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In the 10th edition of Electronic Devices from Thomas L. Floyd Example 6-1 provides a way to find the AC model output signal operating range based on the following circuit. I'd like to ask for a clarification about the solution in the book:

schematic

simulate this circuit – Schematic created using CircuitLab

The provided solution to calculate the AC model \$I_{c(sat)}\$,\$V_{ce(cutoff)}\$ values given the DC model \$V_{CEQ}\$, \$I_{CQ}\$ operating point is:

\$R_c = R_C || R_L\$
\$I_{c(sat)} = V_{CEQ}/R_c + I_{CQ}\$
\$V_{ce(cutoff)} = V_{CEQ} + I_{CQ}R_c\$

The formula for \$I_{c(sat)}\$ and \$V_{ce(cutoff)}\$ make sense to me considering that the maximum delta for \$I_c\$ above and below \$I_{CQ}\$ is \$V_{CEQ}/R_c\$.

However, I don't see how \$R_E\$ can be ignored without a bypass capacitor connected parallel with it. As I understand leaving \$R_E\$ unbypassed would make the \$I_c\$, \$V_{ce}\$ operating range smaller, which is not negligible in general (even if bypassing is the usual arrangement). Is this correct?

Is the effect of the unbypassed \$R_E\$ unimportant in the understanding of the given example?

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    \$\begingroup\$ I don't like the author's approach. But let's follow it, anyway.... Assume the base signal is small. The emitter will "follow" this small base signal. But because the changes are small, the emitter voltage doesn't change much. It just follows the base around. So for the author's purposes of analysis, the emitter voltage is effectively fixed. So this probably explains the lack of \$R_\text{E}\$. Of course, the author is wrong and for any practical temp-stable version here the gain won't be high, so a signal will be large enough to worry over, and larger scale changes bring \$R_\text{E}\$ back. \$\endgroup\$ – jonk Feb 4 at 7:16
  • \$\begingroup\$ Ok, the emitter voltage being practically fixed was something I missed completely. So I think \$R_E\$'s effect can be then ignored here as you pointed out and the formulas are correct. The book has more accurate amplifier models taking into account the transistor internal resistance for instance, so I'll read on and hopefully will get a better understanding. I really like the practical approach of the book btw, ignoring details that are not important for a given model. Here I just thought the model got over-simplified, but I was wrong. \$\endgroup\$ – Imre Deák Feb 4 at 20:55
  • \$\begingroup\$ Okay. In that case, I'm glad I took a moment to write and that it helped. Best wishes! \$\endgroup\$ – jonk Feb 4 at 20:58

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