The Art of Electronics, figure 2.10B

Question

I'm trying to learn electronics and right now I'm reading the book The Art of Electronics.

I'm stuck on this chapter 2 on figure 2.10B.

1. In particular I'm simply confused with the lingo like "pulling" collector to the ground, R3 "hold off" Q3, what does the author really mean? If it is some sort of quantity like voltage or current, why doesn't the author just say something like "voltage/current increases/decreases to X"?

2. What is "sinks the current"?

3. The part where it says "make sure you understand why?" well, no kidding I don't understand... :( I've understood about NPN and PNP thing, like for example to get into saturation mode Vb > Ve and Vb > Vc (NPN) and stuff like that, but I still didn't get how R3 keeps Q3 off when Q2 is off? Well, I used the SPICE simulator, the result is exactly as the book says, but I don't understand why. Many book reviews seems to state the book is great for beginners, but it just skims through so fast on the foundations part.

Here is the text:

Figure 2.10B shows how to do that: NPN switch Q2 accepts the “logic-level” input of 0 V or +3 V, pulling its collector load to ground accordingly. When Q2 is cut off, R3 holds Q3 off; when Q2 is saturated (by a +3 V input), R2 sinks base current from Q3 to bring it into saturation. The “divider” formed by R2 R3 may be confusing: R3's job is to keep Q3 off when Q2 is off; and when Q2 pulls its collector low, most of its collector current comes from Q3's base (because only ∼0.6 mA of the 4.4 mA collector current comes from R3 – make sure you understand why). That is, R3 does not have much effect on Q3's saturation. Another way to say it is that the divider would sit at about +11.6 V (rather than +14.4 V), were it not for Q3's base– emitter diode, which consequently gets most of Q2's collector current. In any case, the value of R3 is not critical and could be made larger; the tradeoff is slower turn-off of Q3, owing to capacitive effects.

Answer 1

1)

• When Q2 is on, it has very low Vce, 'pulling' the collector of Q2 to near ground.
• When Q2 is off, R3 holds Q3 in an "off" state, because it causes the base voltage of Q3 to be almost identical to the emitter of Q3. When Vbe < 0.7V the transistor will be OFF
• OFF = no current

2)

Which current? Q2 sinks current from R3 when it's on. The "Load" sinks current from Q3. It's a resistor or a light bulb or something.

3)

When Vbe is less than the cutoff voltage, the transistor will be off.

It's not the most easy to read passage in the book, but you should be able to get through it by tracing currents and voltages on your own, matching up what he's saying with what you see.

Answer 2

For the terminology...

'Pulling' a node to a voltage or state ('on', 'off' etc.) means drawing a current from it so that it goes that way. The state the node was 'pulled' in can be (or is expected to be) over-rided by another component. A good example is a switch connected to ground in series with a 'pull-up' resistor connected to, say, 5 V. When on, the switch can short the resistor to ground, and when off, the resistor will pull up the switch terminal to 5 V. 'Pulling' is distinct from 'tying' a node to a voltage or state, meaning hard-wired and impossible to override.

'Holding off' in this case means 'keeping it switched off when it could otherwise drift to (partly) on'.

Answer 3

I've been struggling with this one too. For part 2):

• the emitter base ''diode'' of Q3 is in parallel with R3
• typical diode drops are about 0.6V
• so R3 has a voltage of 0.6V

Thus 0.6mA of current flow across R3.

This is why the R2-R3 is described as ''may be confusing'', as the voltage drop across R3 is much smaller than it would be if the emitter base diode weren't in parallel with R3.