I can't come up with any suitable title. I am reading about transistor from a book Here's the 4 properties of an NPN transistor (for PNP, it is reversed)

  1. The collector must be more positive than the emitter.
  2. The base-emitter and base-collector circuits behave like diodes (Fig. 2.2). Normally the base-emitter diode is con- ducting and the base-collector diode is re- verse-biased, i.e., the applied voltage is in the opposite direction to easy current flow.
  3. Any given transistor has maximum values of Ic, IB, and VCE that cannot be exceeded without costing the exceeder the price of a new transistor (for typical values, see Table 2.1). There are also other limits, such as power dissipation (revCE), temperature, VBE, etc., that you must keep in mind.
  4. When rules 1-3 are obeyed, Ic is rough- ly proportional to IB and can be written as \$I_C = h_FEI_B = {\beta}I_B\$. Where \$h_{FE}\$, the current gain (also called beta), is typically about 100. Both \$I_C\$ and \$I_E\$ flow to the emitter. Note: The collector current is not due to forward conduction of the base-collector diode; that diode is reverse-biased. Just think of it as "transistor action."

and then here is the picture

enter image description here

it say when the switch is close then

When the switch is closed, the base rises to 0.6 volt (base-emitter diode is in forward conduction). The drop across the base resistor is 9.4 volts, so the base current is 9.4mA. Blind application of rule 4 gives Ic = 940mA (for a typical beta of 100). That is wrong. Why? Because rule 4 holds only if rule 1 is obeyed; at a collector current of lOOmA the lamp has 10 volts 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 saturation voltages are about 0.05- 0.2V, see Appendix G) and stays there. In this case, the lamp goes on, with its rated 10 volts across it.

What I have understood from this explanation is that here the lamp act as resistor that make the voltage at collector terminal 0 while the voltage at base is 0.6V (because \$V_B=V_E+0.6\$ and \$V_E\$ here equal 0) so it doesn't obey rule 1 and then rule 4 doesn't hold. But the after explanation is unclear. Why the collector must have higher current and what does "pull the collector below ground" mean and then how can collector can have 0.5 or 0.2 voltage while the lamp used up all voltage. So what is the use of the transistor in this case?

  • \$\begingroup\$ It allowed you to turn on a 100 mA lamp with a 10 mA switch. \$\endgroup\$
    – EM Fields
    Commented Aug 14, 2014 at 19:26

4 Answers 4


Rule 1 isn't a "good idea", it isn't a "guideline", it is a fundamental tenet of transistor physics. If for any reason (during normal usage) it is unable to hold then the circuit will not operate.

As for the lamp, it is a purely resistive element. It should have 10V across it, but thanks to the transistor it won't. So the transistor gets 0.2V and the lamp gets 9.8V and reality is saved.

  • \$\begingroup\$ Actually you can swap the collector and emitter, and it will still act like a (really bad) transistor with terrible beta, but it still operates. (Reverse active region.) \$\endgroup\$
    – John D
    Commented Aug 14, 2014 at 19:10
  • \$\begingroup\$ Sure, but I'd be hard-pressed to call that normal usage in the circuit that the asker provides. \$\endgroup\$ Commented Aug 14, 2014 at 19:15
  • \$\begingroup\$ Right, definitely not normal usage. And you could make the case that once you do that the emitter becomes the collector and vice versa, so the rule still holds. In this case the transistor is in saturation, the collector is at 0.2V and the emitter is at 0V. So the rule holds, but the OP still seems confused. \$\endgroup\$
    – John D
    Commented Aug 14, 2014 at 19:28
  • \$\begingroup\$ yes, I am confused, if the the collector and emitter swap so the current will flow from ground which now connect to collector to emitter and the emitter connecto to the lamb but there is 10 voltage there and current will flow from high to low voltage so how can the current flow from 0 voltage at the ground to emitter while there are current flow from 10V to emitter as well? \$\endgroup\$
    – aukxn
    Commented Aug 14, 2014 at 19:34
  • \$\begingroup\$ You can only perform a C-E swap if you bias the emitter higher than the collector. Because rule 1. \$\endgroup\$ Commented Aug 14, 2014 at 19:36

I know that the explanation you are confused by is from the chapter "Bipolar transistors" from the book "The Art of Electronics" by Horowitz and Hill. I too was very confused when I first read this explanation; but since then, I have learned more about circuits and electronics from other resources, and I think I can clear your doubt now.

Step 1: Let's redraw the circuit, but for the sake of simplicity, we will not write any values. Trust me, you will understand the circuit better if you forget about the values for a bit:


simulate this circuit – Schematic created using CircuitLab

Step 2: Now, let us mark the voltages which we are interested in, in this circuit. Really, we are only interested in three voltages: the supply voltage with respect to ground (10 volts), the voltage of the collector with respect to ground (Vc), and the voltage across the lamp (V[lamp]). Also, because the emitter is connected to the ground, thus by convention we take its voltage to be 0 volts.


simulate this circuit

An important observation is this:

V[lamp] = 10 - Vc

That is, the voltage across the lamp is always equal to the supply voltage minus the collector voltage.

Step 3: I know that the Art of Electronics talks a lot about collector currents (Ic), and base currents (Ib), and current gains (β). However, I feel that that is unnecessary when talking about something as simple as a transistor switch! Don't get me wrong, I love the Art of Electronics; it has a lot of intuitive and insightful explanations; however, this particular explanation is, in my opinion, unnecessarily complicated. Here is a simpler model:


simulate this circuit

  • When the switch is open, then we can prove that the collector voltage (Vc) will be pulled up to the supply voltage of 10V. "Pulled up" here is just a fancy way of saying that the collector voltage is "increased" to the high voltage of 10V.
  • When the switch is closed, then we can prove that the collector voltage (Vc) will be pulled down to the ground voltage of 0. "Pulled down" here is just a fancy way of saying that the collector voltage is "decreased" to the low voltage of 0.

Actually, you will hear the terms "pull-up" and "pull-down" a lot in electronics, especially when you're dealing with digital electronics, when a device is either on or off; high or low; up or down; 1 or 0.

So, what is "pulling the collector below ground"?

This is one of the doubts which I had too, when I was reading the Art of Electronics. Let's understand it little by little:

  • In any circuit, the "ground" always has a potential/voltage of 0 volts; this is just a convention which makes circuit analysis easier.
  • Thus, when we say that a voltage is "below ground", this just means that it has a value below 0 volts; i.e., it has a negative voltage. (Don't confuse a voltage being below ground with a voltage being underground!)

Thus, in the Art of Electronics, when the authors said "To get a higher current you would have to pull the collector below ground", this translates into is "To get a higher current you would have to take the collector voltage (Vc) below the ground voltage (0)".


simulate this circuit

Actually, it is impossible to even pull the collector voltage (Vc) to zero/ground, let alone below zero/ground! The authors very correctly mention that the most that you can do is pull the collector voltage to 0.2V. This voltage is known as the saturation voltage of the bipolar transistor, and for a typical transistor it as a value of 0.05V to 0.2V (which is pretty close to 0, wouldn't you say?)

But it doesn't matter. Once you have pulled the collector voltage (Vc) to 0.2V, then the voltage across the lamp will just be 10 - 0.2 = 9.8V. And look, I know that the lamp is rated for 10V; but don't worry, if a lamp is rated for 10V, it will work just fine with 9.8V too (lamps aren't very fastidious).


The point of the circuit appears to be to demonstrate that you can switch something (the lamp) on and off by using a transistor. In this case, they just use a mechanical switch to trigger the transistor, but in the case of most electronics, one switch may be triggered by other transistors (logic gates). In that way, you can have a small electrical signal switch a transistor on and the transistor allows usually much greater amounts of current or voltage than the controlling signal.

In this case, you have 0.01 Amps switching on 0.1 Amps so you still have some gain in your circuit.

It obeys Rule 1 because the collector is somewhere between 0.05-0.2 volts. This means that the amount that drops over the lamp is 10V - 0.2 Volts or only 9.8 volts drops over the lamp.

If the lamp draws about 0.1 Amp then you can assume that the resistance (once it's hot) is 9.8V/0.1 = 98 ohms. All the maths work out.

  • \$\begingroup\$ So you're saying that voltage drop due to base is 0.6 (having 9.4V passing) and for the lamp it is 0.2V passing - so it should be something like: Vb = Ve + 0.6, where Ve = 0.2? giving Vb = 0.8, though its not true because Vb should be 9.4? Im confused here. And what is meant then by rule 1: The collector must be more positive than the emitter? \$\endgroup\$
    – maximus
    Commented Jul 12, 2016 at 6:34
  • \$\begingroup\$ @maximus Ve=0V because it's at ground. Which means Vbe=0.6V or just Vb = 0.6V. Rule one just ensures the BJT operates with high gain and in normal operation. If you were to vertically flip the NPN, you'd reverse the collector and emitter and now the emitter is more positive than the collector and your beta/gain would drop to a very low amount. Let me know if I missed the mark in your question. \$\endgroup\$
    – horta
    Commented Jul 12, 2016 at 15:56

Not to be confused with potential and potential difference......

Simply telling, "pulling below the ground" means taking the potential of collector below 0V. When the collector is at 0V there is 10V potential difference across the bulb. And this 10V allows the current of 100mA(indicated by the stamp in bulb) to flow through the collector (i.e Ic=100mA). But the implication of Ic=BIb gives you the value of 940mA which is higher than the current that we found when the potential difference is 10V. So, by the application of Ohm's Law, for higher current to flow through the component, either its resistance has to change, or potential difference must be increases. Since the resistance here is constant, the only way is to increase the potential difference. And you can see that the +ve terminal is fixed at 10V. So the transistor has to take its collector potential below 0V. This process of taking the potential below 0V is referred as 'taking below the ground'*

But the transistor is not capable of doing this,resulting in the saturation of transistor.

Hope this helps.....


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.