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A part of lecture notes from ee 320, discussing the use of a bjt as an electronic switch. It covers the cutoff and saturation regions of bjt operation, providing diagrams and equations to illustrate the concepts. The document also includes examples to help students understand how to calculate the required base current to saturate a bjt.
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Typology: Lecture notes
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© 2009 Keith W. Whites
The transistor can be used as an electronic switch, in addition to an amplifier. As a switch, we use the cutoff and saturation regions of BJT operation.
(Fig. 5.74)
Consequently, iB ≈ 0 , iC ≈ 0 , and iE ≈ 0 (1) which means v O = VCC (2) These are the cutoff conditions and the BJT is in the “off” state.
Due to asymmetries in the device fabrication, the voltage drops are different for these two forward-biased junctions.
VCE (^) sat ≈ 0.2 V VEC (^) sat ≈0.2 V
These are only approximate values for saturated BJTs. The actual values of VCE (^) sat and VEC (^) sat depend heavily on iC.
Equivalent circuit models for these saturated BJTs are:
VBE ≈ 0.7 V VCE^ sat ≈0.2 V VEB ≈ 0.7 V VEC^ sat ≈0.2 V
So, with vI “large,” then
RC
VCC iC
vO ≈ VCE sat vI iE
RB
Saturation is a bit more difficult to design. We need vI sufficiently large so that the collector current becomes large enough for the CBJ to become forward biased.
For this problem, assume the BJT is saturated so that VCE (^) sat = 0.2V. Therefore,
sat
I = I = − = mA.
sat min
9.8 mA (^) 0. 50
C B
= = = mA
This is I (^) B just on the edge of saturation (EOS). For an “overdrive factor (ODF)” of 10 means we want to force 10 times this current into the base of the BJT: I (^) B = ODF⋅ IB EOS (6)
or I (^) B = 10 0.196 mA⋅ = 1.96 mA.
Therefore, since 5 0. B B
B
= = kΩ
Now, with this design and the transistor saturated, what is the
forced
9.8 mA 5 1.96 mA
C sat B
sat sat forced ODF sat
C C B B
such that
forced (^) ODF
Using (7) for this example,
forced
the problem? Will the transistor stay saturated? Yes, it will.
becoming more saturated.