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BJT as an Electronic Switch: Understanding Cutoff and Saturation Regions, Lecture notes of Electronics engineering

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.

What you will learn

  • What are the cutoff conditions for a BJT?
  • What happens to a BJT when it enters the saturation region?
  • How does the overdrive factor affect the current in a BJT?

Typology: Lecture notes

2014/2015

Uploaded on 08/18/2015

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muhammad.muheeb 🇬🇧

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Whites, EE 320 Lecture 24 Page 1 of 5
© 2009 Keith W. Whites
Lecture 24: BJT as an Electronic Switch.
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)
Cutoff Region. If 0.5
I
v3 or so, the EBJ will conduct
negligible current. Also, the CBJ will be reversed biased with
a large VCC.
Consequently,
0
B
i, 0
C
i
, and 0
E
i
(1)
which means
OCC
vV
=
(2)
These are the cutoff conditions and the BJT is in the “off
state.
Saturation Region. For the “on” state of the switch, we
increase vI until the BJT saturates. This occurs when the EBJ
and the CBJ are both forward biased.
pf3
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© 2009 Keith W. Whites

Lecture 24: BJT as an Electronic Switch.

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)

  • Cutoff Region. If vI 3 0.5 or so, the EBJ will conduct negligible current. Also, the CBJ will be reversed biased with a large V CC.

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.

  • Saturation Region. For the “on” state of the switch, we increase vI until the BJT saturates. This occurs when the EBJ and the CBJ are both forward biased.

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

vOVCE sat vI iE

RB

  • i (^) B

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

C C 1,

I = I = − = mA.

To saturate the BJT with the smallest β we need to provide

sat min

9.8 mA (^) 0. 50

C B

I

I

= = = 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

I

R

B 2.

B

R

I

= = kΩ

Now, with this design and the transistor saturated, what is the

“forced” β?

forced

9.8 mA 5 1.96 mA

C sat B

I

I

This value is much smaller that βmin =50, as expected. Another

way to compute βforced is to notice:

sat sat forced ODF sat

C C B B

I I

I I

such that

forced (^) ODF

Using (7) for this example,

forced

Lastly, what happens if β is increased from 50 to 150 as stated in

the problem? Will the transistor stay saturated? Yes, it will.

Actually, nothing changes in our saturated circuit as β varies.

However, βforced becomes smaller indicating that the transistor is

becoming more saturated.