Art of Electronics 2.5 - Transistor saturation

It's not as complex as you imagine.

Let's look at a slightly re-drawn, but equivalent circuit diagram:

^{simulate this circuit – Schematic created using CircuitLab}

There's no need for the switch, since we can just assume the switch is active and connecting the \$+10\:\text{V}\$ power rail to resistor \$R_1\$. That, plus separating the power supply rails as I do above keeps things about as simple to read, as possible. I've used a \$\frac{10\:\text{V}}{100\:\text{mA}}=100\:\Omega\$ lamp resistance, since that's what your problem statement also implies.

First thing to do is to notice that there is only \$0\:\text{V}\$ (ground) and \$+10\:\text{V}\$ (the other rail voltage.) These are the only voltages available to the circuit. Technically, these might instead be \$+1087\:\text{V}\$ and \$+1097\:\text{V}\$. It really doesn't matter. The point is that there is a \$10\:\text{V}\$ difference. And that's all you need to know.

Since the entire circuit sits between these two rails, all of the voltages within the circuit must also be between these voltages. Assuming \$0\:\text{V}\$ and \$+10\:\text{V}\$, this means that everything inside the circuit must be a value between those two because there's no access to anything outside that range. It's really as simple as that.

Once you accept this reality, then the rest isn't that complex. There are two nodes in the circuit other than the \$0\:\text{V}\$ node and the \$+10\:\text{V}\$ node. This is the base, which will have some voltage between the two, and the collector, which also will have some voltage between the same two.

Since the base-emitter junction, forward-biased, is just a diode-junction it will have about the \$600\:\text{mV}\$ (up to perhaps a few hundred millivolts more, in extreme circumstances) across it. This pretty much means that the base voltage is relatively accurately knowable. It will be about \$+600\:\text{mV}\$. Note that this is, in fact, between the two rails. It's not impossible, at all.

Now, the collector itself must also be between the two rails. Somewhere between \$0\:\text{V}\$ and \$+10\:\text{V}\$. It cannot be negative. It's not possible. Sorry. So why do the authors even bother to mention the idea of a negative collector voltage???

It's because of the idea of \$\beta\$. It turns out that \$\beta\$ is a useful thing to know if, and only if, the BJT is in active mode and is not saturated. It represents the ratio of collector current to base current, in that case. And so long as the collector voltage is not forced to be close to the emitter voltage, you can apply this idea well. So, for the example here, so long as the collector remains between about \$+600\:\text{mV}\$ and \$+10\:\text{V}\$, then the BJT is in active mode and the \$\beta\$ value matters more. But if the collector voltage gets closer to the emitter voltage, here if it goes below \$+600\:\text{mV}\$, then the idea of \$\beta\$ no longer applies and you need to stop using it. In this case, you need to instead just "set" the collector voltage to something below the base voltage, often this is something like \$+200\:\text{mV}\$, and call it good. But this also means you've given up on the active mode of the BJT and have decided that it is now saturated in-mind.

If you are curious about how to tell, it's not hard. In this case, as the authors point out, the base current is \$\frac{10\:\text{V}-600\:\text{mV}}{1\:\text{k}\Omega}=9.4\:\text{mA}\$. Assuming a typical \$\beta=100\$ is still valid (and we don't know, just yet), then we'd assume also that the collector current would be \$\beta\cdot I_\text{B}=100\cdot 9.4\:\text{mA}=940\:\text{mA}\$. That's the rule if and only if this doesn't also imply an impossible collector voltage.

So let's test it. If the lamp is \$100\:\Omega\$ (see above calculation with respect to this value), then the voltage drop across the lamp would have to be \$100\:\Omega\cdot 940\:\text{mA}=94\:\text{V}\$. But, if you buy that argument, then this means the collector voltage has to be \$+10\:\text{V}-94\:\text{V}=-84\:\text{V}\$. Now, we know that this is simply impossible. There's no source around anywhere that has that kind of negative voltage. It's just not in the circuit. And there is no way a simple BJT can manufacture, out of thin air, this kind of voltage. It just doesn't happen.

Instead, the collector voltage is pushed as close as possible to the emitter voltage and then the process stops. So this means the collector voltage is likely very close to \$0\:\text{V}\$. Maybe \$100\:\text{mV}\$, maybe less, maybe a little more. But it will be very, very close to the emitter voltage, which as you can see is exactly \$0\:\text{V}\$.

So, we can say that the current in the lamp is, at most, about \$\frac{10\:\text{V}}{100\:\Omega}\approx 100\:\text{mA}\$. Since the collector cannot get exactly to the emitter voltage, it's more likely something like \$\frac{10\:\text{V}-100\:\text{mV}}{100\:\Omega}\approx 99\:\text{mA}\$. As you can see, that's not much of a difference to worry over.

So this only means that \$\beta=\frac{99\:\text{mA}}{9.4\:\text{mA}}\approx 10.5\$. Which tells us that the prior idea of a fixed and exact \$\beta\ge 100\$ no longer applies because the BJT is now saturated.

Why is it saturated? Because the lamp load forces the BJT into saturation mode. The BJT itself doesn't determine this. For example, if you shorted out the lamp, then the BJT would not be saturated and the idea of \$\beta\$ would then apply and the collector current would be calculated to be \$940\:\text{mA}\$ and this would be about right, too. But with the added lamp, which then acts to force down the collector voltage with increasing collector current, it is entirely possible now for the BJT to become saturated. In short, it is the entire circuit and not the BJT by itself that determines whether or not a BJT can be, or is, saturated.

But if you used a different lamp, say one with only \$5\:\Omega\$ resistance, then the BJT's collector would be \$10\:\text{V}-\left(\beta=100\right)\cdot 9.4\:\text{mA}\cdot 5\:\Omega=5.3\:\text{V}\$ and the collector voltage would be quite a bit above the emitter voltage and in active mode. So, here, \$\beta=100\$ would still apply and you could calculate the value just fine.