Bipolar transistor switch base current calculation example from PEFI seems wrong?

I'm reading through Practical Electronics for Inventors, 3rd edition, on bipolar transistors, and they provide this example of a transistor switch:

I'm a bit confused by the calculation at the bottom for the base current:

$I_{B} = \frac{V_{E} + 0.6V}{R_{1}} = \frac{0V + 0.6V}{R_{1}}$

Shouldn't the base current be calculated as follows?

$+V_{CC} = V_{R_{1}} + V_{BE} + V_{E}$

$+V_{CC} = I_{B}*R_{1} + V_{BE} + V_{E}$

$+V_{CC} = I_{B}*R_{1} + 0.6V + 0V$

$I_{B} = \frac{+V_{CC} - 0.6V}{R_{1}}$

• You sound correct to me. – Ignacio Vazquez-Abrams Aug 3 '14 at 21:38
• FYI for those also reading PEFI, there is an extensive unofficial errata for the Third Edition can be found at eevblog.com/forum/beginners/… – cdwilson Aug 3 '14 at 22:39
• @cdwilson I'm not too keen on the idea of having [relatively obscure and not very widely used] abbreviations for book titles as tags. You're welcome to contest this in meta. – Nick Alexeev Aug 3 '14 at 23:46
• @NickAlexeev do I just start a new topic in the meta SE and link this question? I've never used meta before, just curious what the normal procedure is for these types of things – cdwilson Aug 4 '14 at 0:39
• @NickAlexeev FYI, looks like this question has already been asked/answered meta.electronics.stackexchange.com/questions/2813/… – cdwilson Aug 5 '14 at 2:30

I'd agree with you. When the switch is on, all the charge flowing through the resistor can only flow through the base-emitter junction of the transistor. The voltage across $R_1$ is $V_{CC}-V_{BE} = V_{CC} - 0.6\mathrm V$, making the current $\frac{V_{CC}-0.6\mathrm{V}}{R_1}$.