EE 334Midterm Review

Diode: Why we need to understand diode?

• The base emitter junction of the BJT behaves as

a forward bias diode in amplifying applications.

• The behavior of the diode when reverse bias is the key to the fabrication of the integrated circuits.

• The diode is used in many important nonamplifer applications.

Departure from ideal behavior

The four major reason why the actual diode do not correspond exactly to the ideal.1. Ohmic resistance and contact resistance in series with the diode cause the VI curve to become linear at high forward current.2. Avalanche or Zener breakdown take place at high reverse voltage, causing an abrupt increase in reverse current.3. Surface contaminants cause an ohmic layer to form across the junction, which isIncreasing the reverse current as reverse voltage is increased.4. Recombination of current carrier in the depletion region take place due to traps.

The purpose of modeling

• Nonlinear problems are much more difficult than linear ones. These problems could be impossible to solve manually and could require huge amount of time if solved on a computer.

• One possible solution of the above mentioned problem is to approximate the nonlinear relationship with a model that has a linear relationship.

• The trust of nonlinear modeling is direct towards this end.

• The modeling not only simplifies the solution, it also allows the designer to understand how the circuit behaves. Modeling often increases the conceptual understanding of the circuit operation.

Schottky Barrier Diode

One semiconductor region of the pn junction diode can be replaced by a non-ohmic rectifying metal contact.A Schottky contact is easily formed on n-type silicon. The metal region becomes the anode. An n+ region is added to ensure that the cathode contact is ohmic.

Schottky diodes turn on at a lower voltage than pn junction diodes and have significantly reduced internal charge storage under forward bias.

Reverse Breakdown

Increased reverse bias eventually results in the diode entering the breakdown region, resulting in a sharp increase in the diode current. The voltage at which this occurs is the breakdown voltage, VZ.

2 V < VZ < 2000 V

Half Wave Rectification

Figure 2.7

A full-wave bridge rectifier: (a) circuit showing the current direction for a positive input cycle, (b) current direction for a negative input cycle, and (c) input and output voltage waveforms

Half-wave rectifier with filter

% regulation is used to measure how well the regulator isPerforming its function.

Voltage regulation is the measure of circuit’s ability to maintained a constant output even when input voltage or load current varies.

Demonstration of Zener diode as a voltage regulator

The large value of VCE decreases the effective base width W. Since IS is inversely propositional to W, which cause increase in IC.

Bipolar NOR logic gateExample 3.11 Determine current and voltage in the circuit 3.43(b)

Rc=1KRB=20KVBE(on)=0.7VVCE(sat)=0.2Vβ=50

Lecture #3

The process by which the quiescent output voltage is caused to fall somewhere the cutoff and saturated values is referred to as biasing.

Example 3.13

Q-point has shifted Substantially.Q-point is not stabilizedAgainst the variation.

Chapter 4Small-Signal Modeling and

Linear Amplification

DC and AC Analysis• DC analysis:

– Find dc equivalent circuit by replacing all capacitors by open circuits and inductors by short circuits.

– Find Q-point from dc equivalent circuit by using appropriate large-signal transistor model.

• AC analysis:– Find ac equivalent circuit by replacing all capacitors by short

circuits, inductors by open circuits, dc voltage sources by ground connections and dc current sources by open circuits.

– Replace transistor by small-signal model– Use small-signal ac equivalent to analyze ac characteristics of

amplifier.– Combine end results of dc and ac analysis to yield total

voltages and currents in the network.

DC Equivalent for BJT Amplifier

• All capacitors in original amplifier circuits are replaced by open circuits, disconnecting vI, RI, and R3 from circuit.

AC Equivalent for BJT Amplifier

kΩ100kΩ3.43

kΩ30kΩ1021

RC

RR

RRBR

•Find ac equivalent circuit by replacing all capacitors by short circuits,

Hybrid parameter I: diffusion resistance/input impedance

• The diffusion resistance r is define as the reciprocal of the iB-vBE curve, which can be find as,

Output terminal characteristics of the bipolar transistor: transconductance

• If we assume constant collector-emitter voltage the,•

As we know

• By using the two hybrid parameters (r, gm), we can develop a simplified small signal hybrid-- equivalent circuit for the npn transistor.

•Voltage -controlled current source gmvbe can be transformed into current-controlled current source,

From equivalent circuit we can write as Voltage gain

• If we include the early effect then collector current in terms of early voltage as,

•

then

Summary of hybrid--model parameters

Diffusion resistance

transconductance

Current gain

Output resistance

Characteristics of a CE amplifier• It has moderately low input impedance (1K to 2K)• Its output impedance is moderately large(50K or so)• Its current gain is high• It has very high voltage gain of the order of 1500 or so• It produce very high power gain of the order of 10,000 times

or 40dB• It produce phase reversal of input signal

Uses: many applications because of Large gain in voltage, current and power

(by voltage dividerRule)

Small-Signal Analysis of Complete C-E Amplifier: AC Equivalent

• Ac equivalent circuit is constructed by assuming that all capacitances have zero impedance at signal frequency and dc voltage source is ac ground.

• Assume that Q-point is already known.

21RRBR

Small-Signal Analysis of Complete C-E Amplifier: Small-Signal Equivalent

3R

CRorLR

If we include an emitter resistance in the circuit, the Q-point of the circuit will be less dependant on the transistor current gain .

In order to determine the input impedance Rib, which is the resistance looking into the base of the transistor. We can write the following loop equation

The overall input impedance to the amplifier is now

Voltage gain is less dependant on

The voltage gain is Substantially reducedWhen an emitter resistoris included!!

How can we Improve the voltage gain ?

A common collector amplifier has following chracteristics:1 High input impedance (20-500K)2 Low output impedance (50-2000 Ohms)3 High current gain (50-300)4 Voltage gain of less than 15 Power gain of 20 to 20dB6 No phase reversal between input and output signals

Apply KVL around the base emitter loopUsing above equations we can

write voltage gain as

Input impedance

Clearly the voltage gain is less than 1 and no phase reversal

Input and output impedance for EF

Very large compared to CE

• Common base amplifier has• Very low input impedance (30-150 )• Very high output resistance (500K)• Current gain <1• Large voltage gain of about 1500• Power gain of upto 30dB• No phase reversal between input and output• Uses: for matching low impedance circuit to high impedance circuit

Chapter 5Field-Effect Transistors

The MOS Transistor

Polysilicon

Aluminum

• It is to be noted that the VDS measured relative to the source increases from 0 to VDS as we travel along the channel from source to drain. This is because the voltage between the gate and points along the channel decreases from VGS at the source end to VGS-VDS.

• When VDS is increased to the value that reduces the voltage between the gate and channel at the drain end to Vt that is ,

• VGS-VDS=Vt or VDS= VGS-Vt or VDS(sat) ≥ VGS-Vt

Concept of Asymmetric Channel

NMOS Transistor: Saturation Region

TNGSDSTNGSDVvvVv

L

WnKi

for

2

'2

vDSAT

vGS

VTN is called the saturation or pinch-off voltage

Common source circuit with coupling capacitance Cc, which act an an open circuit to the dc

DC equivalent circuit.

Gate

PMOS common source circuit

If the device is biased in saturation region

If ID=0 VDS=5V

If VDS=0 ID=VDD/RD=0.25mA

Pinchoff at the drain terminal

We see

VDS=VDD-ID(RS+RD)2=3-(0.040)(10+RDRD=15KID=K(VGS-VTN)2

40=250(VGS-0.20)2

VGS=0.6VVG=VGS+VS=0.60+(0.040)(10)=1.0VVG=(R2)/R1+R2)VDD1=(R2)/150(3)R2=50KR1=100k