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I do not want to charge the battery but make a long video. I want to use the Canon D30 camera without battery, but with a PSP charger which has the same plug but 5.3V. The Canon D30 needs 4.3V.

is that possbile?

Found a solution: a $5 DC to DC Converter like this

https://i.imgur.com/MJhHuM1.png

I cut off a normal USB and a PSP loading cable. Tested and works fine. Voltage is around 3.8.

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marked as duplicate by scottbb, inkista, mattdm, Dan Wolfgang, Saaru Lindestøkke Dec 2 '16 at 22:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    That is 23% over-voltage. For the little that they cost, I would buy a PSU with the correct ratings. – Mick Nov 2 '16 at 15:04
  • Please edit your question to ask an actual question. Google search won't index your image as a question... – scottbb Nov 2 '16 at 20:30
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    I'm voting to close this question as off-topic because this question is powering an electronic device with a power supply for a different electronic device. This is fundamentally an electronics design question, not a photography question. – scottbb Nov 2 '16 at 22:35
  • Can the camera be charged from USB? USB is 5V, so that would be pretty close. – vclaw Nov 3 '16 at 23:24
  • The Camera cannot be charged from USB. I tried it for a few seconds with the PSP cable and it worked, but I didn't dare to try it out longer. – Jimmy Jon Nov 10 '16 at 23:41
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If the camera has overvoltage protection based on voltage stabiliser, there is no risk. The stabiliser will just heat up wasting the extra volt.

If there is no protection, or a breaker, there is a hazzard of burning some crucial circuit in the camera. The voltage 4.3V is equal to voltage of three NiMH AA(A) cells in series.

If you are able to build some simple circuits, buy the plugs, 4.3V stabiliser or switching regulator and you are sure. You can also build voltage divider using two resistors or one potentiometer.

The chapest way is using the voltage divider. Choose the resistors parameters (Resistance, wattage limits) according to peak values displayed on camera and the charger. Peak current available from the charger shall be higher than then the peak current shown on the camera whatever solution you choose.

Another option is to build battery stack using battery cells, M cells in series and N cells in parallel. M=4.3/Uc, where Uc is single cell voltage, and N=C/Cc, where C is power capacity you want and Cc is single cell power capacity (measured in mA.h).

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    The voltage divider will be very finicky to set the right voltage, because the camera won't likely present as a fixed load (fixed resistance). That means the voltage to the camera won't be constant. Voltage divider circuits are not the same as voltage supplies. – scottbb Nov 2 '16 at 20:27
  • @scottbb with V-meter attached to the camera input side one can tune the potentiometer to match the voltage. Regulators will do this jobs automatically. – Crowley Nov 2 '16 at 20:48
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4.3V makes me suspicious. That sounds a lot like the charging voltage of a lithium battery, suggesting that the charger controls the current into the battery. I could easily be wrong but if I'm right there's a risk of your cantata catching fire if it's got the battery in. Lithium batteries are fussy about charging and if you get it wrong can explode.

So if you really want to try it and risk your camera, make sure the battery is out first. It's quite possible that the logic circuits run at 3.3V, magic that the regulator has to drop 2V instead of 1V for a fixed current. It probably doesn't turn all this wasted power into heat as that would waste battery life and more efficient regulators are available, but if it does, it could damage something.

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