The James-Baxandall Passive Tone-Control Network

Ramon Vargas Patron rvargas@inictel.gob.pe

INICTEL Independent adjustment of bass and treble frequencies in high fidelity audio amplifiers is usually accomplished utilizing specially designed tone-control networks. There are versions for these tone controls based only on passive components, such as the ground-referenced James network shown in fig. 1. Among those versions using active devices we must make mention of P.J. Baxandall´s proposal, in which the tone control was devised as a feedback amplifier (refs. 1 and 2). In this article we shall analyze the James network (also known as the passive Baxandall tone control), obtaining its design equations.

Let´s begin studying the bass control section (fig.2), which has influence over the frequencies below the designed center frequency of the overall James network.

Here, R4 offers some isolation between this stage and the treble control section (frequencies above the center frequency are affected by this control). R5 represents the input resistance of the amplifier connected to the output of the James network and should be selected such that it imposes no appreciable load on the network. Here we assume that C3 and C4 are open circuits at the bass frequencies. With full bass boost (R2´s slider at the upper end), the equivalent circuit for the tone control is as depicted by fig.3.

From this figure and knowing that “s” is Laplace´s variable, we obtain:

2231

2230 //

//

C

Cg XRRR

XRRV

++V

+=

where:

22

22

22 1

1

//

sCR

sCR

XR C

+

×=

122

2

+CsRR

=

Then:

1

1

22

231

22

23

0

+++

++

=

CsRRRR

CsRRR

VgV

After some algebraic work we arrive to:

31

222

3

222

31

30

1

1

RRRCsR

RRCsR

RRR

VV

g

+++

++⋅

+= ...(1)

The gain for high bass frequencies is:

31

31 RR

RA

+=

The gain at low frequencies is:

321

322 RRR

RRA

+++

=

Expression (1) has a zero given by:

22

3

2

01

1

CRRR

s+

−=

and a pole:

22

31

2

2

1

CRRR

R

s p+

+−=

With the tone control adjusted for maximum bass cut (R2´s slider at the lower end), the equivalent circuit changes to that of fig.4. Again, we neglect the loading effect of R4 and R5.

Now the following holds:

1231

30 // C

g XRRRR

V++

=V

112

231

3

+++

CsRRRR

RVg=

A simple algebraic manipulation takes us to:

31

212

12

31

30

1

1

RRRCsR

CsRRR

RVV

g

+++

+⋅

+= ...(2)

The gain at high bass frequencies is now given by:

31

33 RR

RA

+=

For low frequencies the gain is:

321

34 RRR

RA

++=

Expression (2) has a zero given by:

12

031CR

s −=

and a pole:

12

31

2

4

1

CRRR

R

s p+

+−=

The gain ratio at low frequencies is:

3

2

4

2 1RR

AA

+=

For a 40dB control range the following relationship must be satisfied:

1003

2 =+RR1 ...(3)

Then: 332 10099 RRR ≈= On the other hand, at maximum bass boost, the ratio of the gains for low and high bass frequencies is:

31

3

321

32

1

2

RRR

RRRRR

AA

+

+++

=

321

32

3

31

RRRRR

RRR

+++

⋅+

=

According to expression (3): 332 100RRR =+ Then:

31

3

3

31

1

2

100100

RRR

RRR

AA

+⋅

+=

( )

31

31

100100

RRRR

++

=

For a 20dB bass boost:

( )

10100 31

31 =+

100 +RRRR

Solving for R1 we arrive to the following relationship: 31 10RR = We may check that:

3

321

31

3

4

3

RRRR

RRR

AA ++

⋅+

=

10=

this is, the ratio of the high-frequency and low-frequency bass gains is also 20dB at maximum bass cut. For best symmetry in the response curves we must choose s01 = sp4. Therefore:

1

31

2

2

3

2 11

CRR

R

CRR

++

=+

Substituting the already obtained relationships between the resistances yields:

1

3

2

2

111

CR

R

C

+=

100

1

11991

C

+=

1

10C

=

Then, C2 = 10C1. The corresponding Bode plot of the bass response at full boost and full cut is shown below.

Next, we will analyze the treble control. We have already stated that the treble frequencies are those above the center frequency for which the James network is designed. We may consider C1 and C2 as short circuits at these frequencies. Therefore, R1 and R3 along with R4 are part of the treble network (fig.6).

In order to facilitate calculations, we shall first find the Thevenin equivalent for Vg, R1 and R3. In fig.7, the equivalent Thevenin voltage source, V1, is given by:

31

31 RR

RVg +

=V

11

gV= ...(4)

The equivalent Thevenin resistance, RTH, is: 31 // RRRTH =

With the treble control adjusted for maximum boost (R6´s slider at the upper end), and assuming the current through resistor R6 is much smaller than that flowing through R4, we may state the following:

( ) 304

01 sCVVRR

VVg

TH

−=+−

if R5>>R4. Substituting the value given by expression (4) for V1:

( ) 304

011 sCVVRR

VV

gTH

g

−=+

−

Rearranging this equation we get:

( ) ( )[ ]43043 1111 RRsCVRRsC THTHg ++=

++V

Then:

( )( )43

430

1111

RRsC

RRsC

VV

TH

TH

g ++

++= ...(5)

At sufficiently low treble-frequencies:

1110 =

gVV

the same as -20.83dB. At sufficiently high treble-frequencies:

10 =gV

V

or 0dB. Expression (5) has a zero given by:

( ) 3405 11

1CRR

sTH +⋅

−=

and a pole:

( ) 346

1CRR

sTH

p +−=

With the treble control adjusted for maximum cut (R6´s slider at the lower end) we may write:

044

01 VsCRR

VV

TH

=+−

Then:

( ) 441

0

11

CRRsV TH ++=

V

From the above equation and bearing in mind (4):

( )[ ]44

0

1111

CRRsVV

THg ++⋅= ...(6)

At sufficiently low treble-frequencies:

1110 =

gVV

or -20.83dB. At sufficiently high treble-frequencies:

00 →gV

V

Expression (6) has a pole given by:

( ) 447

1CRR

sTH

p +−=

For best symmetry in the response curves we must choose sp7 = s05. Therefore:

( ) ( ) 3444 1111

CRRCRR THTH +⋅=

+

yielding C4 = 11C3. Fig.8 shows the corresponding Bode plot of the treble response at full boost and full cut.

A convenient value for R6 must now be computed. We need some knowledge on the constraints the network imposes before we may neglect the current through R6, as compared to that circulating through R4 (IR6<<IR4). IR4 is given by the following expression when the treble control is adjusted for full cut:

4

0

411

RR

VV

ITH

g

R +

−=

and IR6 by:

36

06 1

sCR

VVI g

R

+

−=

Therefore, the requirement that must be satisfied is:

4

0

36

0 111 RR

VV

sCR

VV

TH

g

g

+

−⟨⟨

+

−

The worst case for the above inequality occurs when s approaches infinity. Translated to our analysis, at sufficiently high treble-frequencies. Then, it is required that:

6

4

0

011R

RRVV

VV

TH

g

g

+⟩⟩

−

−

or equivalently:

6

4

0

0

1

111

RRR

VVVV

TH

g

g +⟩⟩

−

−

At sufficiently high frequencies:

00 →gV

V

It is clear then that the following must be satisfied:

6

4

111

RRRTH +

⟩⟩

Then: ( )46 11 RRR TH +⟩⟩ ...(7) If we conduct a similar analysis when the treble control is at full boost, we would find that the condition to be met is R6>>1.1R4. Hence, expression (7) prevails. We would like to arrive at easy-to-use design formulae, so, before presenting a design example for the James network we shall try to simplify the expressions for f01, fp2, f03, fp4, f05, fp6 and fp7. First, for convenience, the relationships between the components’ values will be repeated. These are: R1 = 10R3

R2 = 99R3

C2 = 10C1 C4 = 11C3

f01 is given by:

22

3

2

01

1

21

CRRR

f+

⋅=π

22

10021

CR⋅

π=

2399

10021

CR⋅

π=

23

121

CR⋅

π≈ ...(8)

fp2 is given by:

22

31

2

2

1

21

CRRR

R

f p+

+⋅=

π

22

1021

CR⋅

π=

2399

1021

CR⋅

π=

2310

121

CR⋅

π≈ ...(9)

f03 is given by:

12

031

21

CRf ⋅=

π

22

1021

CR⋅

π=

2pf= fp4 = f01, required for the desired symmetry on the response curves. f05 is given by:

( ) 3405 11

121

CRRf

TH +⋅=

π

( ) 3431 //111

21

CRRR +⋅

π=

343111011

121

CRR

+

⋅π

=

( ) 343 11101

21

CRR +⋅

π= ...(10)

fp6 is given by:

( ) 346

121

CRRf

THp +

⋅=π

0511 f= fp7 = f05, required for the desired symmetry on the response curves. The center frequency of the James network is taken as the geometric mean of f01 and f05, this is: 0501 fff c ×= ...(11) which coincides with the frequency of the minimum of the amplitude response curve when the bass and treble controls are at full boost. Agrees also with the frequency of the maximum of the amplitude response curve when both controls are at full cut. Usually, 1kHz is adopted as the center frequency. A Design Example Let us suppose we wish to design a tone control for a transistorized piece of equipment. An adequate value for R1 is 10k ohms. R3 will then be a 1k ohm-resistor and R2 a 100k ohm-potentiometer (standard value). It is convenient that f01 and f05 be separated one decade in frequency. Then, from expression (11) we may obtain that: 0110 ff c = Being fc = 1kHz, we find that f01 must be 316Hz. The value for f05 will then be 3.16kHz. From expression (8) we obtain for C2 a value of 503.65nF. Then, C1 should have a capacitance of 50.36nF. According to (9), fp2 = 31.6Hz. R5 is taken equal to 5 times R2 in order to avoid loading effects on the bass network. Therefore, R5 = 500k ohms. R4 must be chosen such that an affordable value for R6 is obtained when using inequality (7). If we make R4 = 5k ohms, then R6 must satisfy the condition R6>>65k ohms. We may adopt a value of 500k ohms for R6. From expression (10), with the value chosen for R4 we obtain C3 = 774.85pF and C4 = 8.52nF. Finally, in order to avoid additional attenuation in the circuit, the output resistance of generator Vg (source resistance) must be made some 20 times smaller than R1. Simulation of the frequency response of the James network Tone Stack Calculator 1.3 is an excellent software program that can be used for simulation of tone-control networks. It may be downloaded from: http://www.duncanamps.com/tsc/

Two simulations have been made using this program. The first simulation uses the above calculated component values and the second one, standard capacitor values for the circuit. No major variation on frequency response has been observed between them. Simulation results are shown below.

Fig.9 Frequency response of the James network with the bass and treble controls at full boost (upper) and full cut (lower).

Fig.10 Frequency response of the James network when standard capacitance values for C1, C2, C3 and C4 are used.

References 1. Baxandall, P.J. “Negative feedback tone control – independent variation of bass and treble without switches” W.W. 58.10 (Oct. 1952) 402. Correction 58.11 (Nov. 1952) 444. 2. Vargas Patrón, Ramón “Red activa de control de tono” http://www.inictel.gob.pe/publicaciones/rvargas/red-activa.htm

Lima-Peru, South America July 23rd 2004