SME
Electronics
BJTs and MOSFETs
Use this for “A BJT has VCC = 12 V, R₁ = 40 kΩ, R₂ = 10 kΩ, RC = 3 kΩ and RE = 1 kΩ. Find IC and VCE.”
Diagram
VB
=
VCC·
R₂
R₁ + R₂
VB
=
12
V
×
10
kΩ
40 kΩ + 10 kΩ
=
2.4
V
IC
=
VB − 0.7 V
RE
IC
=
2.4 V − 0.7 V
1
kΩ
=
1.7
mA
VCE = VCC − IC(RC + RE)
VCE = 12 V − 1.7 mA × (3 kΩ + 1 kΩ)
VCE = 5.2 V
Active: VCE is above 0.2 V.
Stiff divider (base current ignored), VBE = 0.7 V, IE ≈ IC.
This is an example. Type your own values to start a new problem.
VB (V)
=
VCC (V)
×
R₂ (kΩ)
R₁ (kΩ)
+
R₂ (kΩ)
IC (mA)
=
VB (V)
−
0.7
V
RE (kΩ)
VCE (V)
=
VCC (V)
−
IC (mA)
×
(
RC (kΩ)
+
RE (kΩ)
)
Formulas
VB
=
VCC
×
R₂
R₁
+
R₂
2.4
=
12
×
10
40 + 10
IC
=
VB − 0.7
RE
1.7
=
2.4 − 0.7
1
VCE = VCC − IC × (RC + RE)
5.2 = 12 − 1.7 × (3 + 1)
Assumptions
•
Stiff divider: the base current is too small to load R₁ and R₂ (β is large).
•
The npn transistor is active, with VBE = 0.7 V and IE ≈ IC.
•
Resistances in kΩ give currents in mA (V ÷ kΩ = mA).
Step-by-step
WE KNOW
VCC = 12 V, R₁ = 40 kΩ, R₂ = 10 kΩ, RC = 3 kΩ, RE = 1 kΩ
FIND
base voltage (VB), collector current (IC), collector–emitter voltage (VCE)
Step 1 · Find base voltage (VB)
Use VB = VCC × R₂ ÷ (R₁ + R₂)
The
divider
gives
the
base
its
share
of
the
supply:
R₂
over
the
whole
chain.
VB
=
12
×
10
40 + 10
VB
=
12
×
10
50
VB
=
120
50
VB = 2.4 V
Step 2 · Find collector current (IC)
Use IC = (VB − 0.7) ÷ RE
The
emitter
sits
0.7
V
under
the
base,
so
RE
carries
VB − 0.7
RE
,
and
IC
≈
IE.
IC
=
2.4 − 0.7
1
IC = 1.7 mA
Step 3 · Find collector–emitter voltage (VCE)
Use VCE = VCC − IC × (RC + RE)
Kirchhoff’s voltage law round the output loop: RC and RE take IC(RC + RE), mA × kΩ = V.
VCE = 12 − 1.7 × (3 + 1)
VCE = 12 − 1.7 × 4
VCE = 12 − 6.8
VCE = 5.2 V
Check
Put the numbers back in. Both sides should match.
2.4
=
12
×
10
40 + 10
✓
1.7
=
2.4 − 0.7
1
✓
5.2 = 12 − 1.7 × (3 + 1) ✓
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Current gain: IC, IE and α from β
Topic check
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