MOSFETs and Miller plateau - Page 1
Forum section: electronics / projects

author: TreeOone
author_rank: Newbie
timestamp: April 03, 2019, 09:24:01 am
content: Hello!I'm looking for a suitable MOSFET which could be used to turn ON and OFF coil of electro-mechanical relay (rated current 100mA). I am intending to use +3.3V from MCU's output for charging and discharging MOSFET's gate.I have following doubts:MOSFETs have Miller plateau, which should be completely charged, before Vds can drop to lowest values. If in Vgs-Gate_charge diagram Miller plateau is not reached, Vds stays high.I was happy with this understanding until I stumbled upon following datasheet:https://hr.mouser.com/datasheet/2/308/NTD3055L104-D-114990.pdfNow, by looking at following datasheet, two contradictory informations are presented:1) On page 3 there is "Figure 1. On−Region Characteristics", which clearly states that you can drive this MOSFET with 3V. In fact it says you can also drive my electro-mechanical relay's coil. When gatevoltage is at 3V, I should expect Vds to be less than 500mV @ Id = 100mA. That's great, I found a solution. But....there is also....2) On page 5 there is " Figure 8. Gate−To−Source and Drain−To−SourceVoltage versus Total Charge" which states this MOSFET's gate will not even reach Miller plateau with 3V, i.e. Vds will be high, you need at least Vgs = 5V. Crap....this will not work :/Now, could somebody explain how previously mentioned diagrams can exist in the same datasheet???I also have this MOSFET:https://hr.mouser.com/datasheet/2/268/TN0604%20D080813-965142.pdfThis one has Miller plateau at 1V, but also smaller Id = 0.5A, for Vgs = 3V.Which MOSFET should be used?

author: Zero999
author_rank: Super Contributor
timestamp: April 03, 2019, 09:56:05 am
content: One question: why not use a BJT?It's cheaper and far easier to get parts which will do this. The plain old BC338 can do this with minimal base drive. A forced Beta of1/100, so that's IB= 1mA will give a VCE<1V (probably around 200mV), when IC= 100mA.http://www.mouser.com/ds/2/149/BC337-193546.pdfTo answer the original question, either MOSFET will do, but they're both way overkill and way more costly, than a simple BJT. If it must be a MOSFET, then go with something like the FDN337N, which is specifically designed for 3.3V logic level operation. In any case it will still be more expensive, than a BJT.https://www.onsemi.com/pub/Collateral/FDN337N-D.PDF

author: MagicSmoker
author_rank: Super Contributor
timestamp: April 03, 2019, 10:19:32 am
content: ^this is the correct answer. A MOSFET is the wrong switch for the job; use a BJT instead. If you are really lazy, use a "pre-biased" BJT which already has the base and base to emitter resistors inside such as MMUN2231 or the like.

author: 2N3055
author_rank: Super Contributor
timestamp: April 03, 2019, 11:57:01 am
content: NUD3124

author: TreeOone
author_rank: Newbie
timestamp: April 03, 2019, 01:06:39 pm
content: Ok, you convinced me that BJT is also acceptable option.By looking at the datasheet:https://www.onsemi.com/pub/Collateral/BC337-D.PDF....if I want Ic = 100-200mA (added little redundancy), diagram "Figure 3. DC Current Gain" says I need 1mA in the base for Ic = 200mA. (h = 200 @ Ic = 200mA)Now the question is how to feed that current to base, i.e. how big should be base resistor.Table "ON CHARACTERISTICS" says Base−Emitter On Voltage = 1.2V @ (Ic = 300 mA, Vce = 1.0 V)....so voltage drop on base resistor should equal 3.3V – 1.2V = 2.1V...and resistance equals 2.1V / 1mA = 2100 OhmsIn my case this resistor should insure Vce < 1.0 V @ Ic = 100-200 mAIs this calculation correct?And finally, would please somebody give answer to my initial question regarding Miller plateau?

author: duak
author_rank: Super Contributor
timestamp: April 03, 2019, 08:42:44 pm
content: TreeOone, you asked about the Miller Plateau shown in figure 8 of the NTD3055 datasheet. The reason why the plateau is above 4 V is because the drain current (ID) is 12 amps and this voltage is required to support the drain current. If the drain current is reduced, the plateau voltage will also be reduced as in figure 2. The manufacturer likely chose 12 A because it is the maximum continuous current and is the worst case condition.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 04, 2019, 12:58:23 am
content: BJT isn't really needed here, and I would prefer a MOSFET because the base current requirement is fairly steep (multiple mA).2N7002 is fine for such a load, it would seem. The '3055 is rather overkill, but very jellybean so you can use it if you like. It should be adequate for 3Vgs(on) service, too.Some notes on use of datasheets, and design with both types:- The plateau isn't what you're after (which is how you've phrased this). It's merely a side effect of what you're after.- Vgs(th) is where the transistor begins conducting (specifically, at the Id shown: 0.25mA). This has a spread due to manufacturing tolerance.- All the curves are plotted around the average case, i.e., Vgs(th) istypical. Referring to Fig.1 for example: the spread is approximately equivalent to an off-by-one error on which curve you're looking at. That is, the 3.5V curve could be at 4.0V or 3.0V. They don't give 2.5, 2.0 and so on (which are probably nearly zero on this scale), so it's hard to say how much current it's actually capable of, under worst case conditions (lowest Vgs(on) in your application, highest Vgs(th) of the transistor).- It's also not simply one step lower, because transconductance drops sharply at low Vgs. This is illustrated by Fig.2, where the slope of the curve becomes quite shallow at low Vgs. But it's hard to tell how much, and the tempco is also quite exaggerated down there (it's not given in this datasheet, but a curve of Vgs(th) vs. temperature is sometimes seen). The effect is that the curves in Fig.1 get more closely packed as you go down in voltage, and current doesn't actually go to zero, it tapers off more gradually.The Rds(on) curves also do not show this; they end at Id = 5A. Actually, they are mislabeled, anyway, which is dumb.- Overall, I expect that gain is high enough, that you can be reasonably sure to handle much more than 100mA, at Vgs(on) = 3V. This is only 1V above Vgs(th), but it's probably capable of 0.5A or more under this condition, too.- The plateau occurs when drain voltage is changing. It is due to Miller effect, i.e., Cdg multiplied by effective gain. You only have gain while drain voltage is changing, and that's only when load current is going from somewhere near zero, to somewhere near full load (Vds in voltage saturation). So Fig.8 only applies to the load current specified. It will of course be lower (and shorter length) for lower Vds.Noteworthy that Coss (Cds + Cdg) and Crss (Cdg) is very high at low drain voltages, so the plateau isn't proportionally shorter, at lower voltages.BJTs:- The standard rule of thumb is Ib(on) = Ic(max) / 10 or so. The 10 is the saturated (forced) hFE. It's "forced" because, when Vce is low, putting in more base current isn't going to make it turn on any further (Vce may drop slightly, that's it). So, anywhere from hFE < 50 to hFE > 0.5 maybe, you can use whatever base current. hFE remains defined as the ratio of Ic/Ib, so, if Ic is acting independent of Ib, we canmake-- force -- hFE to be whatever we want it to be.- For real parts with hFE (linear) in the >100 range, I prefer hFE(sat) in the <30 range. This gives reasonable Vce(sat) (usually not the lowest possible, but also definitely not unsaturated), and doesn't store as much excess charge, so is faster.- For low-Vce(sat) types (e.g., PBSS303NX), hFE is fairly high to begin with, but it remains high even at low Vce, and even at very high Ic (several amperes for this example). hFE(sat) = 80 is often found in the datasheets!- Stored charge is the BJT equivalent of diode reverse recovery. It doesn't turn off until the B-E junction has discharged, and the amount of excess charged stored depends on the forward bias current. This charge must be removed before the transistor turns off. (It is for this reason, that a BJT -- like the MOSFET, for reasons just covered!-- can be more accurately considered a charge-controlled device, rather than a current-controlled one.) If left alone, this takes 10+ microseconds on its own. (The B-E junction can be modeled as a very small, very leaky, battery: complete with exponential V(I) and charge(V) dependency, for very similar reasons, actually!)- For this reason, normally the base is driven through a voltage divider, so that there is a path to discharge the base. Design the divider for a Thevenin equivalent output "on" voltage of 1.2 to 2V, "off" of less than 0.3V, and a Thevenin equivalent resistance adequate to deliver the required base current.Or to put that another way, supply say 1.5 Ib of turn-on current, and put a resistor Rbe = (Vbe) / (0.5 Ib) across B-E. That way you get about Ib base current to turn on, and keep it on; and about 0.5 Ib reverse current to turn it off. This will give turn-off times of 50-500ns, a significant improvement.If you aren't pressed for time, and have a low logic voltage to begin with (like 2.5V CMOS), simply using one resistor will suffice. The turn-off current flows through the same resistor, just less compared to the divider.3V CMOS is on the edge of where I'd say "yeah just put in a B-E resistor" versus shrugging it off. YMMV.- If you do follow tradition and use hFE(sat) = 10, then you'll need quite a lot of base current (i.e., 10mA!), which may even need a buffer. (Not all logic is created equally: some pin drivers are only rated for a few mA DC. FPGAs are often limited this way.)- This is also an efficiency hit (10% increased current consumption!), particularly for variable loads that may be lighter (so, not your relay coil, but suppose this were a general switch output that someone might just put a few mA of LEDs on, or a full 100mA coil, or anything inbetween -- such a waste unless it's actually needed, isn't it?).- It is for these reasons that I would also prefer a MOSFET. Basically no DC gate current, so it's fine all around.Tim

author: TreeOone
author_rank: Newbie
timestamp: April 04, 2019, 07:56:07 am
content: Tnx Tim, for your elaborate answer.My knowledge of BJTs in saturation region was poor.So I am back looking at MOSFETs.Before we proceed with discussion, I would like to point out that I am looking only atthrough holeMOSFETs,SMT is not applicable to my project.So far we have concluded that NTD3055 will probably work, we don't know how exactly (Vds @ Vgs = 3.3V is a mystery), because I am using too small currents (<200mA), and that's why you said it is an overkill.Can somebody recommend some other MOSFET (found on mouser.com, with TO-92 case), which could be better alternative than NTD3055 for my aplication (Vgs = 3.3V, Id < 200mA)?Up till now I failed to mention that I intend to supply my coil (rated voltage 12V) from 17V source. So voltage drop Vds can not be high at rated load (max 3-4V). I also have resistor between coil and MOSFET, but it seems I'll ditch it, because there will not be any need for it.

author: David Hess
author_rank: Super Contributor
timestamp: April 04, 2019, 08:11:19 am
content: I would use the bipolar transistor. A forced beta significantly higher than 10 or 20 can be reliable by using a higher current low Vce(sat) part so that the operating point on the hfe curve occurs before the hfe dropoff at high collector current. (1) The base-emitter shunt resistor can be left out in simple switching applications when a CMOS driver is used because the series base resistor will pull current out of the junction and the storage time is not critical anyway. (2)But there are plenty of suitable low threshold power MOSFETs now so use whatever you are comfortable with.(1) Modern low Vce saturation transistors used a distributed emitter like if not the same as RETs (ring emitter transistors) so suffer from less hfe and Ft dropoff at high collector currents anyway.(2) The base-emitter shunt resistor serves a more important purpose which is also not required with CMOS drive; it diverts the Icbo leakage which would otherwise try to turn the transistor on. This can be significant at high temperatures with power transistors. With CMOS drive, the series resistor is more than enough to divert this leakage.

author: GigaJoe
author_rank: Frequent Contributor
timestamp: April 06, 2019, 04:51:31 am
content: just wonder the relay voltage, 12V 100ma ?how about like this : can use any mosfet

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 06, 2019, 06:14:21 am
content: That would be always-on?Tim

author: GigaJoe
author_rank: Frequent Contributor
timestamp: April 06, 2019, 06:27:39 pm
content: definitely, so silly, made me laugh, never do anything in middle of the nightnow instead 1 transistors, we need 3 to manage relayhope it will work, now

author: SiliconWizard
author_rank: Super Contributor
timestamp: April 06, 2019, 07:12:44 pm
content: I haven't had to design a circuit with relays in years, but when I did I used to use BS170's (/MMBF170) and never had any issue driving them with 3.3V logic signals while sinking 100mA or so. I would probably use MMBF0201's instead now for your requirements (of course there are many more available).

author: TreeOone
author_rank: Newbie
timestamp: April 06, 2019, 07:53:15 pm
content: I am going to use TN0606N3-G.

author: GigaJoe
author_rank: Frequent Contributor
timestamp: April 06, 2019, 08:03:06 pm
content: Gate threshold voltage 2.0V max, need some dividerwith 3.3 , i think any logic mosfet will work, what the realy voltage ? 5V? or 12V

author: Zero999
author_rank: Super Contributor
timestamp: April 06, 2019, 08:29:13 pm
content: Quote from: GigaJoe on April 06, 2019, 06:27:39 pmPresumably R2 represents the relay col? It will work, but is a silly idea because it uses three transistors, rather than one and there's no back-EMF protection diode.Use a single BJT.Quote from: GigaJoe on April 06, 2019, 08:03:06 pmI haven't looked at the datasheet, but a voltage divider is only necessary if the gate voltage can exceed the maximum rating, which is normally between 15V and 30V. It has nothing to do with the threshold voltage.

author: GigaJoe
author_rank: Frequent Contributor
timestamp: April 06, 2019, 10:11:05 pm
content: sure,I base * h21 in saturation = Ic that really not enough for relay 100ma,it was discussed abovemosfet , read datasheetVGS (th)Gate threshold voltage 0.6-2.0V , where TC has 3.3V outputyou may find-out a meaning for gate threshold herehttps://en.wikipedia.org/wiki/Threshold_voltage

author: Zero999
author_rank: Super Contributor
timestamp: April 06, 2019, 10:54:49 pm
content: Quote from: GigaJoe on April 06, 2019, 10:11:05 pmYes, it was discussed above. The transistor doesn't have to go into complete saturation to turn the relay on. It just needs to turn on enough. The BC337 has a minimum hFEof 100, when IC= 100mA and VC= 1V. A couple of mA of base drive is more than sufficient to turn on the transistor with <1V of voltage loss, leaving >11V for the relay.QuoteSorry, I don't understand what your point is. It's still a silly idea to use three transistors when one will do. If you needed to control a much larger relay, i.e. a contactor, then ditch the MOSFET and put the relay's coil in place of R3.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: April 07, 2019, 08:51:39 am
content: If you ask me this should not have been posted in the "projects" section, but in the "beginners" forum.I think it's a good idea to get a breadboard and som TO-92 sized MOSfets and BJT's and start experimenting untill you better understand the parameters of of these things.There is no substitute for rolling up our sleeves and building something yourself.I am quite familiar with MOSfet's, but had never heared of the "Miller plateau", but found it a quite intuitive name.Typing it in a search engine confirmed that it is used more often, and that it is exactly what I thought it was.https://duckduckgo.com/?q="Miller+plateau"&ia=web

author: rbola35618
author_rank: Frequent Contributor
timestamp: April 07, 2019, 04:38:43 pm
content: Hi EEVblogger,Here are two videos on the switching characteristic of mosfets and about the miller plateau.It may or may no help but I put up here for your consideration.Robert BolanosMosfet switching characteristics 1Mosfet switching characteristics 2