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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?

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

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  • \$\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 Aug 14 '14 at 19:10
  • \$\begingroup\$ Sure, but I'd be hard-pressed to call that normal usage in the circuit that the asker provides. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 14 '14 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 Aug 14 '14 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 Aug 14 '14 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\$ – Ignacio Vazquez-Abrams Aug 14 '14 at 19:36
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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.

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  • \$\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 Jul 12 '16 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 Jul 12 '16 at 15:56
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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.....

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