Lecture 32
LC Oscillators
Objectives: At the end of this lecture you should be able to:
(a)
(b)
Understand and Design Colpitts Oscillator
Understand and Design a Hartley Oscillator
Op amp based RC oscillators are suitable for low frequency (less than 100KHz)
applications.
For higher frequency application, however, oscillators can be developed based on
transistors and LC tuned circuits or crystals.
The main disadvantage of LC oscillators is the difficulty to tune the frequency of
oscillation. You will see that crystal oscillators generate a single tune signal.
The most commonly used LC tuned oscillators are the Colpitts oscillator and Hartley
oscillator, after their inventors.
The basic circuits without biasing details of these oscillators are shown in Fig. 1. The
required feedback is achieved by capacitive divider in the Colpitts oscillator and by an
inductive divider in the Hartley oscillator.
Fig. 1: (a) Colpitts oscillator (b) Hartley oscillator
In both circuits, the losses of the inductors, the load resistance, and the output resistance
of the transistor are modeled by the resistor R.
In this lecture, we will concentrate on analyzing the Colpitts oscillator. Assuming the
frequency of oscillation will be low in such a way we can neglect the internal
capacitances of the transistor.
The small signal equivalent of the Colpitts oscillator is developed as shown in Fig. 2.
Fig. 2: Signal equivalent of the Colpitts oscillator
(Add rπ, change voltage variables to small)
* Applying nodal analysis, KCL at the collector of the transistor gives:
Change variables to small:
KCL at Vc:
⇒ vc sC1 +
vc
v −v
+ g m vb + c b = 0
R
sL
⇒ vc s 2C1 LR + vc sL + sLg m Rvb + Rvc − Rvb = 0
⇒ vc [ s 2C1 LR + sL + R] + vb [ sLg mb R − R ] = 0"" (1)
(1)
KCL at Vb:
⇒ vb sC2 +
vb vb − vc
+
=0
rπ
sL
⇒ vb s 2C2 Lrπ + vb sL + rπ vb − rπ vC = 0
⇒ vb =
rπ
vc "" (2)
s C2 Lrπ + sL + rπ
2
(2)
Substitute (2) in (1) yields:
⇒ s 2C1 LR + sL + R +
rπ
[ sLg m R − R] = 0
s C2 Lrπ + sL + rπ
2
⇒ [ s 2C1 LR + sL + R][ s 2C2 Lrπ + sL + rπ ] + rπ [ sLg m R − R ] = 0
⇒ s 4C1C2 L2 Rrπ + s 3C1 L2 R + s 2C1 LRrπ + s 3C2 L2 rπ + s 2 L2 + sLrπ + s 2C2 LRrπ + sLR + sLg m Rrπ = 0
The terms Rrπ will cancel. Substituting s=jωo for physical frequency of oscillation and
solving:
Real =0; ω O 4C1C2 L2 Rrπ − ω O 2C1 LRrπ − ω O 2 L2 − ω O 2C2 LRrπ = 0
⇒ ω O 2C1C2 L2 Rrπ = C1 LRrπ + L2 + C2 LRrπ
ω
2
O
=
C1 LRrπ + L2 + C2 LRrπ
⇒ ωO 2 =
C1C2 L2 Rrπ
L2
+ C2 LR
rπ
C1C2 L2 R
C1 LR +
With rπ → ∞ (If MOSFET is used instead of BJT)
ω
2
O
=
C1 LR + C2 LR C1 + C2
1
=
=
"" (3) (3)
2
C1C2 L R
C1C2 L (C1 & C2 ) L
Frequency of oscillation: = ω O =
1
(C1 & C2 ) L
Imaginary=0; − jωo3C1 L2 R − jωo 3C2 L2 rπ + jωo Lrπ + jωo LR + jωo Lg m Rrπ = 0
Dividing by jωo L :
−ωo 2C1 LR − ωo 2C2 Lrπ + rπ + R + g m Rrπ = 0
Dividing by rπ :
−ωo 2C1 LR
R
− ωo 2C2 L + 1 + + g m R = 0
rπ
rπ
As rπ → ∞
⇒
⇒ −ωo 2C2 L + 1 + g m R = 0"" (4)
Substitute (3) in (4):
⇒−
C1 + C2
C2 L + 1 + g m R = 0
C1C2 L
⇒ −1 −
C2
+ 1 + gm R = 0
C1
(4)
To ensure that oscillation will start:
⇒ gm R ≥
C2
Condition of oscillation
C1
For Hartley oscillator, following same procedure you will get the frequency of
oscillation to be:
⇒ ωO =
1
( L1 + L2 )C
A complete Colpitts oscillator circuit is shown in Fig. 3. The radio-frequency choke
(RFC) is used to provide a high reactance (ideally open circuit) at frequency of
oscillation.
Fig. 3: Colpitts oscillator with the biasing circuit
Colpitts oscillator and Hartley oscillator can be built based on MOSFET as shown in Fig.
4 and Fig. 5, respectively.
Fig. 4: MOSFET based Colpitts oscillator: (a) Basic circuit (b) Small signal model
Fig. 5: MOSFET based Hartley oscillator: (a) Basic circuit (b) Small signal model
neglecting internal capacitors
Example 1: Analyze the MOSFET based Hartley oscillator shown in Fig. 5(b) to find the
frequency and condition of oscillation. KCL at Vs:
⇒
vs vs
vs
+
+
− g m vgs = 0
ro sL1 sL2 + 1/ sC
But vgs=vg-vs
⇒
vs vs
v sC
+
+ 2 s
− g m vg + g m vs = 0
ro sL1 s CL2 + 1
Multiply by ( s 2CL2 + 1) sL1ro gives
⇒ vs [ sL1 ( s 2CL2 + 1) + ( s 2CL2 + 1)ro + s 2CL1ro
+ g m sL1 ( s 2CL2 + 1)ro ] − g m vg sL1 ( s 2CL2 + 1)ro = 0"" (1)
By VDR:
⇒ vg =
vs
1/ sC
vs =
"" (2)
1/ sC + sL2
1 + s 2CL2
Substituting (2) in (1) yields:
⇒ vs [ s 3CL1 L2 + sL1 + s 2CL2 ro + ro + s 2CL1ro
+ g m ro s 3CL1 L2 + g m ro sL1 − g m ro sL1 ] = 0
Since vs ≠ 0 , substituting s=jωo for physical frequency of oscillation and solving:
Real =0; −ω O 2CL2 ro + ro − ω O 2CL1ro = 0
⇒ ωO =
1
"" (3)
( L1 + L2 )C
(i.e. Frequency of oscillation)
Imaginary=0; − jωo 3CL1 L2 − jωo 3CL1 L2 g m ro + jωo L1 = 0
Dividing by jωo L :
⇒ −ωo 2CL2 − ωo 2CL2 g m ro + 1 = 0
⇒ ωo 2CL2 (1 + g m ro ) = 1"" (4) (4)
Substitute (3) in (4) yields
⇒ CL2 + g mCL2 ro = CL1 + CL2
⇒ g m ro =
L1
L2
To ensure that oscillation will start:
⇒ g m ro ≥
L1
(i.e. condition of oscillation)
L2
Example 2: This example demonstrates that if you are allowed to use inductors then
designing op amp based oscillators becomes a simple task. Inductors are not
recommended in electronics circuit for several reasons: their size, difficult to
implement in integrated circuits (ICs), and have low quality factors.
Consider, for example, the RLC network shown in Fig. 6. Develop a
sinusoidal oscillator around this circuit using a single op amp and any
number of resistors. Draw a complete circuit of your design. Select your
components to realize frequency of oscillation of 10kHz.
Fig. 6: RLC circuit of Example 2
The circuit of Fig. 6 is a bandpass filter at resonance frequency the capacitive and
inductive reactance will cancel and the network becomes pure resistive. Hence at this
O/P
R
1
=
= . If this loss is compensated by gain of 2 and the loop is
frequency
I /P R+R 2
closed as shown in Fig. 7, a sinusoidal oscillator will be developed with frequency of
1
.
oscillation equal to the resonance frequency ωo =
LC
Fig. 7: RLC op amp based oscillator
For frequency of oscillation of 10KHz, select C=1nF and find L.
ωo =
1
⇒ 2π 104 =
LC
1
L(10−9 )
⇒ L = 0.25H 0.25H. This is an impractical value.
Again select C=1μF, results in L=0.25mH which is accepted.