# 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. Aug 3, 2014 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/… Aug 3, 2014 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. Aug 3, 2014 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 Aug 4, 2014 at 0:39
• @NickAlexeev FYI, looks like this question has already been asked/answered meta.electronics.stackexchange.com/questions/2813/… Aug 5, 2014 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 $\frac{V_{CC}-0.6\mathrm{V}}{R_1}$.