Estimating Battery Drain

[Triffid]

I though for sure I would be able to find something about this in forum search, but my searching came up empty.  

I would just like to estimate how long my 9V will be good in my effect.  I assume that would really depend on the battery I use.  I guess the right thing would to be to see how much current I am using right?  How can I go about doing this?

Thanks in advance...

[petemoore]

you can connect One Pole [say neg and not pos] of the cct's battery clip to the battery's  clip, then run the pos Through the DMM's current meter [A/Ma etc].
 I compare ccts readings, I think 3 ma is 'common' for small ccts, at about 20 ma, I start thinking about a Power supply. Can be used for debugging, I like to try it sometimes as one of the first debug tasks...right after checking for a connection between +V and - Gnd on the cct/battery clip.

[Mark Hammer]

The 12-16ma zone is a sort of inflection point, where 9v battery operation changes from something you can rely on to something that you can turn to if you can't find an adaptor.  Some of the early digital effects pedals could have current drain of 20ma or more.  Under those conditions, and a professional schedule of use (i.e., couple of hours a night, couple of nights a week), the cost of batteries will easily outstrip the cost of the pedal within a year.  Although such pedals WILL work off a decent 9v alkaline for a while, functioning can become undependable in a short period of time (e.g., a week or two of "normal" use).  Ideally, if you want something to provide reliable functioning for any sort of appreciable period, you'll want to keep battery drain below 10ma.

If the effect has high current demands, then you may well also notice an effect of battery type/brand.  One of the differences between better and cheaper brands of batteries is the quality of the connection between the 6 individual cells packed into the 9v battery case.  Think of the connection between them as a resistance that can vary from zero ohms (e.g., in those nice industrial-quality alkaline Duracells, where each cell is spot welded to the other) to something which can have a cumulative effect on how much current the battery can deliver on demand by acting like a current-limiting resistor (e.g., 50 ohms between cells).  Cell shape can have an impact too, although there are obviously only a few possible varieties within a 9v shell.  The "Polapulse" battery which came in every film cartridge for Polaroid instant cameras (and which you can extract from the spent cartridge and re-use) had flat thin cells.  Although it only provided about 6v, it was apparently capable of delivering huge amounts of current on demand because it had a very large surface area of contact between cells (so low intercell resistance).

[Triffid]

I can't seem to get a reading using my dmm with the method explained by petemoore... I have tried the 3 settings on my dmm (2mA, 20mA, and 200mA) in the section called DCA.   I tried put the dmm in series at both the positive and negative ends of the battery, any ideas?  Thanks

[petemoore]

Directional?
 Turn the leads around...[?]
 I'm the one that had it working IIRC, not the one to explain why...

[Triffid]

Ya, I tried both directions too... I can't even get connectivity when I just try to put the leads of my dmm to each of the battery terminals.  Either I don't know how to use my dmm or it's busted

I also tried just putting one resistor in series with the battery and dmm, no luck @#$@#$@

[RickL]

On some multmetes when you measure current you have to move the red/positive lead to a different spot on the meter. Is there a second or third jack on your meter marked for current? Sometimes there are seperate jacks for high and low current.

[Triffid]

It's just a cheapy radio crap dmm, leads are attached and not detachable.

[Triffid]

Ok, turns out my problem was that my fuse had been blown on my dmm
.  

I replaced it and got a reading of 4.64mA for my curcuit, that should be ok then right?

[smoguzbenjamin]

looks fine...