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Analog Electronics

Analog Electronics. Lecture 4:Transistors. Semiconductor material and pn -junction diode. P-type semiconductor N-type semiconductor PN- Junction Diode. Diode and its resistive behavior. R = 0. R = inf. Transformative Resistor. Controlled Resistor.

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Analog Electronics

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  1. Analog Electronics Lecture 4:Transistors

  2. Semiconductor material and pn-junction diode P-type semiconductor N-type semiconductor PN- Junction Diode

  3. Diode and its resistive behavior R = 0 R = inf

  4. Transformative Resistor Controlled Resistor Signal controlled transformative resistor R Control signal Vsup Transformative Resistor

  5. How does it switch? It leads to an electronic switch Controlled Resistor Signal controlled transformative resistor R Control signal Vsup Transformative Resistor

  6. How does it amplify? It leads to signal amplification Controlled Resistor Signal controlled transformative resistor R Control signal Vsup Transformative Resistor

  7. How does it amplify? Constructing amplifying device A semiconductor is doped with penta and tri valent impurities to divide it into three regions. Three regions make 2 PN junctions. A region called ‘Emitter’ is heavily doped as an n-type material. It has an excess of electrons in conduction band. Next to Emitter a lightly doped p-region with fewer holes is called ‘Base’ The 3rd region is n-type and is called ‘Collector’

  8. Constructing amplifying device The heavily doped n-type emitter region has a very high density of conduction-band (free) electrons. These free electrons easily diffuse through the forward biased BE junction into the lightly doped and very thin p-type base region. The base has a low density of holes, which are the majority carriers.

  9. How does it amplify? Note the junction biases for this discussion. Constructing amplifying device A very little free electrons recombine with holes in base and move as valence electrons through the base region and into the emitter region as hole current. The valence electrons leave the crystalline structure of the base and become free electrons in the metallic base lead and produce the external base current. Majority of free electrons move toward the reverse-biased BC junction and swept across into the collector region by the attraction of the positive collector supply voltage. The free electrons move through the collector region, into the external circuit, and then return into the emitter region along with the base current.

  10. How does it amplify? Transistor Currents npn pnp The conventional current flows in the direction of the arrow on the emitter terminal. The emitter current IEis the sum of the collector current ICand the small base current IB. That is, IE = IC + IB. The emitter current is slightly more that collector current. The voltage drop between base and emitter is VBE whereas the voltage drop between collector and base is called VCE.

  11. How does it amplify? Current Relationships The collector current is directly proportional to the base current. IC∝ IB βDC as constant of proportionality. IC= βDC IB This eq explains amplification of current. Ratio of DC collector current and DC base current. βDC == IC/IB Ratio of DC collector current to the DC emitter current. Collector and emitter currents are ca IC= αDCIE, αDC is always less than 1

  12. Bipolar Junction Transistors This is Bipolar Junction Transistor This semiconductor device with three doped regions: Emitter Base Collector And two pn junctions: npn pnp Base-emitter junction. Base-collector junction The device is named as ‘Bipolar Junction Transistor’ or BJT

  13. More about BJTs BJT Biasing In order for a BJT to operate properly , the two pn junctions must be correctly biased with external dc voltages. For the npn type shown, the collector is more positive than the base, which is more positive than the emitter. For the pnp type, the voltages are reversed to maintain the forward-reverse bias. npn pnp

  14. BJT Circuit Analysis Currents and voltages in BJT IB : dc base current IE: dc emitter current IC: dc source current VBE: dc voltage at base wrt. emitter VCE: dc voltage at collector wrt. emitter. VCB: dc voltage at collector wrt. Base. VBE = 0.7V VCE = VCC – IC RC IB = (VBB – VBE ) / RB VCB = VCE – VBE

  15. BJT Collector Characteristics Curve The collector characteristic curves shows three mode of operations of transistor with the variation of collector current ICw.r.tVCE for a specified value of base current IB. VBB is set to produce a certain value of IB and VCC is zero and VCE is zero. As VCE is increased, IC increases until B. Saturation region Both BE and BC junctions are forward biased and the transistor is in Saturation region.

  16. BJT Collector Characteristics Curve - Saturation In saturation, an increase of base current has no effect on the collector circuit and the relation IC = bDCIB is no longer valid. IC(SAT) =VCC –VCE(SAT) /RC At this point, the transistor current is maximum and voltage across collector is minimum, for a given load.

  17. BJT Collector Characteristics Curve - Linear As VCE is increased furthers and exceeds 0.7V the base-collector junction becomes reverse-biased and the transistor goes into the active, or linear, region of its operation. IC levels off and remains essentially constant for a given value of IB as VCE continues to increase. the value of IC is determined only by the relationship expressed as Active region

  18. BJT Collector Characteristics Curve family By setting up other values of base current, a family of collector curves is developed. bDCis the ratio of collector current to base current. It can be read from the curves. The value of bDC is nearly the same wherever it is read in active region.

  19. BJT Collector Characteristics Curve –Cut off In a BJT, cutoff is the condition in which there is no base current, which results in only an extremely small leakage current (ICEO) in the collector circuit. For practical work, this current is assumed to be zero. In cutoff, neither the base-emitter junction, nor the base-collector junction are forward-biased.

  20. BJT Switches A BJT can be used as a switching device in logic circuits to turn on or off current to a load. As a switch, the transistor is normally in either cutoff (load is OFF) or saturation (load is ON). In cutoff, the transistor looks like an open switch. In saturation, the transistor looks like a closed switch.

  21. DC Load Line Here VCE = 0 and IC= IC-Sat = VCC -VCE(Sat) / RC Here IB= 0 and VCE = VCC

  22. The DC Operating Point Bias establishes the operating point (Q-point) of a transistor amplifier; the ac signal moves above and below this point. Improper biasing can cause distortion in the output signal as the transistor may go into the saturation and cutoff region.

  23. The DC Operating Point The point at which the load line intersects a characteristic curve represents the Q-point for that particular value of IB. Point A,Q,B represents the Q-point for IB 400mA. 300mA and 200mA respectively. Load line Assume a sinusoidal Ib is superimposed on VBB varying between 200uA to 400uA. It makes the collector current varies between 20 mA and 40 mA.

  24. The DC Operating Point A signal that swings outside the active region will be clipped. For example, the bias has established a low Q- point. As a result, the signal is will be clipped because it is too close to cutoff.

  25. IB I2 Summary Voltage-Divider Bias A practical way to establish a Q-point is to form a voltage-divider from VCC. R1 and R2 are selected to establish VB. If the divider is stiff, IB is small compared to I2. Then, Example: Determine the base voltage for the circuit. Solution: 4.62 V

  26. Voltage-Divider Bias A practical biasing technique that utilize single biasing sources instead of separate VCC and VBB. A dc bias voltage at the base of the transistor can be developed by a resistive voltage divider that consists of R1 and R2

  27. DC Load Line Example: What is the saturation current and the cutoff voltage for the circuit? Assume VCE = 0.2 V in saturation. Solution: 4.48 mA 15 V Follow-up: Is the transistor saturated? IC = bIB = 200 (10.45 mA) = 2.09 mA Since IC < ISAT, it is not saturated.

  28. DC and AC Quantities The text uses capital letters for both AC and DC currents and voltages with rms values assumed unless stated otherwise. DC Quantities use upper case roman subscripts. Example: VCE. (The second letter in the subscript indicates the reference point.) AC Quantities and time varying signals use lower case italic subscripts. Example: Vce. Internal transistor resistances are indicated as lower case quantities with a prime and an appropriate subscript. Example: re’. External resistances are indicated as capital R with either a capital or lower case subscript depending on if it is a DC or ac resistance. Examples: RC and Rc.

  29. The FET The idea for a field-effect transistor (FET) was first proposed by Julius Lilienthal, a physicist and inventor. In 1930 he was granted a U.S. patent for the device. His ideas were later refined and developed into the FET. Materials were not available at the time to build his device. A practical FET was not constructed until the 1950’s. Today FETs are the most widely used components in integrated circuits.

  30. The JFET The JFET (or Junction Field Effect Transistor) is a normally ON device. The n-channel is connected with two leads i.e. Drain and Source Two p-type regions are diffused in n-type material and both connected to gate.

  31. The JFET Construction For the n-channel device, when the drain is positive with respect to the source and there is no gate-source voltage, there is current in the channel. When a negative gate voltage is applied to the FET, the electric field causes the channel to narrow, which in turn causes current to decrease.

  32. The JFET-Symbol and biasing The symbol for an n-channel JFET is shown, along with the proper polarities of the applied dc voltages. For an n-channel device, the gate is always operated with a negative (or zero) voltage with respect to the source. Drain Gate Source

  33. JFET Biasing Self-bias is simple and effective, so it is the most common biasing method for JFETs. With self bias, the gate is essentially at 0 V. = +12 V An n-channel JFET is illustrated. The current in RS develops the necessary reverse bias that forces the gate to be less than the source. 1.5 kW Example: Assume the resistors are as shown and the drain current is 3.0 mA. What is VGS? Solution: 330 W 1.0 MW VG = 0 V; VS = (3.0 mA)(330 W) = 0.99 V - 0.99 V VGS = 0 – 0.99 V =

  34. The E-MOSFET The metal oxide semiconductor FET uses an insulated gate to isolate the gate from the channel. Two types are the enhancement mode (E-MOSFET) and the depletion mode (D-MOSFET). E-MOSFET An E-MOSFET has no channel until it is induced by a voltage applied to the gate, so it operates only in enhancement mode. An n-channel type is illustrated here; a positive gate voltage induces the channel.

  35. The D-MOSFET The D-MOSFET has a channel that can is controlled by the gate voltage. For an n-channel type, a negative voltage depletes the channel; and a positive voltage enhances the channel. D-MOSFET A D-MOSFET can operate in either mode, depending on the gate voltage. operating in D-mode operating in E-mode

  36. D-MOSFETs D D G G S S n channel p channel The MOSFET MOSFET symbols are shown. Notice the broken line representing the E-MOSFET that has an induced channel. The n channel has an inward pointing arrow. E-MOSFETs D D G G S S n channel p channel

  37. Thank You

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