Transfer Resistor Chapter 9 Bipolar Transistors

Collector Base

Emitter

Two PN junctions joined together Two types available – NPN and PNP The regions (from top to bottom) are called the collector (C), the base (B), and the emitter (E) Operation

• Begin by reverse biasing the CB junction – Here we are showing an NPN transistor as an example • Now we apply a small forward bias on the emitter‐base junction – Electrons are pushed into the base, which then quickly flow to the collector – The result is a large emitter‐collector electron current (conventional current is C‐ E) which is maintained by a small E‐B voltage • Some of the electrons pushed into the base by the forward bias E‐B voltage end up depleting holes in that junction – This would eventually destroy the junction if we didn’t replenish the holes – The electrons that might do this are drawn off as a base current Currents Conventional View Origin of the names

• the Emitter 'emits' the electrons which pass through the device • the Collector 'collects' them again once they've passed through the Base • ...and the Base?... Original Manufacture Base Thickness

• The thickness of the unmodified Base region has to be just right. – Too thin, and the Base would essentially vanish. The Emitter and Collector would then form a continuous piece of semiconductor, so current would flow between them whatever the base potential. – Too thick, and electrons entering the Base from the Emitter wouldn't notice the Collector as it would be too far away. So then, the current would all be between the Emitter and the Base, and there'd be no Emitter‐Collector current. Amplification Properties

• The C‐B voltage junction operates near breakdown. – This ensures that a small E‐B voltage causes avalanche – Large current through the device Common Base NPN Common Emitter NPN Common Collector NPN

How does IC vary with VCE for various IB? Note that both dc sources are variable

Set VBB to establish a certain IB Collector Characteristic Curve

• If VCC = 0, then IC = 0 and VCE = 0

• As VCC ↑ both VCE and IC ↑

• When VCE ≈ 0.7 V, base-collector becomes reverse-biased and IC reaches full value (IC = βIB)

• IC ~ constant as VCE ↑. There is a slight increase of IC due to the widening of the depletion zone (BC) giving fewer holes for recombinations with e¯ in base.

• Since IC = βIB, different base currents produce different IC plateaus. NPN Characteristic Curves PNP Characteristic Curves Load Line

Slope of the load

line is 1/RL

For a constant load, stepping IB gives different currents (IC) predicted by where the load line crosses the characteristic curve. IC = βIBworks so long as the load line intersects on the plateau region of the curve. Saturation and Cut‐off

Cut-off

Note that the load line intersects the 75 mA curve below the plateau region. This is saturation and IC = βIB doesn’t work in this region. Example

• We adjust the base current to 200 μA and note that this transistor has a β = 100 -6 – Then IC = βIB = 100(200 X 10 A) = 20 mA • Notice that we can use Kirchhoff’s voltage law around the right side of the circuit

– VCE = VCC –ICRC = 10 V – (20 mA)(220 Ω) = 10 V – 4.4 V = 5.6 V Example

• Now adjust IB to 300 μA

– Now we get IC = 30 mA

– And VCE = 10 V – (30 mA)(220 Ω) = 3.4 V

• Finally, adjust IB = 400 μA

– IB = 40 mA and VCE = 1.2 V Plot the load line

VCE IC 5.6 V 20 mA

3.4 V 30 mA 1.2 V 40 mA Gain as a function of IC

As temperature increases, the gain increases for all current values. Operating Limits

• There will be a limit on the dissipated power

– PD(max) = VCEIC

– VCE and IC were the parameters plotted on the characteristic curve.

• If there is a voltage limit (VCE(max)), then you can compute the IC that results

• If there is a current limit (IC(max)), then you can compute the VCE that results Example

• Assume P = 0.5 W D(max) PD(max) VCE IC VCE(max) = 20 V 0.5 W 5 V 100 mA IC(max) = 50 mA 10 50

15 33

20 25 Operating Range

Operating Range Voltage Amplifiers

Common Base PNP

Now we have added an ac source The biasing of the junctions are:

BE is forward biased by VBB - thus a small resistance BC is reverse biased by VCC – and a large resistance

Since IB is small, IC ≈ IE Equivalent ac Circuit

rE = internal ac emitter resistance

IE = Vin/rE (Ohm’s Law)

Vout = ICRC ≈ IERC

Vout AV == voltage gain Vin RI R CE C Recall the name – transfer resistor AV == rI EE rE Current Gains

• Common Base

– α = IC/IE < 1 • Common Emitter

– β = IC/IB From Kirchhoff' s Current Law 11 + β += III BCE = α β I E I B 1+= αβ+= αβ IC IC 1 1 += βαβ)1( 1+= β βα α = 1+ β Example

• If β = 50, then α = 50/51 = 0.98 – Recall α < 1

• Rearranging, β = α + αβ β(1‐α) = α β = α/(1‐α) Transistors as Switches The operating points

We can control the base current using VBB (we don’t actually use a physical switch). The circuit then acts as a high speed switch. Details

• In Cut-off

– All currents are zero and VCE = VCC • In Saturation

– IB big enough to produce IC(sat) ≈ βIB • Using Kirchhoff’s Voltage Law through the ground loop

– VCC = VCE(sat) + IC(sat)RC – but VCE(sat) is very small (few tenths), so – IC(sat) ≈VCC/RC Example

a) What is VCE when Vin = 0 V?

Ans. VCE = VCC = 10 V

b) What minimum value of IB is required to saturate the transistor if

β = 200? Take VCE(sat) = 0 V

IC(sat) ≈ VCC/RC = 10 V/1000 Ω = 10 mA

Then, IB = IC(sat)/β = 10 mA/200 = 0.05mA Example

If a square wave is input for VBB, LED then the LED will be on when the input is high, and off when the input is low. Transistors with ac Input

Assume that β is such that

IC varies between 20 and 40 mA. The transistor is constantly changing curves along the load line. Pt. A corresponds to the positive peak. Pt. B corresponds to the negative peak. This graph shows ideal operation. Distortion

• The location of the point Q (size of the dc source on input) may cause an operating point to lie outside of the active range.

Driven to saturation Driven into Cutoff Base Biasing

• It is usually not necessary to provide two sources for biasing the transistor.

The red arrows follow the base-emitter part of the circuit, which contains the

resistor RB. The voltage drop across RB is VCC –VBE (Kirchhoff’s Voltage Law). The base current is then… −VV I = CC BE B and IC = βIB RC Base Biasing

• Use Kirchhoff’s Voltage Law on the black arrowed loop of the circuit

VCC = ICRC + VCE

So, VCE = VCC –ICRC

VCE = VCC – βIBRC

• Disadvantge

– β occurs in the equation for both VCE and IC

– But β varies – thus so do VCE and IC – This shifts the Q‐point (β‐dpendent) Example

• Let RC = 560 Ω @ 25 °C β = 100 RB = 100 kΩ @ 75 °C β = 150 VCC = +12 V @ 25 °C @ 75 °C −VV 12 V - 0.7 V CC BE IB is the same IB = = = 113μA RB 100,000 Ω IC = 16.95 mA

I I C β I B == (100)(113 μA) = 11.3 mA VCE = 2.51 V RI V VCE V CC −= β RI CB IC increases by 50% - V 12 = 12 V - (100)(113 A)(560 Ωμ) V decreases by 56% = 5.67 V CE Transistor Amplifiers

• Amplification – The process of increasing the strength of a signal. – The result of controlling a relatively large quantity of current (output) with a small quantity of current (input). • Amplifier – Device use to increase the current, voltage, or power of the input signal without appreciably altering the essential quality. Class A

• Entire input waveform is faithfully reproduced. • Transistor spends its entire time in the active mode – Never reaches either cutoff or saturation. – Drive the transistor exactly halfway between cutoff and saturation. – Transistor is always on –always dissipating power –can be quite inefficient Class A Class B

• No DC bias voltage – The transistor spends half its time in active mode and the other half in cutoff Push‐pull Pair

Transistor Q1 "pushes" (drives the output voltage in a positive direction with respect to ground), while transistor Q2 "pulls" the output voltage (in a negative direction, toward 0 volts with respect to ground). Individually, each of these transistors is operating in class B mode, active only for one-half of the input waveform cycle. Together, however, they function as a team to produce an output waveform identical in shape to the input waveform. Class AB

• Between Class A (100% operation) and Class B (50% operation). Class C

IC flows for less than half then cycle. Usually get more gain in Class B and C, but more distortion Common Emitter Transistor Amplifier

Notice that VBB forward biases the emitter-base junction and dc current flows through the circuit at all times

The class of the amplifier is determined by VBB with respect to the input signal.

Signal that adds to VBB causes transistor current to increase Signal that subtracts from VBB causes transistor current to decrease Details

• At positive peak of input, VBB is adding to the input • Resistance in the transistor is reduced • Current in the circuit increases

• Larger current means more voltage drop across RC (VRC = IRC) • Larger voltage drop across RC leaves less voltage to be dropped across the transistor

• We take the output VCE –as input increases, VCE decreases. More details

• As the input goes to the negative peak – Transistor resistance increases – Less current flows

– Less voltage is dropped across RC – More voltage can be dropped across C‐E • The result is a phase reversal – Feature of the common emitter amplifier

• The closer VBB is to VCC, the larger the transistor current. PNP Common Emitter Amplifier NPN Common Base Transistor Amplifier

Signal that adds to VBB causes transistor current to increase Signal that subtracts from VBB causes transistor current to decrease

• At positive peak of input, VBB is adding to the input • Resistance in the transistor is reduced • Current in the circuit increases

• Larger current means more voltage drop across RC (VRC = IRC) • Collector current increases • No phase reversal PNP Common Base Amplifier NPN Common Collector Transistor Amplifier

Also called an Emitter Follower circuit – output on emitter is almost a replica of the input

Input is across the C-B junction – this is reversed biased and the impedance is high Output is across the B-E junction – this is forward biased and the impedance is low. Current gain is high but voltage gain is low. PNP Common Collector Transistor Amplifier Gain Factors

I α = C Usually given for common base amplifier I E

I β = C Usually given for common emitter amplifier I B

I γ = E Usually given for common collector amplifier I B Gamma

• Recall from Kirchhoff’s Current Law α – IB + IC = IE And since β = -1 α IC IE α I I B 1 =+÷ 1 =+ γ IB IB -1 α -1 +αα 1 1 =+ γβ LCD γ == -1 α -1 α

Ex. For β = 100 α = β/(1+β) = 0.99 γ = 1 + β = 101 Bringing it Together Type Common Common Common Base Emitter Collector Relation between input/output 0° 180° 0° phase Voltage Gain High Medium Low

Current Gain Low (α) Medium (β) High (γ)

Power Gain Low High Medium

Input Z Low Medium High

Output Z High Medium Low Hybrid Parameters

Condition

hi Input resistance Output shorted

hr Voltage feedback ratio Input open

hf Forward current gain Output shorted

ho Output conductance Input open

Second subscript indicates common base (b), common emitter (e), or common collector (c) Hybrid Parameters

= β

= Slope of curve Hybrid Parameters

hie = VB/IB Ohm’s Law

hie =input impedance

hre = VB/VC Hybrid Parameters

hfe = IC/IB

Equivalent of β

hoe = IC/VC Various Forms

Common Common Common Emitter (e) Base (b) Collector (c)

hi (ohms) VB/IB VE/IB VB/IB hr (unitless) VB/VC VE/VC VB/VE hf (unitless) IC/IB IC/IE IE/IB

ho (watts) ICVC ICVC IEVE Pin‐outs

No standard – look at the spec sheet or the case Loudness

• When the energy (intensity) of the sound increases by a factor of 10, the loudness increases by 1 bel – Named for A. G. Bell – One bel is a large unit and we use 1/10th bel, or decibels • When the energy (intensity) of the sound increases by a factor of 10, the loudness increases by 10 dB Decibel Scale

• For intensities

– L = 10 log(I/Io) • For energies

– L = 10 log(E/Eo) • For amplitudes

– L = 20 log(A/Ao) Threshold of Hearing

• The Io or Eo or Ao refers to the intensity, energy, or amplitude of the sound wave for the threshold of hearing -12 2 – Io = 10 W/m – Loudness levels always compared to threshold • Relative measure Common Loud Sounds

160

Jet engine - close up Snare drums played hard at 6 inches away 150 Trumpet peaks at 5 inches away 140 Rock singer screaming in microphone (lips on mic)

130

Pneumatic (jack) hammer Cymbal crash

Planes on airport runway 120 Threshold of pain - Piccolo strongly played

Fender guitar amplifier, full volume at 10 inches away

Power tools 110 Flute in players right ear - Violin in players left Subway (not the sandwich shop) 100 ear Common Quieter Sounds

90 Heavy truck traffic Typical home stereo listening level Chamber music 80 Acoustic guitar, played with finger at 1 foot away Average factory 70 Busy street Small orchestra 60 Conversational speech at 1 foot away Average office noise 50 Quiet conversation 40 Quiet office 30 Quiet living room 20 10 Quiet recording studio 0 Threshold of hearing for healthy youths The Math

l1 = 10 log(I1/Io) l2 = 10 log(I2/Io) l2 –l1 = Δl = 10(log I2 – log Io – log I1 + log Io) = 10(log I2 – log I1) l2 –l1 = Δl = 10 log(I2/I1)

Threshold of Hearing when I = Io l = 0 dB 12 Threshold of Pain when I ≈ 10 Io l = 120 dB Example

• A loudspeaker produces loudness rated at 90 dB (l1) at a distance of 4 ft (d1). How far can the sound travel (d2) and still give a loudness at the listener’s ear of 40 dB (l2 ‐ conversation at 3 ft.)?

2 2 Sound follows the inverse square law I1/I2 = d2 /d1

Δl = 50 dB = 10 log(I2/I1) 5 log(I2/I1) = 5 which means I2/I1 = 10 2 2 5 2 If d1 = 4 ft, then d2 = (I1/I2) d1 = 10 (4 ft)

d2 = 1260 ft (about ¼ mile) Common Emitter Current Gain

0 dB -3 dB

hfe

Frequency • For the ‐3 dB point

– Δl = 3 dB = 10 log (I1/I2)

• I1/I2 = 2 = P1/P2 • so 3 dB below initial level mean half the power Why do Frequency limits occur?

• It takes a certain time for e‐ to travel from emitter to collector (transit time) • If frequency is too high, applied current varies too rapidly • Electrons may be unable to dislodge rapidly enough to move from E to C before current surges in the other direction.

Making the base thinner reduces transit time and improves frequency response Interelement Capacitance

• As reverse bias increases on the C‐B junction, the depletion zone increases and C decreases (C = εA/d and d increasing). • As emitter current increases, C increases (d decreasing). • If capacitance changes, so does capacitive reactance 1 X = C 2π f C – Increasing C decreases XC Feedback

• Small base current provides a path back to input – If the feedback voltage aids the input voltage, then it is positive (regenerative) feedback – If the feedback is too large, the amplifier will oscillate Superheterodyne Receiver