Given a Collector -Feedback Biased transistor circuit with voltage at common collector is +10v ,base resistor is 100k ohms, Collector resistor is 10k ohms and Base current is 8.38 micro ampere, ,Voltage at Base-emitter junction  is 0.7v. Determine Beta DC , Collector current and Voltage at collector-emitter junction. These might help as a guide to answer the problem...

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2.) Given a Collector -Feedback Biased transistor circuit with voltage at common collector is +10v ,base resistor is 100k ohms, Collector resistor is 10k ohms and Base current is 8.38 micro ampere, ,Voltage at Base-emitter junction  is 0.7v. Determine Beta DC , Collector current and Voltage at collector-emitter junction.

These might help as a guide to answer the problem...

 

Collector Feedback
This self-biasing collector feedback
configuration
dependent biasing method
requires two resistors to provide the
necessary DC bias for the transistor.
The
Vcc
is
another
beta
which
Re
Vcc – VBe
Rf + B(Rc + Re)
Ic = BlB
VCE = Vcc - Ic(Rc + RE)
IB
Ry
o Vo
le
collector
to
base
feedback
Bpc
configuration
transistor is always biased in the active
region regardless of the value of Beta
(B). The DC base bias voltage is derived
from the collector voltage Vc, thus
providing good stability.
ensures
that
the
VIN VN
RE
GND
Transcribed Image Text:Collector Feedback This self-biasing collector feedback configuration dependent biasing method requires two resistors to provide the necessary DC bias for the transistor. The Vcc is another beta which Re Vcc – VBe Rf + B(Rc + Re) Ic = BlB VCE = Vcc - Ic(Rc + RE) IB Ry o Vo le collector to base feedback Bpc configuration transistor is always biased in the active region regardless of the value of Beta (B). The DC base bias voltage is derived from the collector voltage Vc, thus providing good stability. ensures that the VIN VN RE GND
Determine IR, Ic and VCE in
the following circuit.
Step 1: Solve for Ig.
Step 2: Solve for Ic.
Vcc – VBE
Ic = BIB
Ic = 90(11.91µA)
Ic = 1.07MA
+10v
IR =
RF + B(Rc + RE)
10v – 0.7v
4.7kN
IB =
250kN + 90(4.7kN + 1.2kN)
IB = 11.91HA
250kN
o Vo
B = 90
Step 3: Solve for VCE-
VIN
VCE = Vcc - Ic(Rc + Rg)
VCE = 10v – 1.07mA(4.7kN + 1.2kN)
Vce = 10v – 6.31v
Vce = 3.69v
1.2kn
GND
Transcribed Image Text:Determine IR, Ic and VCE in the following circuit. Step 1: Solve for Ig. Step 2: Solve for Ic. Vcc – VBE Ic = BIB Ic = 90(11.91µA) Ic = 1.07MA +10v IR = RF + B(Rc + RE) 10v – 0.7v 4.7kN IB = 250kN + 90(4.7kN + 1.2kN) IB = 11.91HA 250kN o Vo B = 90 Step 3: Solve for VCE- VIN VCE = Vcc - Ic(Rc + Rg) VCE = 10v – 1.07mA(4.7kN + 1.2kN) Vce = 10v – 6.31v Vce = 3.69v 1.2kn GND
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