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I have been working my way through The Art of Electronics by Horowitz and Hill and I've just started the chapter on Transistors. I am having some difficulty in understanding this transistor circuit:

enter image description here

The text explains that "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." What I don't understand is why the voltage drop across the base resistor is 9.4 volts? Surely the voltage across the resistor should be 10 volts (0.01 amps times 1000 ohms) and this would mean that the entirety of the source voltage is dropped?

The text also mentions that this is an example of transistor saturation.

Many Thanks in advance

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  • \$\begingroup\$ How can it be 10, if 0.6 is already on the transistor? 9.4=10-0.6 \$\endgroup\$
    – Eugene Sh.
    Commented Aug 12, 2015 at 16:43
  • \$\begingroup\$ Where did you get 0.01 amperes from? \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Aug 12, 2015 at 16:49
  • \$\begingroup\$ I'm assuming the current is 10 volts divided by 1000 ohms. I'm probably making a stupid mistake. Thanks for the fast reply. \$\endgroup\$
    – q4758u
    Commented Aug 12, 2015 at 16:52
  • \$\begingroup\$ Hm.. egg or chicken? You are calculating voltage based on an assumed current, while you assume current based on that voltage. See a problem? \$\endgroup\$
    – Eugene Sh.
    Commented Aug 12, 2015 at 16:53
  • \$\begingroup\$ Ok so a simple application of Ohm's Law is clearly not a good way to understand this circuit? Thanks for your help btw. \$\endgroup\$
    – q4758u
    Commented Aug 12, 2015 at 16:58

3 Answers 3

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Here are the voltages as they are described:

enter image description here

Now, if you want to calculate the resistor current, you can see that it is 9.4V / 1000-Ohm = 9.4 mA. Since the base is in series with the resistor, the resistor current = the base current.

Does this help?

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  • \$\begingroup\$ Yes that's definitely clearer. Do we ignore any voltage dropped across the lamp because it's not part of the "closed loop"? Thanks for helping me out. \$\endgroup\$
    – q4758u
    Commented Aug 12, 2015 at 17:20
  • \$\begingroup\$ @q4758u Well, so far we don't know what the voltage across the lamp is. The trick to transistor analysis is that often the current is the given value and is used to determine voltage. Have you read about the transistor's "beta"? It is like a current multiplier. For example, in this circuit, there is 9.4mA flowing into the base. If the beta is 100, then you will have 940mA (9.4mA * 100) flowing from collector to emitter. Now you know the lamp current. If you know the lamp's resistance, you can use the current to determine the lamp's voltage drop. It's like a puzzle :) \$\endgroup\$
    – bitsmack
    Commented Aug 12, 2015 at 17:30
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I think you have a mis-understanding of what Ohm's law means:

Ohm's law states the voltage across a resistor is proportional to the current through it.

schematic

simulate this circuit – Schematic created using CircuitLab

\begin{gather*} V_A = 10V\\ V_B = 0.6V\\ I_{R1} = \frac{V_A - V_B}{R_1} = \frac{10V - 0.6V}{1000 \Omega} = 9.4 mA \end{gather*}

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What you have missed is that there is a voltage drop between the base and the emitter of the transitor. As you can see on the figure below, this voltage is between 0.6 - 0.7V with an approximately large base current. In books and theoretical excersises it is typically assumed as 0.6V.

enter image description here

As the others have already pointed out the voltage drop on the resistor is 10V - VBE where VBE is the base-emitter voltage of the transistor and it is equal 0.6V.

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