yes thanks this is perfect:
BJTs have greater transconductance, meaning you are able to get more current output per unit power applied. The transconductance of FETs is much lower. So if you use the same amount of power at the input for both a BJT and FET transistor, the BJT transistor will produce more gain. This is why BJTs are more popular for amplifier circuits. They produce gain than a FET can. This is why in the case of simple amplifier circuits, the use of a BJT is preferred and FETs are rarely used. For simple amplifiers, FETs are really only used only when it is desired for there to be extremely high input impedance.
Heh 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 amplifier
Strictly 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 dissipate
real power, they only consume
reactive 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 [power] 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 get
too much and 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
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