RC Phase-Shift Oscillators

Dr.Sanjeev Mani Yadav

SRM University AP

Oscillators generate periodic signals in the time domain. They convert DC power into
AC signal power. Signal generation implies production of self-sustained oscillations.
According to the types of waveforms produced oscillators can be classified into one of
four generic types:

-Harmonic oscillators: used for sine-wave generation.

-Sawtooth oscillators: used for the generation of exponential or linear sawtooth waves.
-Relaxation oscillators: used for current or voltage pulse generation with negative
resistance devices.
-Astable multivibrators: used for the generation of rectangular or square waves.

Each type of oscillator resorts to either a positive-feedback or negative-resistance

principle to operate. On the other hand, very complex waveforms can be generated
mixing analog and digital techniques.

The Positive Feedback Approach for a Harmonic Oscillator

As a system using positive feedback, the harmonic oscillator can be represented in
block-diagram form as depicted by Fig.1, where G(s) is the Laplace transform of the
open-loop voltage-gain function of the amplifier stage and H(s) is the transfer function
of the passive feedback network.

Fig.1 An oscillator as a feedback system

The closed-loop gain function of the system is found to be:

V0
(s ) = G (s )
V1 1 − G (s )H (s )

Mathematically, it represents the response of the system to an impulse function, i.e., a

function for which V1 (s ) = 1 . In the real-world, impulse functions are approximated by
narrow noise impulses.

-1-
The expression:
1 − G (s )H (s ) = 0 … (1)

is known as the characteristic equation and contains important information regarding

the system’s stability. Generally speaking, the roots of the equation are of the type
s = α ± jω . When α = 0 the system is marginally stable and s = ± jω . If certain
conditions are met, this situation will describe an oscillator in the steady state delivering
a constant-amplitude sine wave of radian frequency ω. A real-world oscillator requires
that α > 0 for oscillations to start and build up. The power supply start-up impulse or
circuit thermal noise will generally trigger the response. Then, some type of amplitude
stabilization mechanism in the system will gradually reduce α and stabilize the
oscillations. An amplitude-stabilized oscillator will usually exhibit good frequency
stability too. One final note here is that once the circuit bursts into oscillation no
external perturbation v1 (t ) is needed for sustained operation.

Equation (1) tells us that after amplitude stabilization:

G ( jω ) ⋅ H ( jω ) = 1 … (2)

This is Barkhausen’s criterion for oscillations. The left hand member is a complex
number and consists of a real and an imaginary part. Thus, we may write the equation:

Re(ω ) + j Im(ω ) = 1 + j 0
Re(ω ) = 1
Im(ω ) = 0

Im(ω) equated to zero will generally give the frequency of oscillation. Re(ω) equated to
unity will yield the conditions to be met for oscillations.

If we open the loop at the input of block G(s) in Fig.1 and inject a probe signal VT(s) at
this point, the output from block H(s) would be:

VL (s ) = VT (s ) ⋅ G (s )H (s )

We now define the gain function:

V L (s )
AL (s ) = = G (s )H (s )
VT (s )

This is the loop gain of our system. Barkhausen’s criterion then states that the
oscillator’s loop gain must be unity for oscillations to develop in the circuit.

-2-
Some Phase-Shift Oscillators
The most simple RC phase-shift oscillator configuration uses three buffered RC cells
and a voltage amplifier with very high input impedance and very low output impedance.
Fig.2 shows a typical schematic.

Because the RC cells won’t load each other, the loop gain may be found to be:

3
 sRC 
AL (s ) = G (s )H (s ) = A ⋅  
 sRC + 1 
 s 3 R 3C 3 

= A⋅ 3 3 3 2 2 2

 s R C + 3 s R C + 3 sRC + 1 

Fig.2 Simple RC Phase-Shift Oscillator

For sine-wave steady-state operation, s = jω , and the following may be written:

− jω 3 R 3 C 3 1
= … (3)
( 2 2 2
) (
2 2 2
1 − 3ω R C + jωRC 3 − ω R C A )
Equating the real part of the denominator to zero will yield the frequency of oscillation:

1
1 − 3ω 2 R 2 C 2 = 0 ⇒ ω 0 =
3RC

This result fed back into expression (3) will render information on amplifier gain A.
Accordingly:
A = −8

The phase shift introduced by each RC cell can be obtained from its transfer function:

jωRC
F ( jω ) =
jωRC + 1

-3-
 j
3
F ( jω 0 ) =
 j 
 + 1
 3 
Then:
 1 
φ = 90º − tan −1   = 90º −30º = 60º
 3

The total phase shift introduced by the RC network at ω = ω 0 will be three times φ , or
180º.

The BJT RC Phase-Shift Oscillator

Alike the preceding configuration, the BJT RC Phase-Shift Oscillator is a popular
configuration for the generation of low-frequency sine waves, starting at a few Hertz
and up to about 100 kHz. A schematic diagram of a basic implementation can be seen in
Fig.3. If the RC cells were isolated from each other, the phase shift per cell would be
60º. However, not being it the case, we need to perform a detailed analysis considering
loading effects.

Fig.3 BJT-based RC Phase-Shift Oscillator

The bipolar transistor is a current amplifier. Therefore, it is convenient to express the

loop gain as a current ratio:
I
AL (s ) = b'
Ib
where I b' is a test signal current injected at the transistor’s base and I b is the
corresponding base current following loop excitation. Please see Fig.4.

-4-
 Fig.4 Equivalent circuit for loop-gain calculation

The node-potential set of equations in the Laplace domain can be written as:

VA
− h fe I b' = + (V A − V B )sC
RC
VB
0= + (V B − V A )sC + (V B − VC )sC
R
V
0 = C + (VC − V B )sC + (VC − V D )sC
R
V
0 = D + (V D − VC )sC
R

A small simplification leads to:

 1 
− h fe I b' = V A  + sC  − V B ⋅ sC ... (4.a )
 RC 
1 
0 = −V A ⋅ sC + V B  + 2 sC  − VC ⋅ sC ... (4.b )
R 
1 
0 = −V B ⋅ sC + VC  + 2 sC  − V D ⋅ sC ... (4.c )
R 
1 
0 = −VC ⋅ sC + V D  + sC  ... (4.d )
R 

On the other hand, usually, Rb >> hie . Then:

( )
V D = RI b ≈ R ' + hie I b ... (4.e )

From Eq.(4.d):
 1 
VC = V D  + 1
 sRC 

-5-
Substituting into Eq.(4.c):
 1  1 
0 = −V B ⋅ sC + V D  + 1 + 2 sC  − V D ⋅ sC
 sRC  R 
which simplifies to:
 1 3 
VB = VD  2 2 2 + + 1
s R C sRC 

Substituting for VB and VC in Eq.(4.b) we get:

 1 3  1   1 
0 = −V A sC + V D  2 2 2 + + 1 + 2 sC  − V D  + 1 sC
s R C sRC  R   sRC 

yielding:
 1 5 6 
V A = VD  3 3 3 + 2 2 2 + + 1
s R C s R C sRC 

Substituting for V A and VB in Eq.(4.a):

 1 5 6  1   1 3 
− h fe I b' = VD  3 3 3 + 2 2 2 + + 1 + sC  − VD  2 2 2 + + 1 sC
s R C s R C sRC  RC  s R C sRC 

After some simplification:

 1 5 6 1 1 4 3
− h fe I b' = VD  3 3 3
+ 2 2 2
+ + + 2 3 2 + 2 + 
 s R RC C s R RC C sRRC C RC s R C sR C R 

Bearing in mind Eq.(4.e), the last expression transforms to:

 1 5 6 R 1 4 
− h fe I b' = I b  3 2 3
+ 2 2
+ + + 2 2 2 + + 3  ... (5)
 s R RC C s RRC C sRC C RC s R C sRC 

If AL (s ) = 1 , then I b = I b' . For sine wave operation, s = jω and Eq.(5) is rewritten as:

1 6 4 5 1 R
− h fe = j 3 2 3
−j −j − 2 2
− 2 2 2 + +3 ... (6 )
ω R RC C ω RC C ωRC ω RRC C ω R C RC

The frequency of oscillation is obtained equating the imaginary part of the right-hand
member to zero:
1 6 4
0= 3 2 − −
ω 0 R RC C 3 ω 0 RC C ω 0 RC

-6-
Knowing that ω 0 can not be zero we may write:

1 6 4
0= 2 2 2
− −
ω 0 R RC C RC R

and solve for ω 0 :

1
ω0 =
 6 4
RC RC  + 
 RC R 
or in a more suitable form:
1
ω0 =
4 RC
RC 6 +
R

in radians per second. The condition for oscillation is obtained equating to − h fe the real
part of the right-hand member of Eq.(6) while making ω = ω 0 :

R 5 1
− h fe = 3 + − 2 2
− 2 2 2
RC ω 0 RRC C ω0 R C
R  6R   4R 
= 3+ − 5 + 4  −  6 + C 
RC  RC   R 
R R
= −23 − 29 −4 C
RC R

Then:
R R
h fe = 23 + 29 +4 C ... (7)
RC R

The minimum value of h fe required for oscillations is h fe min = 44.5 , and occurs when
R
= 0.37 . If h fe is less than the said value the circuit won’t oscillate, because AL ( jω )
RC
would be less than unity. We can write Eq.(7) in the alternate form:

2
R h fe − 23  h fe − 23  4
= +   −
RC 58  58  29

-7-
 R
Again, h fe must be greater than 44.5 for to be a real number. If h fe = 44.5 , then
RC
R R
= 0.37 . The design process would be then: Given h fe and ω 0 , find . Then
RC RC
compute the RC product and select convenient values for C and R. Design the DC bias
network for Class-A operation and symmetrical signal excursion. Amplitude distortion
at the output may be reduced by introducing negative feedback in the emitter branch
with a series added small resistor. Of course, a little more gain would be needed to
compensate for the reduction in the effective transconductance g m' of the circuit, i.e.:
gm
g m' =
1 + g m Re
where g m is the transconductance of the transistor and Re is the series added resistor.

The JFET RC Phase-Shift Oscillator

Another approach uses a discrete high input-impedance voltage amplifier stage. The
JFET is a high input-impedance semiconductor device and is very well suited for
voltage amplification. Fig.5 shows a schematic diagram of a typical JFET-based RC
Phase-Shift oscillator, yet another popular alternative for low-frequency sine wave
generation. As may be noted, resistor R of the far right RC cell substitutes for the gate-
source bias return resistor commonly used in biasing schemes.

A general approach for the passive phase-shift network used in this type of oscillator
can be seen in Fig.6, where Z1 and Z2 are a capacitor C and a resistor R. Either
impedance can be selected to be the capacitor. However, the preferred configuration is
that depicted by Fig.5. Given the JFET’s high input impedance, loading effects
occurring in the gate circuit are virtually eliminated. The amplifying device is biased for
Class-A operation and minimum signal distortion at the output. If needed, negative
feedback may be used for wave shape correction, as in the bipolar transistor case.

Fig.5 JFET-based RC Phase-Shift Oscillator

-8-
 Fig.6 The three-cell RC network general approach

Design work requires some knowledge of the JFET’s small-signal parameters. These
are defined by:
v
rds = ds when v gs = 0
id
id
gm = when v ds = 0
v gs
v ds
µ=− when i d = 0
v gs

where v ds , id , v gs are small-signal variations about a quiescent point (please see Fig.7
below).

Fig.7 Basic JFET amplifier and small-signal equations

Here, gm is the low-frequency forward transconductance, rds is the drain’s dynamic

output resistance and µ is the amplification factor. The latter equals the product g m rds .

Sine wave operation permits us to perform calculations in the frequency domain.

Accordingly, we can write the mesh equations for the network of Fig.6 in phasor form
as:
V1 = I 1 (Z 1 + Z 2 ) − I 2 Z 2
0 = − I 1 Z 2 + I 2 (Z 1 + 2 Z 2 ) − I 3 Z 2
0 = − I 2 Z 2 + I 3 (Z 1 + 2 Z 2 )

The output voltage is given by:

V2 = I 3 Z 2 ...(8)

-9-
By Cramer’s rule:
∆3
I3 =
∆

The determinant of the coefficient matrix is calculated as:

(Z1 + Z 2 ) − Z2 0
∆= − Z2 (Z 1 + 2 Z 2 ) − Z2
0 − Z2 (Z 1 + 2 Z 2 )
which reduces to:

(Z 1 + 2Z 2 ) − Z2
∆ = (Z 1 + Z 2 ) + Z [(− Z 2 )(Z 1 + 2Z 2 )]
− Z2 (Z 1 + 2 Z 2 ) 2
After performing the indicated algebraic operations we obtain:

3 2 2 3
∆ = Z 1 + 5Z 1 Z 2 + 6 Z 1 Z 2 + Z 2 ... (9)

The determinant ∆ 3 is defined by:

(Z1 + Z 2 ) − Z2 V1
∆3 = − Z2 (Z 1 + 2 Z 2 ) 0
0 − Z2 0
2
= V1 Z 2
Then:
2
Z2
I3 = 3 2 2 3
V1
Z 1 + 5Z 1 Z 2 + 6 Z 1 Z 2 + Z 2

Substituting into Eq.(8):

3
Z2
V2 = 3 2 2 3
V1
Z 1 + 5Z 1 Z 2 + 6 Z 1 Z 2 + Z 2

The RC network’s transfer function is then:

3
V2 Z2
= 3 ...(10)
V1 Z 1 + 5Z 1 Z 2 + 6Z 1 Z 2 2 + Z 2 3
2

The network’s input impedance is given by:

V1
Z IN = ...(11)
I1

-10-
Observing that:
∆1
I1 = ...(12)
∆
where:
V1 − Z2 0
∆1 = 0 (Z 1 + 2 Z 2 ) − Z2
0 − Z2 (Z 1 + 2 Z 2 )

[
= V1 (Z 1 + 2 Z 2 ) − Z 2
2 2
]
( 2
= V1 Z 1 + 4 Z 1 Z 2 + 3Z 2
2
) ...(13)

we get, from equations (9), (11), (12) and (13):

3 2 2 3
V1 ∆ Z 1 + 5Z 1 Z 2 + 6Z 1 Z 2 + Z 2
Z IN = = 2 2
...(14)
∆1 Z 1 + 4Z 1 Z 2 + 3Z 2

Next, we choose Z1 and Z 2 to be:

1
Z1 =
sC
Z2 = R

Substituting in equations (10) and (14) we arrive to:

V2 s 3 R 3C 3
= 3 3 3 ...(15)
V1 s R C + 6 s 2 R 2 C 2 + 5sRC + 1
and:
s 3 R 3C 3 + 6 s 2 R 2 C 2 + 5sRC + 1
Z IN =
3s 3 R 2 C 3 + 4 s 2 RC 2 + sC
s 3 R 3C 3 + 6 s 2 R 2 C 2 + 5sRC + 1
=R ...(16)
3s 3 R 3 C 3 + 4 s 2 R 2 C 2 + sRC

From Eq.(15), and for s = jω :

V2 jω 3 R 3 C 3
H ( jω ) = ( jω ) = −
V1 ( )
1 − 6ω 2 R 2 C 2 + jωRC 5 − ω 2 R 2 C 2 ( )

-11-
From Eq.(16), and for s = jω :

Z IN = R
(1 − 6ω 2
) (
R 2 C 2 + jωRC 5 − ω 2 R 2 C 2 ) ...(17)
(
− 4ω 2 R 2 C 2 + jωRC 1 − 3ω 2 R 2 C 2 )
With regards to Fig.5, the expression for the voltage gain of the amplifying device is:

AV ( jω ) = − g m (R0 // Z IN )

where R0 = rds // R D , and Z IN is the RC network’s input impedance. Barkhausen’s

criterion for the oscillator’s loop gain, Eq.(2), states that:

AV ( jω ) ⋅ H ( jω ) = − g m (R0 // Z IN ) ⋅ H ( jω ) = 1 + j 0
V2
where H ( jω ) = ( jω ) is the passive network’s transfer function already calculated.
V1
If we assume that the RC network doesn’t load the JFET’s output, a situation that can be
met with an adequate selection of circuit values, Barkhausen’s criterion will read as:

AV ( jω ) ⋅ H ( jω ) = − g m R0 ⋅ H ( jω ) = 1 + j 0

Substituting for H ( jω ) in the loop-gain condition yields:

jω 3 R 3 C 3
g m R0 ⋅ = 1 + j0
( ) (
1 − 6ω 2 R 2 C 2 + jωRC 5 − ω 2 R 2 C 2 )
which is satisfied when:
1 − 6ω 2 R 2 C 2 = 0

This sets the radian frequency of oscillation at:

1
ω0 =
6 RC
The condition for oscillation is:
ω0 2 R 2C 2
g m R0 ⋅ 2
=1
5 − ω0 R 2C 2
which shows that the minimum value required of voltage gain is g m R0 = 29 . At the
frequency of oscillation the input impedance of the RC network is given by Eq.(17) as:

j 29
Z IN = R ohms
− 4 6 + j3
= (0.83 − j 2.70 )R ohms
= 2.82 R exp(− j 72.9º ) ohms

-12-
An interesting result is noticed here, and is that Z IN is independent of the frequency of
oscillation, depending only upon R.

In order to avoid loading sensibly the JFET’s output, the following should be satisfied
as a rule of thumb:
2.82 R >> R0 ⇒ R >> 0.355 R0 ⇒ R > 3.55 R0

We would now like to consider the effect of the RC network loading the JFET’s output.
Observing Fig.4 we can write the following circuit analogies:

V gs' ⇔ Vbe'
V gs ⇔ Vbe = I b R
for the case in which R ' = 0 . Then:

V gs Vbe I R
'
⇔ '
⇔ b '
g m JFET V gs g m BJT Vbe h fe I b

Barkhausen’s criterion stablishes that:

V gs Ib
'
⇔ =1
V gs I b'
We may write then:
1 R
⇔
g m JFET h fe
and:
h fe ⇔ g m JFET R ...(18)

We already arrived, in the BJT case to:

29 R0
h fe > 23 + 4k + with k =
k R
Using analogy (18):
29
g m JFET R > 23 + 4k +
k
R0 29
g m JFET ⋅ > 23 + 4k +
k k
2
g m JFET R0 > 23k + 4k + 29

Finally, the minimum voltage gain required from the JFET stage considering loading
effects is:
2
R  R 
Av = g m R0 = 4 0  + 23 0  + 29
 R  R

-13-
where g m , R0 and R are quantities previously defined during the analysis process of the
JFET oscillator.

The Five-Cell RC Phase-Shift Oscillator

The five-cell RC phase-shift oscillator is an extension of the three-cell case of the
preceding section, and will also require a high input-impedance voltage amplifier for
ease of analysis and design. A medium input-impedance amplifier could also be used,
but Z2 would need to be considered part of the amplifier’s input circuitry, requiring
further analysis prior to design work. There exists also some concern with regards to the
loading effect of the RC network on the amplifier’s output. This issue will be given
detailed consideration in our study. As will be demonstrated, the use of five RC cells
reduces considerably the gain threshold for oscillation, as compared to the three-cell
circuit. The amplifying device will be considered to be a vacuum triode. Fig.8 shows a
five-cell RC network as used in a phase-shift oscillator.

Fig.8 Five-cell RC network

The mesh equations for the passive network with input V1 may be written in phasor
form as:

V1 = I 1 (Z 1 + Z 2 ) − I 2 Z 2
0 = − I 1 Z 2 + I 2 (Z 1 + 2 Z 2 ) − I 3 Z 2
0 = − I 2 Z 2 + I 3 (Z 1 + 2 Z 2 ) − I 4 Z 2
0 = − I 3 Z 2 + I 4 (Z 1 + 2 Z 2 ) − I 5 Z 2
0 = − I 4 Z 2 + I 5 (Z 1 + 2 Z 2 )

and V2 = I 5 Z 2 . The system of five equations with five unknowns will be solved using,
again, Cramer’s rule. The determinant of the coefficient matrix is:

(Z 1 + Z 2 ) − Z2 0 0 0
− Z2 (Z 1 + 2Z 2 ) − Z2 0 0
∆= 0 − Z2 (Z 1 + 2 Z 2 ) − Z2 0
0 0 − Z2 (Z 1 + 2 Z 2 ) − Z2
0 0 0 − Z2 (Z 1 + 2 Z 2 )

-14-
 which can be reduced to:

(Z1 + 2Z 2 ) − Z2 0 0
(Z 1 + 2 Z 2 ) − Z2 0
− Z2 (Z 1 + 2 Z 2 ) − Z2 0 2
∆ = (Z 1 + Z 2 ) − Z2 − Z2 (Z 1 + 2 Z 2 ) − Z2
0 − Z2 (Z 1 + 2 Z 2 ) − Z2
0 − Z2 (Z 1 + 2 Z 2 )
0 0 − Z2 (Z 1 + 2 Z 2 )
After calculating the determinants and performing the remaining algebraic operations
we arrive to:
5 4 3 2 2 3 4 5
∆ = Z 1 + 9 Z 1 Z 2 + 28Z 1 Z 2 + 35Z 1 Z 2 + 15Z 1 Z 2 + Z 2
∆
Mesh current I5 is calculated from I 5 = 5 . The determinant ∆5 is given by:
∆

(Z 1 + Z 2 ) − Z2 0 0 V1
− Z2 (Z 1 + 2Z 2 ) − Z2 0 0
∆5 = 0 − Z2 (Z 1 + 2 Z 2 ) − Z2 0
0 0 − Z2 (Z 1 + 2 Z 2 ) 0
0 0 0 − Z2 0

which readily simplifies to:

(Z 1 + Z 2 ) − Z2 0 V1
− Z2 (Z 1 + 2 Z 2 ) − Z2 0
∆5 = Z 2
0 − Z2 (Z1 + 2Z 2 ) 0
0 0 − Z2 0
(Z 1 + Z 2 ) − Z2 V1
2
= Z2 − Z2 (Z 1 + 2 Z 2 ) 0
0 − Z2 0
3 (Z 1 + Z 2 ) V1
= Z2
− Z2 0
4
= V1 Z 2

Then:
5
∆5 Z
V2 = I 5 Z 2 = Z 2 = 2 V1
∆ ∆

5
V2 Z 2
The RC network’s transfer function is defined as H ( jω ) = = , yielding:
V1 ∆

-15-
 5
Z2
H ( jω ) = 5 4 3 2 2 3 4 5
Z 1 + 9Z 1 Z 2 + 28Z 1 Z 2 + 35Z 1 Z 2 + 15Z 1 Z 2 + Z 2

Next, we need to compute the RC network’s input impedance. It is given by:

V1
Z IN =
I1
∆1
with current I 1 obtained from I 1 = .
∆

V1 − Z2 0 0 0
0 (Z 1 + 2 Z 2 ) − Z2 0 0
∆1 = 0 − Z2 (Z 1 + 2 Z 2 ) − Z2 0
0 0 − Z2 (Z 1 + 2 Z 2 ) − Z2
0 0 0 − Z2 (Z 1 + 2 Z 2 )
which can be shown to reduce to:

(Z 1 + 2 Z 2 ) − Z2 0
2 − (Z 1 + 2 Z 2 ) Z2
∆ 1 = V1 (Z 1 + 2Z 2 ) − Z2 (Z 1 + 2 Z 2 ) − Z2 − V1 Z 2
Z2 − (Z 1 + 2 Z 2 )
0 − Z2 (Z 1 + 2 Z 2 )
It further simplifies to:

( 3 2 2 3
∆ 1 = −V1 (Z 1 + 2 Z 2 ) − Z 1 − 6 Z 1 Z 2 − 10 Z 1 Z 2 − 4 Z 2 − V1 Z 1 Z 2 + 4 Z 1 Z 2 + 3Z 2 ) ( 2 2 3 4
)
giving:

( 4 3 2
∆ 1 = V1 Z 1 + 8Z 1 Z 2 + 21Z 1 Z 2 + 20 Z 1 Z 2 + 5Z 2
2 3 4
)
Then:
5 4 3 2 2 3 4 5
V V∆ ∆ Z + 9 Z 1 Z 2 + 28Z 1 Z 2 + 35Z 1 Z 2 + 15Z 1 Z 2 + Z 2
Z IN = 1 = 1 = = 1
I1 ∆1  ∆1  4 3 2 2 3
Z 1 + 8Z 1 Z 2 + 21Z 1 Z 2 + 20Z 1 Z 2 + 5Z 2
4
 
 V1 
1
When Z1 is a capacitor C and Z 2 a resistor R, the impedances are Z 1 = and
jω C
Z 2 = R . Hence, substituting for Z1 and Z 2 we obtain:

-16-
 1 9R 28 R 2 R3 15 R 4
−j + + j − 35 − j + R5
Z IN = ω 5C 5 ω 4C 4 ω 3C 3 ω 2C 2 ωC
1 8R R2 20 R 3
+ j − 21 − j + 5R 4
ω 4C 4 ω 3C 3 ω 2C 2 ωC

Multiplying the numerator and denominator of the above expression by ω 5 RC 5 we get:

ω 5 R 5 C 5 − j15ω 4 R 4 C 4 − 35ω 3 R 3C 3 + j 28ω 2 R 2 C 2 + 9ωRC − j

Z IN =R
5ω 5 R 5 C 5 − j 20ω 4 R 4 C 4 − 21ω 3 R 3 C 3 + j8ω 2 R 2 C 2 + ωRC

Rearranging real and imaginary terms:

ω 5 R 5 C 5 − 35ω 3 R 3C 3 + 9ωRC + j (− 15ω 4 R 4 C 4 + 28ω 2 R 2 C 2 − 1)

Z IN =R …(19)
5ω 5 R 5 C 5 − 21ω 3 R 3C 3 + ωRC + j (− 20ω 4 R 4 C 4 + 8ω 2 R 2 C 2 )

Considering the loading effect of the RC network upon the triode’s output yields the
familiar expression for the voltage gain of the vacuum tube:

AV ( jω ) = − g m (R0 // Z IN )

where R0 = rP // RP . The two paralleled resistances here are the plate’s dynamic output
resistance and the external plate bias resistor, respectively. Barkhausen’s criterion for
the oscillator’s loop gain, Eq.(2), gives:

AV ( jω ) ⋅ H ( jω ) = − g m (rP // RP // Z IN ) ⋅ H ( jω ) = 1 + j 0

V2
where H ( jω ) = ( jω ) is the passive network’s transfer function. Substituting for
V1
AV ( jω ) and H ( jω ) their individual expressions we obtain for the loop gain:
5
Z2
AV ( jω ) ⋅ H ( jω ) = − g m (R0 // Z IN ) ⋅ 5 4 3 2 2 3 4 5
Z1 + 9 Z1 Z 2 + 28Z1 Z 2 + 35Z1 Z 2 + 15Z1Z 2 + Z 2

 RZ  5
− g m  0 IN  Z 2
= 5  R0 + Z IN 
4 3 2 2 3 4 5
Z1 + 9 Z1 Z 2 + 28Z1 Z 2 + 35Z1 Z 2 + 15Z1Z 2 + Z 2

 RZ  5
− gm 0 IN  Z 2
=  R0 + Z IN 
Z IN ( 4 3 2 2 3
Z1 + 8Z1 Z 2 + 21Z1 Z 2 + 20 Z1Z 2 + 5Z 2
4
)

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 After some simple algebraic work and simplification we arrive to:

− g m R0 Z 25
AV ( jω ) ⋅ H ( jω ) =
( ) (
R0 Z 14 + 8Z 13 Z 2 + 21Z 12 Z 22 + 20 Z 1 Z 23 + 5Z 24 + Z 15 + 9 Z 14 Z 2 + 28Z 13 Z 22 + 35Z 12 Z 23 + 15Z 1 Z 24 + Z 25 )
Substituting for Z1 and Z 2 their respective expressions, we get, after some manipulation:

− g m R0 R 5
AV ( jω ) ⋅ H ( jω ) =
−
(
 (R0 + 9 R ) 21R0 R 2 + 35 R 3 ) ( 4 5 
)  1
+ 5 R0 R + R  + j − 5 5 +
(
8 R0 R + 28 R 2
−
) (
20 R0 R 3 + 15 R 4  )
 4 4 
 ω C ω 2C 2   ω C ω 3C 3 ωC 

= 1 + j0

The denominator of the fraction above must be a real negative quantity for the loop gain
equation to hold. Then:

−
1
+
(
8 R0 R + 28 R 2
−
) (
20 R0 R 3 + 15 R 4
=0
)
ω 5C 5 ω 3C 3 ωC

Knowing that ωC can not be equal to zero, the above equation simplifies to:

1
− 4 4 +
(
8 R0 R + 28 R 2 ) (
− 20 R0 R 3 + 15 R 4 = 0 )
2 2
ω C ω C

which may be rewritten as:

(20R R 0
3
+ 15 R 4 )ω 4 C 4 − (8 R0 R + 28 R 2 )ω 2 C 2 + 1 = 0

or in the alternate form:

 R0   R 
 20 + 15  R 4ω 4 C 4 −  8 0 + 28  R 2ω 2 C 2 + 1 = 0
 R   R 

Solving for ω 2 we get:

2
 R0  R  R 
8 + 28  ± 64 0  + 368 0  + 724
 R   R  R K2
ω2 = = 2 2 …(20)
 R0  R C
 40 + 30  R 2 C 2
 R 

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This equation gives the radian frequency of oscillation. We keep the positive sign for the
square root above because, as can be shown, the negative sign would not permit
satisfaction of the loop-gain condition.

We now need to calculate the minimum gain for oscillations to occur. The loop-gain
equation dictates that:

− g m R0 R 5
=1
−
(
(R0 + 9 R ) 21R0 R 2 + 35R 3 ) ( 4
+ 5 R0 R + R 5
)
ω 4C 4 ω 2C 2

This expression can be rewritten as:

− g m R0
=1
 R0   R0 
 + 9   21 + 35 
 R − R  +  5 R 0 + 1
4 4 4 2 2 2
 
ω R C ω R C  R 

and solved for g m R0 , giving:

 R0   R 
 + 9   21 0 + 35 
 R 
g m R0 = −  4 4 4 +  2 2 2  −  5 0 + 1
R R
…(21)
ω R C ω R C  R 

This is the minimum value of the product g m R0 for sustained oscillations in the circuit.
R0
Tables I and II below show how the ratio influences results for the frequency of
R
K
oscillation fosc = and the triode’s unloaded-case small-signal voltage gain,
2πRC
R
g m R0 . The particular case 0 → 0 corresponds to situations where the RC network
R
won’t load the vacuum tube’s output.

-19-