(materials covered in UWM ECE340)

Text: Microelectronic Circuits 6th Edition Ade

Chapter 1 Signals and Amplifiers

Transducers: devices used to convert signal into an electrical signal(a voltage or a current).

Root-mean-square(rms) value: equal to the peak value divided by rad 2. common to express amplitude of a sine-wave signal.

Analog signal: analogous to the physical signal that it represents, continuous variation over its range of activity

signal amplification: signal-processing, make the "weak" signals provided by transducers have more energy

distortion: any change in waveform in signal amplification process

voltage amplifiers: make the signal magnitude larger, voltage gain

power amplifiers: current gain

voltage gain (Av) : Vo / VI //output voltage / input voltage

power gain (Ap) : load power(PL) / input power(PI) = Vo*io / Vi*Ii

Transformer can have voltage gain, but not power gain.

current gain (Ai) : io / iI //output current / input current

Ap = Av*Ai

Voltage gain in decibels (dB) = 20 log |Av|

Current gain in decibels (dB) = 20 log |Ai|

Power gain in decibels (dB) = 10 log Ap

The Amplifier Power Supply

图1

The power delivered to load is greater than power drawn from signal source -> amplifiers need two dc power supplies for their operation, some require only one power supply.(图1)

Pdc : dc power delivered to amplifier

Vcc: dc source's voltage (connect to V+ of amplifier)

VEE: the other dc source's voltage (connect to V- of amplifier)

Icc: current drawn from positive supply

IEE: current drawn from negative supply

Pdc = Vcc*Icc +VEE*IEE

Pdissipated: power dissipated in the amplifier circuit

PL: power delivered to the load

PI: power drawn from the signal source(usually small)

Pdc + PI = PL + Pdissipated

amplifier power efficiency: n = (PL/Pdc)*100

Vpeaktopeak: max signal magnitude value and min value

图2

Ri: amplifier draws an input current from the signal source

Ro: change in output voltage, should be much smaller than RL

RL: load resistance

Vs: signal voltage source

Rs: signal resistance

图3

Av: voltage gain of the amplifier

Avo: voltage gain of the unloaded amplifier, or the open-circuit voltage gain. When RL = infinity, Av = Avo.

Chapter 3 Semiconductors

3.1 Intrinsic Semiconductors

Two kinds of semiconductors: 1.single-element semiconductors 2.compound semiconductors

Intrinsic silicon: silicon has 4 valance electrons, they form bonds with 4 neighboring atoms and form a covalent bond.

At room temperature, thermal energy break some of the bonds, thus generating free electrons, electron leave its parent atom thus generate a hole. The increase in numbers of electron and holes results in increase in conductivity of silicon.

recombination: the process that electrons fill some of the holes, recombination rate depends on the generation rate. In thermal equilibrium, these two rates are equal. thus, concentration of free electrons(n) is equal to concentration of holes p. n = p = ni. ni: the number of free electrons and holes in a unit volume of intrinsic silicon at a given temperature.

ni = BT^(3/2)e^(-Eg/2kT)

B: material dependent parameter, 7.3*10^15 for silicon

Eg: bandgap energy, the minimum energy required to break a covalent bond to generate electron-hole pair, 1.12 eV for silicon

k: Boltzmann's constant

p*n = ni^2

3.2 Doped Semiconductors

We doped the silicon for two reasons: 1. the concentrations of n and p is too small to carry appreciable current 2. the concentration and thus conductivity are strong function of temperature, behavior will change sensitive to temperature

Doping: introducing impurity atoms into the silicon crystal to increase the concentration of either electrons or holes but with little or no change in the crystal properties of silicon.

n type: increase concentration of electron, n, doped with element with a valance of 5(ex phosphorous). $nn\cong ND$ , where nn is the concentration of free electron in the n-type silicon, and ND is concentration of donor atoms. pn*nn = ni^2, $pn\cong ni^2/ND$ . Electrons are the majority charge carriers.

p type: increase p, doped with element with a valance of 3(ex boron). $pp\cong NA$ ,where pp is the concentration of holes in the p-type silicon, and NA is the concentration of acceptor atoms. pn*nn = ni^2, $np\cong ni^2/NA$ . Holes are the majority charge carriers.

Notice: A piece of n-type or p-type silicon is electrically neutral. The charge of the majority free carriers are neutralized by the bound charges associated with the impurity atoms.

3.3 Current Flow in Semiconductors

Drift Current

$v_{p-drift} =\mu_{p}E$

$v_{p-drift}$ : holes' velocity, holes move in the direction of E

$\mu_{p}$ : hole mobility

E: electrical field established in a semiconductor crystal

$v_{n-drift} = - \mu_{n}E$

$v_{n-drift}$ : free electrons drift velocity, electrons move in the opposite direction to E

$\mu_{n}$ : electron mobility

Diffusion Current : The net flow of charge

hole diffusion -> concentration profile p(x) -> magnitude of current is proportional to the slope of the concentration profile

$J_{p}=-qD_{p}\frac{dp(x)}{dx} , J_{n}=qD_{n}\frac{dn(x)}{dx}$

$D_{p}=12cm^2/s,D_{n}=35cm^2/s$ : typical diffusion constant of holes and electrons, respectfully.

Einstein relationship

3.4 The pn Junction with Open-Circuit Terminals

Concentration is high in the p region and low in n region, holes diffuse across the junction from the p side to n side. and same for electrons diffuse from n side to p side. These two current components(majority carrier diffusion) add together to form the diffusion current $I_{D}$ , whose direction is from p side to n side.

holes diffuse to n side and recombine with free electron close to the junction. Thus some of the bound positive charge will no longer be neutralized by free electrons, this charge is called uncovered.(depleted of free electrons) same for the p side.(depleted of holes) . All these create carrier-depletion region, or space-charge region.

The charges on both sides of the depletion region cause Electric filed, hence a potential difference, with n side positive relative to p side. The resulting electric field opposes the diffusion of holes into the n region and electrons into the p region. The voltage drop across the depletion region act as a barrier. The larger the barrier voltage, the smaller the lower the magnitude of diffusion current.

$I_{S}$ is due to minority carrier drift exists across the junction. $I_{S}$ = electrons moved by drift(thermally generated electrons) from p to n and holes moved by drift(thermally generated holes) from n to p. $I_{S}$ is strongly dependent on temperature, independent of the depletion-layer voltage Vo.

This equilibrium condition4 is maintained by the barrier voltage V0. Thus, if for some reason ID exceeds IS , then more bound charge will be uncovered on both sides of the junction, the deple- tion layer will widen, and the voltage across it (V0 ) will increase. This in turn causes ID to decrease until equilibrium is achieved with ID = IS. On the other hand, if IS exceeds ID, then the amount of uncovered charge will decrease, the depletion layer will narrow, and the voltage across it (V0) will decrease. This causes ID to increase until equilibrium is achieved with ID = IS.

The junction Built-In voltage: 0.7V. $V_{0}=V_{T}ln(\frac{N_{A}N_{D}}{n_{i}^2} )$ , $N_{A},N_{D}$ are the doping concentrations of the p side and n side of the junction.

3.5 The pn Junction with an Applied Voltage

Apply a dc voltage between the two terminals, if the voltage applied so that the p side is made more positive than the n side, it's called forward-bias voltage, conversely, reverse-bias voltage.

Circuit Analysis

Circuit Analysis

Chapter 1 Signals and Amplifiers

Chapter 3 Semiconductors

Chapter 4 Diodes

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