Hi,

I have a smart lock that runs of 4 double A batteries in series (I think its series). I want to increase the length of life of the smart lock as it only last 5 days! I was thinking of running a car battery to power pack and remove the double A batteries. My impression is that 4 double A's is 6v. Therefore if i get a 12v car battery and use a cigarette lighter that converts the 12v into 5v then i can then run the 5v usb cigarette lighter into the 6v battery pack that has the batteries removed. I would think the smart lock could urn of 5v since it so close to 6v (also when the batteries start to die it will end up providing 5v anyway so it should work).

Is this logic correct? Would this work or would i break the smart lock. Is amps an issue here? Thanks

 

You can do, or just use a Lm2596 pcb and set it to 6v output, a car adapter can be modified to give 6v out.


http://www.ebay.co.uk/itm/like/1713...3D710-134428-41853-0%26rvr_id%3D1117235132702

 

Thanks! but i think the higher amps may burn out some circuitry in the smart lock (cost me $500 so i want to be careful). Do I also need to scale down he amps?

 

was thinking i need a resistor to reduce the amps down to what ever 4 double A batteries in series would be

 

Current is determined by the load, not the source. You need to reduce the voltage and let the current take care of itself.
Basic Ohm's Law, I = V/R.

I suggest you try to figure out why a smart lock only runs for 5 days on batteries. You have something seriously wrong there. Otherwise it's not really a "smart" lock.

 

But won't it only be safe if there are resistors in the lock? The manufacturers may not have thought people would play around with amp variations and not included resistors? For example, won't a LED light burn out if amps are too high because there is no resistor. hmm should i just buy a regulator - bring it down to 6v and 2.4 amp?

The lock dies quickly because I changed the settings to be always online

 

If the lock draws, say, 10mA when running off 6V from batteries, then it will still draw the same 10mA (or less) if run from the 5V car adapter, regardless of the 1A (or whatever) current the adapter is rated at. The 'smart' lock isn't smart enough to figure out where it gets its supply from .

 

Isn't that only true if there are resistors? Wouldn't, for example, an LED light be brighter running on 5v at 2.4 amps apposed to 5v and 1 amps? If thats true, then the components are at risk of overheating?

 

LED lights run on current, not voltage. So your statement would be true, IF THE LED LAMP IS RATED FOR 2.4 AMPS.

Without a resistor or current source, the lamp will certainly burn out.

 

Isn't that only true if there are resistors? Wouldn't, for example, an LED light be brighter running on 5v at 2.4 amps apposed to 5v and 1 amps? If thats true, then the components are at risk of overheating?

No. You are comparing horses with bananas.

A red LED runs on 2.3V. If you pump 5V into an LED you will kill it. An LED does not obey Ohm's Law.

A 6V lock will run on a 6V 1A battery or a 6V 200A battery. No resistor is required.

 

As MrChips noted, you are comparing two different types of devices, voltage operated and current operated.

Most electronic devices are voltage operated.
The are designed to operate from a specific voltage (or range of voltages) and will take only the current they need.
It doesn't matter how much current the source can supply (even a million amps) as long as it can supply the current the device requires at the rated voltage.

A few devices, such as an LED, are basically current operated.
They operate on a specific current (or range of currents), which must be regulated to that value
They will also drop a voltage across them but that is from the power they use (V x I)
That voltage doesn't change much with a change in current (i.e. a small change in voltage across them can generate a large change in current).
Thus there current must be controlled by a series resistor or some form of current source, or the current can exceed the device's rating with only a small change in applied voltage.

 

The lock dies quickly because I changed the settings to be always online

How often was it online before you changed the settings.
Why do you need it always online?