Transistors. 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.
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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)
How does IC vary with VCE for various IB?
Note that both dc sources are variable
Set VBB to establish a certain IB
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 = bIBworks so long as the load line intersects on the plateau region of the curve.
Note that the load line intersects the 75 mA curve below the plateau region. This is saturation and IC = bIB doesn’t work in this region.
= 10 V – 4.4 V = 5.6 V
As temperature increases, the gain increases
for all current values.
VCE(max) = 20 V
IC(max) = 50 mA
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
rE = internal ac emitter
IE = Vin/rE (Ohm’s Law)
Vout = ICRC IERC
Recall the name – transfer resistor
b = a + ab
b(1-a) = a
b = a/(1-a)
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.
IC(sat) VCC/RC = 10 V/1000 W
= 10 mA
Then, IB = IC(sat)/b = 10 mA/200 = 0.05mA
If a square wave is input for VBB, then the LED will be on when the input is high, and off when the input is low.
Assume that b 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.
Driven to saturation to the negative peak. This graph shows ideal operation.
Driven into CutoffDistortion
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…
and IC = bIB
VCC = ICRC + VCE
So, VCE = VCC – ICRC
VCE = VCC – bIBRC
RB = 100 kW @ 75 °C b = 150
VCC = +12 V
@ 75 °C
IB is the same
IC = 16.95 mA
VCE = 2.51 V
IC increases by 50%
VCE decreases by 56%
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.
IC flows for less than half then cycle. Usually get more gain in Class B and C, but more distortion
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
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.
Usually given for common base amplifier to the negative peak. This graph shows ideal operation.
Usually given for common emitter amplifier
Usually given for common collector amplifierGain Factors
Ex. For b = 100 a = b/(1+b) = 0.99
g = 1 + b = 101
Second subscript indicates common base (b), common emitter (e), or common collector (c)
= to the negative peak. This graph shows ideal operation.b
= Slope of curveHybrid Parameters
hie = VB/IB Ohm’s Law
hie =input impedance
hre = VB/VC
hfe = IC/IB
Equivalent of b
hoe = IC/VC
No standard – look at the spec sheet or the case
160 to the negative peak. This graph shows ideal operation.
Jet engine - close up
Snare drums played hard at 6 inches awayTrumpet peaks at 5 inches away
Rock singer screaming in microphone (lips on mic)
Pneumatic (jack) hammer
Planes on airport runway
Threshold of pain - Piccolo strongly played
Fender guitar amplifier, full volume at 10 inches away
Subway (not the sandwich shop)
Flute in players right ear - Violin in players left earCommon Loud Sounds
90 to the negative peak. This graph shows ideal operation.
Heavy truck traffic
Typical home stereo listening levelAcoustic guitar, played with finger at 1 foot away
Conversational speech at 1 foot away
Average office noise
Quiet living room
Quiet recording studio
Threshold of hearing for healthy youthsCommon Quieter Sounds
l1 = 10 log(I1/Io)
l2 = 10 log(I2/Io)
l2 – l1= Dl = 10(log I2 – log Io – log I1 + log Io)
= 10(log I2 – log I1)
l2 – l1 = Dl = 10 log(I2/I1)
Threshold of Hearing when I = Iol = 0 dB
Threshold of Pain when I 1012 Iol = 120 dB
Sound follows the inverse square law I1/I2 = d22/d12
Dl = 50 dB = 10 log(I2/I1)
log(I2/I1) = 5 which means I2/I1 = 105
If d1 = 4 ft, then d22 = (I1/I2) d12 = 105 (4 ft)2
d2 = 1260 ft (about ¼ mile)
0 dB to the negative peak. This graph shows ideal operation.
FrequencyCommon Emitter Current Gain
Making the base thinner reduces transit time and improves frequency response