If I understand you here, you're saying that it's easier to find the hFE at the current you're looking for, then take 1/5th of it instead of dividing the resistance by two? (I see it essentially does the same thing.) In the case of the 2N3904, the table shows a list of hFE values at different currents. At 10 mA hFE = 100; 100mA hFE = 30 etc. I realize now that the voltage specified next to those currents, Vce, is the voltage drop, not the supply voltage. Anyway if I wanted to use the 3904 to switch a 10 mA current, I could assume a hFE value of 20? Similarly if I wanted to pass 100 mA through that poor transistor I could assume a hFE value of 6?
Yes, that would be fine.
Of course, in the already-low-hFE region at high currents, Vce(sat) is high and hFE is low (to begin with), so it's not very favorable; better to avoid that region by choosing a larger device. So a 2N4401 (600mA) would be a good choice for that.
There are low-Vce(sat) types with amazingly high Ic and hFE figures; these are often specified for saturation at several hFE figures, and a typical 5A device might do Vce(sat) < 0.4V at Ic = 3A and hFE = 80! (They also usually have good inverted hFE -- that is, swapping collector and emitter -- it
is N-P-N after all, and on a simple enough level, it shouldn't matter which N you use for what. Most real devices are too asymmetrical for this to be of much value, but it's a neat gimmick on the rare occasion it's useful.)
I like to call them "high gain MOSFETs with really leaky gates". With high hFE, it's really easy to appreciate how a BJT is, in fact, a voltage-controlled device.
You lost me a bit here. I only partially see why the 100mA at 12V thing is bad. I think I'm a just a bit confused. (I think you may have been joking... but I'm a oblivious at this point.)
Yes, that was going on assuming your literal words..
We supply the collector with 12V, it has, say a 1V drop from C->E, therefore we're left with 11V. Assuming I want a 100mA current to flow, I'd need a 110 Ohm resistor. Correct? What that means is that the power dissipated in the transistor would be 1*.1 = .1 W. Correct? (EDIT: I believe this is wrong, but I don't know where.) Well... then I'd be dissipating 1.1W of power in that poor resistor.
Yes, this is the intended situation, and assuming the resistor is rated to handle that watt, not a problem.
Ok, so when we drive it into saturation, we make hFE low, which lowers the power dissipated in the transistor, which makes it able to act as a switch without a heatsink. We're not really using the amplification properties of the transistor at this point. I mean, we are, but not to its full potential. Now I'm afraid to ask what would be the calculation if you wanted to use the transistor as an amplifier (and therefore using the full gain available to you), but seeing as I haven't googled it at all or read about it, we'll leave that for another day.
Well.. you're amplifying, in a gross sense. The peak voltage amplifier gain is about 20 in this configuration*, but obviously, it's zero when fully off (the base is not forward biased = a small change in base voltage gives no change in output), and nearly zero when fully saturated (being the definition of saturation). If you look at the average, you're still doing maybe 0.8V into the base for 10V change in output, which isn't too bad.
*This appears as a homework question in AoE2. I will similarly leave it as a mystery for the 'student' to solve for... er, when sufficient knowledge has developed, I guess. Or to put it a better way, I'm not going to try and overwhelm you with more intricacies of transistors.
The gain is real. You can take a weak, slow signal, pass it through a chain of switches, and each stage sharpens it up. After a few, it's about as fast as it's going to get. One way to think of it is, maybe you have a switch that's designed for a 0-5V input, but it does most of the switching over the 1.5-2.5V range. You can think of it as magnifying that short voltage range, zooming in vertically, and clipping what's outside that range. Plus a little bit of slow, due to the device's own limitations (only an infinite bandwidth transistor could zoom in forever, making a DC-to-light switching edge
).
You can "set" hFE by changing Ib... correct? So, what you're saying here is that the on the "linear" parts of the Ic vs. Vce graphs, the transistor is not yet saturated?
Yes, if Ic is fixed (or limited, as is the case of a fixed (or bounded) resistive load).
How do you change Vce?
Put a battery across it.
I keep thinking of it as a voltage drop (like across a diode,) is that correct? The voltage drop across a diode doesn't change that much with varying voltage or current. Definitely not on the order of 15 volts!
You can reasonably form this concept of "hardness" and "squishiness" in your mind: diodes are pretty hard, resistors are proportional, constant currents (like transistors in the linear range) are very soft, squishy, compliant.
But everything has a component to it. The diode is not infinitely hard, just as diamond is not. Partly, it depends on how hard you look: out of a thousand volts, a 1V diode drop is pretty hard to argue with. Down around the level of a single AA cell (1.5V), that 1V starts looking a lot more squishy -- you wouldn't want to use a plain silicon junction diode in an LED flashlight!
Nor would you want to use a diode drop in a precision circuit: a signal diode is notionally 0.6V, but it's more like 0.5-0.7V depending on current (over some range). So if you're looking at hundreds of millivolts, it's a big deal again. And if you include temperature and manufacturing variations, more like 0.4-0.9V. So you could have an uncertainty of 0.5V in a circuit -- hardly precision work!
Now, when a BJT is saturated -- as long as it is fully saturated, at or below some given hFE, and as long as the current stays within a limited range (0 < Ic < Ib*hFE), the collector voltage will indeed stay pretty low. To the tune of, well, depends on type and range, but for the case of a 2N3904, at hFE <= 10 and 0 < Ic < 100mA (i.e., Ib = 10mA fixed), you will have Vce(sat) < 0.4V. Which is better than a diode drop, by all means!
But what happens if we push Ic even higher? If Ib stays constant, hFE will rise, until it falls out of saturation (probably around hFE = 16, Ic = 160mA). Vce rises and Ic remains fairly steady: suddenly, what used to be a good solid switch is a loose and slippery current sink!
For a mechanical analogy, consider a spring preloaded against a backstop. As you push against the spring, trying to compress it, a whole lot of nothing happens, because you're effectively pushing against the backstop -- that is, until you exceed the preload, and you push the spring off the backstop. Now the stiffness (change in force per change in displacement, dF/dx) is equal to the spring constant -- if it's a very long spring with a small constant, it could be a good mechanical analogy of a current source (which has a high incremental resistance, dV/dI).
I guess I'm confused about this part. I guess if you put a resistor between the source and the collector, and a resistor between the emitter and ground you could set the voltage across the CE junction.
If you fix the base (not B-E) voltage, and place a resistor from emitter to ground, either that emitter resistor gets powered (in part or full) by the base (it's just a series diode), or if you have collector current available (through a relatively low resistance), you do indeed get a limited current flow.
Congratulations, you just discovered linear amplifier bias.
(Pedant alert: there's no "C-E junction". Current actually physically flows through the base layer, across both C-B and B-E junctions... which is probably kind of weird to think about, but hey, it works.)
I'm still confused. Even if you do this with a diode the diode still only has a voltage drop that doesn't change much. Maybe I can't think of this junction as a diode that has a set voltage drop...
If only you could control how much the diode turns on by...
...But wait, you can!....sort of.
If you imagine a BJT as being the fundamental element, somehow (that is, an input voltage on one side makes a load current on the other -- technically, a two port transconductance device), you can easily construct a diode by tying the output to the input: indeed, a diode-strapped (C tied to B) BJT is often used in ICs to make matched Vbe voltage drops.
Well, if you think of a regular (two terminal, and two-layer,
single junction..) diode as this, you're controlling its current flow by... well, putting current on it in the first place.
There are two places I'm going with this, though. One is speed, the other is saturation.
Speed. A junction diode doesn't turn on instantly (FWIW, a schottky diode
does!). It takes time. During that time, current rises gradually, and until current catches up with what the source is capable of, the voltage drop can be quite large. The largest I've heard is a 40V transient across a poor unsuspecting 1N4007, which was achieved by switching a reserve of up to 30A, at a rate on the order of 10s of ns. (A diode turning on looks like an inductor, so Vf ~= L * dI/dt, where L depends on construction, but is relatively constant with respect to what you're doing in the circuit. These numbers suggest L ~= 40V * 10ns / 30A, or ~13nH, which is noticeably larger than the stray inductance of the diode package and wiring.) In effect, you can apply some juicy voltage source to a diode, and either turn it on by waiting some period of time, or leave it off by not waiting (it remains a high impedance for high frequency AC). This is known as forward recovery.
Saturation. A diode-strapped BJT will operate in the linear range at low currents. This is obvious because Vbe > Vce(sat) over most of the range. But as Vce(sat) grows at higher currents, hFE will drop, until at some point, Ib may be as large as Ic, if not larger (hFE < 1)! I don't know what the division of currents actually is in this region, but it's apparent that the transistor isn't transisting very well up here, and even if the B-E junction alone keeps acting like a diode, if we're talking amperes through a poor 2N3904 or something, we're also talking a relatively large voltage drop across internal resistances. A normal diode has the same behavior, where the voltage drop rises relatively quickly with current (that is, proportionally, rather than logarithmically). The resistance is due to the reality of physical materials that aren't superconductors, obviously enough, but if you compare the graph to the Vce(sat) graph, they're kind of similar: pretty flat for a while, then ticking up at very high currents.
As far as I know, the current density in the actual silicon is comparable in this comparison (i.e., Vce(sat) rising, versus the ohmic region of a similarly sized diode).
A diode won't really become constant current, at least not silicon, at currents that don't also cause other kinds of destruction (electromigration?). But there is a limited number of electrons available in the stuff... it's a
semiconductor after all... it should seem intuitively possible to cause "saturation" in some way. As a matter of fact... this is exactly the case, though to a somewhat different effect, and only in other materials. Namely, gallium arsenide (GaAs).
You can take a block of GaAs (pure, obviously), doesn't need a diode junction, just singly doped -- if you pull enough current density / voltage drop across the block, it runs out of electrons and becomes saturated. Electrons are thermally generated, but instead of diffusing about at thermal velocities, they're actually pulled along faster by the strong field. Because of certain solid-state shenanigans that I forget, the response is actually negative [incremental] resistance -- over a range, current drops as voltage rises. This effect happens with very little delay, so it can oscillate at very high frequencies: they're called Gunn diodes, and aren't actually diodes at all (except in the having of two...'odes'), but are commonly used in the 10s of GHz.
Or to put it a better way, I'm not going to try and overwhelm you with more intricacies of transistors.
Scumbag engineer...
Says he's not going to overload with info
Overloads anyway
In my defense, I just kind of get get explain-ey and drifty when I'm sleepy... so, yeah
Nanoseconds are slow. People in my lab work on the femtosecond timescale. But I see your point regardless.
Freakin LAZERS MAN!
Tim