In the Art of Electronics (Second edition), Section 2.02 (Transistor Switch), the book gives a circuit diagram:

Transistor switch example
Figure 2.3. Transistor switch example.

A bulb rated 10V, 0.1 A is connected to the collector of a NPN transistor. The other end of the bulb is connected to constant voltage +10V. Then when current flows into the base, it says that we cannot blindly multiply \$I_B\$ by \$\beta\$, as a current of 0.1A would drive the voltage through the bulb to 10V, making collector voltage 0V. It also says that the collector voltage goes as low as possible, but still stays something around 0.05 - 0.2V above ground.

But then it comments that the bulb would glow in this case. But how? The voltage drop across the bulb is 9.95V at max, which is less than the rated 10V. Why does the bulb glow up?

Also, even if I assume that such low voltage differences do not affect the temperature of the filament, how does the book say that 0.1A would cause a 10V drop? I take the rating of the bulb to mean that when the bulb is glowing, a 10V drop would cause 0.1A (as resistance is almost fixed by the fixed temperature at that time). But how can this be stated even before the bulb is glowing?

  • \$\begingroup\$ The bulb rating is a collection of nominal values. I.e. the ones for typical usage, while one can deviate from in the limits of absolute maximums/minimums. \$\endgroup\$
    – Eugene Sh.
    Commented Sep 29, 2017 at 17:46
  • \$\begingroup\$ It's great that you are reading The Art of Electronics. (The 2nd edition is quite good, by the way.) The light bulb will light up with less than 10 V because that's just a recommended rating -- not some law in physics. It will light, after a fashion, with 6 V, too. (Not well, necessarily.) It's just a tungsten wire. So it just heats up, more with higher voltages, less with lower voltages. Also the wire's resistance will also change with temperature. It's complex, in reality. But the basic idea is that it will still light. \$\endgroup\$
    – jonk
    Commented Sep 29, 2017 at 17:46
  • \$\begingroup\$ @jonk : I kind of guessed that, as mentioned in the post. But my second question still remains, as to how he can use those values to calculate the maximum possible current. \$\endgroup\$ Commented Sep 29, 2017 at 17:49
  • \$\begingroup\$ @Koyomi-chan How is your math? \$\endgroup\$
    – jonk
    Commented Sep 29, 2017 at 17:51
  • \$\begingroup\$ @jonk: Pretty okay for basic electronics I guess. I'm facing some troubles with the book, because it skips a lot of math. :) But please do not assume anything high level including multivariable calc. \$\endgroup\$ Commented Sep 29, 2017 at 17:53

1 Answer 1


In any transistor circuit the collector current will only be \$I_B\$ by \$\beta\$ IF whatever it is connected to can supply that much current.

In this case the lamp is rated at .1A at 10V (1W), so, once in a steady state, the collector current could never exceed that even if Vce could be set to zero volts. If we assume a \$\beta\$ of say 50 and the base current is \$9.3mA\$ the theoretical max collector current would be \$0.465A\$, which is well above the \$0.1A\$ the lamp can supply at 10V Vcc.

Of course, initially, when cold, the lamp resistance will be a lot smaller and more than 0.1A will briefly flow up to a maximum of \$I_B\beta\$. If you know the \$\beta\$ value, you can therefore use the base resistor to limit the inrush current in the lamp.

Lamps will emit heat and light at pretty much all voltages. At 9.95V the bulb will still be glowing almost at full brightness. The current will not, in fact, be quite as high as 0.1A. However it is still a lot less than the \$I_B\beta\$ value which was the authors original point.

  • \$\begingroup\$ But since the bulb is not necessarily glowing at that time, we can not use the values. Because these values necessarily imply a specific resistance, (ie, 0.01 ohms), which is attained only when the bulb is glowing. Prior to that, the not-glowing bulb has a lower resistance, which can cause a greater current to flow at the 10 V voltage drop. \$\endgroup\$ Commented Sep 29, 2017 at 17:52
  • \$\begingroup\$ @Koyomi-chan I'm still writing.. \$\endgroup\$
    – Trevor_G
    Commented Sep 29, 2017 at 17:55
  • \$\begingroup\$ Ok, so you intend to say that the brief high current would soon heat up the bulb. And after a while, those values would become valid as the bulb glows up? \$\endgroup\$ Commented Sep 29, 2017 at 17:57
  • \$\begingroup\$ @Koyomi-chan that is correct \$\endgroup\$
    – Trevor_G
    Commented Sep 29, 2017 at 17:57

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