Sinusoidal Oscillators
• Signal generators: sinusoidal, rectangular, triangular, TLV, etc.
• Obtaining a sine wave:
triangle functional transf.
sine
sine wave generation: frequency selective network in a
feedback loop of a PF amplifier: sinusoidal oscillator
-Oscillation frequency: f0
-Oscillation amplitude: Vˆo
- Oscillation criterion
- Frequency stability
- Amplitude stability
- Distortion coefficient
1/14
Oscillator
feedback loop
PF amplifier:
xi = xs + xr
a
A=
1 − ar
xs = 0; x0 − finit
a ( jω )
A( jω ) =
1 − a ( j ω ) r ( jω )
1 − a ( jω 0 ) r ( jω 0 ) = 0
2/14
Oscillation criterion
Barkhausen’s criterion
a ( jω 0 ) r ( jω 0 ) = 1
Signal reconstruction on
the feedback loop
a ( j ω ) = a ( j ω ) e jϕ a
r ( j ω ) = r ( j ω ) e jϕ r
a ( jω0 ) r ( jω0 ) = a ( jω0 ) r ( jω0 ) e j (ϕ a +ϕ r ) = 1
module condition: a( jω0 ) r ( jω0 ) = 1
phase condition:
Who sets Vˆo ?
ϕ a + ϕ r = 2 kπ
gives a0
gives f0
Nonlinearity of the gain
3/14
a ( j ω ) r ( jω ) < 1
oscillations are attenuated - zero
a ( j ω ) r ( jω ) > 1
oscillations are amplified - saturation
Stability of the oscillation amplitude
Automatic gain control
vo = Vˆo sin 2πf 0t
r ( jω ) = cst
Vˆo ↑, a( jω0 ) ↓, Vˆ0 ↓
4/14
RC Oscillators
Basic amplifier independent of frequency:
o
• inverting ϕ r = −180
• noninverting ϕ r = 0
Frequency selective network
BPF with one zero and
two poles
Phase shift of the network
lies in the range of
[+90o; -90o].
We want ϕ r = 0 just for a
single frequency f0
It is necessary to bring
closer the two poles.
5/14
WIEN Bridge
For only one
single frequency,
f0 we have
ϕr = 0
v r ( jω )
F ( jω ) =
vo ( j ω )
output
input
6/14
WIEN Bridge–cont.
v r ( jω )
=
r ( jω ) =
vo ( jω )
ϕr = 0
1
ω0 Rs Cs −
=0
ω0 R p C p
1
f0 =
2π Rs R p C s C p
Rs = R p = R
Cs = C p = C
1

Rs C p
1

+
+ j ω Rs C s −
1+

Rp Cs
ωR pC p

r ( jω 0 ) =
1
f0 =
2πRC




1
Rs C p
1+
+
R p Cs
1
r ( jω0 ) =
3
7/14
7/14
Op amp and WIEN bridge oscillator
vo (t ) = Vˆo sin 2πf 0t
Rs = R p = R
Cs = C p = C
1
a ( jω 0 ) =
=3
r ( jω 0 )
R4
a =1+
R3
Vˆo = ?
R4
1+
=3
R3
1
f0 =
2πRC
1
r ( jω0 ) =
3
R4 = 2R3
Nonlinearity on the gain, close to saturation
8/14
Automatic control of the amplitude
1. Using diodes
for vo (t ) small, D1, D2 – (off)
R4' + R4''
a =1+
R3
a ( jω 0 ) r ( jω 0 ) > 1
vo (t ) increases,
D1 – (on) on the
positive half-cycle
D2 – (on) on the
negative half-cycle
R4' + R4'' || rd
a =1+
R3
a ( jω 0 ) r ( jω 0 ) = 1
to maintain oscillations
Vˆo is given by the value of rd
9/14
2. Using n-channel
depletion-type MOSFET
R4
a = 1+ '
R3 + rDS
rDS ≈
1
2β (vGS − VTh )
vGS<0
| vGS |↑, rDS ↓
10/14
Op amp and RC ladder network oscillator
• High pass band
• Low pass band
• the phase-shift is in the
range of [0o; -90o]
• inverting basic amplifier
• how many identical RC cells
are necessary to build an
oscillator?
11/14
low pass RC
ladder with
3 cells
r ( jω ) =
1
2
3
1 − 5(ωRC ) + j 6ωRC − (ωRC )
[
]
ϕr = 0
6ω0 RC − (ω0 RC ) = 0
3
6
f0 =
2πRC
1
r ( jω 0 ) = −
29
12/14
The circuit of RC ladder network oscillator
Why the basic amplifier does not contain
only one inverting op-amp amplifier?
13/14
Illustration
How does the voltages vo(t) and v+(t)
look like in the steady-state
regime? What is the oscillation
frequency?
Size R4 so that the circuit will
maintain the oscillation. In
conduction assume the
equivalent diode resistance
rD1=rD2=0,5 KΩ. Verify if the
oscillation can start.
How does the voltage vo(t) look like
in the steady-state regime if D2
diode is mising in the circuit?
14/14