[NotJeffBuckley]

A quick question for you gentlemen who've more experience with amplifier circuits than I:

How much current output is a typical 9V alkaline battery capable of? I understand that NiMH 9V rechargables can output up to 200mA, and I've even seen one lithium polymer able to output 400mA peak (not necessarily continuous, though), but I can't find any figures on, say, a 9V Duracell.

If this is better in DIY, please feel free to move it, moderators; I asked it here because it has to do with an amplifier modification I'm performing to ease up on battery costs

 

[sxr71]

Quote:

Originally Posted by NotJeffBuckley A quick question for you gentlemen who've more experience with amplifier circuits than I: How much current output is a typical 9V alkaline battery capable of? I understand that NiMH 9V rechargables can output up to 200mA, and I've even seen one lithium polymer able to output 400mA peak (not necessarily continuous, though), but I can't find any figures on, say, a 9V Duracell. If this is better in DIY, please feel free to move it, moderators; I asked it here because it has to do with an amplifier modification I'm performing to ease up on battery costs

If you are talking about current output as in mA and not capacity as in mAH, then the order as I have researched is:

Special high current NiMH > NiCd >> regular NiMH >>> Alkaline

I think Li-Ion batteries are roughly around where regular NiMHs are. Alkalines are horrible but for the purposes of an amp they are probably okay, but they suffer terribly when powering switching circuits such as those in XIN's Supermini and Supermicro (and digital cell phones among other things).

Start here:

http://www.batterystore.com/Sanyo/SanyoSeriesNicad.htm


The lower the IR the higher the current delivery capability. You can compare different series of Cadnicas and different chemistries. You will notice that larger batteries have much lower IR than smaller batteries and this makes sense as the electrode surface areas are much larger in larger cylindrical batteries. BTW, you numbers seem low as some of these NiCd AA cells are designed to be used in RC racing battery packs and those motors can drain current at the rate of 15amps under race conditions. No headphone amp will ever need that kind of current so don't overspend on batteries as the really low IR batteries are very expensive.

 

[atx]

I think you're confusing current with capacity. Typical NiMH 9V are rated at 200 mAh, which means at a current drain of 200 mA, the battery would be completely dead in 1 hour. Current output is driven by the load between the two terminals, not by the capacity of the battery.

 

[sxr71]

Quote:

Originally Posted by atx I think you're confusing current with capacity. Typical NiMH 9V are rated at 200 mAh, which means at a current drain of 200 mA, the battery would be completely dead in 1 hour. Current output is driven by the load between the two terminals, not by the capacity of the battery.

..and that one hour is only in theory at a constant drain of C/20 (or 10mA in this example). Not all applications cause that rate of drain and hence the actual usable capacity can be different.

 

[sxr71]

In response to what the OP is probably really asking according to this a 9v alkaline can provide 600 mAH while a Lithium Iron DiSulphide can provide 1200 mAH.

Links:

http://data.energizer.com/PDFs/9vlithium_appman.pdf

http://thomas-distributing.com/9v-ul...ies-9-volt.htm

 

[NotJeffBuckley]

Thanks for all the help, guys - I was looking for instantaneous current maximum, rather than mah, but I think my solution is working just fine.

To phrase the question differently to be sure, though:

Let's say I've got an amplifier running off of a 9V battery. It's good for 20-30 hours of use. If I have an inverter capable of outputting a continuous 300ma current at 9VDC and I wire it in to replace the battery, will I be sufficiently powering the amplifier? Forget noise or any other non-power related question, this is strictly to make sure I'm not underpowering an amplifier.

Edit: It appears that at the beginning of its charge cycle, a 9V alkaline Duracell battery is capable of outputting about 600mA current continuously (which of course lessens very quickly over its life). So, if I'm using a 300mA inverter into my amplifier, am I potentially underpowering it? A headphone amplifier, to my understanding, isn't a high current draw device, is it? How much would a CMoy, for example, draw powering HD650 (300ohm) at 95db? Rough figures are appreciated, as I'm not expecting anyone to bust out the calculators for my sake.

 

[sxr71]

Well this is something I wonder about as well. I think the continuous current demands of a CMOY are small, but I wonder if the instantaneous requirement are higher. One way of testing I guess is connecting a sensitive ammeter to the circuit and powering with batteries that can provide huge amounts of current and look for spikes in current draw.

Also keep in mind that your power supply has a peak current delivery that exceeds its continous power delivery (by how much and for how long I have no idea).


Another way to look at it is that at a worst case 16 ohms on the output, 9v can only drive 0.56A through the circuit in theory (and practically less because we're not even including resistances in the circuit). So if your power supply can generate 563mA then you are covered in a theoretical worst case situation. In reality it should be more than enough.

I might be wrong in my analysis of this so someone correct me if I am wrong.

 

[fewtch]

Wouldn't some good fast power supply capacitors take care of the instantaneous current demands? Or is that only voltage... my electronics knowledge is rusty.