Electronics Questions

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Chapter 1

No. $R_{O}$ is in series with $R_{L}$ , so if $R_{L}$ gets smaller, there will be a smaller voltage drop over $R_{L}$ and a larger voltage drop over $R_{O}$

21, F1.18: Why does it have to have an source impedance? How big is this impedance? Loading effects? Could we get rid of this if we had superconducting wires?

The source impedance is simply to reflect the real world internal losses of the system. If you want to use these models, you will need to make a Thevenin or Norton equivalent.
When you make the Thevnin or Norton equivalent of a circuit, you will find out how big these impedances are.
Loading effects are the impedances that reduce the voltage or current to the intended source. For F1.18, these are $R_{s}$ and $R_{o}$ , for F1.28, it is $R_{o}$
Yes! It depends on what part of the project you get to design. If you get to work on the pre-amplifier section, you can have small resistances (for a Thevenin equivalent) or high resistances (for a Norton equivalent). If you are designing the amplifier section, you can stick in a buffer amplifier in so that there is no voltage drop (due to no current) flowing into the amplifier. This ensures the voltage drop is completely across the load/input resistor.

31, F1.25: Why are the output resistances placed where they are? It seems there would be a voltage drop over the voltage amplifier and extra current not being accounted for in the current amplifier?

The resistor is there because it is a Norton equivalent. Loading effects will play a part, but if you are designing this section of the amplifier, your design can take this into account to provide minimum losses.

If $V_{BE}$ is 0, and $V_{CE}$ is very large, will you get any current flow?
1/22/2010,3: Explain the protection diode so that the transistor doesn't arc over.
Capacitor_L acts like an open circuit for DC and a short circuit for AC
AC coupling goes through a blocking capacitor, while DC coupling doesn't

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