Advice on lowering voltage

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greasemonkey1978

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Hello everyone,

I have a digital camera which drains batteries so quickly its getting very expensive even with rechargeable. The camera has a 5 volt dc socket on it and i want to run the camera from this via an external battery. I have a 6volt 4.5Ah lead acid battery that i would like to use. I asked a few people for advice who suggested a voltage regulator 7805 to reduce the voltage. This worked well until i wanted to take pictures. I powers the screen and i can look at the pictures but as soon as i tried to take a picture the camera shuts off. I think that the voltage regulator as reduced the amperage. The camera details are 5 volt dc, 5 watt which i believe to be 1 amp.

If anyone has any ideas or suggestions i would be very grateful as i am lost now.

Thanks

Mick
 
Just put two power diodes in series, between the battery and camera.
 
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The problem is: as soon as you try to get noticable current out of a 7805, the chip eats 2 or 3 volts for its own purposes.

The solution is to put a single diode in series between the battery and the camera. The diode will eat about 7 tenths of a volt and the camera will get 5.3 volts. You need a 3 amp rectifier diode of no particular wonderfulness. Even Radio Shack should have them.
 
A 6V lead acid battery has an output of about 6.3V when charged, so one diode drop may not be enough. If so you could put two diodes in series. A standard and a schottky diode in series should drop a little over a volt.
 
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hi all thanks for the replies i was just wondering if the diode would give a guaranteed voltage as i cannot chance frying the camera. Someone else suggested using a variable resistor (potentiometer) used as a voltage divider, i was wondering what you thought.

My knowledge is not much better than basic only really used to car electrics.
 
bychon said:
The problem is: as soon as you try to get noticable current out of a 7805, the chip eats 2 or 3 volts for its own purposes.
Click to expand...

So if i used a 12v lead acid would this solve my problem as i already have the regulator wired up in a box with all connectors ready to go.
 
Yes. The only thing yet to consider is that if you're a pro photographer, firing 32 shots at a time, the 7805 will get hot. It will need a piece of metal to help it radiate heat. If you're a normal person, don't worry about it.
 
crutschow said:
A 6V lead acid battery has an output of about 6.3V when charged, so one diode drop may not be enough. If so you could put two diodes in series. A standard and a schottky diode in series should drop a little over a volt.
Click to expand...

Thanks could you suggest what ratings to use for each and if they need to go in a particular configuration please as i am unsure where to start with this.
 
A potentiometer would give poor regulation and dissipate a lot of power.

You could use a 12V battery, but you then have the problem that the regulator would have to dissipate about 6W of power, so it would have to be mounted on a good heat sink.
 
bychon said:
Yes. The only thing yet to consider is that if you're a pro photographer, firing 32 shots at a time, the 7805 will get hot. It will need a piece of metal to help it radiate heat. If you're a normal person, don't worry about it.
Click to expand...

crutschow said:
A potentiometer would give poor regulation and dissipate a lot of power.

You could use a 12V battery, but you then have the problem that the regulator would have to dissipate about 6W of power, so it would have to be mounted on a good heat sink.
Click to expand...

At the moment it has a clip on heatsink on the regulator and the box it is mounted in has ventilation holes, i am not a pro photographer but am intending to shoot quite a few photos.
 
I think it's all been said. You can buy a better regulator, use a 12 volt battery with a heat sink (which you already have), or add a couple of 3 amp diodes in series. Pick one.
 
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