alifie Posted June 2, 2009 Share #1 Posted June 2, 2009 Advertisement (gone after registration) In a fit of pique I decided to measure the current taken by the M7 from its battery (2 cells). In idle, on but no display less then 0.005 mA, that is as low as my meter goes. Turn on 10mA for 2 seconds then 6mA pulsing to 8mA at about 3Hz, changing with light. Shutter on 4s short unmeasurable pulse when opening shutter, same short pulse closing, 10mA while open. Shutter open on 'B' 10mA Disconnecting power when on B shuts the shutter. Summises (in the air conclusions) Most of the current is taken by the display The photocell uses a charge metering system, i.e. fill a capacitor and see how long it takes. Given the 160mA/h capacity of the cells maximum B is 16 hours, unless the display goes out when it reaches maximum (999 seconds?) then it may be more. Couldn't be bothered to try this. It uses a one magnet to open and another to close the shutter. I have no idea how or why the shutter closes when the cells are removed, it used to be an old trick on Canon SLRs for astrophotography where the max B time was limited, the shutter closed when you re-installed the battery. Leicas expectations as to battery life are about right. Link to post Share on other sites More sharing options...
Advertisement Posted June 2, 2009 Posted June 2, 2009 Hi alifie, Take a look here Current consumpotion of M7 . I'm sure you'll find what you were looking for!
alifie Posted June 2, 2009 Author Share #2 Posted June 2, 2009 And the M6 12mA on, less than 0.005mA off and it uses 3V not 6. If the standby current is 0.005mA (less than that in reality) battery life would be 4 years. Link to post Share on other sites More sharing options...
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