Oscillators Open Loop Gain (Forward Complex Gain) Attenuation (Feedback Complex Gain)

1. BSECEIV ECE413F PCOMM
PRELIM ASSIGNMENT #1
1. Referto the circuit of example 1:
a. Determine the total feedback gain and the total phase shift of the signal after the second RC section if
the signal frequencyis equal to the frequencyof oscillation.
If f = 𝒇 𝒐 then 𝑋𝑐 = (6)(𝑅2
)
( 𝑋𝑐)2
= (6) ∗ 𝑅2
𝑋𝑐 = 𝑅 ∗ √6
𝑉1
𝑉2
=
( 𝑅) ∗ ( 𝑅 − 𝑗𝑋𝑐)
𝑅2 − ( 𝑋𝑐)2 − 𝑗3𝑅𝑋𝑐
𝑉1
𝑉2
=
(𝑅) ∗ 𝑅 − 𝑗𝑅√6
𝑅2(𝑅√6)
2
− 𝑗(3)(𝑅)(𝑅√6)
𝑉1
𝑉2
=
(1− 𝑗√6)( 𝑅)2
( 𝑅)2(1− 6 − 𝑗3√6
|
𝑉1
𝑉2
| = 𝟎. 𝟐𝟗𝟕𝟔𝟕∠𝟓𝟔. 𝟒𝟑𝟗𝟔°
b. Determine the total feedback gain and the total phase shift of the signal after the second RC section if
the signal frequencyis equal to 1 kHz.
If f=1kHz
X 𝑐 =
1
(2)()(𝑓)(𝐶)
𝑋𝑐 =
1
(2)()(1000𝐻𝑧)(0.001 𝐹)
Xc = 159.155 k
𝑉1
𝑉2
=
𝑅( 𝑅 − 𝑗𝑋𝑐)
𝑅2 − ( 𝑋𝑐)2 − 𝑗3(𝑅)(𝑋𝑐)
𝑉1
𝑉2
=
(100𝑘)(100𝑘 − j159.155k)
(100𝑘)2 − (159.155𝑘)2 − 𝑗(3)(100𝑘)(159.155k)
|
𝑽 𝟏
𝑽 𝟐
| = 𝟎. 𝟑𝟕𝟒𝟖𝟐∠𝟒𝟗. 𝟗𝟒𝟐𝟓°
c. Determine the total feedback gain and the total phase shift of the signal after the third RC section if the
signal frequencyis equal to the frequencyof oscillation.
𝑓 = 𝑓𝑜
𝑋𝑐 = 𝑅√6

2. 𝑉2
𝑉𝑜
=
( 𝑅)[𝑅2
− ( 𝑋𝑐)2
− 𝑗(3)(𝑅)(𝑋𝑐)]
𝑅3 + ( 𝑗𝑋𝑐)3 − 5( 𝑅)( 𝑋𝑐)2 − 𝑗(6)(𝑅)2(𝑋𝑐)
𝑉2
𝑉𝑜
=
( 𝑅)[𝑅2
− 6( 𝑅)2
− 𝑗(3)( 𝑅)(𝑅√6)]
𝑅3 + 𝑗(𝑅√6)
3
− 5( 𝑅)(𝑅√6)
2
− 𝑗(6)( 𝑅)2(𝑅√6)
𝑉2
𝑉𝑜
=
𝑅3
− 6( 𝑅3) − 𝑗(3)( 𝑅)3
√6
𝑅3 + 𝑗( 𝑅)3(6√6) − 30( 𝑅)3 − 𝑗( 𝑅)3(6√6)
𝑉2
𝑉𝑜
=
( 𝑅)3
(1− 6 − 𝑗(3√6)
( 𝑅)3[1+ 𝑗(6√6)− 30 − 𝑗(6√6)]
|
𝑽 𝟐
𝑽 𝟏
| = 𝟎. 𝟑𝟎𝟔𝟒𝟗∠𝟓𝟓. 𝟕𝟔𝟖𝟏°
d. Determine the total impedance of the feedbackcircuit if the frequencyis 1 kHz.
𝑍𝑡𝑜𝑡 =
( 𝑅−𝑗𝑋𝑐)(𝑅−𝑗𝑋𝑐)
( 𝑅−𝑗𝑋𝑐)+(𝑅−𝑗𝑋𝑐)
// ( 𝑅 − 𝑗𝑋 𝑐 )
At 𝑿 𝒄 = 𝟏𝟓𝟗. 𝟏𝟓𝟓𝐤
𝑍𝑡𝑜𝑡 =
(100𝑘−𝑗(159.155𝑘)(100𝑘−𝑗(159.155𝑘)
((100𝑘−𝑗(159.155𝑘))+((100𝑘−𝑗(159.155𝑘))
//((100𝑘 − 𝑗(159.155𝑘))
| 𝒁𝒕𝒐𝒕| = 𝟐𝟎𝟐. 𝟏𝟓𝟖𝒌 ∠− 𝟔𝟖. 𝟖𝟖𝟎𝟏°
2. Derive the feedbackgain and the formula for the frequencyof oscillation of a circuit with a Common Source
Amplifier and 4 RC graded section feedbackcircuit.
=
𝑉 𝑓
𝑉 𝑜
= (
𝑅
𝑅−𝑗𝑋𝑐
)
4
= (
𝑅
𝑅−𝑗𝑋 𝑐
)(
𝑅
𝑅−𝑗𝑋 𝑐
)(
𝑅
𝑅−𝑗𝑋 𝑐
)(
𝑅
𝑅−𝑗𝑋 𝑐
)
=[
𝑅
√𝑅2+( 𝑋 𝑐)2
∠ tan−1
(
0
𝑅
) − tan−1
(
−𝑋 𝑐
𝑅
)]
4
But − 𝐭𝐚𝐧−𝟏
(
−𝑿 𝒄
𝑹
) = 𝐭𝐚𝐧−𝟏
(
𝑿 𝒄
𝑹
)
=[
𝑅
√ 𝑅2+( 𝑋 𝑐)2
]
4
∠4∗ tan−1
(
𝑋 𝑐
𝑅
)
4 ∗ tan−1
(
𝑋 𝑐
𝑅
) = 
𝑋 𝑐
𝑅
= tan (

4
)  𝑋𝑐 = 𝑅
1
(2)()(𝑓𝑜)(𝐶)
= R  𝒇 𝒐 =
𝟏
( 𝟐)( )(𝑹)(𝑪)

3.  =
𝑅4
(√𝑅2+( 𝑅)2)
4 =
𝑅4
(√(2)(𝑅)2)
4 =
𝑅4
((𝑅)√2)
4 =
𝑅4
( 𝑅4)(4)
=
1
4
= 0.25
3. If the Op-Amp of a Wien Bridge Oscillator is replaced with a Single-stage Common Emitter amplifier, will the
circuit oscillate? Why or why not?
In order for an oscillator circuit to maintain oscillation it must satisfy two important conditions of the
Barkhausen Criterion. Which states that the total phase shift must be a multiple of  or 360 and that
| 𝐴𝐵| must not be less than 1. Since the phase shift produced of a Wien Bridge Oscillator is 360°, the
amplifier used to replace the op-amp must also have a total phase shift of 360°. However, the phase
shift of a Common Emitter Transistor is only 180° which does not satisfy one of the 2 conditions of
the Barkhausen Criterion whichwillmean that the circuitwillnot oscillate.In order to use a
Common Emitter Transistor in the Wien Bridge oscillator setup, it must be cascadedwith another
common emitter transistor so that the total phase shift produced by the two transistors would be
180°+180°=360 ° and only then would the oscillator circuit oscillate.