transistors (both bjt and fet). The differences and my confusion - Page 1
Forum section: electronics / beginners

author: Moriambar
author_rank: Unknown
timestamp: May 11, 2021, 06:02:12 pm
content: Hello.First of all let me say that most of my experience and knowledge come either from AoE or from tinkering and experimenting.Second of all my goal: I mainly pursue knowledge about this stuff but practically I've always been tinkering in the digital world, and definitely cannot afford anything RF (not that that interests me much), but I'm trying to learn as much as I can: you never know when it can be usefulI still am confused a bit from transistors, especially by the differences from BJT and FETs (mostly mosfet) circuits.First question: are JFET that important? I see that the AoE talks a lot about them, but in tutorials on circuits and arduino and such I've never seen a jfet used, with the possible exception of some (at least one) w2aew's videos.Anyhow I kind of understand the basic bjt circuit on a beginner level, I mean the emitter follower, the common emitter amplifier, switch and current source (common base is basically ignored in tAoE). But are MOSFET only used as switches as tAoE seems to imply? I mean, why can't I create a MOSFET common source amplifier or a current source?I understand that FETs are Voltage driven instead of current driven and Vgs determines how much of the channel is open, but I cannot quite grasp why some circuits do not work or are not used. Or perhaps the authors did not want to discuss them?For anyone still reading: thank for your time, I hope that made sense.Cheers

author: Grandchuck
author_rank: Super Contributor
timestamp: May 11, 2021, 06:12:23 pm
content: http://www.learningaboutelectronics.com/Articles/BJT-vs-FET.phpand there are many other sites that cover the difference

author: rstofer
author_rank: Super Contributor
timestamp: May 11, 2021, 06:44:59 pm
content: This is the place where a breadboard and a handful of parts comes into play. Breadboard your BJT CE circuit and, using about 3 DMMs, measure Ib, Ic, Vce. Put the numbers in an Excel spreadsheet and plot the results (all versus Ib).Similar experimental circuits will exist for the MOSFET and, presumably, the JFET. w2aew has pretty good videos about this stuff.

author: Moriambar
author_rank: Unknown
timestamp: May 11, 2021, 07:20:15 pm
content: Quote from: rstofer on May 11, 2021, 06:44:59 pmWell, apart from having only 2 DMMs (my bad), I will try. Too bad I won't understand why things are happening or why probably, but I can ask with more info later thanks.One thing though: isn't Ig in the mosfet practically zero? How should I set up for the FETs?As for w2aew's videos on tranny they're kinda all on bjt, so far I've only seen one using a JFET and 1 using a MOSFET (actually just connecting it to the dmm).thanks anyhow

author: Moriambar
author_rank: Unknown
timestamp: May 11, 2021, 07:22:13 pm
content: Quote from: Grandchuck on May 11, 2021, 06:12:23 pmyes thanks this is perfect:Quote

author: james_s
author_rank: Super Contributor
timestamp: May 11, 2021, 07:23:59 pm
content: You don't need multiple DMMs, you can easily take the measurements with just one of them. You can also get really cheap DMMs for just a few dollars that are sufficient for basic measurements on low voltage circuits.Yes Ig on a mosfet will be effectively zero, except for a very brief instant as the capacitance of the gate is charging or discharging.

author: Manul
author_rank: Super Contributor
timestamp: May 11, 2021, 07:57:32 pm
content: You may also try simulation software for the experiments. It does let you observe a lot of things easy, save time and resources. It is not without a catch, but could be very useful.

author: magic
author_rank: Super Contributor
timestamp: May 11, 2021, 11:13:08 pm
content: Quote from: Moriambar on May 11, 2021, 06:02:12 pmYou absolutely can have those things and it's all routinely done in CMOS analog ICs like the simple opamp below.FET current source can even be more accurate, because drain current is exactly equal source current. For example, you can take the "precision current sink" example in TL431 datasheet (which isn't really precise at all) and improve its accuracy just by replacing the NPN with a FET.

author: rstofer
author_rank: Super Contributor
timestamp: May 12, 2021, 01:51:47 am
content: You can take the measurements separately, one DMM can do the job. It isn't as fun as having all 3 meters running at the same time but it still comes up with the right answers.Pick nice numbers: The base resistor can be 10k and the collector resistor can be 1k. Then just measure the voltage across them and use Ohm's Law I = E / RI would be working in the area of Vcc = 5V or higher. You will need to vary the base resistor voltage to get some particular value of Ib. A 1K potentiometer might be helpful. One end on Vcc, one end on ground and the wiper to the base resistor.If you assume the base end of the base resistor is 0.7V, you can get some idea of how the base resistor current should behave as you rotate the pot. But what really counts is Ib and you get that by measuring the voltage drop across the base resistor or by actually measuring the current. I would prefer to have a meter measuring the current permanently part of the circuit. Same with the collector current and Vce. Then I could watch all 3 values change as I changed the pot.

author: rstofer
author_rank: Super Contributor
timestamp: May 12, 2021, 01:59:52 am
content: Quote from: Manul on May 11, 2021, 07:57:32 pm^^^ This ^^^If you're going to learn electronics, LTspice will be a very helpful tool. There are many videos and most of the contributors around here know how to use it. If you have a problem, just ask.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 12, 2021, 04:51:00 am
content: Quote from: Moriambar on May 11, 2021, 07:22:13 pmHeh well... not quite... but as you'll see in reading about these things, people aren't always clear about what they mean, or necessarily know what they're talking about in the first place...Take for example, the argument of "BJTs are voltage/current controlled", which you all are welcome to resume now...(it's voltage, by the way).(...Actually, the above quote is already contradictory as, how could the FET take as much power as a BJT if it has a "high" input impedance? See what I mean...)So, one needs to understand the impedances at work here. And impedance depends on frequency, so you see, there is more to it.We should look at the definitions first.Prerequisites:- DC and AC steady state circuit analysis- Basic transistor terminology (terminals, current flow conventions, signal conventions, AC coupling or incremental changes, etc.)- Common emitter/source amplifierStrictly at DC, we have the BJT:Ic ~= Is * exp(Vbe / Vth) (according to the Ebers-Moll model in the linear operating range)and FET:Id ~= (Vgs - Vgs(th))^2 * k_p (Vgs > Vgs(th), 0 otherwise) (part of Shichman-Hodges model)Note that Vth, Is and k_p are just parameters, and don't say anything about a particular device. They vary with device dimensions, temperature, etc.There's already one very obvious lie here, in that Id doesn't go to zero at Vgs(th) -- indeed, it's measured at some defined Id (check the datasheet!). What really happens there?Like all amplifying devices(?*), there's still an exponential slope in MOSFETs, it just doesn't extend to high currents as it does for BJTs. This is called the subthreshold region. So for a more honest model, we should write some kind of interpolation between an exponential response (Id ~ exp(Vgs - Vgs(th))) and the above quadratic response.Which, I think offhand is going to be a transcendental formula (meaning, an exponential expression that references its own parameter, that has no closed form solution), so I'm not even going to attempt to write it out, it wouldn't be useful... Just try to imagine a curve which is exponential at low currents, and quadratic at high currents. It just gets shallower at high currents -- somewhere around Id @ Vgs(th).*As it happens, the three major types of amplifying devices all depend upon thermal statistics. It's ultimately the exponential "tail" of thermal energies, which creates the exponential transfer function. In vacuum tubes, it's the energy of electrons "boiled" off the thermionic cathode; in BJTs and MOSFETs, the thermal energy of electrons within the semiconductor. Tubes and FETs have their own properties in the high-current regime; arguably, so do BJTs, as they are ultimately limited by device resistances. The different mechanisms in these devices give different (power-law) curves in this regime.Anyway, it does happen that, for given operating conditions, and comparably sized devices (e.g., 2N4401 NPN vs. 2N7000 N-ch MOSFET), transconductance is higher for the NPN. Okay, then...What is transconductance?Transconductance has units of conductance: amps per volt. It's "trans" because the voltage and current aren't measured in the same place. Namely, it's how much output current, for how much input voltage.Input current and output voltage do not come into it at all!Just from transconductance, we cannot make any statement of power input or output. We need to know the input conductance or resistance as well, and we need a finite, nonzero load impedance. These will fill in the missing variables, and then we can solve for power.Say we apply a 1mV step change to the base of a BJT. Emitter common (grounded). Say we measure that the input current rises by 2uA, and the output current rises by 0.4mA. Evidently, we have gm = (0.4mA/1mV) = 0.4S transconductance, and h_ie = (1mV/2uA) = 500Ω input resistance. (Which corresponds to an h_fe ~ (0.4mA/2uA) = 200 -- pretty normal for a BJT.) That's on the order of (1mV)(2uA) = 2nW applied power (depends on waveform; this is exact for a symmetrical square wave).(The 'h' (hybrid) parameters are a traditional way to express BJTs in linear circuits. There's also h_oe (output conductance) and h_re (feedback voltage gain). Note the symmetry: input resistance, output conductance; forward current gain, reverse voltage gain. They happen to be reciprocal or complementary pairs. There are many other possible pairings to describe the same thing, and find use in various situations. Anyway, 2-port models are beyond the scope of this topic.)Now, what's the actual power gain? Depends on load. If we have a short circuit, that 0.4mA change develops zero voltage, and output power is zero. We can't have infinite impedance either, because only so much supply voltage is available (the collector voltage can only pull down nearly to the emitter; if it pulled negative, that would be producing power out of nowhere!*). If our supply is 12V, and bias current 10mA, then we could use a load resistor around (12V)/2 / (10mA) = 600Ω, and we would get a change of (600Ω)(0.4mA) = 0.24V, for an output power on the order of 96uW -- a handsome gain of 48k times, or about 47dB. Not bad, huh?*There is actually an operating condition where you can get very slight negative collector voltage from an NPN, but it's more of a parlor trick than anything useful. And, by "operating condition", I mean there's a (in fact vastly larger) power source connected -- it's not getting something from nothing, so the universe is fine with that.(Note that the input and output impedances aren't exactly equal, although they happen to be pretty close in this example. Often, we need to match to a specific source or load, such as RF amps to 50 ohm transmission lines, or audio amps to 8 ohm speakers, etc. In that case, the power gain can still be as high, but we need transformers or networks to match the resistances properly. To some extent, we can adjust bias current and supply voltage to suit things better. Or we intentionally discard some gain, in a compromise for directly coupling the transistor to the source or load -- and with so much gain to spare, we can afford to do quite a bit of that. An example of both is the operational amplifier, a circuit that uses dozens of transistors to reduce input current nearly to zero, while making output current essentially independent of input voltage. The resulting gm and Av (voltage gain) are extremely high, in bulk terms -- but nowhere near as high as 47dB * (dozens of transistors) might imply!)Whereas for the FET case, we might have similar parameters, with lower change-in-drain-current, but most notably, gate current ~= 0. The transconductance might be lower, but the input power is almost nil. So the power gain can be quite high indeed.Now, that's just at DC. At AC, we have to consider device capacitances. I won't go into detail, but a basic introduction is to apply steady-state sine waves instead of DC step changes. Suppose there are capacitors at the input and output, shunting some signal current from each path -- obviously, impedances drop, and gain falls. Although, power gain doesn't drop, by itself -- the capacitors don't dissipatereal power, they only consumereactive power. If our source and load are purely resistive, we don't have a choice on that, we have to drive both regardless, and then it's the total (apparent power) that matters, so power gain will drop. (When we're doing RF amps, we can use inductance to cancel out that capacitance and maintain the power gain -- at the expense of bandwidth / frequency range. Actually, bandwidth and gain are linked, being approximately inversely proportional to each other -- but a proof of this also goes beyond scope.)There's also the complication of feedback capacitance (Miller effect). This further changes the apparent input and output capacitances. We also desire a model for the fundamental device characteristics -- transistors are not instantaneous in operation, but take some time to respond; in effect, instead of h_fe or gm being some fixed value, they are a function of frequency, h_fe(\$\omega\$) and gm(\$\omega\$), generally flat at low frequencies, dropping smoothly above some cutoff point (typically in the MHz). Like there's an internal RC time constant (or something messier than that). When this fact is combined with feedback (C-B or D-G capacitance, source/emitter degeneration, etc.), it will be found that  gain or bandwidth drops at high frequencies, as well as some more spooky effects (the input and output impedance get weird, putting a limit on maximum stable gain -- meaning, if you attempt to get too much power gain out of it, it's not that you run out of gain, it's that you gettoo muchand it oscillates instead! Aren't amplifiers great?..).So, when AC is included, it gets harder to say what's what. MOS is clearly the winner for integrated circuits -- we have transistors with fT in the low 100s GHz, put together in huge logic chains to produce powerful CPUs running at multiple GHz.For modest power applications, we also have for example, old standard MOSFETs like the IRF510, or low-Vce(sat) BJTs like ZXTN2011G:IRF510 is rated 100V 5A, about 800pF input (180pF when off, but this figure includes Miller effect), and about 80pF output. The 800pF input, means it takes about 20mA of drive current to push around at 2.5MHz, with 10V of swing -- 200mW required.ZXTN2011G is rated 100V 6A, and about 26pF output; at 5A, it will need a generous >20mA of base current, and a bit more (peak) to turn on and off quickly. With Vbe(on) ~ 1V, it should be able to run about as fast, yet need maybe 1/5th as much drive power. And its lower output capacitance means the load voltage can potentially swing faster. (And as a check, its fT is rated at 130MHz, so we expect it does have enough gain down at 2.5MHz to actually do this.)(fT is the frequency where h_fe drops to 1; assuming a first-order drop, we can solve for the h_fe at the operating frequency. In this case, 130/2.5 = 52. If we're doing switching, we want much faster edges than a sine wave, so we'll need to drive it harder than that, hence some additional peak base current.)(MOSFETs don't really have an fT, they tend to kind of peter out slowly -- at higher and higher frequencies, the available gain just becomes so small that it's not worth using. IRF510 is effective to the low 10s of MHz. Newer power transistors are good to a few times higher, to maybe 100MHz. RF FETs are optimized differently from power switching types, and regularly clock in with good power gain in the low GHz; special materials (GaAs, SiGe, GaN, etc.) push discrete transistors into the 10s of GHz. They have very, very low capacitances!)Tim

author: ledtester
author_rank: Super Contributor
timestamp: May 12, 2021, 06:03:03 am
content: QuoteFound this article comparing the uses of JFETs and MOSFETs:https://sound-au.com/articles/fet-applications.htmI only have the 2nd edition, but the 3rd edition of AoE has a new section entitled "Why JFET (versus MOSFET) amplifiers?" (section 3.3.6) which would be interesting to read.Also note that the first edition was written in 1980 which is around the time when the JFET op amp TL072 was introduced. From what I read the TL072 made a huge impact in the audio amplifier space, so perhaps JFETs were considered an important technology for designers to know about at the time.

author: Zero999
author_rank: Super Contributor
timestamp: May 12, 2021, 08:50:20 am
content: Discrete J-FETs aren't used so much nowadays, especially P-channel which are becoming increasingly harder to get, but they're still commonly use as input stages inside op-amp ICs. Don't worry about J-FETs too much. It's a good idea to understand the basics, because it will help you realise why an op-amp might not be behaving because its input stage isn't properly biased, but there's little need for discrete J-FETs in modern circuits.

author: exe
author_rank: Unknown
timestamp: May 12, 2021, 10:30:43 am
content: Quote from: Moriambar on May 11, 2021, 06:02:12 pmHa! This reminded me myself >20 years ago when I opened a book and got lost in all that types of fets. All that enhanced mode, depletion mode, jfets, p-type, n-type, with separate body terminal, vertical mosfets, lateral mosfets... Each type had its own symbol, and I couldn't memorize them. Now it's very rare to see anything but n-/p-fets.I'd think of JFET as a depletion-mode mosfet. There is are differences (like, reverse-biased gate will conduct as essentially it's a diode), but it's not that important, imho. As discrete devices they are phasing out. But there are many opamps having at the input stage.I used to like using jfets as current regulators, although they have huge spread of parameters, so you have to trim resistor individually for each fet if you need precise current. If you are lucky, within one batch the spread won't be big. Now I'm annoyed of trimming resistors. But they have a relatively better regulation and compliance voltage, comparing to a single-bjt current source.

author: Moriambar
author_rank: Unknown
timestamp: May 12, 2021, 07:02:52 pm
content: Thank you all for the responses.In particular rstofer to pushing me to experiment and T3sl4co1l for the detail explanation that I'm gonna read again and again until I have extracted all of its meaning.

author: TimFox
author_rank: Super Contributor
timestamp: May 12, 2021, 07:48:30 pm
content: Yes, MOSFETs are much more common now than JFETs. One place where a detailed comparison is useful is low noise analog applications. Semiconductor amplifiers can be characterized for noise as a voltage generator in series with the input and a current generator in parallel with the input. Generally, and this depends on operating conditions and device, BJTs can have lower voltage noise (but higher current noise) than JFETs, so BJTs are preferred for low source impedance, but JFETs are better with higher source impedance. Small-signal MOSFETs have higher noise voltage, but very low noise current, but their “excess” noise (“pink”, “1/f”, or “flicker”) noise is worse at low frequencies. These statements are gross generalizations for roughly audio frequencies, but are intended to suggest what you need to consider for low-noise design.For higher-power, JFETs are almost unknown, and the choice between MOSFETs and BJTs is interesting. MOSFETs dominate over BJTs for switching applications.