Transistor Circuits VI
Two-Transistor Direct-Coupled CE
Amplifier / Some basics of
troubleshooting CE Amps
The Two-Transistor Direct-Coupled CE
circuit configuration
Formulas for same
• 𝐼𝐶1 =
• 𝐼𝐶2 =
•
•
•
•
𝑉𝐶𝐶 −2𝑉𝐵𝐸
𝑅𝐹
𝑅𝐶1 + +𝑅𝐸1
𝛽
𝑉𝐶𝐶 −𝑉𝐵𝐸 −𝑅𝐶1 𝐼𝐶1
𝑅𝐸2
𝑉𝐶𝐸1 = 𝑉𝐶𝐶 − 𝑅𝐶1 + 𝑅𝐸1 𝐼𝐶1
𝑉𝐶𝐸2 = 𝑉𝐶𝐶 − 𝑅𝐶2 + 𝑅𝐸2 𝐼𝐶2
𝑉𝐶1 = 𝑉𝐶𝐶 − 𝑅𝐶1 𝐼𝐶1
𝑉𝐶2 = 𝑉𝐶𝐶 − 𝑅𝐶2 𝐼𝐶2
First example
12 V
2.7kΩ
68kΩ
Q2
Q1
470kΩ
330Ω
1kΩ
• Referring to the figure
shown, find the
collector current,
collector-to-emitter
voltage, and collectorto-ground voltage for
each BJT. Each BJT has
an hFE = 50 and VBE =
0.7V.
Work for first example
• 𝐼𝐶1 =
𝑉𝐶𝐶 −2𝑉𝐵𝐸
𝑅
𝑅𝐶1 + 𝛽𝐹 +𝑅𝐸1
10.6V
77.73kΩ
=
12−2 0.7
470k
68k+ 50 +330
=
12−1.4
68k+9.4k+330
=
= 136.37μA
• 𝑉𝐶𝐸1 = 𝑉𝐶𝐶 − 𝑅𝐶1 + 𝑅𝐸1 𝐼𝐶1 =
12 − 68k + 330 136.37μA =
12 − 68.33kΩ 136.37μA = 12 − 9.318 = 2.682V
• 𝑉𝐶1 = 𝑉𝐶𝐶 − 𝑅𝐶1 𝐼𝐶1 = 12 − 68kΩ 136.37μA =
12 − 9.273 = 2.727V
Work for first example (cont.)
•
𝑉𝐶𝐶 −𝑉𝐵𝐸 −𝑅𝐶1 𝐼𝐶1
12−0.7− 68k 136.37μA
𝐼𝐶2 =
=
𝑅𝐸2
1k
12−0.7−9.273
2.027V
=
= 2.027mA
1k
1kΩ
=
• 𝑉𝐶𝐸2 = 𝑉𝐶𝐶 − 𝑅𝐶2 + 𝑅𝐸2 𝐼𝐶2 =
12 − 2.7k + 1k 2.027mA =
12 − 3.7kΩ 2.027mA = 12 − 7.499 = 4.501V
• 𝑉𝐶2 = 𝑉𝐶𝐶 − 𝑅𝐶2 𝐼𝐶2 = 12 − 2.7kΩ 2.027mA =
12 − 5.473 = 6.527V
Second example
• Rework the previous problem using a 680-kΩ
resistor in place of the 470-kΩ resistor.
Work for second example
• 𝐼𝐶1 =
𝑉𝐶𝐶 −2𝑉𝐵𝐸
𝑅
𝑅𝐶1 + 𝛽𝐹 +𝑅𝐸1
10.6V
81.93kΩ
=
12−2 0.7
680k
68k+ 50 +330
=
12−1.4
68k+13.6k+330
=
= 129.379μA
• 𝑉𝐶𝐸1 = 𝑉𝐶𝐶 − 𝑅𝐶1 + 𝑅𝐸1 𝐼𝐶1 =
12 − 68k + 330 129.379μA =
12 − 68.33kΩ 129.379μA = 12 − 8.84 = 3.16V
• 𝑉𝐶1 = 𝑉𝐶𝐶 − 𝑅𝐶1 𝐼𝐶1 = 12 − 68kΩ 129.379μA =
12 − 8.798 = 3.202V
Work for second example (cont.)
•
𝑉𝐶𝐶 −𝑉𝐵𝐸 −𝑅𝐶1 𝐼𝐶1
12−0.7− 68k 129.379μA
𝐼𝐶2 =
=
𝑅𝐸2
1k
12−0.7−8.798
2.502V
=
= 2.502mA
1k
1kΩ
=
• 𝑉𝐶𝐸2 = 𝑉𝐶𝐶 − 𝑅𝐶2 + 𝑅𝐸2 𝐼𝐶2 =
12 − 2.7k + 1k 2.502mA =
12 − 3.7kΩ 2.502mA = 12 − 9.528 = 2.472V
• 𝑉𝐶2 = 𝑉𝐶𝐶 − 𝑅𝐶2 𝐼𝐶2 = 12 − 2.7kΩ 2.502mA =
12 − 6.756 = 5.224V
TROUBLESHOOTING CE CIRCUITS
First example
10kΩ
33kΩ
180kΩ
10.7V
2.2kΩ
12V
12V
• If the 180-kΩ (R2)
resistor became open in
the circuit shown, what
would the collector-toground voltage be?
Normal operation
• Normal operation (Q point):
• 𝑉𝐵 = 𝑉𝑅2 =
180k
213k
𝑅2
𝑉𝐶𝐶
𝑅1 +𝑅2
=
180k
33k+180k
12 =
12 = 0.845 12V = 10.14V
• 𝑉𝐵𝐸 = 𝑉𝐵 − 𝑉𝐸 = 10.17 − 10.7 = −0.56V
• 𝐼𝐶 = 𝐼𝐸 =
𝑉𝐶𝐶 −𝑉𝐸
𝑅𝐸
=
12−10.7
2.2k
=
1.3V
2.2kΩ
= 590.91μA
• 𝑉𝐶 = 𝐼𝐶 𝑅𝐶 = 590.91μA 10kΩ = 5.909V
Circuit evaluation
• If R2 opens, VB ≈12V ∴ VBE is reverse biased.
• If VBE is reverse biased, transistor is cutoff (IC =
0mA).This means VC = 0V as VE = VCC = VCE.
Second example
10kΩ
33kΩ
180kΩ
10.7V
2.2kΩ
12V
12V
• If the 33-kΩ (R1) resistor
became open instead in
the figure shown, what
would the collector-toground voltage be?
Circuit evaluation
• If R1 opens, VB = 0V ∴ transistor is biased full
on (saturation)
• 𝐼𝐶 = 𝐼𝐸 =
𝑉𝐶𝐶
𝑅𝐶 +𝑅𝐸
=
12
2.2k+10k
=
12
12.2kΩ
=
983.607μA
• 𝑉𝐶 = 𝐼𝐶 𝑅𝐶 = 983.607μA 10kΩ = 9.836V
Third example
VCC = 20V
RC = 5.6k
RB = 390k
𝐼𝐶 =
𝑉𝐶𝐶
𝑅𝐸 + 𝑅𝐶 +
𝑅𝐵
ℎ𝐹𝐸
If VBE ≈ 0V and ICEO ≈ 0A
RE = 1k
• Referring to the figure
shown, if the BJT’s hFE =
80, find its IC and VCE.
Assume that VBE and
ICEO are negligible. Hint:
The equation for VCE is
the same as the
voltage-divider-biased
circuits.
Work for third example
• 𝐼𝐶 =
𝑉𝐶𝐶
𝑅𝐵
𝑅𝐸 +𝑅𝐶 +
ℎ𝐹𝐸
20
1k+5.6k+4.875k
=
=
20
390k
1k+5.6k+
80
20V
11.475kΩ
=
= 1.743mA
• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 + 𝑅𝐶 + 𝑅𝐸 𝐼𝐶 =
20 − 5.6k + 1k 1.743mA = 20 −
6.6kΩ 1.743mA = 20 − 11.503 =
8.497V
Circuit revision
• Work the previous problem using hFE = 160
instead of 80.
Work for revision
• 𝐼𝐶 =
𝑉𝐶𝐶
𝑅𝐵
𝑅𝐸 +𝑅𝐶 +
ℎ𝐹𝐸
20
1k+5.6k+2.438k
=
=
20
390k
1k+5.6k+
160
20V
9.038kΩ
=
= 2.213mA
• 𝑉𝐶𝐸 = 𝑉𝐶𝐶 + 𝑅𝐶 + 𝑅𝐸 𝐼𝐶 =
20 − 5.6k + 1k 2.213mA = 20 −
6.6kΩ 2.213mA = 20 − 14.606 =
5.394V
Common issues
(CE Amp – 2 Supply biased)
12V
RC
4.7kΩ
6.5V
DC collector-to-ground
voltage
-0.3V
DC emitter-to-ground
voltage
RB
33kΩ
RE
10kΩ
-12V
Chart of changes/outcomes
Change
in
Value
IC
VC
VE
(1)
VCC
↑
↔
↑
↔
(2)
VCC
↓
↔
↓
↔
(3)
RB
RB
RB
↑
↔
↔
↔
↓
↔
↔
↔
∞
↓
↑
↓
↑
↔
↓
↔
↓
↔
↑
↔
∞
↓
↓
↓
RE
RE
↑
↓
↑
↔
↓
↑
↓
↔
(11)
RE
∞
↓
↑
↑
(12)
VEE
↑
↑
↓
↔
(13)
VEE
↓
↓
↑
↔
(4)
(5)
(6)
(7)
(8)
(9)
(10)
RC
RC
RC
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