Mosfet overheating in my circuit - can't figure out why - Page 1
Forum section: electronics / projects

author: Whoadie
author_rank: Newbie
timestamp: October 11, 2014, 05:55:21 am
content: Hi everyone, my first post after being turned onto this site.So I am trying to build a circuit to drive some LED's. I am using an AP8803 LED driver which uses a switching method with an inductor and sense resistor. The driver has an internal switch, along with a Vin, sense and ground connection... and has a maximum switching frequency of 700kHz.I am trying to use a couple of MOSFET's to switch two different LED's into the circuit - one is warm white and the other is cool white. They are high power (350mA) LED's. The FET's are DMN3404L and I am driving them with a microcontroller which is running at 3V. The FET has an Rds of 82 micro ohms with a Vgs of 3V, and is rated for current of 2A at that level.My LED part of the schematic is attached. The problem is that the FET's get REALLY hot almost instantly - like 150C! And they shortly burn out.Looking at the block diagram of the AP8803, the switch pin is just connected to a FET internally to ground. I think what is happening is that the inductor is obviously building up a magnetic field and when the LED driver switch is off - this causes a voltage spike which bleeds back into Vin... but that causes Vgs to be negative, turning off the FET. And the FET switching at 700kHz with an 82 micro ohm Rds is causing it to heat up.I'm trying to think how I could solve this easily. I realize the ideal solution would be to use a separate LED driver for each LED, but I don't have the room to do that - not to mention I already ordered a few PCB's and am hoping to be able to use them. I thought about changing to a solid state relay but they are very expensive, and they appear to just use optically activated FET's anyway. I tried using an NPN transistor instead of a FET thinking that it would dissipate perhaps less power and heat up less, but it was just as bad.Can anyone guess why the FET is heating up so much and what I might be able to do to fix the problem? I'm hoping there is something simple and obvious I might be overlooking.Thanks!

author: valvedude
author_rank: Contributor
timestamp: October 11, 2014, 06:09:55 am
content: hi have you checked the currant of your circuit and or the voltage and current handling of your fet

author: valvedude
author_rank: Contributor
timestamp: October 11, 2014, 06:15:52 am
content: i checked the data may be the fet is faulty

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 11, 2014, 06:16:02 am
content: The FET1 / FET2 "on" voltage has to be several volts above the SENSE voltage for this to work.Or, try P channel MOSFETs above the LEDs, with some sort of level translation to drive the gates between V(SENSE) and V(SENSE) - 5V or so.Tim

author: Whoadie
author_rank: Newbie
timestamp: October 11, 2014, 06:56:48 am
content: Quote from: T3sl4co1l on October 11, 2014, 06:16:02 amI'm not sure I understand this... when the LED driver switch is on, the line is tied to ground, so shouldn't Vgs by 3V which is in the working range of the FET? And then when the LED driver switch is off, the FET1/FET2 lines will also be off, but that should be OK (I think?) because the inductor's magnetic field will break down and the voltage will rise to Vin + 0.6V and then bleed through D1 back to Vin, no? In that state, Vd and Vs on the FET would be the same and Vgs would be -9V (if running from 12V). Would that be a problem for the FET if Vd and Vs were both the same voltage which was higher than Vg?Do you think the FET is overheating because it's spending too much time switching and its Rds is causing the excess heat?The thing is, the circuit works - well - I can switch between the two LED's no problem with 3V on the "SWITCH" line from the MCU, just the FET gets very very hot. I am not sure why it's getting so hot... but perhaps I am misunderstanding something.Thanks for the help!

author: IconicPCB
author_rank: Super Contributor
timestamp: October 11, 2014, 06:57:20 am
content: LEDs are current driven devices once You exceed the forward voltage drop of an LED it will not limit the current of itself.The sense resistor in Your circuit is ment to keep the current constant. What is the current setpoint for Your circuit?Is it really? To check measure current the driver circuit takes from the powersupply; it will be close enough.Rather than attempting to drive the two fest inside the switching path and hoping for best at 3V gate source voltage measure the voltagedrop across the inductor to ensure that You are in faact operating the fets at 3Vgs.The voltage across the inductor will tend to work so as to reduce gate drive voltage. It will cause the FET to operate in less switched on mode ( back into linear mode where the drain source resistance will increase from 28 miliohms to something more substantial and cause the fet to heat up.Better option might be to get two LED drivers and use the control pin on the LED driver chips to select / enable the desired LED.

author: Phoenix
author_rank: Frequent Contributor
timestamp: October 11, 2014, 07:21:06 am
content: There might be a couple of problems here:Firstly if current is flowing in L1 and both FETs are turned off there is no freewheel current path (D1 can not free wheel in this condition).Secondly the Vgs of the FETs will be Vgs = V(FET) - V(L1), as the previous poster aluded to this may put it in a linear condition.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 11, 2014, 09:26:12 am
content: Quote from: Whoadie on October 11, 2014, 06:56:48 amNo, because the source is not tied to the LED driver thing, it's after the inductor.You'vecascodedthe current source, so that the MOSFET drops almost all the supply voltage, and the LED drops whatever it needs (2-3V?).QuoteSo in addition, you're switching the FETx signals synchronously? So the inductor current has nowhere to go?Yes, that would be a problem, even more so than the first...QuoteIt's not switching at all...Consider what must be true if the MOSFET is saturated. I'm going to assume SENSE is, like, 10V or something. If an LED is on, it's dropping, say, 3V (which would be a blue or white LED, but whatever it is). So the FET drain has to be 7V. The source isn't above 7V (there's nothing to pull it up, only down; and it has a body diode anyway, which means source voltage won't be more than about 0.7V more positive than drain voltage regardless), it must be below. What is the source voltage referenced to? Gate voltage. If you're driving FETx from +5V, the source is below gate, by Vgs(th) and a little more, maybe a few volts. Then, while the driver is going on and off, it's only drawing current from those few volts, not from the 7V at the LED.To saturate the MOSFET, you have to get Vgs up somewhere reasonable, say 5V. But source voltage follows gate voltage, so you have to keep raising gate voltage until it sticks. That happens when source gets just below 7V, which means FETx needs to be more like 12V, and preferably, maybe even 15-17V.Or, use a PMOS above the LED, so the gate voltage needs to be 10V (off) or < 5V (on), giving a Vgs of -5 to -10V which is in the saturation range. Source is solidly referenced to a supply (or something near a supply, I assume SENSE has a low resistance to some other supply), so you don't have to worry about big gate voltage swings. You do still need level translation if you want to switch this from a logic level.Tim

author: mikerj
author_rank: Super Contributor
timestamp: October 11, 2014, 10:01:02 am
content: Quote from: Phoenix on October 11, 2014, 07:21:06 amThat is the primary issue IMO, the FET, inductor and diodes are in the wrong places (which should be obvious by from the AP8803 datasheet). I guess OP has tried this to avoid using two inductors and diodes, but it's not going to work.

author: wraper
author_rank: Unknown
timestamp: October 11, 2014, 10:22:37 am
content: I don't understand how did you figure out that you can put mosfets in the schematic in a such a way. Looking in the datasheethttp://diodes.com/datasheets/AP8803.pdfI just cannot understand how can you attach mosfets in between of the LEDs and the inductor when mosfet source is floating all the time. You can use P channel mosfets on the high side between SENSE and LEDS though.

author: Whoadie
author_rank: Newbie
timestamp: October 12, 2014, 12:49:56 am
content: Quote from: T3sl4co1l on October 11, 2014, 09:26:12 amThank you for taking the time to spell it all out for me, Tim!I am going to go with the P-MOS high-side solution. I'll just tie the gate to Vin and use an NPN on the gate to pull it to ground using the MCU output. I'll double check that but I think that should work OK - thanks again!

author: Whoadie
author_rank: Newbie
timestamp: October 12, 2014, 12:53:01 am
content: Quote from: wraper on October 11, 2014, 10:22:37 amIt was a dumb mistake. I have used N-FET's lots of times for low side switching of LED's, and it was (as we say in the USA) a "brain fart" to use it again in this situation. I had breadboarded it up as I usually do just to confirm my circuit, and it worked fine. Of course, I had breadboarded it with some big TO220 FET's, and they didn't heat up much so I didn't think about it. It was only when my PCB's came back from the fabricator that I saw they were heating up massively. I figured it was a good learning opportunity and wanted to confirm with the more knowledgeable members what my thoughts were on why the FET was heating up so bad.I didn't consider the lack of a current path for the inductor when my FET turns off... I figured it was an Rds issue since the source would be high when the fet turned off but the inductor still was charged up.