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My confusion stems from a an led that shows: Input(DC): 600mA-700mA / 3V-3.4V. the product suggests a constant current led driver (same brand as LED) that shows: Output: Current 600mA (Constant) ; Voltage 18-34V

In researching I have read that excessive current will burn out the LED and over time as the battery drains the amps drawn gets smaller resulting in the need for a constant current led driver.

I am having trouble finding information for my question: Can LEDs receive a voltage above their recommended value if the current is constant?

Also, side question, if the current is constant at 600mA and the led's current takes 600mA, then I don't need a resistor?

update:

Thank you for all of your input. @mkeith was correct when saying "I think the point of confusion is that you don't understand the relationship between the power source and the load."

I was under the impression that only the value of the Amps through an led mattered. I thought that since the company suggested the driver with a minimum output voltage of 18V, when the LED that it was referenced from, had a voltage of 3-3.3V, that that meant the constant current is what mattered. I was asking if the higher voltage (18V from the driver vs. 3.3V of the LED) with the correct Amps would be okay.

I have com to realize why this driver was suggested. wiring the 3V LED in series to get about 18V maybe have been the intended use. additionally, I better understand the difference between constant current and constant voltage and how the LED will control the amps (when constant voltage is applied) and the Volts (when constant current is applied)

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    \$\begingroup\$ voltage in excess of the guidelines practically guarantees more current than suggested. the voltage guidelines lay out the space from barely turning on to almost too hot, whereas current can be about anything under max. in normal operation, max current equates with max voltage. \$\endgroup\$ – dandavis Jan 2 '18 at 4:23
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    \$\begingroup\$ @dandavis Maybe so. But if it is a production circuit, I don't think driving the LED with a voltage will work very well. You will get too much unit-to-unit variation in brightness (I suspect). And you would need reasonably tight voltage control. \$\endgroup\$ – mkeith Jan 2 '18 at 4:59
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    \$\begingroup\$ You cannot have it both ways. If you drive with a current source, then the current is fixed and controlled. It does not change. The voltage is free to adjust to whatever it needs to be to make the LED happy. In fact, it will settle at a constant voltage (more or less). \$\endgroup\$ – mkeith Jan 2 '18 at 5:27
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    \$\begingroup\$ @JordanKlaers, all the comment put together are quite confusing ..... here is a simple explanation .... if you connect an LED (any load actually) to a constant current source that is adjusted for 100mA, the constant current source will adjust the voltage at the LED until 100mA flows through the LED. ... if you put 2 LEDs in series then the constant current source will crank up the voltage to get 100mA flowing through the 2 LEDs. \$\endgroup\$ – jsotola Jan 2 '18 at 5:31
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    \$\begingroup\$ Can LED's receive a voltage above their recommended value if the current is constant? ... yes, when the constant current is set above the LED recommended value ... do not confuse constant current supply with safety ... you could set the current to 10A (if your supply would allow) and blow up most LEDs connected to it \$\endgroup\$ – jsotola Jan 2 '18 at 5:39
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The driver is inappropriate for the LED because the minimum voltage from the driver (18V) is greater than the minimum LED voltage at 600mA (3V). The driver is likely designed for LED arrays that have at least 6 dice in series, so 18V.

When you feed the particular LED die you mention with a constant current between 600 and 700mA you will get a voltage (assuming you have not destroyed the LED) that will be between 3V and 3.4V (or maybe the voltage is specified at a particular current).

If you do not exceed the recommended current, the LED voltage should not exceed the range given (it will actually drop a bit as the LED heats up).

You only get to pick either the voltage or the current. With an LED, you are expected to pick the current and the voltage across the LED will be a result of that current. If you tried to run the LED from a constant voltage supply you would have to find the voltage experimentally and it would not be stable (and could kill the LED).

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    \$\begingroup\$ Correction: The driver is inappropriate for the LED because the minimum voltage from the driver is MORE than the LED voltage at 600 mA. If the OP put several LEDs in series to get the total LED voltage above 18 V he could use that driver. \$\endgroup\$ – Peter Bennett Jan 2 '18 at 5:12
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An LED is a very simple device. It behaves according to:

$$I_{LED}=I_{SAT}\cdot\left(e^\frac{V_{LED}}{n \cdot V_T}-1\right)$$

Or, alternately,

$$V_{LED}=n\cdot V_T\cdot \operatorname{ln}\left(\frac{I_{LED}}{I_{SAT}}+1\right)$$

In the above examples, \$n\$ is the emission coefficient (some number that is 1 or larger, but probably not much larger than 10), \$V_T\$ is the thermal voltage (which is \$\frac{k\cdot T}{q}=26\:\textrm{mV}\$ at room temperatures), and \$I_{SAT}\$ is the saturation current (which is the apparent y-axis intercept on a log scale chart based on the slope of the curve representing the voltage vs current of the LED) and is often quite small -- usually much smaller than \$10^{-9}\:\textrm{A}\$.

Suppose, in your case, that the LED is best modeled by \$n=5\$, \$I_{SAT}=1\times 10^{-11}\:\textrm{A}\$ (\$10\:\textrm{pA}\$) and \$V_T=26\:\textrm{mV}\$. Then you could compute:

$$V_{LED}=5\cdot 26\:\textrm{mV}\cdot \operatorname{ln}\left(\frac{600\:\textrm{mA}}{10\:\textrm{pA}}+1\right)\approx 3.226\:\textrm{V}$$

Now, you do NOT get to simultaneously force both the voltage and the current. You can have a power supply that maintains a fixed voltage and simply "complies" with whatever current is needed (up to the specified compliance limits of the power supply.) Or you can have a power supply that maintains a fixed current and simply "complies" with whatever voltage is needed (up to the specified compliance limits.) The LED itself will respond, either way.

I mentioned some "parameter" values above for a hypothetical LED. But LEDs vary all over the place. So let's say that if you grab out a bunch of LEDs and have special equipment that simply prints out the right values whenever you plug in a different LED. Using it you get the following table for six LEDs from the same manufacturer:

$$\begin{array}{r|lr} \text{LED} \# & n & I_{SAT}\\ \hline 1 & 5 & 10\:\text{pA} \\ 2 & 4.8 & 30\:\text{pA} \\ 3 & 4.6 & 15\:\text{pA} \\ 4 & 5.7 & 18\:\text{pA} \\ 5 & 5.3 & 22\:\text{pA} \\ 6 & 4.9 & 27\:\text{pA} \end{array}$$

Let's say you have a power supply that supplies a fixed voltage of \$3.2\:\textrm{V}\$ and does it perfectly. What will be the currents for each of these different LEDs that you hook up? Well, let's look:

$$\begin{array}{r|r} \text{LED} \# & I_{LED}\\ \hline 1 & 490\:\text{mA} \\ 2 & 4100\:\text{mA} \\ 3 & 6250\:\text{mA} \\ 4 & 43\:\text{mA} \\ 5 & 268\:\text{mA} \\ 6 & 2190\:\text{mA} \end{array}$$

Wow! That's bad. All these supposedly similar LEDs produce huge differences in their current using this exact same voltage power supply. And not a single one of them very close to the assumed \$600\:\text{mA}\$, either. Assuming that the power supply can actually deliver over six amps, you could do some serious damage to the LEDs.

Now let's switch over and use a constant current supply designed to provide a fixed \$600\:\textrm{mA}\$ and see what happens with the LED voltage, instead:

$$\begin{array}{r|r} \text{LED} \# & V_{LED}\\ \hline 1 & 3.23\:\text{V} \\ 2 & 2.96\:\text{V} \\ 3 & 2.92\:\text{V} \\ 4 & 3.59\:\text{V} \\ 5 & 3.31\:\text{V} \\ 6 & 3.04\:\text{V} \end{array}$$

Note here that the range of voltages is much smaller! All you need to do is find a constant current power supply that can handle at least \$5\:\textrm{V}\$ or so and you are fine.

Yes, I provided some "clinkers" in the LEDs above. Your specifications said that the LEDs went from \$3\:\textrm{V}\$ to \$3.4\:\textrm{V}\$ at \$600\:\textrm{mA}\$. But that's also the point. While the specifications tell you that it is statistically unlikely to see LEDs out of that range, the fact is that you will still encounter some that are just outside of it from time to time.


This very small variation in voltage is a big reason why "current limiting" resistors work as well as they do. Since the differences in voltage hug a small range, it's very easy to estimate what voltage remains (within a small error range) for a resistor's voltage drop.

If you have a power supply voltage of \$6\:\textrm{V}\$ (not a constant current source, but now a constant voltage source again), then you can be pretty sure that the resistor needs what remains after the LED drop of about \$3.2\pm 0.2\:\text{V}\$. The remainder voltage is then \$2.8\pm 0.2\:\text{V}\$. So if you compute a resistor that will generate the right current given that remaining voltage drop, then the actual current in practice won't vary that much because the remaining voltage drop for the resistor also doesn't vary that much.

(As a note, you can also see here that if you used a constant voltage power supply of \$4\:\textrm{V}\$, that the remainder voltage of \$0.8\pm 0.2\:\text{V}\$ has a much wider variation, percentage wise. And this means that there would be far less consistency in the LED current as a result of that fact. So here, you find that higher voltages for the constant voltage power supply improve current regulation. But this benefit comes at the expense of added wasted dissipation as useless heat.)

A constant current source is often quite similar to a voltage source with an added variable resistor that can adjust itself to drop just the right amount of voltage to keep the current constant. This is done with transistors and/or ICs. But the effect is that instead of a fixed resistor, some added circuitry allows the power supply to vary the resistor automatically, instead. Otherwise, not so different.

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  • \$\begingroup\$ would the CV current diffs be so dramatic if you wired the LEDs in series and fed them 3.2vX6? \$\endgroup\$ – dandavis Jan 2 '18 at 5:13
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    \$\begingroup\$ @dandavis Not so bad, when in series, unless you get really unlucky. The idea then is that you are playing a statistics game. The more you put in series, the tighter the standard deviation becomes. Assuming random distributions in the batch, of course. It's also possible for manufacturers to "bin" their parts and sell binned as well as unbinned parts. If so, the unbinned ones might be the cast-offs of some unknown binning process for high-valued sales elsewhere. This would very much mess up your statistics, then, while still being "in spec." \$\endgroup\$ – jonk Jan 2 '18 at 5:17
  • \$\begingroup\$ thanks for the info. i think of series units as sharing equal current, so that makes sense. \$\endgroup\$ – dandavis Jan 2 '18 at 5:32
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    \$\begingroup\$ +1 nice numerical analysis. I suppose there is an ohmic term in there too. \$\endgroup\$ – Spehro Pefhany Jan 2 '18 at 5:41
  • \$\begingroup\$ @jonk although Theory might support your discussion of picoamps and saturation voltage , this is not only impossible for mortals to measure but more importantly, is never done in industiy rating diodes of this type. I suggest you use more practical models for teaching above some visible threshold or the voltage of a dim LED which is more consistent threshold voltage. \$\endgroup\$ – Sunnyskyguy EE75 Jan 2 '18 at 7:06
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A constant current LED driver being specified as "18-24V output" means the output voltage compliance range (where CC operation will be maintained) in constant current mode is 18-24V. Using it with a LED string with a combined threshold voltage far outside that range can be expected to cause it to shutdown (making the setup useless), overheat (damaging the driver), or lose constant current properties (damaging the LEDs with overcurrent).

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