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03-04-2014, 12:42 PM   #1
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K-3 failure - couldn't stop the mirror flapping up and down

Hi
Today I was taking pictures outdoors. When pressing the shutter the mirror went bananas. It sounded and felt like the camera took 100 shots, cause the mirror kept flapping up and down. Turning off the camera didn't stop the camera.
I began trying to get the battery out, but before I managed, the camera stoped. Afterwards I checked to see all the images - but none were recorded!

I certaily hope this will never happen again - it could be a big problem, if shooting something, that could not be done all over again!

Have anyone had similar experiences?

Regards
Jens

03-04-2014, 12:45 PM   #2
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I've had it happen on my K-x several times. I have to take the battery out to stop it. But this started after it was several years old, not when new.
03-04-2014, 12:55 PM   #3
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QuoteOriginally posted by Photodana Quote
Hi
Today I was taking pictures outdoors. When pressing the shutter the mirror went bananas. It sounded and felt like the camera took 100 shots, cause the mirror kept flapping up and down. Turning off the camera didn't stop the camera.
I began trying to get the battery out, but before I managed, the camera stoped. Afterwards I checked to see all the images - but none were recorded!

I certaily hope this will never happen again - it could be a big problem, if shooting something, that could not be done all over again!

Have anyone had similar experiences?

Regards
Jens
see here..you're not the only one

https://www.pentaxforums.com/forums/172-pentax-k-3/242257-odd-shutter-issue-f...0-problem.html
03-04-2014, 01:13 PM   #4
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The dreaded "mirror flop" problem. I thought that was history after being solved in the early K5 days. The (temporary) fix back then was to be sure the battery was not fully charged all the way up. Users who had the problem got in the habit of ripping off a few dozen throw-away shots on high-speed continuous immediately upon replacing the battery with a fresh one, just to drain the charge a little. Starting with a battery at 98% charge or so seemed to solve the problem.

You might also make sure you have the latest firmware installed.

EDIT: Oops. In the other thread, it seems that the current problems with the K3 may not be related to voltage. Suggest you read through that thread.

03-04-2014, 01:16 PM   #5
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K-5 had that problem for a while (but only after 1-1/2 years) and eventually died. $274 - mirror box motor...
03-04-2014, 02:04 PM   #6
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I have encountered exactly the same problem last week on the K3. It happens only one time. I have the latest firmware 1.02. I suspect a firmware bug. I have encountered the "Mirror flop" issue on my old K5, that needed a Mirror box motor replacement, but it wasn't the same symptoms. We shall see....
03-04-2014, 02:10 PM   #7
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Ditto. Had this happen a couple of times when first purchased (& yes, battery removal is only option), but not since firmware upgrades....and have shot many thousands of shots since upgrade.

03-05-2014, 11:39 AM   #8
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I AM usingFirmware 1.02 - the latest, right?
Pehaps I used a K-5 battery, which as slightly more mA...

Jens
03-05-2014, 04:02 PM   #9
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The K-5 Battery is exactly the same as the K-3 battery and having bought 3rd party batteries as well I don't have a problem. The mirror flopping happened to me on the first day but only once and never since.
03-06-2014, 05:11 AM   #10
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See here: https://www.pentaxforums.com/forums/172-pentax-k-3/250865-k3-locking-up-while...ime-lapse.html
and also my update on Page 4 of that thread.


It's an issue. I've had it happen a lot when shooting time-lapse, but others are reporting it when not shooting time-lapse too. I've used old batteries (k5) and new, and Pentax even sent me a brand shiny new one to try and it happened with that as well. Different brands of SD cards. Different weather conditions, indoors and outdoors. I even got a new K3 body and right out of the box it still happened. Unfortunately
03-06-2014, 01:39 PM   #11
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My K-5 bat]teries are 2000 mAh. The K-3 battery (from the K-3 box) is 1860 mAh - that is a little less powerful.

Regards
Jens
03-06-2014, 01:48 PM - 1 Like   #12
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QuoteOriginally posted by Photodana Quote
My K-5 bat]teries are 2000 mAh. The K-3 battery (from the K-3 box) is 1860 mAh - that is a little less powerful. Regards Jens
The mAh rating has to do with how long he battery will last, not how much voltage it will put out.
That being said I wonder how exact the voltage levels are actually matched on third party batteries.
Most devices tolerate a range of voltage, but there may be something wrong with a component on some of these cameras.
I hope folks are recording this info in the serial number database here, we may be able to see a trend such as date of manufacture or serial number relationships. https://www.pentaxforums.com/forums/pentax-serial-number-database/?do=viewserials&id=495

Last edited by crewl1; 03-11-2014 at 10:38 PM.
03-13-2014, 05:16 AM   #13
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Anyone who has had this issue, could you please report it here in this thread ...

Go here: https://www.pentaxforums.com/forums/172-pentax-k-3/254464-k3-crazy-mirror-sic...ours-here.html

There are quite a few threads on the issue, and we're trying to get everyone all in one spot, so we can get a good sense of how many people it's happening to. Would be so helpful

Thanks!
03-13-2014, 05:47 AM   #14
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QuoteOriginally posted by crewl1 Quote
The mAh rating has to do with how long he battery will last, not how much voltage it will put out.
]
The mAh rating (if true) will reflect how much the battery voltage sags under load.

Because the maximum current is determined by the internal resistance of the cells.
therefore
Voltage drop = rating * cell resistance
06-05-2014, 07:28 AM   #15
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QuoteOriginally posted by awaldram Quote
.... therefore
Voltage drop = rating * cell resistance
With all due respect, this is not true. The actual voltage under load has absolutely nothing to do with the battery's amp-hour rating. Nothing at all.

The voltage under (VL) load of a battery, is determined by 3 parameters

Vo = No load voltage
Ri = Internal resistance of battery
I = load current

Such that VL = Vo - I * R

The voltage drop is then the load current times the internal resistance or I * Ri
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