background is in R&D and I am formally trained in electronics. I appreciate and respect the cautions McSleepy extends, but I don't agree with the magnitude of risk.

[Mr Bill]

After reading all the posts. I thought of the similarity to heart transplants.
(1) One my have difficult finding a donor heart with suitable tissue match.
(2) The operation may be difficult resulting in the death of the heart or even the patient.
(3) Even after a successful operation, there is still a chance of rejection and/or heart failure.

[McSleepy]

Heat and current are both inversely porportional to resistance.

No. Generated heat is proportional to dissipated power, and electrical power is equal to current squared times resistance. In an electrical circuit, you want as much of the voltage distributed to the load, not to the source (battery), wires, control elements, etc., which is why you want the resistance of those to be as low as possible. Here is a simple enough example: thicker cables are used in high-current applications (e.g., car jumpers) because they have lower resistance; if a thinner, high-resistance cable were used to jump a dead car battery, it would catch on fire. A battery with high internal resistance is like an ideal battery in series with a resistor - the higher it is, the more of the power is dissipated in it (and not in the load), and thus, the more heat is generated inside the battery.

Again, the risk is not so much with hurting himself but with the whole thing not working out and him wasting his money and time.

McSleepy

[-SWS]

Heat and current are both inversely porportional to resistance.

No. Generated heat is proportional to dissipated power, and electrical power is equal to current squared times resistance.

Well, we were talking about varying battery resistance relative to a constant charging voltage on Uncle_Bob's oximeter. When we maintain voltage as a constant that way, but vary the resistance, then power dissipated is the square of that constant voltage divided by changing resistance. So at any given fixed/charging voltage, lowering resistance increases power dissipated. I don't think the current and dissipated heat in this circuit pose a significant risk of explosion.


A very fruitful discussion IMHO. And this part of the manual suggests either lenient tolerance or remiss documentation regarding replacement-battery requirments:

The user's manual just says to replace with a 3.7v lithium battery.... no part # per se.

It sure would be nice if we could get more details from the manufacturer.

[Emilia]

I noticed that the CMS50D model uses 2 AAA batteries... then they moved to these lithium batteries with the E model. Built in money maker.....

[-SWS]

Well they still offer the CMS-50D Plus, which uses ordinary alkaline batteries. Since I rarely use an oximeter, I bought the CMS-50D Plus because I didn't want to worry about maintaining and storing a lithium battery. The downside, however, is that the CMS-50D Plus doesn't maintain time-of-day in the data----only transpired time.

As a side note the CMS-50D does not record and store data, whereas the CMS-50D Plus does. The CMS 50D is for spot-checking only.

[McSleepy]

Well, we were talking about varying battery resistance relative to a constant charging voltage on Uncle_Bob's oximeter. When we maintain voltage as a constant that way, but vary the resistance, then power dissipated is the square of that constant voltage divided by changing resistance. So at any given fixed/charging voltage, lowering resistance increases power dissipated. I don't think the current and dissipated heat in this circuit pose a significant risk of explosion.

It isn't as simple as that, you are assuming the internal resistance of the battery is the only resistance in the circuit while that is not true - in this case the charger plays a significant role. You also need to understand that the charger doesn't keep a constant voltage. If you maintained 4.2V (the target voltage), you would smoke the battery; if you applied less, it wouldn't charge fully. The current is regulated according to various strategies (different for Li-Ion and Li-Po) and the voltage is controlled according to supplementary criteria. But what you really need to understand is that the only thing that matters for heat generation in the battery is the power dissipated inside the battery, not the whole amount of power in the circuit. Given that temperature (which is what causes ignition) rises by accumulating heat, as long as there is more power dissipated inside the battery, even if the total power consumed by the entire circuit were lesser, heat would still be accumulated at a higher rate and if that exceeded the thermal dissipation of the assembly (i.e., cooling), the temperature will eventually become high enough to ignite the battery.
Let's take the example of a circuit with a 10V battery with high resistance (1 Ohm) and current of 1A and let's assume the thermal dissipation is equal to the accumulation and the temperature is kept at a safe level (albeit, barely). Let's instead put a battery with the same thermal dissipation characteristic but with lower (0.1 Ohm) internal resistance and increase the current to 2A (say, because of charger strategy). The total power in the circuit has increased but which battery will heat up more? Battery #1 dissipates 1A*1A*1Ohm = 1W while battery #2 dissipates 2A*2A*0.1Ohm = 0.4W. The difference is the first circuit has a load (e.g., the charging circuit) whose equivalent resistance is 9 Ohms while the second - 4.9 Ohms, the remaining power is dissipated in the charger. The second battery is operating well under its thermal limits, so if we pushed things a bit, this battery would have no problem, while the first battery would smoke (remember, it was at the limit of its thermal capacity already). Clearer now?
McSleepy

[-SWS]

It isn't as simple as that, you are assuming the internal resistance of the battery is the only resistance in the circuit while that is not true - in this case the charger plays a significant role. You also need to understand that the charger doesn't keep a constant voltage. If you maintained 4.2V (the target voltage), you would smoke the battery; if you applied less, it wouldn't charge fully. The current is regulated according to various strategies (different for Li-Ion and Li-Po) and the voltage is controlled according to supplementary criteria. But what you really need to understand is that the only thing that matters for heat generation in the battery is the power dissipated inside the battery, not the whole amount of power in the circuit

No I'm not assuming that at all. The CMS oximeter likely employs an inexpensive IC-based linear charger rather than switching type. If/when a circuit/battery combination comes close to risking battery volatility (based on thermodynamics) then the IC-based charger selected is also going to employ analogue linear or digital loop thermal feedback measures to reduce the charging current as ambient temperature rises.

When a typical IC based linear charger like the CMS's IC administers its first two charging phases, it will initially trickle charge; then it will more quickly charge at higher current; during the third stage voltage-monitor type charging occurs before the charging process is then terminated. You can place a 3.7v Li-Ion or Li-Po in the CMS's modest little application and thermodynamics are not going to result in an explosion. Regardless, the IC charging circuit's resulatant voltage impressed across the battery terminals is primary and salient regarding electrical power dissipation inside that battery---which was the tangential hair-splitting at-hand I believe. Accumulative thermodynamics from both battery and supporting/ambient circuits certainly enter the equation. This device's far-end or primary load (oximetry) is essentially open-terminal or high-impedance during charging.

Built in over/under voltage protection circuit

Li-Ion are more tolerant, that kind of protection would not help - in this case it's the regulator in the CMS that needs protecting. Looking at the trip points: high is 4.275 and low is 2.30V, that is well wider than the requirements for a Li-Po, whose useful operating voltage is 4.2V to 3.75V. If the original battery is Li-Po and has the regulating circuitry in the device (almost certain), and the difference doesn't end up otherwise killing it, that Li-Ion battery will not end up being charged/used properly.

The protective trip points are inside the battery and are typically N-channel MOSFET switches. Those cut-off MOSFET switches are for the benefit of the battery's safety. In this case the Li-Ion battery will simply not power the oximeter if battery voltage drops to 2.3. The battery needs that low-end protection---not the oximeter's charging circuit.

Here, the Li-Ion substitution would safely withdraw from charging if it reached 4.275, which it will never see. Although built-in battery protection trip points typically consume about 100mW for either battery type, they do not demand more or less current from the regulating IC on the CMS itself based on battery type. The latter IC is going to deliver its pre-programmed current during the first two stages and deliver its pre-programmed voltage during the third charging stage. Dropping a MORE voltage-tolerant Li-Ion battery into a Li-Po charging circuit means the Li-Ion battery will charge with suboptimal efficiency.

The Li-Ion would likely not fry the CMS's on-board charging IC in this case. It would probably charge less efficiently, but I suspect it would run the oximeter just fine. That might be WHY the manual simply calls for a 3.7 volt litium battery. Because it will probably actually work...

[McSleepy]

No I'm not assuming that at all. The CMS oximeter likely employs an inexpensive IC-based linear charger rather than switching type. If/when a circuit/battery combination comes close to risking battery volatility (based on thermodynamics) then the IC-based charger selected is also going to employ analogue linear or digital loop thermal feedback measures to reduce the charging current as ambient temperature rises.
When a typical IC based linear charger like the CMS's IC administers its first two charging phases, it will initially trickle charge; then it will more quickly charge at higher current; during the third stage voltage-monitor type charging occurs before the charging process is then terminated. You can place a 3.7v Li-Ion or Li-Po in the CMS's modest little application and thermodynamics are not going to result in an explosion. Regardless, the IC charging circuit's resulatant voltage impressed across the battery terminals is primary and salient regarding electrical power dissipation inside that battery---which was the tangential hair-splitting at-hand I believe. Accumulative thermodynamics from both battery and supporting/ambient circuits certainly enter the equation. This device's far-end or primary load (oximetry) is essentially open-terminal or high-impedance during charging.

What was that? Reads like a Sci-Gen generated text. Anyway, you don't know what kind of charger the CMS employs; it is highly unlikely to waste resources on trickle-charging as... Lithium batteries simply don't need it; the sequence you describe is nowhere near any of the charging strategies I've seen; the rest is a sequence of terms that make little sense together and do not address the issues that I brought up but are rather unrelated, chaotic statements.

The protective trip points are inside the battery and are typically N-channel MOSFET switches. Those cut-off MOSFET switches are for the benefit of the battery's safety. In this case the Li-Ion battery will simply not power the oximeter if battery voltage drops to 2.3. The battery needs that low-end protection---not the oximeter's charging circuit.
Here, the Li-Ion substitution would safely withdraw from charging if it reached 4.275, which it will never see. Although built-in battery protection trip points typically consume about 100mW for either battery type, they do not demand more or less current from the regulating IC on the CMS itself. The latter IC is going to deliver its pre-programmed current during the first two stages and deliver its pre-programmed voltage during the third charging stage. Dropping a MORE voltage-tolerant Li-Ion battery into a Li-Po charging circuit means the Li-Ion battery will charge with suboptimal efficiency.
The Li-Ion would likely not fry the CMS's on-board charging IC in this case. It would probably charge less efficiently, but I suspect it would run the oximeter just fine. That might be WHY the manual simply calls for a 3.7 volt litium battery. Because it will probably actually work...

The only point I see here is you agreeing that the CMS charger is unlikely to charge the battery properly, which is definitely a problem (longevity, ability to power the device as designed, etc.) Again, you can't know what protective circuits are used in the CMS or that battery, but either way the charging strategy is quite certain to include high-current stages if the original battery is Li-Po as that is how those work. The typical Li-Po charges at 1C, although many go to 2C or even 3C. That means, at 1C, the current the CMS charger will try to deliver will be 0.62A for a 620mA/H battery, which means the entire 0.5A of the USB port is likely to be utilized, and if the battery refuses to take it, either by "tripping" or by virtue of higher internal resistance, the charger would try to keep raising voltage, and in the best case, the battery will disconnect and not charge at all.
As far as theoretical possibility of ignition, none of the protections listed on that Li-Ion battery can prevent the problems I described - the battery itself has enough energy to ignite itself and being connected to a circuit that wasn't designed for it can certainly trigger that (by damaging the flimsy circuits), even if the charging current itself were never higher than 100mA (which, as I described above, is highly unlikely). There are examples of battery recalls when battery ignition was caused by damage to the battery's own circuits, long after being disconnected from its charger.
The point is, there is no way you can be sure that the substitution is safe, and even if it is likely safe, it is still highly unlikely to work to the satisfaction of the user. Also, this last post of yours starts to suspiciously feel like you are responding more for the sake of arguing rather than in substance, and I don't really intend to continue in that direction.
McSleepy