# LEDs are burning. Cannot figure out why

I have a lamp that has LED driver producing 35 V (actually lamp driver says 16 - 28 V, but I've measured 35 V). On the lamp I have 7 LEDs (originally 5 + 2 I've added) with heatsinks that have burned and replaced a couple of times now. All LEDs are connected in sequence.

LEDs specs:

• LED Emitter: 3W
• Output Lumens: 180-210 Lumens
• DC Forward Voltage (VF) : 3.6-3.8Vdc
• DC Forward Current (IF) : 700mA
• Color Temp: 6000~6500K (white)
• Beam Angle: 120 degrees
• LifeSpan Time : > 50,000 hours

So I've figured out:

$$R=\frac{V_{Driver}-V_{LED} \times 7}{I_{LED}} = \frac{35\ V-3.7\ V \times 7}{0.7\ A} = 13\ \Omega\$$

which is nothing. At the same time my LEDs are burned down within a month.
What am I doing wrong?

• How much current is the driver putting out? – Ignacio Vazquez-Abrams Aug 21 '14 at 0:38
• What is the resistance calculation for? Do you have a resistor in series? – venny Aug 21 '14 at 0:44
• @IgnacioVazquez-Abrams I don't see the driver's current in the formula. Does it also matter? – Alex Okrushko Aug 21 '14 at 0:45
• @venny I was calculating if I need one, because of LEDs failing. – Alex Okrushko Aug 21 '14 at 0:45
• Well, yes. If your driver is putting out too much current then your LEDs will burn out. – Ignacio Vazquez-Abrams Aug 21 '14 at 0:47

The driver is clearly out of spec and mostly likely internally damaged. From you photos it isn't clear how the rest of your lights and power sources are connected, but you may have made an error here (I've even seen people connect the AC power line to the DC output side of the supply).

## You have a current source, not a voltage source

If you look at the label you will notice that current is specified precisely (600mA) and voltage is specified as a range (16v-28v). You will also notice that the drawings on the label show a single current loop and specify 7 3-Watt LED's in series.

That you provided this equation indicates that you are confused about the difference between the two types of sources:

$$R=\frac{V_{Driver}-V_{LED} \times 7}{I_{LED}} = \frac{35\ V-3.7\ V \times 7}{0.7\ A} = 13\ \Omega\$$

In your equation, you cannot know the value of $$V_{Driver}$$ as it is determined by the network (it's somewhere between 16 and 28 volts if an acceptable load is attached). Only the current value is constant in the normal operating condition.

## Some background

A voltage source presents a single output voltage, no matter what you connect to it. To make that a true statement it will output any current the load wants up until it is incapable of outputting any more (current-limit or failure). Most people are familiar with this behavior as it is intuitive and commonly encountered.

A current source will attempt to output a constant current no matter the load attached to it. It will do this by adjusting the output voltage until it either can raise it no further (limit of it's upward adjustment range) or lower it no further (it will produce insufficient voltage to operate itself and shutdown).

This works via Ohm's law (V = I R) such that increasing the voltage will increase the current flowing and decreasing the voltage will decrease the current. The system is active and senses it's output current (while adjusting its output voltage) until the output current equals the number printed on the label (in this case 600mA).

If nothing (or too little load) is attached, it will output it's maximum voltage as it keeps trying to increase voltage to get increased current... and vice versa if too much load is attached.

## Driving LED's

If your LED's are connected as parallel strings of series lights, they will need to be driven by a voltage source. This configuration is cheaper to design and build, but more difficult to install and subject to greater line losses as the LED strings get bigger (since you need to bring the full voltage of the power supply all the way to the end of the line).

If your LED's are connected in series (one to the next), then the same current that drives one light will drive the next. This configuration is used in most higher-end architectural lighting. You use a current source to ensure that no matter how many lights are on the string, the current output remains the same. The advantages are that you can easily add lights to existing strings without worry. The fact that current flows through all lights ensures that voltage losses in the line are minimal (most efficient power distribution). And, LED light output is proportional to current so a current driven approach best ensures uniformity of light output.

There are some limitations however. Current drivers are more expensive as they are produced in lower quantities and have to be matched to the exact fixtures being used. The fixtures must all be the same so that they have the same light-to-current relationship and will not be too dim (or burn up) under the constant current value applied. Series wiring of light fixtures is inconvenient in some installations.

• This is a really good answer. – Dave Van den Eynde Aug 21 '14 at 7:26
• LED's in series can also burn out due to unequal voltage drops due to variance in manufacturing, it is easy to get one "weak" one which then means the others are sharing too much voltage and hence operating outside their rating. – John U Aug 21 '14 at 11:39
• @JohnU Initially the lamp had 5 LEDs, but they first first melted the plastic light spreaders (don't know how they are called) and then eventually one of the LEDs was burned breaking the chain. I've replaced all 5 LEDs with the new ones and added 2 more thinking that I would "unload" the system. That didn't help as LEDs started to burn again (within a month). – Alex Okrushko Aug 21 '14 at 12:47
• @DrFriedParts thanks for explaining everything in details. Really appreciate it! Before this I've only been playing around with battery-powered circuits, and looks like battery is a voltage source. – Alex Okrushko Aug 21 '14 at 12:51

According to the data printed on the driver, it's a constant-current supply and it's rated to supply about 600mA into a load that can drop anywhere between 16 and 28 volts with 600mA through it, which is a load resistance range of from about 27 to 47 ohms.

Your LEDs are spec'ed to drop a nominal 3.7 volts each with 700mA through them, which makes them equivalent to about 5.3 ohms each, but that's with 700mA through them.

With 600mA through them - which is what your driver is designed to provide - the drop will be a little less, say 3.6 volts each, so they'll look like about 6 ohms each.

Seven in series, then, will look like about 42 ohms, which is pretty close to the 47 ohm ceiling for the driver and, if you add up all the tolerances, may well climb past it and overtax (fry) the driver.

Certainly with a 35 volt output pushing 1 ampere through your LED string, the driver has lost its magic smoke and is convincing the LEDs to lose theirs as well...

A solution, if you replace the driver with an equivalent unit, is not to add a series resistor to the string, but to drive only six lamps, bringing the resistance of the string down to around 36 ohms, which the new driver should be able to handle.

• Thank you for the answer, it summaries things nicely. I think @DrFriedParts pretty much has the same opinion and so I'd accept his answer (as it came first). Also, the driver burned 5 LEDs at first so I've replaced all LEDs and added 2 more thinking that I'd "unload" the system. Again, my inexperience. Thanks again! – Alex Okrushko Aug 21 '14 at 12:40

If you don't have any series resistor, then the answer is clear: 7 x 3.8 = 26.6V, and your driver is producing 35V.

If you DO have a series resistor (YOU SHOULD!), then it should be dropping at least 8.4V (10.5V would be better) at 700mA total current. Your ideal series resistor would be 15 ohms, but your LEDs may possibly (if run at absolute maximum ratings) be able to deal with a series resistor of only 12 ohms. Your 13-ohm resistor is running painfully close to absolute maximum.

Are your heat sinks dissipating enough power? Is it well thermally (BUT NOT ELECTRICALLY!) connected to the LEDs with a generous smear of heat sink compound? If it's a single heat sink and it's ELECTRICALLY connected to them all, then you're bypassing all but one and getting an AWFULLY high voltage across that one.

Also... since your driver is already known to be delivering SOME voltage higher than it says it SHOULD be delivering, what's the possibility that it's delivering pulses of still-higher voltage?

• Well, all we know is that 35V is no-load voltage. It is a 'LED driver', it would be strange to have constant voltage driver for such high power setup. – venny Aug 21 '14 at 1:02
• Thanks. I don't have resistor in the series at the moment. I thought 13 ohms is very minor and wouldn't affect anything. I guess I'll have to add 15 ohms resistor – Alex Okrushko Aug 21 '14 at 1:03
• @Alex, that resistor is extremely important and is almost certainly the reason your LEDs have been burning up. Also, note the power dissipation of the resistor. With a 15 Ohm resistor, the current will be 560mA. So the power dissipation is 0.56^2*15 = 4.7W. Make sure the resistor you pick can handle that. – Dan Laks Aug 21 '14 at 1:25
• @DanLaks thanks! the driver also states that current output is 600mA +- 5%, but I've measured it to be up to 1A. Could it also be a factor. I still cannot figure out how driver's current output fits into the formula. – Alex Okrushko Aug 21 '14 at 1:28
• All that matters is that the current you choose for the LEDs (which is based on the resistor) is less than the maximum current the driver is spec'ed to source. That said, I think you've got a bigger problem if your driver is behaving that far outside of its spec. My guess is something has gone wrong and I wouldn't trust it any longer to power the LEDs (or anything else). – Dan Laks Aug 21 '14 at 1:42

When a driver rated at 600mA is pushing out over 1000mA, the answer is simple: it is toasted.