Definition of Engineering Phrases
"A" "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M",
"N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"

Audio Base Control

Minimum Base Control

This circuit represents the minimum component design for a Base control, using two resistors, one capacitor and a potentiometer. Normally the resistors all have the same value; R1 = R2 = R3.
Ignoring the capacitor the three resistors form a voltage divider with the output resistance ranging from R3 [low end] to R2 + R3 [high end].

The output voltage would than range from
Vo = Vi x [R3 / (R1 + R2 + R3)] Minimum value for R2
Vo = Vi x [(R3 + R2) / (R1 + R2 + R3)] Maximum value for R2.

With the output amplitude increasing as R2 is turned up.
Vo = 1 x [1k / (1K + 1K + 1K)] = 0.33 volts
Vo = 1 x [(1k + 1K) / (1K + 1K + 1K)] = 0.66 volts

Minimum Component Audio Base Control
Base Control

Attaching a shunt capacitor to the potentiometer changes its linear output into a frequency dependent output. As the frequency of the audio signal increases the impedance of the capacitor [Xc] decreases.
Using a 1uF capacitor the capacitance reactance is equal to
Xc = 1/ (2 x pie x f x C)
Xc = 1 / (2 x 3.14 x 20 x 1uF) = 7962 ohms
Xc = 1 / (2 x 3.14 x 500 x 1uF) = 318 ohms
[using the Woofer Frequency Range from this site]

Recalculating Z2 with Xc in parallel with R2 gives;
Z2 = 1 (1/R1 + 1/ Xc1) = 888 ohms
Z2 = 1 (1/R1 + 1/ Xc1) = 214 ohms

The chart to the right shows the impedance change Z2 over increasing frequency.
As the frequency increases the impedance across Z2 decreased [from 20 to 500Hz].

Replacing R2 with Z2 in the original equation gives;
Vo = 1 x [1k / (1K + 888 + 1K)] = 0.35 volts
Vo = 1 x [(1k + 888) / (1K + 888 + 1K)] = 0.65 volts
Vo = 1 x [1k / (1K + 214 + 1K)] = 0.45 volts
Vo = 1 x [(1k + 214) / (1K + 214 + 1K)] = 0.54 volts

Basically, as the volume is turned up it has a decreasing effect as the frequency is increased. The higher the audio frequency the lower the output volume. However because this is a minimum component design the volume is effected regardless of frequency.

Common Base Control

Normal Audio Base Control with shunt capacitors
Base Control

Component equations:
R1 = R2 = R3

F1 = 1/ (2 x 3.14 x R1 x C2) = 1/ (2 x 3.14 x R2 x C1)
F2 = 1/ (2 x 3.14 x R3 x C2) = 1/ (2 x 3.14 x R1 x C1)

R1/R2 = R3/R1 = C1/C2

C1 = 1/ (2 x 3.14 x F2 x R1)
C2 = 1/ (2 x 3.14 x F2 x R3)

A passive Base-Boost circuit is shown to the right. The Base-Boost circuit is designed to only emphasize the base or lower frequencies, relative to the higher frequencies. The resistors R1 and R2 form a voltage divider; however capacitor C1 and R2 form the filter portion of the circuit. As the frequency increase the impedance [Xc] of capacitor C1 decreases, decreasing the parallel resistance of R1 and C1. So higher frequencies are attenuated, while lower frequencies are not.

Each of these circuit segments show a single channel, or mono circuit. However the potentiometer could be ganged together with another channel forming a dual channel base control [Dual-Ganged Control]. In fact that works for any of the other audio controls liked off this topic. A combo Potentiometer and SPST Switch could be used to add the loudness control.

PC motherboard

Distributor rolodex Electronic Components Electronic Equipment EDA CDROM Software Engineering Standards, BOB card Cabled Computer Bus Electronic Engineering Design Table Conversion DB9-to-DB25.
DistributorsComponents Equipment Software Standards Buses Design Reference