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Understanding Voltage Drop
This thread has 39 replies. Displaying posts 1 through 15.
Post 1 made on Sunday January 20, 2019 at 00:14
tca
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We constantly run into situations where we only have 1 cat5 cable at a gate and need to provide power for a door strike, keypad, and request to exit button. Most situations we need to use 12v dc to accommodate all 3 things. Runs can be anywhere for 100 feet to 300 feet. If we can run new wires, we do, but I want to understand voltage drop.

12vdc 1 amp power supply should be enough to power a keypad (typically draws not more than 200ma), a strike (typically draws no more than 300ma), and a request to exit (typically draws no more than 100ma for led light).

I've seen all the voltage drop calculators, but I don't understand fully. I know the voltage drops, but what about the amps? Can you install a 13 or 14 VDC power supply to make up for the drop? Does that also help with the amps?

Typically we would quadruple up the pairs of a cat5 and connect to a 12vdc 1 amp power supply. As an example, if we do this for a run of around 200 feet, and connect it to the strike, keypad and request to exit button using the above current draws, it may work, but is it mathematically appropriate? If not, what would we need to do to make it correct when we can't run new wires and only have 1 cat5 cable, and no local power?

Thanks for helping me to understand.
Post 2 made on Sunday January 20, 2019 at 00:59
edizzle
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Power it at the gate!
I love supporting product that supports me!
OP | Post 3 made on Sunday January 20, 2019 at 01:00
tca
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NO LOCAL POWER
Post 4 made on Sunday January 20, 2019 at 01:29
davidcasemore
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On January 20, 2019 at 00:14, tca said...
I've seen all the voltage drop calculators, but I don't understand fully. I know the voltage drops, but what about the amps? Can you install a 13 or 14 VDC power supply to make up for the drop? Does that also help with the amps?

Without any load at all there won't be any voltage drop measured at the far end. As you start piling on devices, and adding to the current flow, the voltage will start to drop in a predictable ratio (Ohms Law).

You can increase the voltage at the power supply but you have to be careful. If one thing is drawing current all the time and the others are only momentary then you could eventually burn out the full-time device by running it at a higher voltage than it's rated for.

As far as doubling up conductors, here is good rule of thumb:
When you double up two conductors of equal size the resistance is cut in half.

You can search online for the resistance of a single conductor of Category cable. I think it's around 0.026 Ohms per foot. As long as you know how Ohms law works this should be fairly simple.

Many power supplies for CCTV cameras and low-voltage lighting have multiple secondary taps just for this reason. But you can't just jack up the voltage as high as you want without calculating the parallel conductor resistance, the continuous current draw, the momentary current draw, the low and high voltage range maximums of the devices, the length of the cable run, and the ambient temperature of the cables in the raceway.

Aren't you glad you asked?

ERNIE, As always, please correct any misinformation or missing information that might be contained in this answer.
Fins: Still Slamming' His Trunk on pilgrim's Small Weenie - One Trunk at a Time!
Post 5 made on Sunday January 20, 2019 at 05:58
buzz
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Voltage drop across a single conductor is always I x R where I is the current in the wire and R is its resistance. It doesn't matter what voltage is measured at one end of the wire with respect to ground or other common reference.

"Voltage" is a "potential" relative to some reference point and the reference point must always be stated.

For a similar example, if we ask how high buildings are in Denver, Colorado, and our zero reference is sea level, then every building, even a shack out back, is at least a mile high. For most cases we are interested in height from the base of the building, "ground" level if you like. An aviator, however, usually works with a sea level reference and would typically fly at a mile or more altitude.

A useful circuit technique is to establish a "high" reference and a "low" reference. Usually the "low" is called "ground" and "high" would be "power supply". Now replace each snip of wire with a resistor and current.

The perfect power supply's voltage never sags and it can supply infinite current to maintain this voltage. This voltage is reduced by every snip of wire (I x R) in the external circuit until it reaches the target device. The low side ("ground" if you like) at the target device is sneaky to calculate because its measured voltage (relative to our low reference point) rises due to the loss in each snip of wire along the way and these losses add to the voltage measured at the device low side.

Now, lets look at our hypothetical DC device that has a (+) and (-) terminal. For whatever reason lets assume that the sum of the I x R drops along the positive wire is 0.5V, while the sum of the I x R drops along the negative wire is 0.4V. In this case the voltage measured between the (+) and (-) terminals is 0.9V lower than the supply.

Note that I have deliberately chosen different drops for the (+) and (-). This is very common in real circuits and one must be prepared to recognize and deal with this, but it is a subject for a different day.
Post 6 made on Sunday January 20, 2019 at 10:29
Ernie Gilman
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I'll throw some things in here but I've got several hours of other stuff to do first.
A good answer is easier with a clear question giving the make and model of everything.
"The biggest problem in communication is the illusion that it has taken place." -- G. “Bernie” Shaw
Post 7 made on Sunday January 20, 2019 at 10:36
kwkshift
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This is something you could mock up in your garage to test out. If all the gear runs on 12v and you have potentially very long cable runs, you can purchase larger (more amperage) and adjustable voltage power supplies.

Most gear like this can take some variance in voltage, so if you have a standard 12v/1a ps and you try actuating the strike, it might not work at all since the voltage at the far end may only be 10v, for example. If you put a larger, adjustable ps, say one with variable voltage up to 18v and 2amps of available current, then you could probably set it to 14v and everything should be fine. At that point, you might be reading 12-13v at the far end and when you try actuating the strike, there will be a large enough "bucket" of current available to make it work.

The most basic way to explain it, which is how it was explained to me is that voltage is pushed and current is drawn.

You can have a 12v device that draws 1 amp to do something. If we hook it up to a huge, adjustable ps that is rated at 12v/ 50 amps, it will work perfectly fine.
However, if we connect it to the ps set at 100v/100ma, you'll see smoke instantly, even before we try actuating it.

Hopefully that makes sense.
Post 8 made on Sunday January 20, 2019 at 12:48
highfigh
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On January 20, 2019 at 00:14, tca said...
We constantly run into situations where we only have 1 cat5 cable at a gate and need to provide power for a door strike, keypad, and request to exit button. Most situations we need to use 12v dc to accommodate all 3 things. Runs can be anywhere for 100 feet to 300 feet. If we can run new wires, we do, but I want to understand voltage drop.

12vdc 1 amp power supply should be enough to power a keypad (typically draws not more than 200ma), a strike (typically draws no more than 300ma), and a request to exit (typically draws no more than 100ma for led light).

I've seen all the voltage drop calculators, but I don't understand fully. I know the voltage drops, but what about the amps? Can you install a 13 or 14 VDC power supply to make up for the drop? Does that also help with the amps?

Typically we would quadruple up the pairs of a cat5 and connect to a 12vdc 1 amp power supply. As an example, if we do this for a run of around 200 feet, and connect it to the strike, keypad and request to exit button using the above current draws, it may work, but is it mathematically appropriate? If not, what would we need to do to make it correct when we can't run new wires and only have 1 cat5 cable, and no local power?

Thanks for helping me to understand.

Here's a chart for wire gauge, with current limits. It shows the resistance for 24ga wire at about 25 Ohms/1000 feet, so you would cut that to ~25% by combining the wires and further to 20% of that number if it's 200', so it would be around 1.2 Ohms. You would be better off using a regulated power supply rated for more current than trying to send higher voltage.

[Link: powerstream.com]

The thing that's creating/being a load will cause heat- if the wire has no resistance, it will cause no heat and therefore, no voltage drop. Also, if the temperature of a conductor increases, the resistance increases, too. It's counter-intuitive, but that also increases the current if the power is to remain constant. It's what causes a weak car battery to kill the starter.
My mechanic told me, "I couldn't repair your brakes, so I made your horn louder."
Post 9 made on Sunday January 20, 2019 at 21:59
Ernie Gilman
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On January 20, 2019 at 12:48, highfigh said...
Here's a chart for wire gauge, with current limits. It shows the resistance for 24ga wire at about 25 Ohms/1000 feet, so you would cut that to ~25% by combining the wires...

The resistance of two of the same conductor in parallel roughly equals the resistance of a wire three numbers larger on the chart. Frinstance, two #4 in parallel are about the same as a #21. Take two such setups and parallel them, and you'll effectively have 18 gauge. It's worth sticking in your brain somewhere that a CAT5 cable, using all conductors, is roughly equivalent to an 18 gauge pair.
... and further to 20% of that number if it's 200', so it would be around 1.2 Ohms.

Yes, roughly. But...
You would be better off using a regulated power supply rated for more current than trying to send higher voltage.

Oops, that last assertion doesn't pan out.
A power supply rated for more current will be able to output more current than one rated for a lower current. But you're talking about somehow getting more current out of a supply at 12 volts than out of another supply at 12 volts, with the same load each time. That won't happen. 12 volts divided by the same load in two different setups will be the same current.

What WILL happen, that might influence things, is that if you have an unregulated supply, its output will float up to about 16 volts when there's no load on it, and it will.... it should... output nice clean 12 volts DC when the supply's rated load is being drawn. Important: an unregulated supply will output more volts when less than its rated load is drawn. When more than its rated load is drawn, it will drop slightly from its rated voltage and hum will be introduced into the output voltage.

The thing that's creating/being a load will cause heat- if the wire has no resistance, it will cause no heat and therefore, no voltage drop. Also, if the temperature of a conductor increases, the resistance increases, too. It's counter-intuitive, but that also increases the current if the power is to remain constant. It's what causes a weak car battery to kill the starter.

While that might be true, it's like mentioning the food coloring you'd use to make gummi bears while telling someone how to make boeuf bourguignon. It's the same field, and it's a fact, but it doesn't help and might even confuse someone who's thinking the answers are all related things.

And I still haven't had the time to actually write my response to the original question.
A good answer is easier with a clear question giving the make and model of everything.
"The biggest problem in communication is the illusion that it has taken place." -- G. “Bernie” Shaw
Post 10 made on Sunday January 20, 2019 at 23:51
edizzle
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On January 20, 2019 at 01:00, tca said...
NO LOCAL POWER

How does the gate operate?
I love supporting product that supports me!
Post 11 made on Monday January 21, 2019 at 02:31
Ernie Gilman
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Here are some ideas. I could go on. This is not well organized.

As for a strike or other device far away from your power, you can simulate having power out there by using a capacitor.

What you do is feed power through a couple of conductors (just to get decent current flow), and feed a capacitor in the range of 100,000 to 470,000 mfd. It should take a bit of time, maybe a half second or two, to charge this capacitor.

But then you have a MASS of stored electrical power. A control voltage is then sent to a relay at the gate. That relay closes, connecting the charged capacitor to the load.

Now, about the load. If it's to unlock or open a gate, it has to be able to respond and release, or better yet spring open, the gate. A puny amount of current, just enough to open a relay, can thus make several amperes of current flow into a load, a latch release. (In fact this is the definition of a RELAY: it relays the control of and release of a high current using a low current. It's a relay. Use a diode across the coil.

EDIT START
With the large capacitor, you have a HUGE amount of current for a small amount of time. That's why you want the opener to spring the door open; the voltage on the capacitor rapidly lowers to whatever you'd get with the wire all by itself, and the current lowers to whatever the wire would deliver to the load if the capacitor were not there.
EDIT END

About cameras.
The one manufacturer I've spoken with did not even understand my question when I asked them what the minimum voltage was that the camera could function with. So it's up to you to find out.

Meanwhile, when power flows through a wire, the resistance of the wire lowers the voltage that is delivered at the load. The hot and the ground EACH reduce the voltage. I did the math a bit this morning and came up with a few things.

Four 24 gauge wires, wired in parallel, have about the same resistance as one 18 gauge wire. If you had a 100 foot run of 18 gauge wire (which is 200 feet of wire because there's a hundred feet of hot lead and a hundred feet of ground lead), and a camera (PLUS IR LEDs) that draws 500 mA, you'd have a voltage drop of .6385 volts, so a 12 volt supply would deliver 11.3615 volts. Your 200 foot run will have twice the drop, so you'd have 10.723 volts at the load.

To figure voltage drop, determine the current your load requires. Multiply that current by the ohms per foot that the wire has. Putting wires in parallel lowers the resistance. A helpful rule of thumb is never to beat your wife with anything larger in diameter than your thumb. No, wait, that's the origin of the term... rule of thumb that doubling up wires results in a conductor that is nearly equal in resistance to a wire size three numbers smaller. For instance, two 24s in parallel have the resistance of a 21 gauge wire. Two of those 2x24 in parallel have the resistance of an 18 gauge wire.

Once you have the voltage drop, subtract that from the supply voltage. That's the voltage at the load.

Okay, you know that the voltage drops, but what about the amps....
There are some ways that envisioning delivering water through a pipe can help you. If you have some water (electrons) that can be pumped at a particular pressure (voltage) through a pipe that limits the flow (resistance), some things can make sense.

Your regulated supply is your pump. The size of the pipe limits how much water you can push through it. And if you're trying to use the flow and the pressure of the water to do some kind of work at the far end, the flow allowed by the pipe is crucial. The pipe limits the water flow (current) and reduces the push from the pump (the voltage) available to do the work.

The electrical current is the voltage divided by the resistance, so it will simply be what it will be depending on those amounts.

There's sometimes a misunderstanding about current specs of power supplies. A 12 volt 1 am power supply will supply, say, 500 mA to a camera when its LEDs are on. On the other hand, a 12 volt 500 amp power supply will supply 500 mA to the same camera. Yup, that's right -- the same voltage and the same load mean the same current. What's different between those two cases is that the second power supply could power a crapload more cameras, and the second supply could probably start a fire if its output is shorted, since a short would make it deliver 500 amps until protective circuits or devices kick in. But even this follows the same rules about current -- a short circuit is a circuit with zero ohms, so doing the math, the current would be infinite. The power supply can't do that, so it will put out its maximum current of 500 amps.

You end your post asking if something is mathematically correct. Anything you do is mathematically correct because you can't make it mathemaically incorrect. 3+1 is never 5... Will enough power get to the load? That's the question.

Last edited by Ernie Gilman on January 21, 2019 02:47.
A good answer is easier with a clear question giving the make and model of everything.
"The biggest problem in communication is the illusion that it has taken place." -- G. “Bernie” Shaw
Post 12 made on Monday January 21, 2019 at 06:54
Mario
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1. How does the gate operate? What powers the gate?
2. a capacitor or a battery at the gate can be a great source of power. Long cable run can trickle charge it and plenty of Watts will be available for the local devices.
Post 13 made on Monday January 21, 2019 at 09:45
highfigh
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On January 20, 2019 at 21:59, Ernie Gilman said...
The resistance of two of the same conductor in parallel roughly equals the resistance of a wire three numbers larger on the chart. Frinstance, two #4 in parallel are about the same as a #21. Take two such setups and parallel them, and you'll effectively have 18 gauge. It's worth sticking in your brain somewhere that a CAT5 cable, using all conductors, is roughly equivalent to an 18 gauge pair.
Yes, roughly. But...

Really? Did you look at the chart for the resistance of 24ga vs 18ga? I did and that's why I wrote that the resistance would be ~25%. Jesus! Read, don't skim! I have to tell myself to do the same but if you doubt my statements, the least you could do is look at the source!

Oops, that last assertion doesn't pan out.
A power supply rated for more current will be able to output more current than one rated for a lower current. But you're talking about somehow getting more current out of a supply at 12 volts than out of another supply at 12 volts, with the same load each time. That won't happen. 12 volts divided by the same load in two different setups will be the same current.

Will two power amps produce the same output into low impedance loads? Why not?

Power supply. If one power supply is more robust, the output will be more similar to connecting to a higher impedance load and if not, we'll see them rated for output @1KHz into 8 Ohms. Also, did you NOT see my comment about a regulated supply?

What WILL happen, that might influence things, is that if you have an unregulated supply, its output will float up to about 16 volts when there's no load on it, and it will.... it should... output nice clean 12 volts DC when the supply's rated load is being drawn. Important: an unregulated supply will output more volts when less than its rated load is drawn. When more than its rated load is drawn, it will drop slightly from its rated voltage and hum will be introduced into the output voltage.

While that might be true, it's like mentioning the food coloring you'd use to make gummi bears while telling someone how to make boeuf bourguignon. It's the same field, and it's a fact, but it doesn't help and might even confuse someone who's thinking the answers are all related things.

And I still haven't had the time to actually write my response to the original question.

Maybe you should have responded to the original- you wasted time with your first comment about not having time and here you are, not responding to it again.
My mechanic told me, "I couldn't repair your brakes, so I made your horn louder."
Post 14 made on Monday January 21, 2019 at 09:47
buzz
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With respect to using a capacitor to augment available current for operating the gate lock:

First, there is the assumption that the lock operates from a DC drive. I have no depth of installer experience here, but just listening to the action as I enter various spaces, many locks operate from AC. Of course, as the installer you would have control over this point.

Second, capacitor charging and support times are a consideration. One should not assume that the door operator (the human) will instantly yank the door open after the lock is enabled. My guess is that several seconds of support would be appropriate. Now, what happens if the person failed to meet the timing window and must then wait until the capacitor recharges before the next attempt?
Post 15 made on Monday January 21, 2019 at 09:53
highfigh
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On January 21, 2019 at 06:54, Mario said...
1. How does the gate operate? What powers the gate?
2. a capacitor or a battery at the gate can be a great source of power. Long cable run can trickle charge it and plenty of Watts will be available for the local devices.

Honestly, I think it would be best to use a battery at the gate with trickle charge on two pairs and use the other two pairs to trigger a relay. Unless the battery goes totally dead and the charger is suddenly activated, the two pairs of wire should be enough but I would also make sure their capacity can't be exceeded. For that reason, I think I would try to use a solar cell to charge the battery. I would be more comfortable if I could use the Cat5e for one purpose, rather than two.
My mechanic told me, "I couldn't repair your brakes, so I made your horn louder."
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