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Voltage Drop in a Simple DC Circuit Path

11/10/2012 2:40 PM

I am building a simple low voltage LED lighting system where my 12 volt source is on one end of a dock and my last parallel assembly is 200 feet away.

I am using good 18-2 thermostat wire that has a single wire resistance of about 3.6 ohms per 500 feet.

My sum circuit load should be about .150 amp. (30 mA X 5 in parallel)

Is my circuit length 200 feet or 400 feet? (200 down on the positive wire and 200 back on the negative wire.)

Can I calculate my voltage drop as Vdrop = circuit current X Wire Resistance ?

There will be about 10 wire nuts used to parallel my assemblies. Are the nutted connections going to contribute significantly to the voltage drop?

I am a retired railroader; obviously NOT an electronics technician.

Any information that could me afforded me in this matter would be greatly appreciated.

Thank You Guys and Gals.

Gav

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#1

Re: Voltage drop in a simple DC circuit path

11/10/2012 3:10 PM

You have the correct concept. You should account for the wire resistance for both directions of travel. The resistance for 18 AWG copper wire is a little bit better than you listed at 6.38 ohms/1000ft. You will likely find that your connection resistance will bring you back to close to your original number of 7.2 ohms/1000ft.

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#35
In reply to #1

Re: Voltage drop in a simple DC circuit path

11/17/2012 9:21 PM

You could also use a multimeter (some are very inexpensive), measure the actual resistance of components and then do some quick model in a simulator.

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#36
In reply to #35

Re: Voltage drop in a simple DC circuit path

11/18/2012 2:23 AM

"You could also use a multimeter (some are very inexpensive), measure the actual resistance of components"

First off we're talking about LEDs which are nonlinear devices as I explained in post #28 so you can't just measure the resistance you need to be able to measure the Vf voltage for the LEDs.

Most multimeters do in fact have a diode test function that will give you Vf for a given diode. However, white and blue LEDs typically have Vf values of 3.2 to 3.6 V and from my experience the cheaper multimeters just can't generate diode test voltages high enough to turn them on. Also older multimeters that were manufactured before blue and white LEDs became readily available often weren't designed to test diodes with such high Vf values so they can't turn on white and blue LEDs making them useless in this situation.

Mind you, I'm not certain that we are talking about LEDs here because from Gavilan's description of how the devices he is using function they aren't acting like typical white LEDs so testing them with a multimeter could possibly give us a better idea of what is going on.

Actually given Gavilan has quoted voltages it would seem that he may already have a multimeter, so Gavilan do you have access to a multimeter and if so does it have a diode test function on it?

PS: I haven't had a chance to play with your link but it looks interesting and I will definitely have a play with it when I get a chance so thanks for posting it.

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#38
In reply to #36

Re: Voltage drop in a simple DC circuit path

11/18/2012 4:02 AM

I think you will enjoy the link, it is a good tool.

.

My reply was aimed at the questions the OP was asking about specifically. The questions had to dowith resistance of a length of wire and a question about contribution of wire nuts to resistance. (I rechecked before I typed this reply and those are the only places that any question marks are placed).

(But re-reading for content a little slower this time.....it looks like you have a point...moreis being asked than is at first apparent...running out of time to edit this post.... so leaving this as is.... Checkout the link I provided in the previous comment. it is a great simulator ..free)

.

The OP provided information on the LEDs leading me to believe a spec sheet is being utilized and that those bases were covered.

.

I think anyone inexperienced working on a project like this would better off with a multimeter, a simulator, and some time to gain a working knowledge of both (not that either is in the least bit challenging)

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#2

Re: Voltage drop in a simple DC circuit path

11/10/2012 4:53 PM

I am a retired railroader; obviously NOT an electronics technician.
Yeah, but you write a good concise question with all the relevant information which makes a refreshing change.
If I could a award a 'good question' rather than a 'good answer' I'd do so!The nutted connections are an interesting question, very old computers used wire wrap back planes because the wire wrapping was a better joint than solder, so my guess is they would be good low resistance connections, but I'd be interested to hear the experiences of others.
Del

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#3
In reply to #2

Re: Voltage drop in a simple DC circuit path

11/10/2012 5:36 PM

The wire wrapping of old back planes and sockets resulted in lower contact resistance because the wire wrap pins were square cylinders with sharp edges. The wire and pin edges would cut into each other at four points on each wrap rotation. This abrasion would cut through any oxidation on wire or pin for at least a dozen points on a good wrap. This is why any unwrapped wire could not be reused (usually) unless one shortened and re-stripped the unwrapped end.

This phenomena of using sharp edges in connecting is why I prefer using only new wire nuts when I use wire nuts to connect two wires. The thread edges found in the wire nut cut into the conductor of each wire as the two wires twist together.

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#5
In reply to #3

Re: Voltage drop in a simple DC circuit path

11/10/2012 8:15 PM

Hey! I remember wire wrapping!

I remember wire wrapping a Motorola 68020 computer for myself using one of the very first 68020 PGA samples that I coaxed from a very young and beautiful Motorola field rep.

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#6
In reply to #5

Re: Voltage drop in a simple DC circuit path

11/10/2012 11:29 PM

Some backplanes on modern Siemens automation equipment is still wire wrapped.I had to show some newbies how to use the tool.

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#12
In reply to #5

Re: Voltage drop in a simple DC circuit path

11/11/2012 9:38 AM

I still have my old wire wrapping tools. Every prototype was wire wrapped. What a messy looking animal on the back of the perf board !

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#13
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Re: Voltage drop in a simple DC circuit path

11/11/2012 9:42 AM

Mine should be in the attic somewhere, but I have not seen them in years!

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#15
In reply to #13

Re: Voltage drop in a simple DC circuit path

11/12/2012 1:01 AM

I still use mine on a regular basis for building one off circuits that I embed in scale models to do all the lighting and things like that. I used to use 555 timers to do the fancy stuff like strobes, but have since moved on to microcontrollers. That way I can use a simple common board that I have designed, wire all the LEDS up to it using wire wrap wire and programme the microcontroller to do whatever I want. No messing about trying to solder wire on to LEDs or PCB, wire wrap works a treat in situations like this.

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#4
In reply to #2

Re: Voltage drop in a simple DC circuit path

11/10/2012 5:45 PM

I agree that this was a wonderful well thought out and presented question worthy of a "good question" award.

Thank you for pointing out that I had not added any insight to the wire nut question.

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#7

Re: Voltage Drop in a Simple DC Circuit Path

11/10/2012 11:31 PM

First I want to thank all of you for replying to my question.

I am trying to reduce the bug scat on the boats (including mine) by using directional LED lighting at our marina. It is a huge problem here on Lake O the Pines.

Using Red's figure of 7.2 ohm / 1000 feet (factoring the nutting) leaves me with .0072 ohm / foot. At 400 feet of circuit path, this gives me 2.8 ohm of wire resistance.

If my sum circuit load is .150 A, and using Vdrop = IR , gives about .432 v drop from my source voltage at my farthest assembly. Is this correct? Is that the right way to use that equation (E=IR)??

I have noticed that when I charge my small gel cell batteries with my 20 watt panel I get about 12.8 volts of source no load voltage. This is for a very low power system that is WAY overbuilt because I was using what I had available.

If my new gell cell charges that high then I should still have about 12.368 volts to drive my farthest assembly. Is that correct?

My plan, unless I am screwing up, is to charge a 12 volt 12ah gell cell with a 30 watt panel. This should give me enough charging power to add another gell cell for the other dock if this works. The charge controller should keep it from overcharging, I hope.

I think this means the battery can supply about 144 watt hours of energy; is that correct?

I am using a gell cell because I have had good luck with my current one that drives my car port lights, I don't have to check the fluid levels, AND it only costs $55.

A 14 hour night should use about 12v X .15 A X 14 h = 25.2 watt hours which gives me about 5 nights of lights if the charging conditions are poor.

At 50% of rated power/charging efficiency I should still get about 15w X 8hr = 120 watt hours a day out of my charging system, in the winter.

Are there any glaring screw ups here?

Thank You

Gav

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#9
In reply to #7

Re: Voltage Drop in a Simple DC Circuit Path

11/11/2012 4:56 AM

I think that your system should work a little better than your factoring since your LED light are in parrallel and you are only driving 30mA the 200 feet and the 150mA through the first section, with reductions of current after each light. I can't see why it would not work.

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#18
In reply to #9

Re: Voltage Drop in a Simple DC Circuit Path

11/12/2012 3:16 PM

Wow - good point - Thanks!

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#10
In reply to #7

Re: Voltage Drop in a Simple DC Circuit Path

11/11/2012 6:29 AM

Gav, The reasoning in your last post looks sound to me except for the solar production forecast. The simple formula is: 77% x Wattage x 'peak hours'. 'peak hours' is a bizarre concept if you ask me, but it gets the job done. It is measured in kWh per m2 per day, but you can use it as just hours/day in the above formula and it will give you kWh produced. So example, for Lurkin, TX in Dec.: 30 watt panel x 3.6 peak hours x 77% = 83.16Wh I am not sure there is a place on earth, no matter the day of the year, where you get 8 peak hours. If I remember correctly there are a few places like the desert in Northern Chile where they get 7+. To find the peak hour for your specific place and month, go to: http://eosweb.larc.nasa.gov/cgi-bin/sse/grid.cgi You can also get a forecast at PVWatts http://rredc.nrel.gov/solar/calculators/pvwatts/version1/US/Texas/ but it will give you a number for the monthly production. Denis.

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#19
In reply to #10

Re: Voltage Drop in a Simple DC Circuit Path

11/12/2012 3:19 PM

Great info Solarguy - thanks!!

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#8

Re: Voltage Drop in a Simple DC Circuit Path

11/10/2012 11:37 PM

I do not see any problem with the wire nuts as long as you use the proper size for the wiring you use.Some wire nuts are made for high moisture locations and have a gell-fill inside to seal out moisture.A crimp type connector with heat shrink sleeve is a more reliable method,however.

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#11

Re: Voltage Drop in a Simple DC Circuit Path

11/11/2012 6:46 AM

Have you considered running the power to the middle of the LED light "string"? This should have give a more even appearance to the light level by cutting the distance from the "first & last" lights in half.

Common sense is an oxymoron and the world if full of morons! (I am not one of them)!!!

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#20
In reply to #11

Re: Voltage Drop in a Simple DC Circuit Path

11/12/2012 3:23 PM

I really had considered that; but the docks are covered and run N/S. Panel and battery placement is considerably less problematic on the south end of the covered dock.

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#14

Re: Voltage Drop in a Simple DC Circuit Path

11/11/2012 4:03 PM

1. Is my circuit length 200 feet or 400 feet? (200 down on the positive wire and 200 back on the negative wire.)

-- your circuit length is 400 feet.

2. Can I calculate my voltage drop as Vdrop = circuit current X Wire Resistance ?

-- Vdrop = Vsource (12V) - (circuit current X Wire Resistance)

3. There will be about 10 wire nuts used to parallel my assemblies. Are the nutted connections going to contribute significantly to the voltage drop?

-- I doubt it. Their effect should be negligible

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#21
In reply to #14

Re: Voltage Drop in a Simple DC Circuit Path

11/12/2012 3:26 PM

Thanks for letting me know I wasnt using "voltage drop" in proper context; and especially in the tactful way you did it. You would make a great teacher.

Gav

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#16

Re: Voltage Drop in a Simple DC Circuit Path

11/12/2012 7:10 AM

A slight deviation from the original question.

What is the forward voltage drop of your LEDs?

For all these sorts of LED projects you are better off trying to drive as many as possible in series rather than parallel.

For example if your forward voltage drop is 3.6V then two parallel strings of three in series would give you a total voltage across the LEDs of 10.8V you then just need to calculate the the values of the resistors to provide the correct current, and you'd be using two fifths of the current to drive 5 in parallel.

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#22
In reply to #16

Re: Voltage Drop in a Simple DC Circuit Path

11/12/2012 3:38 PM

There are 2 parallel branch circuits in each assembly where each branch is 4 - series LEDs where the spec sheet says nominal voltage is 3.2 volts. I have run 4 from a 12 volt source in the past and they seem to be plenty bright; but never from a solar powered system where the voltage path is this long.

I try to avoid using resistors because I think they waste power - right?

Gav

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#24
In reply to #22

Re: Voltage Drop in a Simple DC Circuit Path

11/12/2012 4:53 PM

Technically resistors do waste power. However, you should have at least one resistor in series with each series of LEDs to limit the current flowing through the LED as the temperature of the LED and voltage fluctuations change. A LED is a diode remember and follows the very non-linear Shockley diode equation. Putting a "power robbing" linear resistor in series will allow voltage changes from the primary supply, or other induced voltage sources and even the diode voltage drop change with temperature to not over drive the diode.

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#25
In reply to #22

Re: Voltage Drop in a Simple DC Circuit Path

11/13/2012 12:58 AM

"I have run 4 from a 12 volt source in the past and they seem to be plenty bright"

Please, please, please tell me that you are not just connecting four LEDs in series to a 12V source without a current limiting resistor. Because if you are you will blow the living daylights out of the LEDs when the charging circuit tries to charge the Sealed Lead Acid Battery you are using to power the system.

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#26
In reply to #25

Re: Voltage Drop in a Simple DC Circuit Path

11/14/2012 7:21 PM

The automatic dusk/dawn switch isolates the LED's when charging.

When connecting 4 - 3.2 volt LED's in series and running it from a 12 volt source what would be the series resistance I would need to add to protect that branch?

My Radio Shack book says the equation for the series resistance is R= Supply Voltage - LED Voltage / LED Current.

The Supply voltage at peak battery charge should be about 12.8 volts. 4 series 3.2 volt LED's sum to 12.8 volts.

Plugging those numbers into the equation show that no series resistance is required.

Although the forward continuous current is given as 30 mA they will hand a pulse current of 100 mA.

I also protect each circuit with fuse between the source and automatic switch.

On the plug in systems (8) I have operating here in East Texas and the Kingdom of Tonga I haven't used current limiting resistors. 5 of these systems have been operating for up to a year.

The single solar system I have operating has been operating since March without series resistance and so far nothing has happened.

Gav

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#27
In reply to #26

Re: Voltage Drop in a Simple DC Circuit Path

11/14/2012 11:55 PM

You've assembled precisely the combination that implies to me possibly the most problems for you by this being a not well controlled scenario. Having four series 3.2V LED (assuming that this is a version of white LED) implies that only semiconductors achieves this LED voltage drop. Thus any supply surge voltage or a junction voltage change can cause a significant current change that significantly dims (flickers) the lights or burns out the LED closest to the supply and eventually all of them.

Putting groups of three 3.2V 30mA LED in series with a single 80 ohm resistor will mean that each resistor will need to dissipate 0.048 watts of power while each of the three LED will turn to light 0.096 watts. (This assumes that the rating of your LED is 20mA @ 25°C.)This will not be as efficient as having no resistors in series but if you walk through the Schokley diode equation earlier you'll find that the briefest of voltage surges to 15V across the whole package of three or four LED can quickly exceed the surge rating of 100 mA if no current limiting resistor but cable resistance is utilized.

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#29
In reply to #27

Re: Voltage Drop in a Simple DC Circuit Path

11/15/2012 9:45 AM

Red and Masu;

Thank you for your thoughtful replies. Your time and energy is much appreciated.

Would it be possible to use one resistor ahead of the first assembly to protect all of the assemblies?

Would it be possible to use capacitance in parallel with the power circuit to act like a shock absorber in the case of voltage spikes?

What type of events could cause a voltage surge in the battery powered circuit?

Thank You

Gav

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#31
In reply to #29

Re: Voltage Drop in a Simple DC Circuit Path

11/15/2012 10:44 AM

It would be possible to use one series resistor but I would not recommend it. Again going back to the Schokley diode equation:

The temperature of the PN junction will change the voltage drop and current draw with each diode location. This is why diodes are used as precise devices for the measurement of temperature. So in your application you may get the brightness of lamps changing with the local temperature swing. A more practical problem of using just one resistor becomes apparent when one set of LED fail to draw current for any reason. (bad contact, junction failure, boat impact,etc.) The current flowing through the resistor will drop. The voltage across the resistor will also drop making the voltage across the remaining LED higher. So a break in the middle of your wiring can fry the half of your LED array that is still connected to your supply. Putting a current limiting resistor in each LED series will mean that a break anywhere will not damage the rest of your circuit. Now a local array of six LED in two series with one current limiting resistor in a lamp location may provide more local light and wire break protection.

Now as for the possible voltage spikes across a battery. Your several hundred foot wire array will still act as an antenna capable of picking up interference. I admit that a very strong EMP spike will be needed to damage a LED but lightning does produce significant EMP without having to be a direct strike. Capacitors can provide ways to deposit this energy but predicting where the capacitor should be placed and what value should be used gets complicated very quickly.

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#28
In reply to #26

Re: Voltage Drop in a Simple DC Circuit Path

11/15/2012 4:11 AM

The problem with diodes is that they are nonlinear devices which means that the current flowing through them is not related to the voltage across them in a linear manner.

To explain I will be using the diagram below which shows the voltage applied across the device on the horizontal axis with positive to the right and the current flowing through the device on the vertical axis with positive being up.

The BLUE line show what happens when we apply a voltage to a resistor. If we start at the origin with no voltage across the resistor we have no current flowing through it. Now if we increase the voltage in the positive direction then the current flowing through the resistor increases in a linear relationship to the increase in voltage. The exact opposite happens if we apply a negative voltage, the current flows in the opposite direction but the relationship between the voltage and current stays on the straight BLUE line. We therefore say that resistors are linear devices because the relationship between the voltage across them is directly related to the current flowing through them in a linear manner. In other words it's called Ohm's law and can be expressed mathematically as:

Now let's look at the RED line which represents the voltage across and current through a typical diode. If we start again at zero and increase the voltage slowly at first nothing happens. Then when we get to a point called Vf which in your case is 3.2 V the current suddenly starts to rise extremely rapidly. This is because the diode will not pass any current through it until it reaches a certain point at which time the diode suddenly turns on and starts to conduct current. Now the important thing here is to note how steep the RED line is after Vf. This means that with very little increase in the voltage across the diode you will have a dramatic increase in the current flowing through the diode. Because of the steepness of the RED line after Vf it means that it is very easy to burn out a diode should the voltage across it go much past Vf and to prevent this damage you have to place a resistor in series with the diode.

Redfred has worked out that with three of your diodes in series with an 80 Ω resistor will give you 30 mA through each of the LEDs while limiting the current to 68 mA which your LEDs have shown to be capable of handling should the voltage rise to as much as 15 V.

On the other hand if the voltage were to rise to 15 V without the resistor the current through four diodes in series would likely be well over the LEDs maximum rated current and result in at least one if not all the LEDs burning out.

Another point worth noting is that with four diodes in series should the voltage drop below 12.8 V which is four time Vf then the diodes will not turn on at all because the voltage across each of the diodes is below Vf. On the other had with 3 diodes and a resistor in series you system will operate as low as 9.6 V although at that voltage the light would be very dim.

Something else worth noting is that with sealed lead acid batteries it's not unrealistic to expect voltages as low as 12 V or even less if the battery is heavily discharged which with your 4 diodes in series that requires 12.8 V minimum to operate would result in it shutting down completely. However with 3 LEDs and an 80 Ω resistor the system will still operate albeit at a lower intensity, but that's better than nothing whatsoever.

NOTE: I haven't gone into what happens when a diode or LED is reversed biased as it's not applicable in your application, but if you would like to know what happens just ask and I will explain.

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#30
In reply to #28

Re: Voltage Drop in a Simple DC Circuit Path

11/15/2012 9:59 AM

Masu;

Thank You for your very comprehensive and informative reply.

As info; the forward trigger voltage (the emf at which the diode begins to conduct) is well below 3.2 volts. I don't know exactly where it is at but I have played with these things with my little homemade variable voltage supply and I know it is well below the recommended operating voltage of 3.2 volts. The spec sheet says these things will operate in continuous service up to 4 volts. I have accidentally put 6 volts across them for a few seconds and they still work.

I am about to go back down to my playhouse so I will check what the trigger voltage is.

Thanks again.

Gav

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#32
In reply to #30

Re: Voltage Drop in a Simple DC Circuit Path

11/15/2012 3:08 PM

Can you give us the part number of the LED you are using? It doesn't sound much like the ones I am familiar with.

It almost sounds as though it has a series resistor built in, but, then on the other hand 3.2V is already a very low forward voltage for a white LED.

Ahh. Are they white LEDs?

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#33
In reply to #32

Re: Voltage Drop in a Simple DC Circuit Path

11/15/2012 9:47 PM

ETG-5CGWHT-15

18,000 mcd -

Two in series (6.4 volt) will begin conducting at about 3 volts. I have exposed a two series assembly (6.4 volt) to 12 volts for a couple of seconds without apparent damage.

Individually they are designed to operate at up to 4 volts.

They are great little devices but cost me about $1.60 each. I have found super bright blue, yellow, and green designed for the same 3.2 votlage and have white and blue operating on the same line in one system.

Gav

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#34
In reply to #33

Re: Voltage Drop in a Simple DC Circuit Path

11/16/2012 2:33 AM

ETG-5CGWHT-15 sounds like some after manufacture stock number because I can't find any reference to that part number on any of the data sheet sites I use, so at the moment I haven't got a clue about the type of LED you are using.

I also did some checking and for all the white LEDs I looked at they all had a Vf of 3.2 to 3.6 V.

Having a white LED turn on with a Vf as little as 1.5 V (even a red LED needs about 2.2V) and then survive a Vf of 6 V as stated in post #33 doesn't add up.

One question, do these LEDs work if you connect them backwards? If they do work backwards then they aren't LEDs because LEDs like all diodes will not conduct when reverse biased, at least not until you reach Vr at which point they will turn on but this voltage is usually considerably higher than Vf so in your situation if you wire them up backwards they shouldn't work.

Something fishy is going on here.

Sorry the pun was not intentional.

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#37
In reply to #34

Re: Voltage Drop in a Simple DC Circuit Path

11/18/2012 2:35 AM

Thanks for the reply.

Are the forward voltages referred to the nominal operating V or the trigger voltages?

I think common diodes start conducting at about .6 volts.

What I did was connect my VOM in series with a 2- LED subassembly and began to slowly increase my supply voltage from 0 until my meter indicated current flow or I could see the LED begin to glow very faintly. I am confident these are indeed LEDs; but after an off thread message with Randal I think I may be paying WAY to much for the LED's I am using. I will order some of the Cree devices from digi-key he linked me to and do some destructive testing to see which I think are the most robust.

The ETG-5CGWHT-15 may be an obsolete device, I don't know; but they are tough as nails. You can get the data sheet at allelectronics.com part number LED-127. They cost me $1.60 each compared to the 23 cent devices that Randall linked me to.

Gav

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#39
In reply to #37

Re: Voltage Drop in a Simple DC Circuit Path

11/18/2012 8:21 AM

According to the data sheet the LED you are using has a typical Vf of 3.2 V so it shouldn't be conducting anything at all until the voltage across it reaches 3.2 V.

However, if you look at the graphs on the last page you will one that details If against Vf (top left hand corner of the page)

This graph indicates that the LEDs will start conducting current when the forward bias voltage Vf reaches as little as 2.8 V, however the current remains fairly small until the Vf reaches about 3.0 V after which it rises rapidly with a small increase in Vf. Also from the slope of the curve after a Vf of 3.0 V tells us that the LED has an internal resistance of approximately 8 Ω. It doesn't show it on the graph but we can calculate that if a Vf of approximately 3.72 V will result in the maximum If of 100 mA being achieved and the diode can only tolerate this for 100 µs after which it must be off for 900 µs to cool down. That means a sustained Vf or 3.72 V or above will almost certainly damage the junction and either degrade the performance of the LED or destroy the junction completely.

What the data sheet tells us is that the LED is designed to have a maximum continuous If of 30 ma and that at a and If or 20 mA the typical Vf will be 3.2 Volts. That tells us that when you design the circuit you should be aiming for a Vf of approximately 3.2 V if the LEDs are to operate continuously. By the way continuously in electronic terms means for more than a couple of milliseconds not 8 to 16 out of 24 hours.

You said you had a VOM or in other words multimeter. If it's like most multimeters it should have a diode test function, usually indicated by a diode symbol on the function selector switch.

If it does have this feature it may not be able to test white LEDs but it's at least worth a try so could I get you to try something for us.

First off set the meter to the diode test function then connect the red lead to the LEDs anode and the black to the cathode. Your LED should light up (probably very dimly) and you should see a reading on the display of something like 2.8 to 3.0. This is the Vf or forward bias voltage of the diode and it shouldn't conduct any current until you reach this voltage.

If you do this and it does work then try reversing the polarity of the leads (red to cathode and black to anode). In this configuration the diode should not light up and should not give you a reading on the display.

The results of this test will tell us a considerable amount about the devices you have and how they are operating.

Now if I were using the LEDs detailed in the data sheet with a 12 V supply then I would either use three in series with an 80 Ω resistor or if you need four LEDs to give you the intensity you need then two banks of two LEDs in series with a 180 Ω resistor. Both of these configurations will result in 30 mA flowing through you LEDs which is their designed If as well as limit the current to less than 100 ma which is the maximum short term If for the devices should the voltage go over 12 V.

Another point that's worth making is that since you are powering this ultimately from solar cells efficiency is a factor that must be considered. The two options above are designed to limit the current to 30 mA which means they will both draw 360 mW. Now since one option uses three LEDS then you effectively get 54,000 mcd (or 54 cd) for your 360 mW. On the other hand if you need four LEDSs to give you 72,000 mcd (or 72 cd) then the current draw at each lighting point will be increased to 60 mA and therefore your power consumption will increase to 720 mW. Put in other word to get 33% more light you need to double the expenditure of energy which makes your system considerably less efficient and will effectively halve the time the system can operate before exhausting the battery.

Which bring me to another problem. Sealed lead acid batteries should never be allowed to run absolutely flat because if you do you will destroy them and they will not recharge or hold a charge properly. So have you thought about the chance that the battery could become close to being exhausted? If you haven't then you should probably consider including a low battery voltage detector/shutoff circuit that disconnects the load from the battery should the voltage drop below a point where the battery could be damaged. It should then prevent the load from being reconnected until the battery has been sufficiently recharged to at least fully drive the load without the voltage dropping below 12 V.

Sorry for throwing all this at you, but as you are probably getting the idea good design of electronics isn't just a matter of throwing things together and getting them to work. You have to take into account a whole host of other things that may affect the operation of your system particularly when you are designing automatic systems that use rechargeable batteries, solar panels and will have to operate autonomously for lengthy periods of time.

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#40
In reply to #39

Re: Voltage Drop in a Simple DC Circuit Path

11/19/2012 3:45 PM

Although this is a reply to Masu's post most of it is really for Gavilan.

I agree with most of what Masu has said, but, with a one notable exception.

The 30 mA figure is in the absolute maximum ratings: data sheets normally include a disclaimer something like this:-

_________________________________________________________________________

Stresses beyond those listed under "absolute maximum ratings" may cause permanent damage to the device. This are stress ratings only, and functional operation of the device at these or any other conditions beyond those indicated under "recommended operating conditions" is not implied. Exposure to absolute-maximum-rated conditions for extended periods may affect device reliability.

________________________________________________________________________

I would always try to design things so that they fall in the recommended characteristics. In this case that means aiming at 20 mA (but see below).

LEDs have a half life. I would normally expect a normal LED to emit half the light after 10,000 hours of operation at 20mA (I can't offer confirmation of this figure, and, of course it may be different for any make or brand). This life is very dependent on temperature, so running one at 30 mA would reduce the half life considerably. On the other hand running one at 10mA would increase the half life much more than doubling it.

10,000 hours is just over a year or about 4 years at 8 hours per day. I'd guess that after 5 years (at 8 hours per day) a design which aimed to run the LED at 10 mA would be still emitting about the same light as one than aimed at 20mA. Of course the 10mA design would start at the same brightness as the 20mA design would reach after 4 years.

"Twice as much" light is not nearly as much as you might imagine you could probably see the difference if someone turned the current up or down but if someone just turned one on at 10 or 20 mA you'd probably be struggling to see the difference. (I might be destroying my argument here but I think you get the drift).

I think that Gavilan may well have misunderstood the optical/electrical characteristics part of the data sheet:-

In particular the line which reads Forward voltage, Typ: 3.2V, Max: 4V

This does not mean that you can apply 4V across a diode.

All components are subject to variations in their specifications. Manufacturers try to keep the important ones (variations) to a minimum, and allow the not so important ones to spread across their bell curve.

What the above line means is that if you took a statistically significant sample of the LEDs and applied 20mA (see the "conditions" column) to them then on average they would have a forward Voltage of 3.2V but some of them would have a smaller forward voltage and some larger up to a maximum of 4V.

This does make it difficult to design a simple optimum circuit, without deliberately wasting quite a lot of power.

On the whole I think I'd use a constant current design like this:-

D1 and D2 are 1n4148s; Q1 is a 2N3906; R1 = 27KΩ, and, R2 = 62Ω

You might need to "trim" the value of R2 a little, but even if you left the design as it is you wouldn't go far wrong.

Basically D2 matches the voltage drop across the base emitter of the transistor so that D1 then effectively sets the voltage across R2 to be about 0.6V, and 0.6Volts across 62 Ohms gives you approximately 10mA.

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#41
In reply to #40

Re: Voltage Drop in a Simple DC Circuit Path

11/20/2012 10:23 AM

"The 30 mA figure is in the absolute maximum ratings"

The 30 mA maximum If is only for continuous current, the next line in the data sheet says you can go as high as 100 ma provided it's a 10% duty cycle and only on for 100 µs (0.1 ms) then off for a minimum of 900 µs or in other words only on for 10% (1/10) of the time.

However, I wouldn't recommend ever pushing this device that hard and I agree that ideally from the data sheet 20 mA is about as hard as you should drive it.

Your idea for designing it to use 10 mA instead of 20 mA to increase the diodes life expectancy has a lot of merit and you would only lose 40% of the intensity according to the If - Relative Intensity chart on the last page of the data sheet. If a 40% drop in intensity is too much then go for 15 mA. This will still increase your LEDs life expectancy and only result in a 20% drop in intensity which I doubt many people would notice.

Also what do you think about my comment about protecting the battery from being over discharged?

Given the Vf - If chart that indicates that these LEDs will start conducting at a Vf as low as 2.8 V, three in series would be capable of flattening a SLA (Sealed Led Acid) battery well and truly below its minimum safe discharge voltage.

Therefore, there is always the chance that given a long enough spell of bad weather when the battery isn't fully recharged during the day that over time the battery will steadily lose charge till it's discharged below its safe minimum.

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#42
In reply to #41

Re: Voltage Drop in a Simple DC Circuit Path

11/20/2012 11:17 AM

Pulsing current through an LED so that a higher current and maximized light output is done all the time. Remember our eyes perceive light through a chemical process that integrates flashes of light into a continuous image. This is why the monitor we're staring at does not seem to flicker and movies can be made. (There's a single word that describes this. I just cannot come up with that term right now.) I did not bring this up earlier because I was fearful of overwhelming Gavilan with useful but complicating concerns.

Your concern of completely discharging the battery is a valid concern that Gavilan must choose if this is critical concern. I would likely handle this by implementing a similar current sourcing design at each multiple LED lamp like the one Randall offered. I would instead use a FET biased into saturation (constant current) mode for all of the LED. I would also select the FET to have a pinch OFF voltage at about half of the battery voltage. Now an ideal FET with all the right parameters might not be readily available to work. With Gavilan dabbling into electronics I did not want to confuse him more with FET parameter selections and explaining the equations of triode and saturation operation of a FET.

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#43
In reply to #42

Re: Voltage Drop in a Simple DC Circuit Path

11/21/2012 3:41 AM

Hi redfred & Gavilan,

Having brought up the subject of protecting the battery from being over discharged it may not be necessary as Gavilan is using a solar battery charger/control unit and any well designed unit like that would more than likely have a built in cut-off circuit that prevents the battery from being over discharged.

I had a look at the data sheet for the SLA batteries I use for my telescope's control system and for Gavilan's application I would not recommend letting the on‑load voltage drop below 10.5 V. I would then not let the load be reconnected until the on‑charge voltage reached 13.5 V which is the recommended charge voltage for the battery when used in a standby situation. However, in Gavilan's application the battery is being used in a cyclic application so the normal charge voltage for the battery would be between 14.4 and 15.0 V provided the charge current doesn't exceed 1.62 A.

The voltages batteries generate and need to charge is dependent on the chemistry of the battery so in Gavilan's situation the above voltages wouldn't be far off the mark provided he is using a SLA battery, however, the maximum charge current will be dependent on the storage capacity and design of the battery, but that would controlled by the battery charger system provided it has been matched to the battery being used, so there's no need to worry about that.

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#44
In reply to #43

Re: Voltage Drop in a Simple DC Circuit Path

11/21/2012 5:40 AM

'.....I would not recommend letting the on‑load voltage drop below 10.5 V......'

.

I would not recommend the voltage for any 12V SLA (or any other 12V lead acid battery) be allowed to drop anywhere near as low as 10.5V in normal service, without at least a reasonable explanation of what is being gained at the expense of shortened battery life.

.

It isn't hard to undestand that the discharge limits set on the data sheet need to be in line with the advertized amp hour rating of the batteries. Designing to utilizing that full range on a lead acid battery instead of perhaps go with a little additional capacity up front, isn't likely to save money. Regularly discharging batteries 12V lead acid batteries down below 11V is likely toresult in much higher lifetime costs in many cases.

.

Since this application is fairly low current over a fairly long period, the curves representing highcurrent draw aren't really relevant.

Really the upper most discharge curve is relevant in this case. For longevity, not discharging below 12.5 in normal operation would be good., but certainly not below 12.0V, without strong reason.

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#45
In reply to #44

Re: Voltage Drop in a Simple DC Circuit Path

11/21/2012 8:34 AM

The voltages I quoted were with the load connected, without the load the voltage would more than likely be up around the 12 V mark, although it would depend on the load. Nevertheless according to the charts on the data sheets it's what they are rated to, however, having said that erring on the side of caution is always advisable.

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#17

Re: Voltage Drop in a Simple DC Circuit Path

11/12/2012 3:14 PM

You folks are awesome !!! Thanks for all your input and great ideas.

I am going forward with 5 parallel light assemblies consisting of 2 - 4 series - 3.2 volt LEDs in each assembly.

It will be powered by a 30 watt panel and a 12 volt 12 ah Gel Cell.

You guys really helped me to move forward in confidence.

Hopefully next season the bug scat will be a little less a problem for me and my fellow slip renters at Marley's Bullfrog Marina on Lake O The Pines !!!

Thanks Again

Gav

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#23

Re: Voltage Drop in a Simple DC Circuit Path

11/12/2012 3:58 PM

Here is a couple of chicken scratchings showing the basic circuits. The first one omits the battery fuse - but there will be one there. The marina will be using a 30 watt panel instead of a 20 watt.

Here is the automatic dusk/dawn switch CR4 helped me with a few months back. And it works great!!

I once used a shunt to the positive lead of my panel to control the gate but this circuit works well for both plug in and solar powered supplies - without modification.

Gav

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#46

Re: Voltage Drop in a Simple DC Circuit Path

11/22/2012 12:46 PM

I used the battery I will be using at the marina in a charged state of 12.7 volt.

All readings were at the battery terminals.

Each Assembly consists of two parallel branch circuits with 4 series LEDs in each branch circuit.

I put one meter in series at the positive lead of my battery source and measured the circuit current in the "lights off" state. ( Sum parasitic power of the charge controller and Auto switch.) The meter was in the 200mA position and read 5.2 mA. P= .0052 * 12.7 = 0.06604 watts of parasitic power loss in "lights OFF" state. (P channel MOSFET being held in open state.)

I then blacked out the photo-switch circuit and found the parasitic power unchanged.

I then connected one of the assemblies. The assembly drew 40.7 mA - 5.2 mA = 35.5 mA which means that each branch of the two branches was drawing about 17.75 mA.

Using test leads I then went through the process of hooking up additional assemblies in parallel with the following results. (This includes parasitic power)

2 Assemblies = 68.6 mA @ 12.58 volts

3 Assemblies = 90.1 mA @ 12.57 volts

4 Assemblies = 107 mA @12.57 volts

5 Assemblies = 118 mA @ 12.56 volts : P = 12.56 * .118 = 1.428 watts including parasitic losses.

The supply line resistance totaled 3.8 OHM in the positive lead from the battery to the last assembly for a total of 7.6 OHM in the circuit length.

Estimated energy usage 12 hours on and 12 hours off = (12 hrs * 1.428 watts) + (12 hrs * .06604 watts) = 17.93 Watt Hours per day

I am using a 12 volt 12 Amp Hour Gel Cell for storage which should have a capacity of about 144 watt hours. Using 50 percent of capacity should give me 4 days reserve power.

Given the results of earlier systems I have decided to go with what I have found works.

If I blow the assemblies then I will do a redesign and replace them.

I they do blow I will be sure to let you know - it will not be the first "I told you so" event I have experienced and given the worst case scenario, combined with empirical evidence, I think its worth the risk.

Gav

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#47
In reply to #46

Re: Voltage Drop in a Simple DC Circuit Path

11/22/2012 1:45 PM

Excellent! I look forward to the update.

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#48
In reply to #46

Re: Voltage Drop in a Simple DC Circuit Path

11/22/2012 9:45 PM

Your system sound like it will work well, you have done your homework and I think learnt a lot about electronics in the process, so well done.

One final thing that's worth checking is the charge voltage that the battery controller is supplying to the battery when the sun is up and the battery is being charged. For a Sealed Lead Acid battery it should be somewhere between 14.4 and 15 V. I know this sounds high for a 12 V battery but you have to overcome the natural voltage of the battery which is about 12.7 V before you can even start to get current flowing in the opposite direction and charging the 6 internal cells. If the battery controller isn't supplying at least 14.4 V then it's not going to fully charge the battery and you may find yourself overestimating how much run time you have before the battery goes flat.

So, I would definitely recommend checking the charge voltage to make sure it's high enough to fully charge the battery. Oh yes, I almost forgot, you will need to test the charge voltage when the battery is fully charged as the battery controller may reduce the charge voltage to prevent the charge current from exceeding the maximum for the battery.

If you find that the charge voltage is under 14.7 V when the battery is fully charged then I would suggest having a look at the manual for the battery controller to see if there is some way to adjust the charge voltage and maximum charge current. A good battery controller should have some way of adjusting these as the manufacturer has no way of knowing the specific characteristics of the battery the customer decides to use. As a result they usually set them conservatively to prevent batteries from being overcharged and possibly even exploding and starting a fire. Unfortunately this usually means that they never fully charge the battery, so it's definitely something worth looking into.

If you are unsure about the charge voltage and maximum charging current for your battery then consult the data sheet for the battery. If it didn't come with one of doesn't have the data printed on the side of the battery which seems to be the norm these days, then I suggest looking for the data sheet for your battery at ALLDATASHEET.COM . Failing that use this data sheet for a 12 V 12 Ahr SLA Battery which should be close enough.

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#49
In reply to #48

Re: Voltage Drop in a Simple DC Circuit Path

11/26/2012 12:10 PM

Thank you for the information. I will indeed check the charging voltages.

I want to again thank everyone for assisting me in this project.

I have learned a lot in the process.

Gav

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#50
In reply to #49

Re: Voltage Drop in a Simple DC Circuit Path

11/30/2012 1:25 PM

It's a pleasure to think we have helped and I'm sure you have learnt much from you experience. This truly is what the spirit of CR4 is all about, helping others learn and understand. Keep up the good work and we look forward to your next project.

Electronics is certainly complex but if you start with relatively uncomplicated projects then step by step build more complex projects you'll find yourself building things that use microcontrollers in no time.

Well done mate.

PS: I'm in hospital as I write this so my input over the next few weeks will be limited but I will certainly follow how you go so pleas keep us updated with how things work out.

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#51
In reply to #50

Re: Voltage Drop in a Simple DC Circuit Path

12/01/2012 9:39 PM

Masu;

Hospitals are dangerous places; please get well soon so you can get out of there.

Thanks for your input.

Gav

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Location: Port Noarlunga, South Australia, AUSTRALIA (South of Adelaide)
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Good Answers: 75
#52
In reply to #51

Re: Voltage Drop in a Simple DC Circuit Path

12/02/2012 6:47 AM

"please get well soon so you can get out of there"

Thanks mate and I'll certainly do my best to get out of here as soon as possible, but part of the surgery involves a skin graft in about two weeks' time so I'm stuck here for at least a month or so.

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Join Date: Aug 2005
Location: Hemel Hempstead, UK
Posts: 5242
Good Answers: 287
#53
In reply to #52

Re: Voltage Drop in a Simple DC Circuit Path

12/02/2012 11:31 AM

Good luck and best wishes.

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