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Transcript 
Basic Bipolar Transistor Inverter (RTL)
*
*
IB
IC
p
+
VBE
n
n
0.7 V 0.8V
IC
VCC
RC
Active
Ic=βIb
Saturation
IC< β IB
~ 0.2 V
Cutoff IC≈ 0
ECES 352 Winter 2007
Resistor Transistor Logic (RTL)
For low vi input, output vo is high
ïƒ Transistor is off (cutoff) since
iB ≈ 0 because base-emitter junction
is not biased sufficiently (VBE is too
small).
ïƒ Since iB ≈ 0, then iC ≈ 0 because
iC ≈ β iB.
ïƒ So vo = VCC - iC RC ≈ VCC
VBE
* For high vi input, output vo is low
ïƒ Transistor is on since iB > 0
because base-emitter junction is
biased sufficiently (VBE is large).
ïƒ Since VBE is large (~0.8 V), iB >> 0,
then iC >> 0 since iC ≈ β iB.
ïƒ So vo = VCC - iC RC is very small.
ïƒ Transistor driven into saturation
region so . vo = VCE,sat ≈ 0.2V
VCE
Ch 11 Bipolar Digital Pt. 1
9
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