Power/Voltage Drop over cable length!

[chrissfour 0]

I'm starting to realise that this was not thought of before when People Buy a CCTV camera Kit and adaptors and start to find that the Ceiling space may not have nearby Power access.

 

1) When I searched for 12v - 24v Power drop off I found this quite helpful.

 

http://www.cctvcamerapros.com/AC-DC-voltage-drop-cable-distance-s/846.htm

 

Print it out or copy paste etc so if your designing from scratch then this might help.

 

Im currently looking at a FOSCAM brand IP camera that required 5v 2A feed.

http://www.foscam.co.nz

 

2) The company recommended a 3 meter extension the max to the power board. I wanted about 10 -15 meters But then I knew that POE was about and I need not be too restricted in length for our Home use. I also have seen "Injectors" like a splitter carry the power through a Cat5? cable so thought that would be an option too?

 

Question - Am I on the right track or do I need to think of something else?

 

Tia

 

Chris

[almelst 1]

You can't use long power cabling for 5 V supply. Reason: voltage drop does not depend on supply voltage, and for instance 0,5 V is 10% of 5 V , but only 4,2% of 12V and 2,1% of 24 V.

If you need longer power wiring you should use 12V adapter, place a step down DC/DC converter near camera and set output voltage to 5V.

[the toss 0]

I think you should get a better working knowledge of Ohms Law. Voltage drop does indeed depend (to a degree) on supply voltage. Lets look at two examples.

A) DC resistance of a cable run = 2 ohm (this is a fixed value in this example)

DC load (ie camera) = 100 ohm (this is a fixed value for this example)

Total DC resistance = 102 ohm

If DC supply voltage = 12V then current will be 12v/102^ = 117.64 mA

Therefore the voltage drop over the cable will be 117.64 mA X 2^ = 0.235V

 

NOW keeping everything the same but reducing the supply voltage to 5V we get -

 

DC current now will be 5v/ 102^ = 49 mA

therefore the voltage drop over the cable will be 49 mA X 2^ = 0.098V

 

All three operators ( supply voltage , load resistance and current ) will affect the transmission line voltage drop.

[chrissfour 0]

Thankyou for the explanation. I understand what your saying as in the formula.

 

Is it then "feasible" to use an Injector for such a low voltage and hook up to a Cat5 cable? to carry a further distance ... to the power location or simply... the same situation occurs... even with cat5?

 

I understood that a Cat5 would be more efficient in some uses, however is this valid for my use?

 

Thanks

[chrissfour 0]

You can't use long power cabling for 5 V supply. Reason: voltage drop does not depend on supply voltage, and for instance 0,5 V is 10% of 5 V , but only 4,2% of 12V and 2,1% of 24 V.

If you need longer power wiring you should use 12V adapter, place a step down DC/DC converter

near camera and set output voltage to 5V.

 

Thankyou for your thoughts there. I see my idea still needs further thought as to viable options. ie "step down dc/dc" good idea.

 

If the end user then was to consider a 12v system what would the "acceptable distance of power supply" be recommended?

 

Thanks guys.

[almelst 1]

I think you should get a better working knowledge of Ohms Law. Voltage drop does indeed depend (to a degree) on supply voltage. Lets look at two examples.

A) DC resistance of a cable run = 2 ohm (this is a fixed value in this example)

DC load (ie camera) = 100 ohm (this is a fixed value for this example)

Total DC resistance = 102 ohm

If DC supply voltage = 12V then current will be 12v/102^ = 117.64 mA

Therefore the voltage drop over the cable will be 117.64 mA X 2^ = 0.235V

 

NOW keeping everything the same but reducing the supply voltage to 5V we get -

 

DC current now will be 5v/ 102^ = 49 mA

therefore the voltage drop over the cable will be 49 mA X 2^ = 0.098V

 

All three operators ( supply voltage , load resistance and current ) will affect the transmission line voltage drop.

 

The voltage drop depends ONLY on current draw (and cable resistance, of course)!

You made a mistake, electronic device (including camera) is a non-linear load and can't be represented as a resistant load so you can't use Ohm's Law that way. Usually, there is a voltage stabilizer inside (for instance, 12V powered device works on 5V internally with stabilizer at input to allow wider supply voltage range) which makes it draw the same current in allowed supply voltage range ( 10,5 - 13 V or so).

Next, your example is wrong. You can't supply 12V device with 5 V, it won't draw expected current , and if you connect 5V device to 12V source you may burn it.

[the toss 0]

I think you should get a better working knowledge of Ohms Law. Voltage drop does indeed depend (to a degree) on supply voltage. Lets look at two examples.

A) DC resistance of a cable run = 2 ohm (this is a fixed value in this example)

DC load (ie camera) = 100 ohm (this is a fixed value for this example)

Total DC resistance = 102 ohm

If DC supply voltage = 12V then current will be 12v/102^ = 117.64 mA

Therefore the voltage drop over the cable will be 117.64 mA X 2^ = 0.235V

 

NOW keeping everything the same but reducing the supply voltage to 5V we get -

 

DC current now will be 5v/ 102^ = 49 mA

therefore the voltage drop over the cable will be 49 mA X 2^ = 0.098V

 

All three operators ( supply voltage , load resistance and current ) will affect the transmission line voltage drop.

 

The voltage drop depends ONLY on current draw (and cable resistance, of course)!

You made a mistake, electronic device (including camera) is a non-linear load and can't be represented as a resistant load so you can't use Ohm's Law that way. Usually, there is a voltage stabilizer inside (for instance, 12V powered device works on 5V internally with stabilizer at input to allow wider supply voltage range) which makes it draw the same current in allowed supply voltage range ( 10,5 - 13 V or so).

Next, your example is wrong. You can't supply 12V device with 5 V, it won't draw expected current , and if you connect 5V device to 12V source you may burn it.[/quot

 

Lets make this simple.

* As I have show the voltage drop may depend on the current draw BUT the current draw depends on the supply voltage (assuming the total resistance [load + cable doesn't change])

 

* Any load offered up to a DC supply will present as a DC equivelant load. For a given supply voltage it will present as a particular DC load. Things are completely different if the supply is not DC.

 

* The internals of (most) cameras are 12V NOT 5V and will have the normal operating dynamic range of 20%. 24Vac cameras operate in the fashion you describe. Their internals are the same as a 12Vdc camera but have a 24Vac - 12Vdc rectifier/regulator built in. This is why they are used in long (power) cable runs

 

* Your last point seems to contradict everything you have been saying and confirms everything I have been saying. The current depends not only on the load (+cable) but ALSO on the supply voltage.

 

* I have never talked about running a 5V camera on 12V. I have talked about dropping the supply voltage from 12V to 5V and seeing the current change as a result.

 

Hope this is helpful

[almelst 1]

Again: Ohm's Law is not valid in non-linear circuits the way you proposed. Camera can't be represented as same resistor for both 5V and 123V supply (100 mA at 5V would be 50 ohms and 100 mA at 12V would be 120 ohms). Theoretically speaking, camera's equivalent would be a current sink. Current draw is not proportionally dependent on supply voltage. Try measuring camera's current draw throughout full voltage supply range, it will just slightly change. Using the camera outside operational supply voltage is pointless.

And, for the last time: voltage drop is the product of the cable resistance and the current draw and nothing else.

My point was that the same current draw produces much bigger relative voltage drop on 5V than on 12V and the resultant camera's supply voltage goes out of the operational range more easily on 5V rather than on 12V supply. Conclusion: using the same wiring cross-section , maximum cable run is much shorter on 5V than on 12V supply.

[the toss 0]

I'm not going to argue with you, I was trying to enlighten you but you dont want to listen. It will not affect me that you believe what you do.

 

all the best to you

[almelst 1]

Enlighten me !!!! LOL !!! Who do you think you are, Georg Simon Ohm himself ?

[the toss 0]

I'm not entering into a " my dongle is bigger than your dongle" contest. You go your own way.

[almelst 1]

Toss, I suggest we both go our own ways and hope that chrissfour found what he needed.

We should have led the discussion on Ohm's law directly, by email, but it's too late now. Sorry chrissfour, it's your topic after all.

[amy888 0]

Cable is like a pipe. Charges is like the water. Current is like the water flowing rate through the pipe, volume of litres per sec. Voltage is like the water pressure from one point to another inside the pipe. Energy is like the water meter terminal located at the outlet of individual home, counting the accumulating charges that reaches our home. This is how we can have an alternative view on electricity.

Here is a Voltage drop calculator for your refeference:

https://www.elect-spec.com/voltage-drop-calculator.html

[chrissfour 0]

" title="Applause" /> to all who participated! lol

 

Thanks guys