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Revision as of 16:05, 28 January 2010 by Fonggr
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- 20,1,3: Shouldn't the output voltage get larger as gets smaller?
- No. is in series with , so if gets smaller, there will be a smaller voltage drop over and a larger voltage drop over
- 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 and , for F1.28, it is
- 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 is 0, and 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
- How is the BJT amplifier an inverting amplifier?