Hello,
I have a question regarding the LED driver TLC5927, we just produce 50.000 units of a device that utilize 3 of your LED drivers (TLC5927) and we just found a problem that we do not know if it is related to the LED driver itslef.
Our product has a led matrix of 252 leds, the way we arrange such matrix makes that each output of the led drivers handle 6 led in total. We use a PWM signal to ocntrol the power and the TLC5927 in order to reduce the power cosumption. Also we utilize dual color led, blue and red, and the bleu led is getting damaged very often. The way we do the matirx scan is as follows, we have some trnasistors that control when the power is applied over the leds, and if a particular led need to be light up then the proper TLC5927 output is turned ON.
The problem comes when all led are off, the way our matrix scan works is that the power is always applied to the leds so the transistors we use are being turned ON and OFFdepending on which column need is being scanned, we actually control each led by using the driver. If the led need to be ON then the porper information is sent to the drivers order to turn on the proper led, if all led are off then only zeros are sent to the driver so when the OE signal is activatd then all output should be OFF regardless if the power is being applied.
We are using a 8.4V power supply since the brightness is not enough using a lower voltage, when the led are off we have a 7.7V voltage on the TLC pins. Having this voltage on the led cathode makes that it is being reverse biased with a 7.7V and the maximum allowed is 5V. We think that having such high voltage is damaging the led since this leds, InGaN leds, are very suceptible to ESD and reverse voltage stress. My questions are:
1) Why we have such high voltage, 7.7V, when the TLC outputs are OFF?
2) Since all the outputs are OFF I thought there should be no current following through the led and then the led should not be polarized, it seems that what we are seeing is like a diode drop of 0.7V (8.4V - 7.7V = 0.7V).
3) When the outputs are in the OFF state, Does they are a open circuit?
We are about to make a new production of around 10.000 nmore devices, which represents 30.000 TLC5927, and we need to find a suitable solution ASAP.I hope you can provide me any help or guidance on how to solve the problem.
Regards
Mark,
I have a couple of questions to help me understand your system. Can you send me a partial schematic for a single output so I can fully understand the configuration? The TLC5927 outputs will be high impedance when off. The output will be pulled up to whatever voltage is sourcing the LEDs (minus some drop). I was looking at another device yesterday (similar output to the TLC5927). When I used a volt meter to measure the voltage on the driver when the LEDs were OFF, I notice that the LED turned on very slightly. In other words, by probing that node I am actually biasing the LED. In fact, I saw a 2V+ drop across that LED. I was using an RGB LED and saw different voltage drops depending on which LED I was probing. Do you have a high impedance meter?
What current are you using for the outputs? What is the forward voltage range of the LEDs you are using?
Regards,
Dick
Dick,
Sorry for the late response, I was kind of busy this last months. I am attaching the schematic you requested, the output are being configured to source a current of 18mA. The forward voltage is 2.1V for the red LED and 3.3V for the blue one. We are using a higher voltage becuase the brightness we get using 3.3V was not enough, we also have a resistor in series with the led, which actually we should not being using since the TLC controls the current but we need it to lower the voltage.
Here a samll drawing that shows the way we control the leds.
As you explain, if the TLC output will pull high to the source voltage then that explain the phenomenon I am seeing. In that case the voltage at the TLC pin will be around 7.7V which will reverse bias the other leds connected to that particular line (maximum reverse voltage is 5V). My assumption here is that this huge reverse voltage is the reason why the leds are being damaged. We lower the source votlage to 5V but we still have reverse voltage, Do you have any idea to avoid this reverse voltage reach the led?
I have another question regarding your post, you said that when the TLC output is OFF it will be high impedance, having the output in OFF states for you means OE set to logic 1 (all outputs dissable) or OE set to logic zero (all outputs enable) but the corresponding OUTPUT LATCH bit set to logic 0 (no current being drain on that especific output)? I am interested in the output pin state when the OUTPUT LATCH bit is set to logic 0 and OE is set to logic 1, as you can see in the image above the TLC's are almost all the time with the OE pin set to logic 0....the case when OE is set to logic 1 is irrelevant for us.
Best regards
Mark
One note of clarification, the TLC5927 output will not internally drive high, it is high impedance, so it will be pulled to whatever voltage is present. If an output is off, that EXxx will be pulled high through the respective LEDs and their resistors.
The effect is the same if the output is disabled by writing a logic 1 to the output or if OE is high. It is clear in your diagram, that OE is rarely high.
At 18mA, the minimum output voltage at the driver is about 0.5V to 0.6V. The LED voltage needs to be greater than the driver output voltage plus the LED forward voltage. In the case you discuss that would be 3.3V (for the blue) and 0.6V for the driver. The minimum LED voltage would need to be roughly 3.9V. The variation on the forward voltage would need to be added. Any voltage above this will create excess power dissipation. You are dumping some of it into the series resistors. The rest will be seen at the driver output. Since your LED voltage is 8V, the series resistors are a good idea to minimize the power dissipation on the driver.
Can you verify the Reisstors on the C01/C02... lines? I believe they are 80 ohms. This would then give 8V - 80*0.018 - 3.3V = 3.26V at the driver (or 4.46V). Since each off column is tied to ground through a 10k ohm resistor, the result is 3.26V across the off channel LEDs (or 4.46V if the Blue LED is on). If this is an issue, changing the location of the 80 ohm resistor tie directly to the driver output pin might help. That would change the 1.44V resistor drop to the output pin which reduces the reverse voltage across the off LEDs.
Can you give me any more information about the FB components? I would also be concerned about inductive kick-back at that area.
Let me know if this helps.
Thanks for your reply, here my comments:
1) When you say the "each off column is tied to ground through a 10k resistor" you mean that internally in the TLC5927 has a 10k pulldown when the ouput pin has been set to OFF (bit set to zero and OE set to one)? In this case, how much current will flow through the TLC pin?
2) Why you say that changing the resistor position may help, correct me if I am wrong but the way I see it is the following. The TLC output pin voltage will can be calculated by the following equation 8v - Vres - Vfw - Vtlc =0 then Vtlc = 8v - (80 x 0.018) - 3.3V (blue led) Vtlc = 3.26. this is the same calculation you made, so it does not matter where the resistor is located the voltage will remain the same...right? My point is that it does not matter the order the outcome will reamin the same, maybe I am misunderstood all this since you are the expert here your input will be greatly appreciated.
3) I do not have the FB spec file, tomorrow I can upload it. But even the schematic symbol looks like a inductor it is actually jus for EMI filtering.
4) I am attaching 2 pictures, one shows the anode and cathode voltages when the red led is on and the other when the blue is on. We actually lower the voltage to 5V but we still has some voltages spikes, every spike comes when the TLC output is OFF (OE set to one). We need still to lower the voltage a bit to be safe on the reverse voltage side (maximum 5V) and also to be able to use the open circuit and short circuit detection the TLC offers. As you must understand cathode (CATODO) voltage is the TLC output voltage, sorry that the labels are in spanish.
5) We are going to have a new production next February and I am thinking to lower even more the voltage, I think around 4V...do you have any comment about this change?
6) As you can see in the image looks like the reverse voltage, led OFF, is now around 3.2V. Just to undestand how this works I would like to know why it is not more near to 5V. As you said the TLC output will pull high to the supply voltage, minus a small drop, but 1.8V does not seems small to me.
I really appreciate all your help.
1) The 10k resistor I was referring to is from your schematic (R45 for C01/C02). There is no internal pull-up or pull-down. If C01 is off but DBLUNET is on, the Q3 is on but C01 is tied to R45. When the TLC pin is off, the leakage current will be less than 0.5uA at room temperature with 17V on the output pin (I-LEAK from the Electrical Characteristics table).
2) I was referring to the OFF state, not the on state. I agree with your note.
3) That is what I expected, just verifying.
4) I am not sure why the voltage peaks at 5V then drops to 3.2V. It might be interesting to zoom in and view the OE timing as well.
5) I think 4V would work well. Just be aware of the calculation from note 2. If there is not enough voltage at the TLC device, it will not get to the full current level.
6) For an experiment, take one of the LEDs and hook the Anode to the LED voltage and leave the cathode floating. I did this in the lab and found with a 3V LED voltage at the anode, the cathode measures 1.5V. Is this what you mean?