Do MOSFETS turn FULLY off?? - Page 1
Forum section: electronics / projects

author: Dave92F1
author_rank: Contributor
timestamp: August 11, 2021, 04:26:45 pm
content: I'm controlling a LED with a MOSFET.When the LED is off, I need it FULLY off - I don't want it generating even one single photon. (LEDs are inside a low-light camera...)Does anybody know - do MOSFETS, when off, go FULLY off? In other words, is RDS(off) truly infinite, or only very large?If they don't go fully off, will a BJT?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 11, 2021, 04:29:44 pm
content: The figure you're looking for is leakage current.You may find it effective to add a switch in parallel with the LED.Tim

author: magic
author_rank: Super Contributor
timestamp: August 11, 2021, 04:35:29 pm
content: Switch? I think a resistor would do. 100kΩ will sink 10µA before the LED sees a single volt across it.In BJTs the equivalent spec is usually called "collector cutoff current". I think they tend to be better somewhat; old 2N2222 is speced <10nA.

author: ejeffrey
author_rank: Super Contributor
timestamp: August 11, 2021, 04:42:15 pm
content: Quote from: Dave92F1 on August 11, 2021, 04:26:45 pmNeither the MOSFET nor the BJT will completely shut off (zero current), they both have leakage current intrinsic to the semiconductor and may have additional leagage from packaging or the PCB. LEDs are made of high bandgap semiconductors so they will generally have less leakage than silicon devices, so a silicon switch leaking will tend to turn an LED on.The solution to this is to provide an alternate path for the current to go. If you have 1 nanoamp of leakage current then a 10 kohm resistor in parallel with the LED will only have 10 mV voltage drop. There will still be some tiny leakage (picoamps?) through the LED but it will not be enough to cause photoemission. If the extra conduction through the resistor in the "ON" state is no problem you are done. Otherwise you can do as Tim suggests and use a second switching device across the LED to drive the terminal voltage to zero.

author: Ground_Loop
author_rank: Frequent Contributor
timestamp: August 11, 2021, 04:43:20 pm
content: MOSFETs do not have infinite resistance when "off." You should drive your LEDs with a dry contact (relay). If space is a concern, I occasionally use something like this:

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 11, 2021, 04:55:46 pm
content: (The switch may come for free, i.e. as in CMOS output pins -- drive the LEDs from an MCU or logic gate. If not, a resistor is a fine method as well.)

author: Benta
author_rank: Super Contributor
timestamp: August 11, 2021, 05:21:16 pm
content: MOSFETs have significant leakage current. BJTs are better, but not perfect. A reed relay might be the best solution.

author: Dave92F1
author_rank: Contributor
timestamp: August 11, 2021, 05:37:46 pm
content: Quote from: T3sl4co1l on August 11, 2021, 04:55:46 pmThanks, but I don't understand what you have in mind.When the LED is on, it'll be driven in PWM from a microcontroller. The MCU will PWM the gate of the MOSFET (because I need more current than the MCU can provide). My concern is leakage thru the MOSFET when it's off.How can I get the "switch for free"?I can imagine a second MOSFET that can be turned on to bypass the LED when I want the LED off (that's a great idea).Does anybody know if the bypass resistor will really work - I think it'll just reduce the current thru the LED, but not (quite) to zero. Do LEDs have a threshold current below which they don't generate ANY photons at ALL? (My own experience is that if any current at all goes thru a LED you get SOME light; but I can't tell what's happening at really low currents with my eyes.)(For sure the brute-force solution is a relay - but I'm hoping for something cheaper and simpler.)

author: Siwastaja
author_rank: Super Contributor
timestamp: August 11, 2021, 05:44:17 pm
content: Switch for free aka driven from a push-pull stage; for example, a small led directly from the MCU output pin. Then this extra switch shorts out the LED.In your case where you only have the low-side NFET, resistor (in parallel with the LED) is indeed simpler than another switch shorting the LED.Reverse leakage current is specified in almost all MOSFET datasheet but it's always 1µA (I have foundonepart, namely SSM6K361NU, which has a different value - 10µA) so manufacturers clearly don't characterize or even simulate anything, real-world number is very likely many orders of magnitude smaller especially with less than maximum Vds and die temperature.But for a robust design, you must design for the 1µA leakage. Then you need to verify at which Vf the LED does not generate even a single photon. It may be a surprisingly low value, and I don't know if you can get a guarantee for it. Then size the resistor per Ohm's law.If you want to be extra sure you can slightly reverse-bias the LED by pulling it up to a higher aux supply rail by the resistor.

author: Dave92F1
author_rank: Contributor
timestamp: August 11, 2021, 05:54:24 pm
content: Quote from: Siwastaja on August 11, 2021, 05:44:17 pmOh, yeah, that'll do it.But thinking more about this, as long as I keep the voltage across the LED well below the ~0.7v bandgap, the LED shouldn't conduct anything, and no photons (at all).Or am I confused? I'm not so good on semiconductor physics.If so, then I think ~ 10k ohms in parallel with the LED should completely resolve this.

author: razvan784
author_rank: Regular Contributor
timestamp: August 11, 2021, 06:06:43 pm
content: Quote from: Dave92F1 on August 11, 2021, 04:26:45 pmIt will generate a huge lot of photons, mainly in the far infrared, due to its temperature being above absolute zero. So will everything around it. So at some point you have to realize that everything in engineering is about approximations and figuring out at what point something is negligible.For most practical purposes, as others have said, a resistor in parallel with the LED should be enough to limit the efects of leakage. 10kohm should be enough, 1k to be extra-sure. Also make sure you don't use a MOSFET that's much bigger than it needs to be - those will generally have higher leakage. At this point I should probably point out board contamination that can also induce leakage.

author: bdunham7
author_rank: Super Contributor
timestamp: August 11, 2021, 06:21:21 pm
content: Quote from: Dave92F1 on August 11, 2021, 05:54:24 pmIf you really have an application where you actually need zero or single-digit photon levels, you need to do some pretty in-depth research. There is a phenomenon where at very, very low power levels an LED will have greater than unity efficiency as well as non-zero photon production at voltages well below the bandgap voltage. Basically any voltage bias can cause a statistical amplification effect where it converts heat energy into photons, actually cooling the LED in the process. I suppose his could happen even at zero voltage, although from what I know it is typically observed at fairly high temperatures and voltages on the order of a single-digit denominator fraction of the bandgap.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 11, 2021, 06:39:25 pm
content: I assume you're concerned about visible range photons, not "any and all". But that would be an important clarification, we have no idea if this camera is visible, like CMOS or CCD, or PMT(s) or APD(s), or IR or what...LEDs generate light around a characteristic wavelength, which takes energy, because quantum mechanics. That energy has to come from somewhere, and normally it comes from the appliedpotential, literally: a 3eV (blue) photon needs an electron to drop through 3V, to get enough energy to emit one. (Plus other considerations, which are met by the semiconductor, and emission results.) There's a subthreshold region ("endothermic", actually) where, say at 2.98V, 3eV photons can still be emitted, but at a lower probability: thermal energy (~26meV at room temperature) will sometimes provide the extra kick required, and emission occurs. At least if quantum efficiency is 1 (i.e., the photon guaranteed exits the die), this implies over 100% efficiency, and the chip actually getting very slightly cooler in operation. (Crazy, huh?) The thermal distribution has an exponential cutoff, so the probability drops exponentially as the voltage falls below here (and it's no accident that most semiconductors have exponential transfer curves -- they depend on thermal motion to work!). If the emission rate is say 10^18/sec at 3V, and the rate is a decade every 60mV say, then you need 18*0.06 = 1.08V below, or under 2V, to get to about one photon per second.So yeah, some ~mV should be pretty good. If the LED is red or IR, it might simply not be possible, because the thermal tail at room temperature extends to that level. (In which case, it's as good as a black body emitter, and you'll have no better luck from anything else in the surroundings, without cooling the whole thing that is.)Note that reverse bias isn't a slam-dunk either; there are emission mechanisms that way as well. Infamously, silicon has a broad (green-yellow) emission when avalanched (>6V); silicon does not happen to meet the considerations for forward emission (which would be ~1000nm anyway due to its bandgap), but that doesn't prevent other mechanisms from doing the same.Mind, I'm not sure what if any applies to LEDs, and it may be quite safe to use reverse bias. Easy enough to test, right?In any case, leakage current rises with reverse bias, quickly leveling off to the "saturation current" after some volts, until rising steeply, up around the avalanche breakdown voltage. Current flow means electrons crossing the junction, means potential for light emission; keeping current low (zero voltage, or slight reverse bias) should be key.Aside:Every so often you see a press release from some damn lab claiming "overunity LEDs", or a softened but equivalent version of that. This is merely the trick above! There's no practical use for this -- the intensity is perhaps just barely enough to read by, if you tiled every wall of a room solid with LED chips and biased them to this range. And if you don't mind that it's blue-green light (to which the eye is most sensitive, and for which LEDs of current technology are most efficient).More importantly, modern LEDs aren't 100% quantum efficient (i.e., not every photon generated, exits the die), so they'll likely still heat up, even if most of the emission is truly endothermic. We're talking small differences here (like a few points above 100%), so it doesn't take much absorption to cancel it out on the whole. Anyway, I'm quite happy that modern white LEDs are pushing over 50% efficiency -- the phosphor's loss included*, that's a damn fine showing!*AFAIK, no phosphors are used that have high quantum efficiency; and this would only work for yellow-red light anyway (i.e., emitting two 1.5eV photons for every 3eV photon absorbed). What's used is a blend with broadly yellow emission, balancing the remaining narrow blue line from the die. So they're losing a fair amount as heat in phosphors, plus the die's efficiency; to get into the 50% range is no small feat!Put more viscerally: LEDs emit so much light, that they are, not just perceptibly cooler, compared to the raw electrical input power; but indeed they dissipate enough less that smaller heatsinks are possible, i.e. it affects thermal design. Contrast this with the >95% heat output of incandescents, you get a little light out as a side effect and have to dissipate essentially everything you put into it. Very, erm, cool.Tim

author: Dave92F1
author_rank: Contributor
timestamp: August 11, 2021, 06:44:15 pm
content: Quote from: bdunham7 on August 11, 2021, 06:21:21 pmGood to know. I don'tactuallyneed zero (visible band) photons - just very few. It's a Peltier-cooled CCD; we do exposures up to 5 minutes long.Altho these days there really are single-photon imagers:https://wisionlab.cs.wisc.edu/project/quanta-burst-photography/andhttps://www.hamamatsu.com/us/en/product/type/C15550-20UP/index.htmlI'm glad I don't have to worry aboutthat.It sounds like I need to run some tests - build a circuit and put it under the camera. Probably the resistor will do the job. Probably.

author: David Hess
author_rank: Super Contributor
timestamp: August 11, 2021, 06:54:43 pm
content: Drain current has an exponential relationship to Vgs *below* the Vth (threshold voltage). Above Vth, the drain current is proportional to the square of the difference between Vgs and Vth. Note that Vth as specified in datasheets is commonly *not* the real Vth, but is instead just measured at some specified low current for convenience and close enough for most applications.Beyond this I assume there is fixed leakage between the drain and source but it is not well characterized.I like magic's suggestion of adding a resistor to provide the leakage current and that is how I have done it in the past. An improved method is to put a forward biased diode in series with the LED on the power side with another resistor to ground to keep it forward biased. Now with both resistors and the diode, when the MOSFET turns off, the LED will see a slight negative voltage so there is no possibility of forward current through the LED.

author: coromonadalix
author_rank: Super Contributor
timestamp: August 11, 2021, 07:15:22 pm
content: An different aproach would be to short the led when not in use to kill anything that coud "drive it"you could add an inverter on the mcu control pin outputhigh = led mosfet onlow = short the led with added circuitry

author: Kleinstein
author_rank: Super Contributor
timestamp: August 12, 2021, 08:36:05 am
content: With the LED in slight reverse bias, there is a good chance this can reduce the emission below the black body value for the relevant wave lengths. Consider this a very black LED, that is more than off. The reverse effect is measurable: exposing a IR detector LED to low temperature will produce the little bit of reverse voltage. It may even work with Silicon - PV elements producing reverse voltage from a cold dark sky (no moon or bright stars).The resistor in parallel should be sufficient to keep the LED voltage low. Even 1 nA and 10 K would be only 10 mV, less than the thermal kt/e (= 26 mV).The leakage currents found in the MOSFET datasheets are usually test limits actual leakage current is usually lower - way lower. It is just impractical to tests parts in the low cent range for leakage in the sub nA range if hardly anbody needs this.After turn off the LED may show some delayed light from long lived internal states.There is a chance the LEDs (and maybe already the expoxy) will show scintillation and thus convert some cosmic rays to light. Here shortening the LED may help to reduce this.

author: nali
author_rank: Frequent Contributor
timestamp: August 12, 2021, 09:15:39 am
content: A push-pull output stage may help, depending on the current you need, something like a MOSFET gate driver may suffice.Also just a thought - what LEDs are you using? For example white LEDs contain phosphor which will give you afterglow effects...

author: Zero999
author_rank: Super Contributor
timestamp: August 12, 2021, 10:40:09 am
content: I'd just put a current source across the LED. A PNP BJT with a 10M resistor to 0V will do. The current will not be accurate and depend on the power supply voltage and hFE.

author: jonpaul
author_rank: Super Contributor
timestamp: August 12, 2021, 12:38:39 pm
content: Check LED spec. Below a certain threshold of current, no light can be emitted.You can characterize the LED with a curve tracer or DCcurrent source, use a photodiode to detect the output.Jon