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

**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

**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.