Why does voltage drop become more noticeable in longer LED circuits?

[cedricuk]

I've been trying to understand voltage drop in longer low-voltage LED circuits from a physics perspective.

If the supply voltage is constant, increasing the length of the conductor increases the total resistance. With the same current flowing through the circuit, this should produce a larger voltage drop according to Ohm's law.

What I'm less clear about is how this develops along the length of a circuit rather than simply appearing as a single voltage loss. Does the potential decrease continuously along the conductor, and can this be thought of as an electric field driving the current through the resistance of the wire?

I'm also interested in how the wire's cross-sectional area changes the situation. A thicker conductor has lower resistance, but is there a useful way to relate this directly to the potential gradient along the conductor?

I'd appreciate an explanation from the physics side rather than just a practical wiring rule.

 

[Baluncore]

If the supply voltage is constant, increasing the length of the conductor increases the total resistance. With the same current flowing through the circuit, this should produce a larger voltage drop according to Ohm's law.

The same current does not flow in the circuit.

The supply voltage is fixed, Vs.
The LED voltage is fixed, Vd, (by the band gap of the material).
The resistance, R, of the circuit wires is proportional to length.
The current that flows in the circuit will be, I = ( Vs – Vd ) / R .

If you increase R, then I will fall.

 

[berkeman]

What I'm less clear about is how this develops along the length of a circuit rather than simply appearing as a single voltage loss. Does the potential decrease continuously along the conductor,

Yes. It's a little more complicated than that, depending on how the LEDs are driven. If they are simple LEDs each with its own series current-limiting resistor, then the current through each LED depends on the voltage available at the tap point where the LED and its series resistor are connected on the wires. So the LEDs near the end of the string will receive a lower voltage from the string wires, and will be a little dimmer than the first LED in the string.

If each LED has its own switching current source that draws power from the string wires, then the LEDs will be the same brightness even though the string wiring voltage decreases with distance. Each LED circuit will draw the same power from the string wiring, so the LED circuits at the end of the string will be drawing a higher current from the wires due to the lower voltage that they see. This is a non-linear effect that has some strange behaviors, which I can describe in more detail if you are interested.

 

[sophiecentaur]

If the supply voltage is constant, increasing the length of the conductor increases the total resistance.

Yes. That's correct. The resistance of the supply wires is set by the material (copper, mostly) and the dimensions. So far so good.
You mention Ohm's Law but that law only applies to ideal conductors like metals. An LED is far from ohmic so all the basic calculations you can do when adding resistances in series etc etc fail with semiconductor devices.
Also, as mentioned above, your regular DC power supply provides you with a constant voltage over a range of device resistances. You 'can' make 'constant current' power supplies and all the calculations need to be modified but that's what we would call "second year work" lol.

 

[DaveE]

Does the potential decrease continuously along the conductor,

Yes.

can this be thought of as an electric field driving the current through the resistance of the wire?

Sort of. Field descriptions can get complicated. I would stick with the voltage (potential) for now. The current would be determined by the entire loop circuit, so wires plus the devices (LEDs) that the current has to flow through.

Khan academy has some really good tutorials on electronic circuits.

 

[sophiecentaur]

Field descriptions can get complicated. I would stick with the voltage (potential) for now.

And why? It's because Fields are vectors and have direction so, in the end you'd need to consider the actual route of connecting wires and positions of components in your calculations. Using Potential (Volts) you are only concerned with the way the Electrical Potential Energy changes around the circuit. Electric Fields come into it in particle accelerators and Cathode Ray Tubes where the electrons tend to go in the direction of the field.

Khan academy has some really good tutorials on electronic circuits.

It's full of proper Physics, agreed, but the presentation is very pedestrian with writing on a white board etc.. You need to be very determined to get through it. But presenting the same content in a zippy way involves a lot of hard work from the presenter so its often necessary to do a lot of hunting for a more exiting alternative.

 

[Baluncore]

Where you have an end fed strip of many groups of LEDs, each group with a current limiting resistor, the voltage drop along the supply lines will result in some voltage variation along the strip. LEDs will be brighter at the feed end, dimmer at the open end.

One way to reduce that variation is to feed the strip from both ends. Connect the positive supply at one end, with the negative supply at the other end. That places the same series supply resistance in all LED current paths, so the voltage variation should not result in a current or brightness variation along the strip.

All LEDs will then be equally bright, the same brightness as the LED in the middle of the strip would be, if fed with both supplies from one end only.

 

[sophiecentaur]

All LEDs will then be equally bright, the same brightness as the LED in the middle of the strip would be, if fed with both supplies from one end only.

And so the UK RingMain idea was born for domestic supplies in UK homes.

 

[Baluncore]

And so the UK RingMain idea was born for domestic supplies in UK homes.

The UK ring main is a quite different system, one that does NOT balance the voltage drops. The UK ring main supplies current from both ends of the ring through both conductors, that is with four connections at the distribution box. A break in either conductor can double the current in the broken wire, which then has only half the copper needed to meet the voltage droop specification. Overheating of the wires can take place without there being any warning to the occupants of the premises.

LEDs, or loads, on an intact UK ring main configuration would be brightest at the ends, with the dimmest LED being in the middle of the ring. The UK ring main configuration is actually used for most LED strips, because it provides more light, but it only halves the brightness variation, it does not cancel the variation completely.

One pair of conductors, in a ring, with only one conductor fed at each end, equalises the line resistance to all nodes on the ring. Unlike in the UK ring main, any break in either conductor will extinguish all nodes between the break and the open end of that conductor. For that reason, it is easier to diagnose the presence and position of a break, than it is in the UK ring main system, but it doubles the cost of the copper needed in the wires.

 

[Herman Trivilino]

If each LED has its own switching current source that draws power from the string wires, then the LEDs will be the same brightness even though the string wiring voltage decreases with distance.

Could you elaborate on this a bit, please? If the string wires carry dc, I don't understand how these "switching current sources" can do this.

 

[sophiecentaur]

The UK ring main is a quite different system,

Could you clarify this a bit more, please? If you could refer to a diagram of the LED strip arrangement then It could be clear but I would assume that there's parallel connection for each LED unit onto the power leads. The ring main uses a Live ring and a neutral ring. Will a ring supply to a number of LEDs have two conductors, arranged in a ring (+v and 'Earth) or some other arrangement?. There are, of course, some differences in applications and power levels the ring main is designed to cope with the full current without getting hot and the LED system equalises the LED brightnesses but is there really any fundamental difference?

With just one ring main conductor cut, the system works 'fine' except under heavy loads. Checking for a break is not easy without testing for continuity with the whole ring disconnected from the board. I think it will be a matter of differences between power dissipation in the power and the lighting applications.

 

[berkeman]

Could you elaborate on this a bit, please? If the string wires carry dc, I don't understand how these "switching current sources" can do this.

For example, say the DC source power supply outputs 12V, and each LED's forward voltage when ON is about 2.0V. Then each LED can be driven by a buck current source that drops the 12V down to about 2V and regulates by sensing the LED current (usually done with a low-side resistor below the LED, with feedback for the buck circuit tapped off the top of that current-sensing resistor).

Assuming the buck switching circuit has high efficiency, then the power drawn from the distribution wires is just the LED current multiplied by the voltage drop across the LED and the current-sensing resistor, which is the constant power that I mentioned. The LED circuits at the end of the distribution wiring will see an input voltage less than the source 12Vdc, but they will still draw the same power determined by the LED current and the voltage drop of the LED+resistor. Since they are constant power consumption devices, the input current will be proportionally higher since the input voltage is lower.

Does that help?

 

[Baluncore]

Could you clarify this a bit more, please?

In these schematics, a LED with a series resistor is represented by a vertical 1k resistor. The resistance of the conductor is represented by a horizontal 1R resistor. The voltage on a node is proportional to the vertical on the schematic, so hopefully it will clearly illustrate the voltage droop for the different configurations.

 

[sophiecentaur]

Got it - thanks. The cost of copper in a ring main is very relevant (power dissipation costs you). Thin lighting wire is cheap so you can afford to ‘waste’ the extra current carrying capacity in the interest of even lighting.
Smart!

 

[Baluncore]

Got it - thanks.

You are too easily persuaded.
The voltage droop diagrams shown, are true only for one LED turned on at the time.

When all LEDs are on, the variation in the ring-main easily beats the variation in the single-ended, and comes close, but not equal to, the variation in the equal-brightness circuit.

I agree that, the ring-main is a clear win for the economy.

 

[sophiecentaur]

You are too easily persuaded.

I was born yesterday! But, as with all Engineering, it all has to do with cost benefit and practicality. Low current conductors are cheap so you can throw just a little money at the LED circuit and keep the brightness variation as low as you need.
BTW how does the overall voltage reduction decrease by changing from the equal brightness system to the double ring system? The effective source resistance for a double ring would be lower than for the equal brightness system.

 

[Baluncore]

BTW how does the overall voltage reduction decrease by changing from the equal brightness system to the double ring system?

For only one LED on, that LED would be operating at the same voltage.
For all LEDs on, double end fed would have higher voltage than in the equal brightness system.
The analysis is complicated by the type of current regulation employed for the individual groups of LEDs.