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My book (The Art of Electronics by Paul Horowitz and Winfield Hill) is stating something that seems to be the opposite of the example it is using. I am not going to school for my studies so I apologize for having no go-to resources but here. Can anyone shed light on where I'm getting confused?

NPN Transistor circuit

I'll type out the relevant bits: "A transistor is a 3-terminal device available in 2 flavors (npn and pnp), with the properties that meet the following rules for npn transistors: 1. The collector must be more positive than the emitter. 2. .... 3. .... 4. When rules 1-3 are obeyed, IsubC is roughly proportional to IsubB and can be written as IsubC = (hFE)(IsubB) where hFE, the current gain (also called beta), is typically about 100. ... 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 IsubC - 940mA (for a typical beta of 100). This is wrong. Why? Because rule 4 holds only if rule 1 is obeyed; at a collector current of 100mA the lamp has 10 volts across it. To get a higher current you would have to pull the collector below ground."


My question is isn't the collector more positive than the emitter in this case? The collector has 10 volts and 100mA. The emitter is literally connected to ground. The collector is therefore more positive than the emitter. Is it not?

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  • \$\begingroup\$ A lamp voltage is a NOMINAL VALUE for use by a designer, as is the lamp current rating. As a designer, you must use that. \$\endgroup\$ – Whit3rd Nov 15 '17 at 7:09
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What they're saying is that if IC were 940mA then VE would have to be higher than VC. Since the voltage follows rule 1, IC cannot be 940mA.

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  • \$\begingroup\$ Can you quickly explain your rationale behind "if IC were 940mA then VE would have to be higher than VC"? \$\endgroup\$ – ABC DEF Nov 15 '17 at 1:23
  • \$\begingroup\$ 10V / 0.1A = 100ohm ... 940mA * 100ohm = 94V ... 10V - 94V = -84V ... -84V < 0V \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 15 '17 at 1:26
  • \$\begingroup\$ Beautiful. Thank you. Measuring the voltage across an LED I see 0. Maybe I didn't do it right. I would mark this as the correct answer, but I do not have enough reputation. \$\endgroup\$ – ABC DEF Nov 15 '17 at 1:33
  • \$\begingroup\$ You always have enough reputation to accept an answer, you just can't upvote until you have 15. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 15 '17 at 1:33
  • \$\begingroup\$ Indeed. I'm wrong once again. \$\endgroup\$ – ABC DEF Nov 15 '17 at 1:34
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Assuming the switch is closed, the collector is at a few hundred mV in the example, not 10V.

The transistor stops having gain (Ic/Ib = HFE) as the collector voltage approaches zero.

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