transistor circuits v
TRANSCRIPT
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
โข ๐ผ๐ถ1 =๐๐ถ๐ถโ2๐๐ต๐ธ
๐ ๐ถ1+๐ ๐น๐ฝ+๐ ๐ธ1
โข ๐ผ๐ถ2 =๐๐ถ๐ถโ๐๐ต๐ธโ๐ ๐ถ1๐ผ๐ถ1
๐ ๐ธ2
โข ๐๐ถ๐ธ1 = ๐๐ถ๐ถ โ ๐ ๐ถ1 + ๐ ๐ธ1 ๐ผ๐ถ1
โข ๐๐ถ๐ธ2 = ๐๐ถ๐ถ โ ๐ ๐ถ2 + ๐ ๐ธ2 ๐ผ๐ถ2
โข ๐๐ถ1 = ๐๐ถ๐ถ โ ๐ ๐ถ1๐ผ๐ถ1
โข ๐๐ถ2 = ๐๐ถ๐ถ โ ๐ ๐ถ2๐ผ๐ถ2
First example
โข 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
68kฮฉ
330ฮฉ
2.7kฮฉ
1kฮฉ
470kฮฉ
Work for first example
โข ๐ผ๐ถ1 =๐๐ถ๐ถโ2๐๐ต๐ธ
๐ ๐ถ1+๐ ๐น๐ฝ+๐ ๐ธ1
=12โ2 0.7
68k+470k
50+330
=12โ1.4
68k+9.4k+330=
10.6V
77.73kฮฉ= 136.37ฮผA
โข ๐๐ถ๐ธ1 = ๐๐ถ๐ถ โ ๐ ๐ถ1 + ๐ ๐ธ1 ๐ผ๐ถ1 =12 โ 68k + 330 136.37ฮผA =12 โ 68.33kฮฉ 136.37ฮผA = 12 โ 9.318 = 2.682V
โข ๐๐ถ1 = ๐๐ถ๐ถ โ ๐ ๐ถ1๐ผ๐ถ1 = 12 โ 68kฮฉ 136.37ฮผA =12 โ 9.273 = 2.727V
Work for first example (cont.)
โข ๐ผ๐ถ2 =๐๐ถ๐ถโ๐๐ต๐ธโ๐ ๐ถ1๐ผ๐ถ1
๐ ๐ธ2=
12โ0.7โ 68k 136.37ฮผA
1k=
12โ0.7โ9.273
1k=
2.027V
1kฮฉ= 2.027mA
โข ๐๐ถ๐ธ2 = ๐๐ถ๐ถ โ ๐ ๐ถ2 + ๐ ๐ธ2 ๐ผ๐ถ2 =12 โ 2.7k + 1k 2.027mA =12 โ 3.7kฮฉ 2.027mA = 12 โ 7.499 = 4.501V
โข ๐๐ถ2 = ๐๐ถ๐ถ โ ๐ ๐ถ2๐ผ๐ถ2 = 12 โ 2.7kฮฉ 2.027mA =12 โ 5.473 = 6.527V
Second example
โข Rework the previous problem using a 680-kฮฉ resistor in place of the 470-kฮฉ resistor.
Work for second example
โข ๐ผ๐ถ1 =๐๐ถ๐ถโ2๐๐ต๐ธ
๐ ๐ถ1+๐ ๐น๐ฝ+๐ ๐ธ1
=12โ2 0.7
68k+680k
50+330
=12โ1.4
68k+13.6k+330=
10.6V
81.93kฮฉ= 129.379ฮผA
โข ๐๐ถ๐ธ1 = ๐๐ถ๐ถ โ ๐ ๐ถ1 + ๐ ๐ธ1 ๐ผ๐ถ1 =12 โ 68k + 330 129.379ฮผA =12 โ 68.33kฮฉ 129.379ฮผA = 12 โ 8.84 = 3.16V
โข ๐๐ถ1 = ๐๐ถ๐ถ โ ๐ ๐ถ1๐ผ๐ถ1 = 12 โ 68kฮฉ 129.379ฮผA =12 โ 8.798 = 3.202V
Work for second example (cont.)
โข ๐ผ๐ถ2 =๐๐ถ๐ถโ๐๐ต๐ธโ๐ ๐ถ1๐ผ๐ถ1
๐ ๐ธ2=
12โ0.7โ 68k 129.379ฮผA
1k=
12โ0.7โ8.798
1k=
2.502V
1kฮฉ= 2.502mA
โข ๐๐ถ๐ธ2 = ๐๐ถ๐ถ โ ๐ ๐ถ2 + ๐ ๐ธ2 ๐ผ๐ถ2 =12 โ 2.7k + 1k 2.502mA =12 โ 3.7kฮฉ 2.502mA = 12 โ 9.528 = 2.472V
โข ๐๐ถ2 = ๐๐ถ๐ถ โ ๐ ๐ถ2๐ผ๐ถ2 = 12 โ 2.7kฮฉ 2.502mA =12 โ 6.756 = 5.224V
TROUBLESHOOTING CE CIRCUITS
First example
โข If the 180-kฮฉ (R2) resistor became open in the circuit shown, what would the collector-to-ground voltage be?
12V
12V
33kฮฉ 180kฮฉ
2.2kฮฉ
10kฮฉ
10.7V
Normal operation
โข Normal operation (Q point):
โข ๐๐ต = ๐๐ 2 =๐ 2
๐ 1+๐ 2๐๐ถ๐ถ =
180k
33k+180k12 =
180k
213k12 = 0.845 12V = 10.14V
โข ๐๐ต๐ธ = ๐๐ต โ ๐๐ธ = 10.17 โ 10.7 = โ0.56V
โข ๐ผ๐ถ = ๐ผ๐ธ =๐๐ถ๐ถโ๐๐ธ
๐ ๐ธ=
12โ10.7
2.2k=
1.3V
2.2kฮฉ= 590.91ฮผA
โข ๐๐ถ = ๐ผ๐ถ๐ ๐ถ = 590.91ฮผA 10kฮฉ = 5.909V
Circuit evaluation
โข If R2 opens, VB โ12V โด VBE is reverse biased.
โข If VBE is reverse biased, transistor is cutoff (IC = 0mA).This means VC = 0V as VE = VCC = VCE.
Second example
โข If the 33-kฮฉ (R1) resistor became open instead in the figure shown, what would the collector-to-ground voltage be?
12V
12V
33kฮฉ 180kฮฉ
2.2kฮฉ
10kฮฉ
10.7V
Circuit evaluation
โข If R1 opens, VB = 0V โด transistor is biased full on (saturation)
โข ๐ผ๐ถ = ๐ผ๐ธ =๐๐ถ๐ถ
๐ ๐ถ+๐ ๐ธ=
12
2.2k+10k=
12
12.2kฮฉ=
983.607ฮผA
โข ๐๐ถ = ๐ผ๐ถ๐ ๐ถ = 983.607ฮผA 10kฮฉ = 9.836V
Third example
โข Referring to the figure shown, if the BJTโs 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
๐ผ๐ถ =๐๐ถ๐ถ
๐ ๐ธ + ๐ ๐ถ +๐ ๐ตโ๐น๐ธ
If VBE โ 0V and ICEO โ 0A
Work for third example
โข ๐ผ๐ถ =๐๐ถ๐ถ
๐ ๐ธ+๐ ๐ถ+๐ ๐ตโ๐น๐ธ
=20
1k+5.6k+390k
80
=
20
1k+5.6k+4.875k=
20V
11.475kฮฉ= 1.743mA
โข ๐๐ถ๐ธ = ๐๐ถ๐ถ + ๐ ๐ถ + ๐ ๐ธ ๐ผ๐ถ =20 โ 5.6k + 1k 1.743mA = 20 โ6.6kฮฉ 1.743mA = 20 โ 11.503 =8.497V
Circuit revision
โข Work the previous problem using hFE = 160 instead of 80.
Work for revision
โข ๐ผ๐ถ =๐๐ถ๐ถ
๐ ๐ธ+๐ ๐ถ+๐ ๐ตโ๐น๐ธ
=20
1k+5.6k+390k
160
=
20
1k+5.6k+2.438k=
20V
9.038kฮฉ= 2.213mA
โข ๐๐ถ๐ธ = ๐๐ถ๐ถ + ๐ ๐ถ + ๐ ๐ธ ๐ผ๐ถ =20 โ 5.6k + 1k 2.213mA = 20 โ6.6kฮฉ 2.213mA = 20 โ 14.606 =5.394V
Common issues (CE Amp โ 2 Supply biased)
12V
-12V
RB33kฮฉ
RC4.7kฮฉ
RE10kฮฉ
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 โ โ โ โ
(2) VCC โ โ โ โ
(3) RB โ โ โ โ
(4) RB โ โ โ โ
(5) RB โ โ โ โ
(6) RC โ โ โ โ
(7) RC โ โ โ โ
(8) RC โ โ โ โ
(9) RE โ โ โ โ
(10) RE โ โ โ โ
(11) RE โ โ โ โ
(12) VEE โ โ โ โ
(13) VEE โ โ โ โ
Any questions?
โข Contact us at:
โ 1-800-243-6446
โ 1-216-781-9400
โข Email: