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Transcript
```BJT structure
heavily doped ~ 10^15
provides the carriers
lightly doped ~ 10^8
lightly doped ~ 10^6
note: this is a current of electrons (npn case) and so the
conventional current flows from collector to emitter.
BJT characteristics
BJT characteristics
BJT modes of operation
Mode
EBJ
CBJ
Cutoff
Reverse
Reverse
Forward active
Forward
Reverse
Reverse active
Reverse
Forward
Saturation
Forward
Forward
BJT modes of operation
Cutoff: In cutoff, both junctions reverse biased. There is very little current flow, which
corresponds to a logical "off", or an open switch.
Forward-active (or simply, active): The emitter-base junction is forward biased and the
base-collector junction is reverse biased. Most bipolar transistors are designed to afford the
greatest common-emitter current gain, βf in forward-active mode. If this is the case, the
collector-emitter current is approximately proportional to the base current, but many times
larger, for small base current variations.
Reverse-active (or inverse-active or inverted): By reversing the biasing conditions of the
forward-active region, a bipolar transistor goes into reverse-active mode. In this mode, the
emitter and collector regions switch roles. Since most BJTs are designed to maximise
current gain in forward-active mode, the βf in inverted mode is several times smaller. This
transistor mode is seldom used. The reverse bias breakdown voltage to the base may be an
order of magnitude lower in this region.
Saturation: With both junctions forward-biased, a BJT is in saturation mode and facilitates
current conduction from the emitter to the collector. This mode corresponds to a logical
"on", or a closed switch.
BJT structure (active)
IE
-
E
IC
VCE +
C
current of electrons for npn transistor –
conventional current flows from
collector to emitter.
+
VBE
VCB
IB
+
B
BJT equations (active)
 = Common-base current gain (0.9-0.999; typical 0.99)
VBE
iC

iE
iC  I S e
iE 
IS

VT
VBE
e
VT
BJT equations (active)
 = Common-emitter current gain (10-1000; typical 50-200)
VBE
iC

iB
iB 
iC  I S e
IS

VBE
e
VT
VT
BJT equations (active)
 = Common-base current gain (0.9-0.999; typical 0.99)
 = Common-emitter current gain (10-1000; typical 50-200)


 1


1
BJT large signal models (forward active)
   F ~ forward
BJT large signal models (reverse)
 R ~ reverse
Common-base current gain (0.1-0.5) BJT transistor is not a symmetrical device
BJT Ebers-Moll (EM) model
BJT structure
The npn transistor has beta=100 and exhibits an Ic=1mA at
VBE=0.7V. Design the circuit so that a current of 2mA flows
through collector and a voltage of +5V appears at the collector.
BJT equations
The voltage at the emitter was measured and found to be -0.7V.
If beta=50, find IE, IB, IC and VC.
BJT equations
A given npn transistor has beta=100. Determine the region of operation if:
a)
IB=50uA and IC=3mA
b)
IB=50uA and VCE=5V
c)
VBE=-2V and VCE=-1V
BJT equations
RB=200kΩ, RC=1kΩ, VCC=15V, beta=100. Solve for IC and VCE
BJT equations
RB=200kΩ, RC=1kΩ, VCC=15V, beta=100. Solve for IC and VCE
```
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