Art of Electronics 2.5 - Transistor saturation

Question

With reference to the image below the text says

When the switch is closed, the base rises to 0.6 V (base– emitter diode is in forward conduction). The drop across the base resistor is 9.4 V, so the base current is 9.4 mA. Blind application of Rule 4 gives IC = 940 mA (for a typi- cal beta of 100). That is wrong. Why? Because rule 4 holds only if Rule 1 is obeyed: at a collector current of 100 mA the lamp has 10 V across it. To get a higher current you would have to pull the collector below ground. A transistor can’t do this, and the result is what’s called saturation – the collector goes as close to ground as it can (typical sat- uration voltages are about 0.05–0.2 V, see Chapter 2x.) and stays there. In this case, the lamp goes on, with its rated 10 V across it.

Rule 1 is: The collector must be more positive than the emitter.

I’m missing a step - how would a negative collector voltage obey this rule and result in a higher current?

Answer 1

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.

Answer 2

I couldn't understand AoE because when it referred to Rule 1, I thought "OK, collector is 10V and Emitter is 0v, so what is the problem?"

What helped me to understand was that this was impossible as lamp was 100R and would imply 94V voltage drop. Whereas the maximum possible drop is around 9.9V and 9.9v / 100R = 99mA of current.

Answer 3

The intention

With all due respect to jonk's amazing answer I will try to generalize it because, although the saturation phenomenon is observed in transistors, it is much more general; it is a concept.

Basic idea

... how would a negative collector voltage obey this rule and result in a higher current?

Saturation occurs when, in an effort to increase current, the regulating element of an amplifier stage runs out its resistance. The additional negative voltage source added its voltage to the positive voltage of the power supply thus allowing the transistor to later go into saturation. Let's follow step by step how this is done.

Operation

I will illustrate the basic idea with a conceptual circuit of an "amplifier" of two elements in series - a load with a constant resistance of 100 Ω (representing the OP's lamp) and a variable resistor Rce (representing the collector-emitter part of the transistor). To simplify the circuit, I will use an imperfect ammeter LOAD with 100 Ω internal resistance as a load and an imperfect voltmeter Rce with a variable resistance as a transistor.

Single power supply

Cut-off (Rload = 100 Ω, Rce = 1 GΩ, Vcc = 10 V): Initially, the transistor is cut-off so Rce tends to infinity (open circuit) and no current flows. There is no voltage drop across the load; so the collector voltage is equal to the supply voltage.

^{simulate this circuit – Schematic created using CircuitLab}

Active mode (Rload = 100 Ω, Rce = 100 Ω, Vcc = 10 V): The transistor begins conducting and decreases its "resistance" (Rce = Rload = 100 Ω in the schematic below). The voltage divider has a gain of 0.5; so the collector voltage is half of the supply voltage. A 50mA current flows.

^{simulate this circuit}

Saturation (Rload = 100 Ω, Rce = 0 Ω, Vcc = 10 V): The transistor continues decreasing its "resistance" so the current increases and the collector voltage continuously decreases. Finally, it runs out its resistance (Rce = 0), and the voltage drop across Rce becomes zero.

^{simulate this circuit}

There is nothing more the transistor can do to continue increasing the current (Rc is zero and we can not make it less than zero). Simply put, the transistor is not a transistor but just a "piece of wire". Only the load resistance remains in the circuit to determine the load.

Adding a negative power supply

Rload = 100 Ω, Rce = 0 Ω, Vcc = 10 V, Vee = -5 V: Obviously, to get the collector voltage "below zero" we need to "pull" the emitter down using a negative voltage source. Well, let's start with -5V. Now the transistor can increase the current to 150 mA before entering saturation (Rce = 0 Ω). Why?

^{simulate this circuit}

The trick here is that the -5 V negative voltage Vee is actually added to the 10 V positive voltage Vcc, and the voltage divider is supplied by a 15 V voltage source (note that the two supply voltages have opposite polarities in respect to ground but the same polarity when traveling along the loop).

Rload = 100 Ω, Rce = 0 Ω, Vcc = 10 V, Vee = -10 V: Then let's increase the magnitude of the negative voltage Vee up to -10 V. Now -10 V negative voltage Vee is added to the 10 V positive voltage Vcc, and the voltage divider is supplied by a 20 V voltage source.

^{simulate this circuit}

As a result, the transistor can increase the current to 200 mA before entering saturation.

Conclusions

• The regulating element of an amplifier stage saturates when runs out its resistance.

• The negative voltage source adds its voltage to that of the positive one thus allowing the transistor to later go into saturation.

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See also my related question.

Answer 4

If ideal ic = 100*iB = 940mA, then the lamp (collector load) would need... maybe 100V assuming it was a resistive load, and could dissipate that much power (100V * 1A = 1000W, rather a lot)... So if the +10V supply were high enough, we could get the NPN BJT out of saturation.

Transistor current gain (beta) is not a constant, it's an estimate of the best current gain possible for that transistor when not in saturation. With iB=9.4 mA and beta=100 mA/mA, then iC would be 940mA, but that's not possible because of what's attached to the collector in this circuit. The voltage across the lamp is about 10V, so Vc must be close to 0V because of the voltage drop across the lamp. Since Vc<Ve, this NPN BJT is in saturation, and we can't get very much current gain.

If the emitter voltage Ve wasn't ground, but instead Ve=-100V (and the base bias circuit was changed so that Vb=-98.4V, so that still Vbe=0.6V and iB=9.4mA), then the transistor would not be in saturation anymore: iC=940mA causes the lamp to drop 100V(!) and so Vc=10V-100V=-90V. (The transistor doesn't know or care what point we label as 0V ground.) Vc>Ve so the NPN transistor isn't in saturation, and we can use beta to estimate current gain.