# Why did my LED resistor burn while lighting four LEDs in series?

I have tried to create a circuit to switch a large 7-segment LED display (LDS-CD16RI) using a pair of MOSFETs, as follows:

simulate this circuit – Schematic created using CircuitLab

Here I am trying to use a 3.3V logic signal (illustrated as the circled 1) to switch the 24V to drive the LEDs. This circuit is repeated for each of the segments of the display.

The typical forward voltage of each of the LEDs (which are in series inside each segment of the display) is 6.8V, and their max steady forward current is 20mA, so I aimed for 10mA current through the LEDs. Since my supply voltage is only 24V I planned to actually drop about 5.75V across the LEDs to give me some headroom for the voltage dropped across M2 and R2.

I arrived at the value for current-limiting resistor R2 at 100Ω using: $$R = \frac{V_s - V_f}{I} = \frac{24 - (5.75*4)}{0.01} = 100Ω$$

Before building this circuit I calculated the power dissipated by R2 as follows: $$P = \frac{V^2}{R} = \frac{1^2}{100} = 0.01\mathrm{W}$$

0.01W seemed safely below the 0.25W limit of the through-hole resistors I used, so I proceeded with constructing and testing this circuit.

To cut a long story short: R2 burned up shortly after a segment was illuminated. This occurred for each of the separate instances of this circuit driving the various display segments, suggesting that it was a design error rather than a single component failure.

From my calculations and further analysis, I cannot yet understand why this occurred. To check my work, I re-constructed the circuit in a simulator which suggested that power from R2 would in fact be 6.84mW, which is a result I cannot explain but in any case one smaller than what I had calculated above.

I expect I have made an error somewhere in my calculations or my assumptions, but I have been unable to locate it. Assuming the problem is that the resistor is indeed dissipating too much power, can my circuit be adjusted to address this? Is R2 a red herring here and the problem exists elsewhere in my circuit? Is my approach itself flawed?

• Measure the actual voltage across each of the LEDs when running with your 100R resistor and 24V power supply rail. For good measure, also measure the voltage across the resistor. – Dwayne Reid Jun 29 '16 at 5:23
• According to your circuit, you are putting 24 v on the gate of M2. This doesn't answer your resistor question, but once it's working, you don't want to burn M2! I can't turn up data quickly on an FQU13P06L, but presumably you have it. – Neil_UK Jun 29 '16 at 5:41
• your power calculation do not match with the simulation result cause you do not consider the voltage drop across the PMOS M2 . What is the saturation voltage drop across it ? – Anklon Jun 29 '16 at 5:53
• Second @DwayneReid. A multimeter should solve this mystery pretty quickly. – Daniel Jun 29 '16 at 5:56
• Thanks for the tip on the gate voltage on M2... I didn't catch that in my initial working, but I see it now. I expect I will have a separate question about solving that after I think this through a bit more, though I need to decrease my supply voltage anyway (per the answers below) so that is likely to change the problem – Martin Atkins Jun 29 '16 at 14:09

6.8 volts seems awfully high for a single LED. Are you sure that 6.8 is not the number for all four LEDs? That would make it 1.7 volts per LED, which is more reasonable for a red LED. And that would mean that you are currently pushing 172 milliamps, or almost 3 watts through your resistor.

If that is the case, you should lower your power supply to less than 20 volts (maybe 12 volts) to keep from destroying the gate of your MosFET (M2).

• Mark is correct. R2 needs to be about 1800 ohm for 10 mA at 24 V. – winny Jun 29 '16 at 6:42
• Thanks! I had concluded that this was a per-LED number because the decimal point only has two digits and so I expected that it couldn't be for all four LEDs, but your explanation makes sense as to why that is not a valid conclusion. – Martin Atkins Jun 29 '16 at 13:57
• For my own edification as well as re-calculating it I re-simulated it with the LEDs at 1.7V forward voltage and confirmed that indeed ~3W was dissipating at R2 in simulation too. – Martin Atkins Jun 29 '16 at 14:05
• Well, that or add like 8 or 9 more leds. Use the power that would otherwise be wasted as heat. – Passerby Jun 30 '16 at 8:00
• @Passerby Add them for why? If he had a hundred LEDs and was planning on driving them four at a time, then of course he should be running more in series. But since that clearly isn't the case, this isn't much helpful. It's like "oh, they're giving away dogfood today, so I need to buy a dog". :) – Graham Jun 30 '16 at 10:41

I see your problem. Your circuit shows how you're driving a single LED segment. (I presume you then have 7 of these circuits, one for each segment.) The datasheet shows 4 LEDs in series, covering the segment.

Where you've gone wrong is assuming there's 6.8V forward voltage drop per LED. There is no such red LED. Typically a red LED will be around 1.6V-1.8V forward voltage drop, and that's a characteristic of the physics involved so there isn't really much scope for variation. This tells me that you have 6.8V forward voltage drop for all four LEDs in that segment in series.

So with a 6.8V voltage drop and a 24V supply, you're dropping 17.2V across the 100R. As Mark says, this gives you 172mA and 2.96W power dissipation on the resistor. Not healthy for a 0.25W resistor.

In fact you're lucky that the 0.25W resistor basically becomes a fuse under those conditions and burns out almost immediately. If it hadn't, putting 172mA through the display would burn that out pretty quickly, and a large 7-segment display is going to be a fair bit more expensive than a resistor. If you'd used a higher-powered resistor, you'd be wondering why the display briefly flashed very brightly indeed and then went black forever.

• Thanks for this answer! I accepted Mark's answer because it was earlier but both were great answers and I wish I could've accepted both of them. The clarification that the resistor acted as a fuse in this case was helpful to think about what happened here. – Martin Atkins Jun 29 '16 at 13:59
• @MartinAtkins Fair enough. Mark was there first - I just added a bit more detail around the edges. :) Glad it was helpful. – Graham Jun 30 '16 at 10:42
• @MartinAtkins Speaking generally about SE, not this thread specifically: You shouldn't accept an answer purely because it's first. See the many threads across SE where the 1st answer is extremely poor and often deleted once that becomes evident. If one comes along later that better answers your question, then you can - and probably should - change your accepted answer. – underscore_d Jun 30 '16 at 14:42

You got it wrong.

Forward voltage in the datasheet is for series of LEDs, not for individual LED.

Thus voltage drop across LEDs is 6.8V and not 6.8V * 4 or 5.xV *4

Thus resistor has to deal with 17.2V and not 0..2V.

Thus the current is .17A and power dissipated ~4W.

• What exactly does this answer add that other answers did not find out already? – Thomas Weller Jun 30 '16 at 9:53
• Good question. I think this answer is both concise and accurate. – Dima Tisnek Jul 1 '16 at 16:39

depends on the power rating (internal resistance) and color of the led (Vf), the 100 ohm resistor has to bear all the extra voltage of 24 - 4xVf -4Ir... Vf ~ 3V for white, 2V for other color. Most LED has high resistance except power LED like the kinds from Cree... If the Resistor holds 12 V, the power will be ~ v2/r 1.5W watt... add more string of LEDs.