Beginners Transistor Circuit - Page 1
Forum section: electronics / beginners

author: Paul#1966
author_rank: Contributor
timestamp: February 01, 2021, 10:20:30 am
content: Hi AllI apologise in advance if I'm asking "daft" questions but I'm not sure how you learn otherwise.Firstly I'm trying to develop an understanding of schematics and created the simple transistor circuit on a breadboard (see attached transistor.jpg) and then tried to produce a schematic to match it (see attached transistorSchematic.jpg). Do I have at least this part right?I didn't get quite what I expected, I used the 1k ohm & 150 ohm resistors to drop the 5v dc supply to .7v to supply to the base of the 2N3904 and wired a small LED circuit across the emitter and collector as I believed this would just act as a switch; and it does to a point. The bit I didn't foresee was the voltage drop which means that my LED has barely enough to light it up, when activated I get a reading of 3.2v across collector and emitter (when I thought it would be nearer 5v) which only leaves 1.8v for the LED (which is what I get with the meter).Is this expected (such a large drop through the transistor)? Could I have predicted this from the datasheets and if so what would I have been looking for (these things are hard to read).In this scenario would there been a better outcome or is it what it is?Thanks in advance.

author: Kleinstein
author_rank: Super Contributor
timestamp: February 01, 2021, 10:36:12 am
content: The problem with the ciruit is, that there is no good limitation for the LED current. So if the powersupply is powerfull enough it may break the LED and maybe also the transistor from too much current. So normally there should be a resistor in series with the LED.For the understanding of the transistor: the base to emitter votlage is usually not set by an external divider, but more with just a base side series resistor to limit the current. The BJT is considered current controlled. The base emitter voltage is more like a minimum hurdle to overcome before the base current can flow. So the base current is calculated from (5 V - VBE) / R_base. To get a reasonable 1 mA base current for a switch it would thus need some 4.3 K or as the next standard value 4.7 K and no extra resistor to ground, at least non that is so small.

author: Paul#1966
author_rank: Contributor
timestamp: February 01, 2021, 11:01:31 am
content: Thanks for taking the time to respond.So no need for a potential divider at all?I generally do put resistors in my LED tests as I've blown a few up, LOL. I didn't put one in this as there was barely enough getting through to light it up in the first place.So looking at the datasheet I see Vbe (sat) as a min 0.65v. Are you saying then remove potential divider, stick a 4.7k resistor in series to base?

author: Zero999
author_rank: Super Contributor
timestamp: February 01, 2021, 02:00:00 pm
content: Yes, a 4k7 base resistor will do, but add 470R in series with the LED.

author: amyk
author_rank: Super Contributor
timestamp: February 01, 2021, 02:10:39 pm
content: Note that collector (and thus emitter) current is roughlyexponentiallyproportional to Vbe, and also temperature-dependent. Thus trying to control Vbe will almost certainly lead to either nearly no current (as you have luckily experienced), or destruction of the transistor from overcurrent, depending on component tolerances.

author: Paul#1966
author_rank: Contributor
timestamp: February 01, 2021, 02:46:55 pm
content: Thanks again for your time.So I've rebuilt the circuit (transistor1.jpg) and have outputted the values of each component.I see how the 5k reduces the current to the base and how the 470 ohm protects the LED. What I'm not understanding is why do I lose so much V across c & e.I expected the transistor to act like a normal switch on a 5v circuit. As it stands in a practical sense I don't have enough V to illuminate the LED (barely glimmers).I know I've a lot of learning to do, being able to ask these sorts of questions really helps me.

author: Zero999
author_rank: Super Contributor
timestamp: February 01, 2021, 02:57:56 pm
content: IC= 6.8mA, which should be enough to illuminate a modern LED, with decent efficiency bright enough to see, under normal offiice lighting conditions. If it needs to be brighter, reduce the series resistor to 220R, for just over double the current.VCE= 61mV, i.e. 0.061V which is the voltage dropped over the transistor's collector and emitter and is tiny, compared to the 5V supply voltage.

author: Tom45
author_rank: Frequent Contributor
timestamp: February 01, 2021, 03:04:20 pm
content: Quote from: Paul#1966 on February 01, 2021, 02:46:55 pmLook at the voltage drops across each component. Starting with 5 volts, the biggest drop is across the 470 ohm resistor. The second biggest is across the LED. The smallest is across the transistor, less than 0.1 volts.The transistor is saturated, so you can't reduce Vceby much. The LED forward voltage is going to be what it is. What you need to change is the 470 ohm resistor. It is dropping too much.You have 5 volts. Minus 1.7 for the LED and 0.06 volts for Vceleaves 3.24 for the resistor. If you want 20 ma of current for a nice bright LED, then R = 3.24 / 0.02 = 162 ohms. So use 150 or 180 ohms for the resistor.

author: CharlotteSwiss
author_rank: Frequent Contributor
timestamp: February 01, 2021, 03:26:39 pm
content: but this circuit would make sense if the transistor worked as a switch (off / saturation), with a current control on the base

author: Paul#1966
author_rank: Contributor
timestamp: February 01, 2021, 03:31:57 pm
content: Thanks everyone for taking the time to respond, I've made those changes and replicated it on a breadboard. I'll drop those numbers out of the simulator again and get busy measuring and try to understand what all those values mean.

author: rstofer
author_rank: Super Contributor
timestamp: February 01, 2021, 04:33:42 pm
content: Pick a real transistor and find the datasheet. It will specify things like VCE(sat), VBE(sat)and hFE. The datasheet will give you values, Google will give you the specific definition. Here's a datasheet for the popular 2N3904https://www.sparkfun.com/datasheets/Components/2N3904.pdfHere's a definition for hFEhttp://www.learningaboutelectronics.com/Articles/What-is-hfe-of-a-transistorIn your simulation, notice how the various Pd values add up to the original -38 mW figure. From this, you can determine where your circuit is dissipating power. That's the nice thing about simulators, they can give you a guess to several decimal places. The answer is right for the assumptions the simulator is making but when you look at the datasheet, you may find the hFEvaries with load current.The only values that matter are the ones you measure on the breadboard. Simulations may be useful but they aren't real.For a transistor used as a switch, you can guess that VCE(sat)is 0.2V and VBE(sat)is 0.7V and, because we want deep saturation, hFEis 10.

author: Paul#1966
author_rank: Contributor
timestamp: February 01, 2021, 04:42:19 pm
content: Thanks, I have the datasheet for the transistor I'm using (2N3904 hfe 100-300) but at the moment they're a bit like hieroglyphics, but these conversations allow me to look at them and try to understand what each of them mean and where they fit into the example I posted above. I hoping that with enough tenacity I'll eventually get it.

author: rstofer
author_rank: Super Contributor
timestamp: February 01, 2021, 07:03:43 pm
content: There are many sites that discuss transistor switches including:https://learn.sparkfun.com/tutorials/transistors/allIn general, we don't care about the actual value of hFEbecause we're going to assume a low value, like 10, to guarantee saturation. The collector current is 10x the base current, or the other way around, that base current is 1/10 of the collector current.Since hFEis the DC current gain, we can just say that if we want to drive that LED with 10 mA, we need 1 mA of base current.Looking at the base circuit, we have some signal voltage, let's say 3.3V from a uC, and we assume a 0.7V drop from VBE(sat)(the voltage difference between the base and emitter) and, for an NPN switch, the emitter is ground or 0V. So, we need 1 mA to drop (3.3V - 0.7V) or 2.6V across the base resistor. This implies we need a 2600 Ohm resistor and this isn't a common value. We can pick 2700 Ohms and still get the transistor into saturation. If we had 5% resistors, we could pick 2400 Ohms.The uC will drive about 1 mA of base current. Should work...On the collector side, assume the LED has a Vfof 2.2V and VCE(sat)is 0.2V so the collector resistor needs to drop 5.0 - 2.2 - 0.2 or 2.6V at 10 mA. A 260 Ohm resistor works here. I would choose 270 Ohms because I don't care if the LED is not quite as bright. It will still be bright enough.So, this is all of the work to figure the resistors for an LED on a 5V source driven by a 3.3V logic signal.The first couple of times you work this out, it will be a chore. Later on it becomes automatic.

author: rstofer
author_rank: Super Contributor
timestamp: February 01, 2021, 07:16:18 pm
content: Above, I gave you the procedure for calculating the 2 resistors and let's assume 2.7k on the base and 270 Ohms on the emitter with a 2.2V LED (you can change the resistor if your LED has a different voltage).Breadboard the circuit and actually measure the voltages and currents. You can use the DMM current ranges but you have to break into the circuit. Instead, measure the resistance when you build the circuit and just measure the voltage drop across the resistor to get the current.See if the numbers resemble the calculations. They won't be exact due to tolerances but they should be close.Ohm's Law is your friend in this project.

author: tooki
author_rank: Super Contributor
timestamp: February 01, 2021, 09:18:05 pm
content: Quote from: Paul#1966 on February 01, 2021, 02:46:55 pmWhy do you consider the voltage drop large? What kind of voltage drop were you expecting instead? Sharing your assumptions can help us understand what parts of your mental model might be wrong or incomplete.(FWIW, I find BJTs to be irritating little bastards. I still just haven't quite managed to really wrap my head around them, at least compared to the nice, docile MOSFET…My mental model of the BJT is still incomplete. But hey, the semester just ended and at least I got a good grade in that class all the same!)

author: CharlotteSwiss
author_rank: Frequent Contributor
timestamp: February 01, 2021, 10:13:13 pm
content: Quote from: Paul#1966 on February 01, 2021, 04:42:19 pmI have also studied transistors recently; before experimenting I did a whole theoretical part on bjt: interdiction, saturation, active zone, common collector, common base, common emitter. Let's say then I started to understand something (not much, but always better than before)

author: TimFox
author_rank: Super Contributor
timestamp: February 01, 2021, 10:29:08 pm
content: For simple circuits like this, with no need to optimize for speed, it is important to concentrate on which parameters are well-determined, and which cannot be trusted.Untrustworthy:Collector current vs. base-emitter voltage. The current is an exponential function of the voltage, which is bad enough, but the co-efficients are temperature dependent. However, the base-emitter voltage is therefore logarithmic in the collector current, and the possible range ofVBEis small compared with 5 V.Trustworthy:VCE(sat) at sensible LED current will be small compared with 5 V and the LED voltage.Limited trust:LED voltage vs. current is logarithmic, but depends on the LED type/color and temperature. You need this voltage, not negligible with respect to 5 V, to determine the voltage across the current-limiting collector resistor.“DC beta”, the ratioIC/IB, will vary from unit to unit, but should have a well-defined minimum value. Base current above the value implied by the minimum beta (within reason) will cause no harm for a slow switch.

author: Zero999
author_rank: Super Contributor
timestamp: February 01, 2021, 10:39:41 pm
content: Quote from: Paul#1966 on February 01, 2021, 04:42:19 pmhFEis the ratio of the base to collector currents. An hFEof 100 means IC= 100mA, when IB= 1mA. It's normally specified with a relatively high collector-emitter voltage. When the transistor is used as a switch, a much higher base current is required, than the HFEon the data sheet would suggest, which is the bare minimum required for the desired collector current. The general rule of thumb is IB= 10/IC, which is not always ideal, but stick with it for now as it's easy to remember.

author: tooki
author_rank: Super Contributor
timestamp: February 01, 2021, 10:42:52 pm
content: Quote from: CharlotteSwiss on February 01, 2021, 10:13:13 pmFYI, "interdiction" is another linguistic false friend.In English, that's the "cutoff" region.Did you find a good source that explains it all well? I am still looking for something that really makes them "click" in my brain, especially the h-parameters beyond hFE...

author: CharlotteSwiss
author_rank: Frequent Contributor
timestamp: February 01, 2021, 11:29:21 pm
content: let's say I studied several online sources ...and then from the book " The art of Electronics"

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 02, 2021, 12:27:55 am
content: I realize this is a learning exercise, and very useful for understanding the BJT used as a saturated switch, but golly, can't you just use a mosfet? An IRF3205 for example will have negligible on-state resistance even when the gate is underdriven at 5 volts.

author: rstofer
author_rank: Super Contributor
timestamp: February 02, 2021, 01:01:23 am
content: Quote from: alsetalokin4017 on February 02, 2021, 12:27:55 amHow well will it work with 3.3V logic? There are some MOSFETs considered Logic Level but when you really look at the graphs, they aren't all that good at low VGSThe IRF3205 graph of VGSversus RDS(on)doesn't even extend below about 4.5V, 3.3V logic levels won't even tickle it. And it's a monster TO220 package for such a small current. Figure 3 page 3http://www.irf.com/product-info/datasheets/data/irf3205.pdfThe PMV16XN looks like a great candidate for 3.3V interfacing of modest loads:https://www.mouser.com/datasheet/2/916/PMV16XN-1600220.pdfSee Figure 9 page 7I haven't used this device but it sure seems attractive at first glance. Next time I order parts, I'll pick up a few.I realize the OP was using 5V input but I changed the game a few replies back because there is more 3.3V logic around than 5V. If we plan to use MOSFETs (and we should), we have to find something with a really low spec for VGSAlways read the datasheet! All those numbers, every single one, puts a limit on the way you can use the part.BTW, the popular 2N7002 will just barely work at such a low VGSFigure 5 page 5https://www.mouser.com/datasheet/2/308/NDS7002A_D-1522662.pdf

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 02, 2021, 01:54:29 am
content: Well, I do agree with you, but just for grins I dug up my stash of IRF3205s. ((IMHO if you are just an experimenter who likes to use components that you can see and handle with your fingers instead of microscopes and tweezers... these are great for all kinds of switching circuits.)And.... using a single partially depleted CR2016 button cell as power source, and a Radio Shack 273-074 piezo buzzer as load with the 3205 as switch ... and it works fine. In fact I can turn it on and off with just a fingertip, or turn it on hard by shorting gate to drain and off again by shorting gate to source. I didn't even have a breadboard handy, I just stuck stuff together and used a clothespin for a battery holder. Sure, the mosfet's resistance is no longer negligible but at 20 mA current for a LED... it's dissipating a few more microwatts and you need a smaller value current-limiting resistor for the LED. But it does work sufficiently well enough for home experimentation even at less than 3 VGS.

author: Zero999
author_rank: Super Contributor
timestamp: February 02, 2021, 08:40:48 am
content: It seems like a few well-intentioned posts have strayed from the original topic. . .The IRF3205 is way overkill for the original poster's application, which is to light a 10mA LED. A low threshold device isn't required because the drive voltage is 5V. The old 2N7000 would be a more suitable MOSFET.I suspect this is all a learning exercise, rather than a real-world application. A small indicatior LED would normally be directly connected to the microcontroller's output pin.

author: Paul#1966
author_rank: Contributor
timestamp: February 02, 2021, 08:30:28 pm
content: Its funny you say that, I'm working on a practical project which is based around a Basic Stamp microprocessor, i thought i was going to need relays in order to switch a 12v 30A circuit but someone mentioned MOSFETS and looking at the FQP30N06L it would appear to up to the job. So, I started playing around with one.The datasheet says Vgs(th) is between 1 and 2.5V and Rds(on) resistance 0.035ohm so I looked to calculate the resistor value 5V - 2.5 = 2.5V then 2.5/0.035=71.43 ohm. Again its probably my simple thinking at fault but I assumed that if I apply between 1-2.5V to the gate the Drain & Source connection would be made. The problem I'm finding is that even with the power connected but not switched on the connection between Drain & Source is made.I don't understand how?I've attached the schematic I was using and the breadboard implementation,

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 02, 2021, 09:08:18 pm
content: SO your LED lights up even when the switch is not pressed?1. Normally the gate pulldown resistor need only be of some higher value like 10k or even higher. It is there to allow charge to drain off the mosfet Gate so that the mosfet will turn off cleanly, but 100 ohms is far too low. It will waste power whenever the switch is pressed. You want current going into the gate, not running away down the pulldown resistor. This should not affect your main problem though.2. Other than that your circuit should be working as planned, I think, if the mosfet is good. Are you sure it isn't shorted internally?3. (edit) If you are really planning on switching 30 amps with that mosfet... I notice that the datasheet gives the Absolute Maximum drain-source current as 32 amps and that these maximum current and minimum resistance values are taken at a Gate drive of 10 volts. That will be with an adequate heatsink.4. (second edit) Another possibility that occurred to me.... do you perhaps have the pushbutton switch installed on the breadboard incorrectly? Try rotating it 90 degrees. IIRC both pins on one side of those switches are connected. So if the switch is inserted 90 degrees rotated from "correct", the two connected contacts will be making the circuit no matter whether the button is pressed or not.

author: wizard69
author_rank: Super Contributor
timestamp: February 02, 2021, 09:26:42 pm
content: Quote from: Paul#1966 on February 01, 2021, 02:46:55 pmAsking questions is key to learning. It is at the root of science.I'm not sure what you are working from as far as learning goes but a good handbook / electronics text can do wonders. In fact I'd suggest going through all the common transistor configurations one by one. Don't be afraid to experiment with each configuration, do try to avoid blowing up your components though. Even then there is value in what you learn when something goes wrong.For example you can vary the base current a bit in your circuit to move the transistor in and out of the saturation region.

author: wizard69
author_rank: Super Contributor
timestamp: February 02, 2021, 10:19:33 pm
content: Quote from: Paul#1966 on February 01, 2021, 03:31:57 pmThis might set off some people but if you are starting out in electronics drop the simulator. Do the calculations manually which shouldn't be too bad for a simple switch.Also you are suffering from one of the same problems I had when first starting out and that was thinking about things as idealized devices. Almost any switch that you will work with will have a voltage drop across it, this includes mechanical switches. That drop can vary from hardly measurable to very significant (in the context of a circuit it is in). To put it another way, we don't live in a perfect world and the voltages calculated by your simulator are not un-normal. What you will get in real life might vary a bit depending upon all sorts of factors.Back in the day when I was in tech school, they recommended the purchase of a handbook. One of them was "HANDBOOK FOR ELECTRONICS ENGINEERING TECHNICIANS" by Kaufman and Seidman. Mine is copyright 1976 so a bit dated (maybe a lot dated), but the basics don't change. Maybe more up to date people will have a line on a similar book as such a reference comes in very handy over a lifetime of working with electronics (who remembers all of this stuff). I'm even less up to date on educational textbooks, that is texts designed to teach, but it never hurts to download the Navy's old NEETs program.Obviously you will want to return to the simulator at some point but I really see hand cranking through a transistor design, at least the simple parts, as one of the better ways to learn what is going on. This applies to a lot of things in electronics right for the beginning when looking as simple restive parallel and series circuits. Walk through those equations and manually calculate a variety of series parallel circuits to get a feel for what is going on. A simulator can certainly give you answers for all the voltages and currents in a resistive circuit but you should be able to derive them yourself and frankly in simple circuits it should become obvious just my looking what is happening.So yeah I think it is a bad idea to go anywhere near a simulator until you have explored some of the basics with pencil and paper and a good calculator. In the case of your original post if you had learned in this way current flow in the circuit would be pretty obvious even if you did not have the precise numbers delivered from calculation.

author: wizard69
author_rank: Super Contributor
timestamp: February 02, 2021, 10:55:10 pm
content: Quote from: Paul#1966 on February 02, 2021, 08:30:28 pm30 Amps is significant current and as such you have to have your circuit board design and everything else up to snuff. For years I've worked in automation and the way this would be handled in industry is to buy an off the shelf model rated for the input and outputs. Ideally something that snaps on a DIN rail. As long as you are using common signalling voltages found in automation you will have a ready made device that you can count upon. Often these are referred to as Solid State relays and if you get the right device will be optically isolated and have an integrated heat sink. This can significantly cut down on the engineering required.This is not a mass produced solution by any means but if you have less than 6 machines world wide to modify these sorts of modules become amazingly cheap. You will not get the satisfaction of a optimal design that can be had implementing discreet devices so that can be a factor.QuoteMake sure your FET is good.Still I'd be very concerned as to how you expect to get 30 amps onto and then off the board.

author: Paul#1966
author_rank: Contributor
timestamp: February 03, 2021, 09:22:30 am
content: QuoteI replaced it with a 10k and it still behaves the same, I'm curious about the gate pull down resistor, was 5V - 2.5 = 2.5V then 2.5/0.035=71.43 ohm not the right way to calculate this?QuoteI think MOSFET is good:Gateno continuity between drain, source or heatsinkDrainno continuity between gate or source but there is continuity with heatsinkSourceno continuity between drain, gate or heatsink but once tester connected to gate then there is continuity to drain and heatsink.QuoteTo be honest I'm not 100% sure of the 30 amps, it will basically switch a 12V DC circuit powered from leisure battery (clay pigeon trap), I'm a way of getting to that point and am not in a position to get a meter on the switch to check what amps I will be dealing with.QuoteThanks, as it happens the switch was orientated incorrectly (i swearI tested that several times LOL) but the switch doesn't appear to be playing any part in this. If I power up the breadboard the LED lights, if i pull the wire between switch and gate the LED stay lit.I think I read somewhere about a capacitive nature of mosfets, is it there is a small current leaking to the gate which makes drain and source which persists after power is removed from the gate perhaps?

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 03, 2021, 08:36:47 pm
content: I've built your circuit on my breadboard and it works perfectly as desired. (I don't have the same mosfet but any N-ch. mosfet should work in this circuit, and I am using my favorite IRF3205.) The only thing I can figure out is that your breadboard setup must be shorted at the point where the D and S leads connect to the mosfet. Please try simply removing the mosfet and see if the LED still comes on when you apply power.Yes, to answer your question the mosfet gate can be charged by a simple finger touch sometimes and residual charge leaking onto the gate can turn the mosfet on. This is why the pulldown resistor is there, to bleed charge off the gate so it will turn off when desired. You might check your pulldown resistor for continuity -- it is barely possible that this resistor is open or not connecting properly and your gate is staying charged up even when power is cycled.How to calculate this resistor? I just use the largest value that works. Someone up on the theory will no doubt tell us how to calculate it precisely depending on the mosfet's gate capacitance and the circuit's gate drive parameters.Note that this is _not_ the same as the LED current-limiting resistor. These mosfets we are using have essentially negligible on-state resistance in simple DC circuits like this. Just calculate the LED resistor as usual, considering the mosfet as a simple switch with essentially no resistance.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 03, 2021, 08:53:15 pm
content: QuoteI'm confused about this calculation. Are you trying to make a voltage divider to give less than the supply voltage to the gate?Mosfets are generally designed to work as basic switches. When they are operated with low gate drive they can be in the "linear region" of operation where the Drain-Source resistance is linearly proportional to the gate drive. They produce more heat when operated this way as their internal resistance is higher than when operated with full gate drive.If you want to experiment with different gate voltages in your present setup you can simply put a potentiometer in the circuit. One leg to positive rail, other leg to negative rail, wiper to the switch (instead of connecting switch directly to positive rail.) Leave the gate pulldown where it is, on the other side of the switch.EDIT: I've been back to the post where you introduced this calculation and I can't figure out what you are trying to do. The units of your equation don't even match. You have volts divided by ohms and an answer in ohms -- a unit discrepancy, for one thing.Mosfets are different than bipolar junction transistors. BJTs require current flowing from Base to Emitter to enable greater current to flow from Collector to Emitter. Mosfets require no such current flow from Gate to Source, rather they require filling the gate "bucket" with charge to turn on and bleeding charge off by the same connection to turn off.So if you want to supply some reduced voltage to the Gate for some reason, you could use a voltage divider (the potentiometer I mentioned above, or fixed resistors calculated as normal for voltage dividers). And for the output side, you can consider the Drain-Source path through the mosfet as having negligible resistance when Gate "bucket" is full of charge, and more and more resistance as there is less and less charge on the Gate. No gate charge, mosfet is "off" and no current flows between Drain and Source. Full gate charge, mosfet is "ON", fully conducting between Drain and Source.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 03, 2021, 09:20:15 pm
content: To calculate the LED current-limiting resistor you simply subtract the LED's forward voltage from the total supply voltage, and divide the result by the desired current through the LED. Example: 5Vsupply - 2.5 VLED = 2.5 V more to drop and 2.5V / 0.020A = 125 Ohms for the LED current-limiting resistor -- of which the mosfet's Drain-Source junction in the On-state contributes ... what, 0.035 ohms? Fergeddaboudit.Note the unit consistency: Volts divided by Amps equals Ohms.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 03, 2021, 09:34:20 pm
content: While I'm thinking about it -- for those TO220 cases and larger, they do put some stress on the breadboard sockets due to the width of the leads and will eventually cause loose connections in the breadboard. To lessen this problem you can carefully turn the mosfet's legs through 90 degrees with some needlenose pliers; this will present the narrow dimension of the legs to the slots in the breadboard and the breadboard will last longer.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 04, 2021, 06:52:55 pm
content: I just discovered something else about this circuit. If _both_ the breadboard power supply and the LED are "backwards", then the circuit behaves as you are finding. The LED stays lit as long as the mosfet is in place, no matter the state of the switch or pull down etc.I realize you are using a breadboard power supply that has its output polarity clearly marked, and that your wiring on the breadboard is consistent with those markings. But.... check the power supply anyway to make sure it is putting out the right polarity as marked! And that the LED is installed the right way round!

author: Paul#1966
author_rank: Contributor
timestamp: February 06, 2021, 02:15:49 pm
content: Thanks for taking the time, I know my questions and understanding are very basic but I am learning from it. The issue with my MOSFET was I hadn't a pull-down resistor on the gate and when it first powered on there was enough voltage present to saturate (is that the right word) the MOSFET and make the connection then of course once made without that pull-down the connection was set.

author: Paul#1966
author_rank: Contributor
timestamp: February 06, 2021, 02:21:22 pm
content: I think the maths was sound V/I=R it was more the application of it that was badly flawed, I don't think I needed to be too fussy over the value of the resistor I was going to use as my pull-down. Anyway that issue was sorted out.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 06, 2021, 03:25:30 pm
content: Quote from: Paul#1966 on February 06, 2021, 02:15:49 pmYou are welcome of course... but I must point out that both your circuit schematic and your photo of the breadboard show your Gate pulldown resistor in the circuit. And as I confirmed by constructing it, it works fine, whether this resistor is 100R or 10k. But now you are saying that you did not have the gate resistor installed. So we were talking about a different circuit than what you showed on schematic and breadboard. It's really hard to figure out what is going on if the schematic and photo posted aren't the actual circuit that is misbehaving.Have you tried using the potentiometer as I suggested, to put the mosfet into the linear operation region where the pot setting controls the LED brightness? That experiment, coupled with a few current and voltage measurements with your DMM, will help you learn a lot about how mosfets work and how they can be used for all kinds of things in the home lab. It is also how you can determine the actual Gate threshold voltage for the mosfet you are using.

author: Paul#1966
author_rank: Contributor
timestamp: February 07, 2021, 01:27:47 pm
content: QuoteThe first attempt very definitely wasn't working for some reason, when I revisited it it worked fine, see attached (ignore Arduino). The only thing I can see differently was I replaced the original gate resistor (100ohm) with 220ohm. Would that have made a difference?QuoteThe mystery still continues as the pins tested fine with no shorts.QuoteNot tried the potentiometer yet, but I will, again thanks for your input.