For the circuit shown in Figure P5.61, the bias voltages are changed to V + = 3V and V − = − 3V . (a) Design a bias−stable circuit for β = 120 such that V C E Q = 2.8 V . Determine I C Q , R 1 , and R 2 . (b) If the resistors R 1 and R 2 vary by ±5 percent, determine the range in I C Q and V C E Q . Plot the various Q −points on the load line. Figure P5.61
For the circuit shown in Figure P5.61, the bias voltages are changed to V + = 3V and V − = − 3V . (a) Design a bias−stable circuit for β = 120 such that V C E Q = 2.8 V . Determine I C Q , R 1 , and R 2 . (b) If the resistors R 1 and R 2 vary by ±5 percent, determine the range in I C Q and V C E Q . Plot the various Q −points on the load line. Figure P5.61
For the circuit shown in Figure P5.61, the bias voltages are changed to
V
+
=
3V
and
V
−
=
−
3V
. (a) Design a bias−stable circuit for
β
=
120
such that
V
C
E
Q
=
2.8
V
. Determine
I
C
Q
,
R
1
, and
R
2
. (b) If the resistors
R
1
and
R
2
vary by ±5 percent, determine the range in
I
C
Q
and
V
C
E
Q
. Plot the various Q−points on the load line.
Figure P5.61
a.
Expert Solution
To determine
The design parameters of the circuit and the collector current at Q -point.
Answer to Problem D5.62P
ICQ=1.453 mA , R1=14.21 kΩ and R2=2.92 kΩ .
Explanation of Solution
Given Information:
β=120,VCEQ=2.8 V, V+=3 V and V−=−3 V
The given circuit is shown below.
Calculation:
First, redraw the circuit with a Thevenin equivalent circuit in the base. Then find the Thevenin equivalent voltage and resistance. Calculate the transistor currents and voltages. Then find the required resistor values using the equations for Thevenin voltage and Thevenin resistance. The below figure shows the circuit with the Thevenin equivalent circuit at the base of the transistor.
Calculation:
Applying Kirchhoff’s law around C-E loop
V+=VCEQ+IEQRE+ICQRC+V−V+=VCEQ+((1+ββ)RE+RC)ICQ+V−ICQ=V+−VCEQ−V−(1+ββ)RE+RC=3−2.8−(−3)(1+120120)0.2+2 mA
ICQ=1.453 mA
Thevenin equivalent resistance is
RTH=0.1(1+β)RE=0.1×121×0.2 kΩ= 2.42 kΩ
Applying Kirchhoff’s voltage law around the B-E loop,
VTH=VBE(on)+IEQRE+IBQRTH+V−VTH=VBE(on)+(1+ββ)ICQRE+(1β)ICQRTH+V−VTH=V−+VBE(on)+(1+ββ)ICQRE+(1β)IcQRTHVTH=−3+0.7+(1+120120)×1.453×0.2+(1120)×1.453×2.42VTH=−1.978 V
Thevenin resistance is,
RTH=(R1R2R1+R2)(R1R2R1+R2)=2.42→(1)
Thevenin voltage is,
VTH=(R2R1+R2)(V+−V−)+V−
Using equation (1), rewrite the above equation as,
The range of Q -point values, ICQ and VCEQ for the percent change of bias resistors.
To plot: Various Q -pints on the load line.
Answer to Problem D5.62P
The range of values is
1.0906 mA≤ICQ≤1.8452 mA and 1.9375 V≤VCEQ≤3.599 V
Th plot is shown below.
Explanation of Solution
Given Information:
β=120,V+=3 V and V−=−3 VChange of bias resistence is ±5%
The given transistor circuit is shown below.
Calculaion:
First, redraw the circuit with a Thevenin equivalent circuit in the base. Then find the Thevenin equivalent voltage and resistance. Calculate the transistor currents and voltages at Q -point.
Thevenin resistance is,
RTH=(R1R2R1+R2)→(1)
Thevenin voltage is,
VTH=(R2R1+R2)(V+−V−)+V−→(2)
Applying Kirchhoff’s voltage law around the B-E loop,
For full RC charging the constant "n" in the product: "nRC" is "5". Now, which should be the constant "n" in "nRC" such that the voltage in the capacitor is 50% of the supply voltage?
6. A simple ohmmeter can be constructed from an ammeter as shown in Figure P5.6. In
this design the measured resistance is given by the equation R 10/1 1 k2. If this
ohmmeter is used to measure the forward resistance of a signal diode, the measurement
will typically be about 100 2 when Ry is really closer to 1 2. Explain with a current versus
voltage diagram.
Ohmmeter
10 V +
I
1 ΚΩ
I
Figure P5.6
Unknown
A
B
Create a circuit design that will supply an output voltage of 2VDC, having a 6V DC source supply, using a resistor and a diodes. Current that is flowing on the circuit should be 20mA. It will be used as power source for a small project. (With solution)
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