This is a good point.
When someone uses a bench top power supply plugged into mains to test an emitter or driver, they are using constant current. The large bench top device strictly regulates the current as well as the voltage. With the voltage, it’s coming from a much higher source… the mains… and so it has no sag. Constant voltage and constant current. Everything there is managed and calculated, no surprises. This is why djozz can create a chart showing how an emitter works, then I build a light that differs from the chart. He’s using a power supply plugged into mains, I’m using a power supply that is variable in the Li-ion cells.
When we get the flashlight assembled and have a Li-Ion cell in it, or multiples, this is no longer easy to achieve. The cells sag under load and their ability lags as they drop in power level. With multiple emitters the apparent forward voltage is lower, so they can hit the cell harder and pull higher current than a single emitter might. (This is a direct drive scenario, as in the A6 or Bistro drivers) I don’t often build a light with a Buck or Boost driver, so I don’t have as good a grasp on how those work. Most of my builds are direct drive for absolute top end power from the start. I do, however, see the benefit of using a Buck driver to supply a lesser voltage than the supply for a regulated run time. As usual, ones intended usage will define much about how a light should be built.
For small output lights the style of direct drive delivery is less observable. Build up a hot triple and these issues become very critical to the performance of the flashlight. This is when top cells come into play.
Hope that helps to some degree, not sure I’m managing to say what I actually mean.
LEDs don’t require a specific wattage but a specific current at a moderately specific voltage. Small changes in voltage induce greater changes in current so drivers are designed to maintain a certain current and make small frequent adjustments in the voltage to do so but its the current not the wattage or voltage that’s being regulated. The driver draws the current necessary from the cell to maintain that led current for as long as the cell can supply it. For single mode boost and buck drivers the output of the driver is constant, the input current rises as the cell voltage drops. Let me try another way. If you change the voltage to the emitter will not have the same power output. You can’t lower the voltage and raise the current, they don’t work that way. You might need to look at a Vf vs current graph to see what I mean if I’m not being clear for you.
The term Vf is inherently tied to the notion of having a value at a given current. It changes based on the current passing through the emitter and isn’t adjustable relative to that current. It’s not a single number but a curve with current on one axis and voltage on the other. Wattage isn’t constant along the curve but the product of multiplying the two together. A Vf of 3V might draw 1A and use 3W, and a Vf of 3.5 V at 3A use 10.5W. The driver is the means we use to manipulate the power source to achieve the output we want from this curve. The driver can either boost the voltage as with a AA cell or buck the voltage as with multiple lithium ion cells but in either case it controls the current to the led drawing whatever it needs from the cell to do so.
No, no, no. I get what you’re saying and I’ve seen more than one Vf vs Current plot on data sheets.
I wasn’t describing what happens in drivers. I’m asking whether what I’m describing can be done.
Lets pick a point off the Vf vs Current plot of an imaginary LED. Vf = 4V, Current = 3A. Fair enough? Say that combination produces output “X.” Can we also power the LED to produce “X” with 12V and 1A instead? If so, can drivers be designed to that: not bucking voltage above what’s needed, but drawing the current as needed for the required output. Kind of like a DD driver that regulates the current, as opposed to drawing as much as the cell can supply and the LED will take at the current voltage.
The short answer is no.
Each emitter type has it’s own forward voltage and it’s going to remain based on that ~3V or ~6V or ~12V lineage. You’re not going to run an emitter that takes 4V at 3A on 12V at 1A, it’ll fry. Likewise, a 12V emitter won’t even glow at 4V.
The emitter is the determining factor, the driver equation has to serve that final function.
Thanks, Dale. Followup question: why would it fry at 1A at 12V, when it can handle 3A at 4V just fine? I’ve forgotten all my E&M physics, which I was never good at anyways. I thought it’s the resulting wattage that matters, not the voltage alone, assuming the voltage isn’t so high to cause arcing. I understand that if it’s rated for 4V and you hooked it up to 12V and let however much current that can pass at 12V through it then it’ll fry, but in my example, the current is constrained to 1A to produce 12W.
0K; I've just taken a very brief tl;dr overview of this subject, and neither I am an electrical/electronic engineer so, please bear this in mind.
What do the initials CC & CV (Constant Current & Constant Voltage) usually refer to/mean?
Essentially: they refer to non necessarily exclusive different modes of operation for power supplies. A led driver is a specific kind of power supply.
CV: Constant Voltage mode of operation is based on the principle of keeping the supply output voltage constant (within a certain specification) at least up to the rated current. Therefore, output current will match load impedance up to the PSU's limits.
CC: Constant Current mode of operation means the supply will tune the output voltage to whatever is required for such current to flow through the load. Of course, practical power supply designs may have their limitations, and this means their DC/DC converters will only be able to operate within certain voltage ranges (the wider the better). These supplies usually achieve this through sensing the voltage drop across a current shunt (a very low precisely known value resistor set in series with the load), and comparing this value to some sort of reference voltage in their integrated circuits in order to determine to what amount raise or lower the voltage at the output so the voltage drop seen at the shunt remains constant.
0K, wanting some sleep. Suggestions/additions welcome.
Cheers ^:)
Electrical components are designed for a certain voltage range. In your example it is the same wattage. Using a buck driver to the led will work but giving that Power straight to the led will only work on the lower voltage as that was what it was designed for. Giving voltage higher outside what it was designed for can kill it faster than over Amps.
Over Amps = heat > pop!
Over volts = pop!
If you want to use 12V then put 4 LEDS in series so they each get 3V.
OP updated with
Must add Energy to the list, often ignored but extremely important when it comes to batteries: it tells how many watt-hour can a battery supply and it takes into account both capacity and voltage. It is required to understand it in order to chose the correct battery for your flashlight, e.g. a 3,000mAh 18650 can provide longer runtime than a 3,500mAh one under certain conditions.
I think the answer you are looking for is that what you are proposing is not possible, as a result of the current/voltage characteristics of an LED. Using your example: if 1A is being sent through a 3V LED, its forward voltage will be defined by the forward voltage vs current curve. It has to be on the curve, not on or below it. It will be somewhere around 3.2V or so; that is the voltage you would measure across the LED if you measured it with your DMM. It is inherently not possible to send 1A through the LED and measure 12V across the LED.
A buck driver would effectively do what you are proposing. There would be 12V or so across your battery pack, but there would still be 3.2V across the LED if the driver sends 1A through it.
That was just an example I gave to clarify my questions.
Thanks, EasyB. If I understand you correctly, for LEDs there’s a 1-to-1 correspondence between the forward voltage and allowed current. The allowed current at any given forward voltage does NOT represent the upper bound of the range of currents that the LED allows.
That’s all I needed. Don’t know why it took so many rounds of exchanges to finally get a straight answer. I thought I was pretty clear in stating my assumptions in my first post.
My understanding is overvoltage LEADS to overamps, which is what kills. But in the absence of overamps, does overvoltage alone kill. Please provide an explanation if you know. I’ve not found one.
You’re welcome. Yes, that is correct.
This stuff can get confusing and it’s easy to misunderstand questions.
LEDs have a small range of voltages in which they operate in. That’s their forward voltage. Too low a voltage and it won’t light and too high a voltage and they die. That’s where the driver comes in at. They transform the voltage given by the cell or battery and change it to the voltage the LED needs.
If a LED is using a certain number of watts in a combination of volts and amps then the driver has to change the combination of volts and amps provided by the battery to the combination needed by the LED.
There’s a lot to understand with caveats and special conditions everywhere along the way. It’s a language with very specific terminology that doesn’t allow for “youknowwhatimeant” sloppiness in choice of wording. Learn the basics of Ohm’s Law and Kirkhoff’s Law and most of this will naturally fall into place. You don’t need knowledge of quantum mechanics or advanced electricity and magnetism. Just be aware that there will be complications in most circuits and choices governing overall size, run time, output current, input current, input voltage, type of LEDs and how many, reflector vs optics, switch type, etc. all come with consequences and compromises. Understanding how these factors interact is a body of knowledge it takes some time to learn. It’s not organized here in any sort of step by step tutorial, you just have to dive in wherever you happen to be and dive in again somewhere else. The more you play with it, the more it sticks. There are lots of different analogies one can use to describe voltage, current, resistance, and power, just remember, the analogue isn’t the thing, it only resembles it in some fashion so don’t get hung up trying to force reality to fit the analogy. The quantity represented by each of those four words exists in relationship to the others, changing one will result in a change in the others
Voltage is potential energy, how hard is the push, it describes the difference in potential energy between different points in a circuit. The sum of all the differences equals the total voltage in the battery.
V = I x R, the voltage drop across any two points is the product of the resistance between those two points and the current.
Resistance is drag, it’s everywhere because there’s no such thing as a perfect circuit. In the cell, in the switch, in the wires, in every single component it adds up and the voltage drop caused by each of these whether large or small adds up to the total battery voltage available. From the formula above an increase in the supplied voltage will result in an increase in the current. Likewise, that increase in current will result in an increase in each of the voltage drops in the circuit, again adding up to the new voltage supplied.
R = V/I The resistance between any two points is the ratio of the voltage divided by the current. An ohm is an arbitrary value of 1 volt/ 1 amp.
I = V/R Current is flow. It’s a measure of how many amperes( a really big number of small charges) is moving per second. Push harder(raise the voltage) and flow increases, add more resistance and flow decreases.
Power is the rate of energy use. Expressed as energy times flow or P = V x I. For available cell power it’s common to multiply cell voltage times cell capacity but this isn’t very precise since cell voltage drops as it depletes so it’s usual to just use the nominal cell voltage(an average number). Every place there’s resistance there’s a voltage drop and a place where power is consumed. Thicker wires, lower resistance FET’s, better switches, and cleaner mechanical connections all reduce resistance in other areas leaving more power available for use in the led. At least at BLF this is commonly referred to as parasitic resistance.
Capacity for cells is listed in milliAmp hours or how much current for how long. Actual value depends on how much current is drawn from the cell since the cell itself is a circuit and has internal resistance which varies by size and cell chemistry so more current causes a voltage drop within the cell leading to a lower measured capacity. At a certain point there’s enough of a voltage drop(power used inside the cell) to cause the cell to heat up excessively. Somewhat before that is the maximum sustainable current output listed in the specs for any given cell. It’s a bad idea to disregard the limitations of any given cell size or chemistry.
Excessive heat affects circuits, cells, and LEDs so there’s much to be learned and unlearned regarding commonly held notions about controlling and dissapating it. By far the most critical spot is the led itself where the heat is concentrated in the die. Using a copper dtp star immediately creates a high speed thermal path to the larger star bottom surface and its connection to the pill. Improving the heat path from there on is the subject of much debate and depends on the amount of power used but that first thermal coupling is the most important. Non dtp aluminum stars are still fine for lower power uses so it’s not even an issue for many but if you run a light that gets hot to the touch it’s pretty much a given that it will run better, brighter, and longer on copper simply by lowering the junction temperature of the led.
There’s lots more detail and questions to be answered but understanding these basics will go a long way to helping those make sense.
Here’s a question using water pressure = V, Tube ID = Current control, to help visualize.
If a PWM sends a pulse of electrons (H2O molecules) down a wire (pipe), will the absence of electrons both in front and behind the gaggle (or is it horde) of electrons cause the electrons to spread out, thus dropping pressure (voltage)?