Transistor Circuits V
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Transistor Circuits VI Two-Transistor Direct-Coupled CE Amplifier / Some basics of troubleshooting CE Amps
Transcript of Transistor Circuits V
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 =𝑉𝐶𝐶−2𝑉𝐵𝐸
𝑅𝐶1+𝑅𝐹𝛽+𝑅𝐸1
• 𝐼𝐶2 =𝑉𝐶𝐶−𝑉𝐵𝐸−𝑅𝐶1𝐼𝐶1
𝑅𝐸2
• 𝑉𝐶𝐸1 = 𝑉𝐶𝐶 − 𝑅𝐶1 + 𝑅𝐸1 𝐼𝐶1
• 𝑉𝐶𝐸2 = 𝑉𝐶𝐶 − 𝑅𝐶2 + 𝑅𝐸2 𝐼𝐶2
• 𝑉𝐶1 = 𝑉𝐶𝐶 − 𝑅𝐶1𝐼𝐶1
• 𝑉𝐶2 = 𝑉𝐶𝐶 − 𝑅𝐶2𝐼𝐶2
First example
• Referring to the figure shown, find the collector current, collector-to-emitter voltage, and collector-to-ground voltage for each BJT. Each BJT has an hFE = 50 and VBE = 0.7V.
12 V
Q1
Q2
68kΩ
330Ω
2.7kΩ
1kΩ
470kΩ
Work for first example
• 𝐼𝐶1 =𝑉𝐶𝐶−2𝑉𝐵𝐸
𝑅𝐶1+𝑅𝐹𝛽+𝑅𝐸1
=12−2 0.7
68k+470k
50+330
=12−1.4
68k+9.4k+330=
10.6V
77.73kΩ= 136.37μA
• 𝑉𝐶𝐸1 = 𝑉𝐶𝐶 − 𝑅𝐶1 + 𝑅𝐸1 𝐼𝐶1 =12 − 68k + 330 136.37μA =12 − 68.33kΩ 136.37μA = 12 − 9.318 = 2.682V
• 𝑉𝐶1 = 𝑉𝐶𝐶 − 𝑅𝐶1𝐼𝐶1 = 12 − 68kΩ 136.37μA =12 − 9.273 = 2.727V
Work for first example (cont.)
• 𝐼𝐶2 =𝑉𝐶𝐶−𝑉𝐵𝐸−𝑅𝐶1𝐼𝐶1
𝑅𝐸2=
12−0.7− 68k 136.37μA
1k=
12−0.7−9.273
1k=
2.027V
1kΩ= 2.027mA
• 𝑉𝐶𝐸2 = 𝑉𝐶𝐶 − 𝑅𝐶2 + 𝑅𝐸2 𝐼𝐶2 =12 − 2.7k + 1k 2.027mA =12 − 3.7kΩ 2.027mA = 12 − 7.499 = 4.501V
• 𝑉𝐶2 = 𝑉𝐶𝐶 − 𝑅𝐶2𝐼𝐶2 = 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
𝑅𝐶1+𝑅𝐹𝛽+𝑅𝐸1
=12−2 0.7
68k+680k
50+330
=12−1.4
68k+13.6k+330=
10.6V
81.93kΩ= 129.379μA
• 𝑉𝐶𝐸1 = 𝑉𝐶𝐶 − 𝑅𝐶1 + 𝑅𝐸1 𝐼𝐶1 =12 − 68k + 330 129.379μA =12 − 68.33kΩ 129.379μA = 12 − 8.84 = 3.16V
• 𝑉𝐶1 = 𝑉𝐶𝐶 − 𝑅𝐶1𝐼𝐶1 = 12 − 68kΩ 129.379μA =12 − 8.798 = 3.202V
Work for second example (cont.)
• 𝐼𝐶2 =𝑉𝐶𝐶−𝑉𝐵𝐸−𝑅𝐶1𝐼𝐶1
𝑅𝐸2=
12−0.7− 68k 129.379μA
1k=
12−0.7−8.798
1k=
2.502V
1kΩ= 2.502mA
• 𝑉𝐶𝐸2 = 𝑉𝐶𝐶 − 𝑅𝐶2 + 𝑅𝐸2 𝐼𝐶2 =12 − 2.7k + 1k 2.502mA =12 − 3.7kΩ 2.502mA = 12 − 9.528 = 2.472V
• 𝑉𝐶2 = 𝑉𝐶𝐶 − 𝑅𝐶2𝐼𝐶2 = 12 − 2.7kΩ 2.502mA =12 − 6.756 = 5.224V
TROUBLESHOOTING CE CIRCUITS
First example
• If the 180-kΩ (R2) resistor became open in the circuit shown, what would the collector-to-ground voltage be?
12V
12V
33kΩ 180kΩ
2.2kΩ
10kΩ
10.7V
Normal operation
• Normal operation (Q point):
• 𝑉𝐵 = 𝑉𝑅2 =𝑅2
𝑅1+𝑅2𝑉𝐶𝐶 =
180k
33k+180k12 =
180k
213k12 = 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 = 𝐼𝐶𝑅𝐶 = 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
• If the 33-kΩ (R1) resistor became open instead in the figure shown, what would the collector-to-ground voltage be?
12V
12V
33kΩ 180kΩ
2.2kΩ
10kΩ
10.7V
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 = 𝐼𝐶𝑅𝐶 = 983.607μA 10kΩ = 9.836V
Third example
• 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.
VCC = 20V
RC = 5.6k
RB = 390k
RE = 1k
𝐼𝐶 =𝑉𝐶𝐶
𝑅𝐸 + 𝑅𝐶 +𝑅𝐵ℎ𝐹𝐸
If VBE ≈ 0V and ICEO ≈ 0A
Work for third example
• 𝐼𝐶 =𝑉𝐶𝐶
𝑅𝐸+𝑅𝐶+𝑅𝐵ℎ𝐹𝐸
=20
1k+5.6k+390k
80
=
20
1k+5.6k+4.875k=
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
𝑅𝐸+𝑅𝐶+𝑅𝐵ℎ𝐹𝐸
=20
1k+5.6k+390k
160
=
20
1k+5.6k+2.438k=
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
-12V
RB33kΩ
RC4.7kΩ
RE10kΩ
6.5V
-0.3V
DC collector-to-ground
voltage
DC emitter-to-ground
voltage
Chart of changes/outcomes Change
in
Value
IC VC VE
(1) VCC ↑ ↔ ↑ ↔
(2) VCC ↓ ↔ ↓ ↔
(3) RB ↑ ↔ ↔ ↔
(4) RB ↓ ↔ ↔ ↔
(5) RB ∞ ↓ ↑ ↓
(6) RC ↑ ↔ ↓ ↔
(7) RC ↓ ↔ ↑ ↔
(8) RC ∞ ↓ ↓ ↓
(9) RE ↑ ↓ ↑ ↔
(10) RE ↓ ↑ ↓ ↔
(11) RE ∞ ↓ ↑ ↑
(12) VEE ↑ ↑ ↓ ↔
(13) VEE ↓ ↓ ↑ ↔
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