Key Facts from Reading Chapter 1

The following facts are not profound and are possibly very obvious. Nonetheless, they might help cement certain concepts. Please add things you think would be helpful.

Transistors

MOSFET diagram <ref> University of Cambridge </ref>

• Conduction in n-type material is from free electrons.
• Conduction in p-type material is from holes (positive particles).
• The function of metal-oxide-semiconductor field-effect transistors (MOSFETs) depends on the voltage applied to the gate.
  • Certain ranges of voltage allow no current to flow between the drain and the source. In this way, the MOSFET acts like an open switch.
  • Another particular range of voltage allows current to easily flow from the source to the drain.
  • When the voltage is in between the ranges of open and closed switch, the MOSFET can smoothly control the amount of current flowing.
• Bipolar Junction Transistors (BJTs) can act as either switches or current controls as well.

Amplifiers

• An inverting amplifier has a negative voltage gain, $\ {A_{v}}$.
• A noninverting amplifier has a positive voltage gain, $\ {A_{v}}$. (If you get this, you deserve a cookie)
• The power gain,$\ G$, is the ration of the output power to the input power
  • $G={\frac {P_{o}}{P_{i}}}={\frac {V_{o}{I_{o}}}{V_{i}{I_{i}}}}=A_{v}{A_{i}}=(A_{v})^{2}{\frac {R_{i}}{R_{L}}}$
    • $\ R_{i}$ is the amplifier's input resistance and $\ R_{o}$ is the amplifier's output resistance.

Decibel Conversion

• Power gain, $\ G$, can be converted to decibels:

$G=(A_{v})^{2}{\frac {R_{i}}{R_{L}}}$

$\ G_{db}=10log{A_{v}}^{2}+10logR_{i}-10logR_{L}$

$\ G_{db}=10log|A_{v}|+10logR_{i}-10logR_{L}$

• By the equation above, we can say voltage gain in decibels can be found with this equation: $\ A_{vdB}=20log|A_{v}|$, where $\ A_{v}$ is the voltage gain and $\ A_{vdB}$ is the voltage gain is decibels.
• Similarly, we can say current gain in decibels can be found with this equation: $\ A_{idB}=20log|A_{i}|$, where $\ A_{i}$ is the current gain and $\ A_{idB}$ is the current gain in decibels.

Differential Amplifiers

• Differential Amplifiers have two inputs. The terminal marked with a "+" is the noninverting input, and the terminal marked with a "-" is the inverting input.
• The differential input signal is given by $\ v_{id}=v_{i1}-v_{i2}$
• The output of an ideal differential amplifier is given by $\ v_{o}=A_{d}v_{id}$.
• The common-mode input signal is given by the following: $v_{icm}={\frac {1}{2}}(v_{i1}+v_{i2})$.
• A real differential amplifier's output is given by $\ v_{o}=A_{d}v_{id}+A_{cm}v_{icm}$, where $\ A_{cm}$ is the common-mode signal gain.

References

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Contributors

• Christopher Garrison Lau I

Readers

• Shepherd,Victor

Reviewers

• Henry,Ben

• Vier, Michael

• The sentence that says "voltage gain and is the voltage gain is decibels." I think you mean "in decibels." Other than that nice work!