Transistor Circuits V
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
β’ πΌπΆ1 =ππΆπΆβ2ππ΅πΈ
π πΆ1+π πΉπ½+π πΈ1
β’ πΌπΆ2 =ππΆπΆβππ΅πΈβπ πΆ1πΌπΆ1
π πΈ2
β’ ππΆπΈ1 = ππΆπΆ β π πΆ1 + π πΈ1 πΌπΆ1
β’ ππΆπΈ2 = ππΆπΆ β π πΆ2 + π πΈ2 πΌπΆ2
β’ ππΆ1 = ππΆπΆ β π πΆ1πΌπΆ1
β’ ππΆ2 = ππΆπΆ β π πΆ2πΌπΆ2
First example
β’ 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
68kΞ©
330Ξ©
2.7kΞ©
1kΞ©
470kΞ©
Work for first example
β’ πΌπΆ1 =ππΆπΆβ2ππ΅πΈ
π πΆ1+π πΉπ½+π πΈ1
=12β2 0.7
68k+470k
50+330
=12β1.4
68k+9.4k+330=
10.6V
77.73kΞ©= 136.37ΞΌA
β’ ππΆπΈ1 = ππΆπΆ β π πΆ1 + π πΈ1 πΌπΆ1 =12 β 68k + 330 136.37ΞΌA =12 β 68.33kΞ© 136.37ΞΌA = 12 β 9.318 = 2.682V
β’ ππΆ1 = ππΆπΆ β π πΆ1πΌπΆ1 = 12 β 68kΞ© 136.37ΞΌA =12 β 9.273 = 2.727V
Work for first example (cont.)
β’ πΌπΆ2 =ππΆπΆβππ΅πΈβπ πΆ1πΌπΆ1
π πΈ2=
12β0.7β 68k 136.37ΞΌA
1k=
12β0.7β9.273
1k=
2.027V
1kΞ©= 2.027mA
β’ ππΆπΈ2 = ππΆπΆ β π πΆ2 + π πΈ2 πΌπΆ2 =12 β 2.7k + 1k 2.027mA =12 β 3.7kΞ© 2.027mA = 12 β 7.499 = 4.501V
β’ ππΆ2 = ππΆπΆ β π πΆ2πΌπΆ2 = 12 β 2.7kΞ© 2.027mA =12 β 5.473 = 6.527V
Second example
β’ Rework the previous problem using a 680-kΞ© resistor in place of the 470-kΞ© resistor.
Work for second example
β’ πΌπΆ1 =ππΆπΆβ2ππ΅πΈ
π πΆ1+π πΉπ½+π πΈ1
=12β2 0.7
68k+680k
50+330
=12β1.4
68k+13.6k+330=
10.6V
81.93kΞ©= 129.379ΞΌA
β’ ππΆπΈ1 = ππΆπΆ β π πΆ1 + π πΈ1 πΌπΆ1 =12 β 68k + 330 129.379ΞΌA =12 β 68.33kΞ© 129.379ΞΌA = 12 β 8.84 = 3.16V
β’ ππΆ1 = ππΆπΆ β π πΆ1πΌπΆ1 = 12 β 68kΞ© 129.379ΞΌA =12 β 8.798 = 3.202V
Work for second example (cont.)
β’ πΌπΆ2 =ππΆπΆβππ΅πΈβπ πΆ1πΌπΆ1
π πΈ2=
12β0.7β 68k 129.379ΞΌA
1k=
12β0.7β8.798
1k=
2.502V
1kΞ©= 2.502mA
β’ ππΆπΈ2 = ππΆπΆ β π πΆ2 + π πΈ2 πΌπΆ2 =12 β 2.7k + 1k 2.502mA =12 β 3.7kΞ© 2.502mA = 12 β 9.528 = 2.472V
β’ ππΆ2 = ππΆπΆ β π πΆ2πΌπΆ2 = 12 β 2.7kΞ© 2.502mA =12 β 6.756 = 5.224V
TROUBLESHOOTING CE CIRCUITS
First example
β’ If the 180-kΞ© (R2) resistor became open in the circuit shown, what would the collector-to-ground voltage be?
12V
12V
33kΞ© 180kΞ©
2.2kΞ©
10kΞ©
10.7V
Normal operation
β’ Normal operation (Q point):
β’ ππ΅ = ππ 2 =π 2
π 1+π 2ππΆπΆ =
180k
33k+180k12 =
180k
213k12 = 0.845 12V = 10.14V
β’ ππ΅πΈ = ππ΅ β ππΈ = 10.17 β 10.7 = β0.56V
β’ πΌπΆ = πΌπΈ =ππΆπΆβππΈ
π πΈ=
12β10.7
2.2k=
1.3V
2.2kΞ©= 590.91ΞΌA
β’ ππΆ = πΌπΆπ πΆ = 590.91ΞΌA 10kΞ© = 5.909V
Circuit evaluation
β’ If R2 opens, VB β12V β΄ VBE is reverse biased.
β’ If VBE is reverse biased, transistor is cutoff (IC = 0mA).This means VC = 0V as VE = VCC = VCE.
Second example
β’ If the 33-kΞ© (R1) resistor became open instead in the figure shown, what would the collector-to-ground voltage be?
12V
12V
33kΞ© 180kΞ©
2.2kΞ©
10kΞ©
10.7V
Circuit evaluation
β’ If R1 opens, VB = 0V β΄ transistor is biased full on (saturation)
β’ πΌπΆ = πΌπΈ =ππΆπΆ
π πΆ+π πΈ=
12
2.2k+10k=
12
12.2kΞ©=
983.607ΞΌA
β’ ππΆ = πΌπΆπ πΆ = 983.607ΞΌA 10kΞ© = 9.836V
Third example
β’ Referring to the figure shown, if the BJTβs 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
πΌπΆ =ππΆπΆ
π πΈ + π πΆ +π π΅βπΉπΈ
If VBE β 0V and ICEO β 0A
Work for third example
β’ πΌπΆ =ππΆπΆ
π πΈ+π πΆ+π π΅βπΉπΈ
=20
1k+5.6k+390k
80
=
20
1k+5.6k+4.875k=
20V
11.475kΞ©= 1.743mA
β’ ππΆπΈ = ππΆπΆ + π πΆ + π πΈ πΌπΆ =20 β 5.6k + 1k 1.743mA = 20 β6.6kΞ© 1.743mA = 20 β 11.503 =8.497V
Circuit revision
β’ Work the previous problem using hFE = 160 instead of 80.
Work for revision
β’ πΌπΆ =ππΆπΆ
π πΈ+π πΆ+π π΅βπΉπΈ
=20
1k+5.6k+390k
160
=
20
1k+5.6k+2.438k=
20V
9.038kΞ©= 2.213mA
β’ ππΆπΈ = ππΆπΆ + π πΆ + π πΈ πΌπΆ =20 β 5.6k + 1k 2.213mA = 20 β6.6kΞ© 2.213mA = 20 β 14.606 =5.394V
Common issues (CE Amp β 2 Supply biased)
12V
-12V
RB33kΞ©
RC4.7kΞ©
RE10kΞ©
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 β β β β
(2) VCC β β β β
(3) RB β β β β
(4) RB β β β β
(5) RB β β β β
(6) RC β β β β
(7) RC β β β β
(8) RC β β β β
(9) RE β β β β
(10) RE β β β β
(11) RE β β β β
(12) VEE β β β β
(13) VEE β β β β
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