Cmoy damaging rechargeables...?
Jan 28, 2005 at 3:03 PM Thread Starter Post #1 of 10

BavariaBarbarian

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Well, this may be a stupid question, but well: Anybody tries to embarass himself as best he may...
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I'd like to plop two Nimh rechargeables into my Cmoy instead of using Alkalines.
The thing is: I read once that one should never, ever deep-discharge a nimh battery below 0.6 or so Volts per cell or it may get damaged.
So now i figure: One battery has 8.4 V so there's gotta be 7 1.2 V cells in a row in there - correct?

If so, and i deplete them to 0,6 V each in both batteries I'll get 7*2* 0.6V = 8.4V.

So methinks i won't hear any clipping due to low supply voltage and go on listening until i've squeezed the last electron out of those batteries.
(Which can't be long from that point on... but anyway.)

Am i making any sense? At all?

Because, you see, it wouldn't save me much money if i killed my rechargeables by deep-discharge after a few cycles.
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Jan 28, 2005 at 4:43 PM Post #2 of 10
I recharge whenever I can, so I always have a full tank. There's no harm in this. I'm much more worried about being 30,000 feet over Kansas with no music, than being out $16 (Sam's Club) because I tanked my batteries.

There are way too many things one could worry about in life. Why don't you proceed, and see if this is a problem for you? If it is, you're out $15 once, you can then regroup so you don't get fooled twice.

Edit: I should add that in the course of experimenting with charging circuits I've done everything you're not supposed to do to NiMh cells, like my 3am addled brain wondering what that smell is at the same time trying to process why my DMM says I have 6 amps of current, that can't be right? So I replace the cells, and no measurement I make can distinguish the "good" cells from the "bad" cells.

In other words, NiMh cells are more robust than any datasheet suggests.
 
Jan 28, 2005 at 5:25 PM Post #3 of 10
Quote:

Originally Posted by BavariaBarbarian
Well, this may be a stupid question, but well: Anybody tries to embarass himself as best he may...
wink.gif


I'd like to plop two Nimh rechargeables into my Cmoy instead of using Alkalines.
The thing is: I read once that one should never, ever deep-discharge a nimh battery below 0.6 or so Volts per cell or it may get damaged.
So now i figure: One battery has 8.4 V so there's gotta be 7 1.2 V cells in a row in there - correct?

If so, and i deplete them to 0,6 V each in both batteries I'll get 7*2* 0.6V = 8.4V.

So methinks i won't hear any clipping due to low supply voltage and go on listening until i've squeezed the last electron out of those batteries.
(Which can't be long from that point on... but anyway.)

Am i making any sense? At all?

Because, you see, it wouldn't save me much money if i killed my rechargeables by deep-discharge after a few cycles.
wink.gif



If you really want to, you could devise a voltage splitter for the LED so that when you reach say 9 V, the LED will not have enough voltage to emit light. Only design issue that I think of off of my head is that the easiest way to do this is with resistors but they'll draw off current and dissipate power, either use very large resistances or go with maybe diodes or something else.
 
Jan 28, 2005 at 5:47 PM Post #4 of 10
my cmoy doesn't work with rechargables (the ones I have anyway - Rayovac.) The volume gets really low after a few minutes and I have to turn the amp off and on to reset it (and then it happens again.) Considering how many friggin' hours of battery life this cmoy gives me, I'm happy using regular batteries though.
 
Jan 28, 2005 at 10:13 PM Post #5 of 10
Quote:

If you really want to, you could devise a voltage splitter for the LED


That is actually a really good idea. You want the LED to light with a charged battery but turn off at low voltage.
So it should turn off around half of the full voltage, that is 8.4/2 = 4.2 volts.
It wants around 1.7 volt to light by itself, so if you connect a zener diode of around 4.2-1.7=2.5 volts in series with the LED you should see the led go out when it is time to charge your battery.
battry_plus ---¨|_<---->|---R--- battery_minus

Where ¨|_<-- is a zener
and -->| is a LED with 1.7 volt forward drop.
 
Jan 28, 2005 at 10:23 PM Post #6 of 10
Quote:

So methinks i won't hear any clipping due to low supply voltage and go on listening until i've squeezed the last electron out of those batteries.


My Pimeta was originally designed to use 2x9V rechargables, but after class A biasing and a pair of extra BUF634s it's not been very feasible (~16 hours to charge, 3 hours run time
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). However, I do have some first hand experience on this matter. I think (and here's where I potentially embarass myself) NiMH batteries have rapidly increasing internal impedance when they approach depletion; this means while the 8.4V NiMHs can still have ~7V each across open terminals near depletion, once you hook them up to the amp the voltage across the circuit drops precipitously to something like 5V for the two of them in serial. I have measured this myself when my amp started clipping (AD8610), and you can easily verify this yourself. Once you remove the batteries, they rapidly regain their voltage to around 7V where they remain.

I have discharged my batteries beyond the clipping point for more times than I care to remember, usually because I thought the amp's on DC. They still work. While doing this to the batteries can certainly shorten their life, I don't think you can damage them permanently in anyway by doing this a few times. In short, I believe battery internal impedance will protect the individual cells from becoming so weak so as to suffer from things like cell reversal, so there's nothing to worry about as far as damaging the battery goes--as long as you turn the amp off quickly after it starts clipping.
 
Jan 28, 2005 at 11:18 PM Post #7 of 10
Quote:

one should never, ever deep-discharge a nimh battery below 0.6 or so Volts per cell or it may get damaged.


The 0.6V value is pretty arbitrary. Personally, I use 0.9V per cell, based on the impedance and voltage vs. discharge curves I've seen. There just isn't much to be had below that value. You can study this material to see this.

Quote:

One battery has 8.4 V so there's gotta be 7 1.2 V cells in a row in there - correct?


Yes, for that case. There are also 7.2V and 9.6V "9V" rechargeables, with 6 and 8 cells, respectively.

Quote:

methinks i won't hear any clipping due to low supply voltage


That depends on several factors, actually. Have you read this?

Quote:

NiMh cells are more robust than any datasheet suggests.


That's because datasheets are in the business of making guarantess in the face of manufacturing variability. They therefore try to be conservative, to minimize the number of return claims they have to process. Even if only 1% of parts fail to meet all the datasheet's specs during use within the limits the datasheet imposes, that's too much risk.

Quote:

you could devise a voltage splitter for the LED


I prefer something a little more complicated. See page 2 of this schematic. Since you don't need adjustability in your application, you can replace the pots with fixed resistor dividers to save money.

Quote:

NiMH batteries have rapidly increasing internal impedance when they approach depletion


Yes, but it's not nearly as bad as with alkalines. See the links page I linked to above for details. The same companies that offer NiMH datasheets also offer alkaline datasheets, and they're often useful for comparison. Sometimes not, though, sadly.
 
Jan 29, 2005 at 11:23 AM Post #8 of 10
Whoa, folks! Thanks for all the response, your being very informative!

Tangent:
I never cease to be amazed at that site of yours!
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I dug through some of the material you linked - and learned a lot.
- That battery indicator circuit diagram in your tpm is beyond my comprehension, though.

But a zener diode inline with the led - that i get (after looking up what a z-diode actually does...
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).
Seems pretty simple and forward to me - just right for a cmoy then.
(hmm... just discovered something similar in the pimeta schematic...
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)
So i think i'll stick a 11 V zener in there. Together with the 1,8 V drop across the led, that gives me 12,8 V supply-voltage. So the batteries discharge until ~0,91 V (12,8/14) per cell - led goes out - i recharge. Or should i aim for 1,0 V / cell ?
Maybe it'll start sounding bad before that point, i haven't figured yet. But then that'll be my singnal to recharge again.

Also, i'll quit worrying about my batteries too much. I keep sticking them in a tricklecharger without timer anyway...
 
Jan 30, 2005 at 5:57 AM Post #9 of 10
Quote:

That battery indicator circuit diagram in your tpm is beyond my comprehension


Oh, it's not that scary. The only broad concept you need to understand is that those op-amps are being used as comparators. A comparator's output goes high when +IN is higher than -IN, and it goes low when the relative voltages are reversed. (That's why it's called a comparator: it compares two voltages and indicates which one is higher.)

Take the upper comparator first: its -IN is pinned to 2.5V above V- by the LM385 voltage regulator. (R7 is just there to give it a fairly constant current.) Therefore, the comparator's output is low as long as the +IN is presented less than 2.5V by the divider. You adjust that divider such that your trip voltage, divided down, equals 2.5.

The other comparator functions similarly, only in reverse.

Between the two, you have three states: low voltage, where only U6B's output is high, high voltage, where only U6A's voltage is high, and a middle ground between the two where both are high. That gives red, green, and a yellow or amber color; the middle color depends on the LED characteristics, and the R9 and R10 values.

Quote:

just discovered something similar in the pimeta schematic


Yes. The addition of the constant current source improves the zener cut-off behavior, and it keeps the LED's brightness constant while the zener is inactive. It's highly recommended for this circuit.
 
Jan 31, 2005 at 10:05 AM Post #10 of 10
Quote:

Originally Posted by tangent
Oh, it's not that scary.


Says you.
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But I think i get it. - I was actually wondering how those fancy battery-indicators with different colors actually work.

Quote:

Originally Posted by tangent
Yes. The addition of the constant current source improves the zener cut-off behavior, and it keeps the LED's brightness constant while the zener is inactive. It's highly recommended for this circuit.


Ah, understand... My current cmoy is pretty crammed as it is but number two just needs another tweak like that.
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