Battery Power For An Axis Camera?

We have a wi-fi Axis M1031-W that we had used for covert purposes in the past. At some point one of the big brained engineer guys put together a power cable with an in line resister to allow us to use a 6v gel cell battery for it. We are trying to recreate this cable as the old one got lost. We can not get it to work, and the originator is no longer around. Anybody have any experience in making this camera completely wireless?

The specs on the camera read: 5.0-5.1 V 1.5A

Thanks-

Jay T.


Accupack for 12VDC cameras

"This model can be used for all 12VDC models from AXIS. The basic version has a runtime from appx. 8h (depending on the camera model) Version with longer runtime are also available. All models are equipped with Li-Ion access so they have no memory effect. Also we have a overload and shortcut control inside, for longer livetime and security."

Undisclosed, that's a neat alternative though James is just looking for wiring advice in this instance.

Can you furnish Amp Capacity or Amp/Hour specs on the battery?

Ohm's Law is a three legged stool, and you need two variables to solve for the third.

Although we have two factors (V=5.0 and I=1.5A), there are multiple issues here:

  • Is the 1.5A accurate and constant? Such specs are often maximum, not operating requirements.
  • Is the tolerance on the camera that tight (5.0-5.1V)? Often 5V circuits can tolerate at least 5% and often 10% (4.75-5.25V or 4.5V-5.5V)
  • What is the voltage of a gel cell battery at different levels of discharge.
  • Do you intend to charge the battery while it's connected to the camera.

The simple solution becomes complex.

  • Gell cells typically provide 6.3V at full charge, tapering down to 6.0V at 25% of charge.
  • Charging voltage would be in the neighborhood of 6.5-7.0V

Assuming you are not charging the battery while it is connected to the camera and the 5.0-5.1V and 1.5A is accurate, you would need a 0.8 ohm, 5W or greater resistor to drop the voltage. The variables are the killer. To determine the resistance, use the formula R=E/I or 1.2V/1.5A = 0.8 ohms. To figure the wattage dissipated in the resistor, use P=E*I or 1.2V*1.5A = 1.8W. A 2 watt resistor would get extremely hot so use at least a 5 watt resistor (10W would be even better).

It sounded like he already had a 6V battery, so we need to know what the source is capable of producing first. Am I seeing this wrong?

Yes, because you can't trust the accuracy of the camera's specs and selecting a resistor for just the cell (dropping 1.2V) would allow the voltage to get too high when charging (7.0-1.2=5.8V).

By the way, it would be simpler to maintain 5.0-5.1V with a 12V battery and a voltage regulator IC, since the output voltage wouldn't change with changing input voltages. Unfortunately it's impossible to regulate down 1.2V (6.3V-5.1V) without a complex circuit but from 12V (or higher) it is a breeze.

We would not charge the battery while connected, just swap it out. We have a healthy supply of 6v 12ah batteries.

Option "B", although quite a bit heavier, is just to utilize a relatively small UPS. The whole set up will still fit in a backpack. With all of the companies out there making crappy "spy cameras", you would think there would be better options for wireless IP options.

Hmmm.

I wonder how tolerant those input voltage specs are. Adding a diode in the circuit would give you a .7V drop, making your 6V supply 5.3V. Close enough to 5.1v? Probably, but it's up to you to risk a camera trying it ;) What does the 6V battery read when fully charged? If it's 6.4V or more, 2 diodes would give you a 1.4V drop, getting you right to 5V.

Might be easier to use a diode as a regulator than to use a resistor as essentially a passive heater.

Brian,

It's a common misconception that diodes maintain a constant voltage across their junction. In actuality, the voltage drop can vary from 0.6V to 0.85V or more. 6V gel cells are rated at 6.3V at 100% charge and 6.0V at 25% charge but even that depends on a number of factors. And diodes also dissipate the voltage drop as heat.

That said, the simplest suggestion might work, depending on how critical the circuit is. Two diodes in series, rated at at least 3A (5A would be better) would be a good starting point. If the resulting voltage is more than 0.4V too low, perhaps remove one diode and try again.


Usually you would use an LDO for this, like the L4940V5 or MIC29300-5.0WT. These are simple three-legged ICs that only need two additional capacitors. Like a linear regulator, they burn everything above the output voltage, but only need about 0.5V of "overhead" (LDO=low drop-out). Unlike a resistor, diode or Z-diode, a regulator will keep the specified output voltage even while the battery discharges or is charged.

As with resistors or diodes, take care to get rid of the excess heat. Unless you are using a more efficient switched DC/DC converters, 1V dropped at 1.5A load means 1.5W of waste heat. For little electronic parts without access to a heat dump this is quite a lot.