Bass Booster Circuits

Bass Boost is today’s sound… whether it’s the driving, gut-vibration pulsation of disco, or the solid bass line of soft, hard, or laid-back rock. One way to get the modern bass-boost sound without running out and buying an all-new expensive piece of equipment is to use a Bass Booster between your guitar, electronic organ or what-have-you, and the instrument amplifier.

A bass booster strips the highs from the instrument’s output signal and amplifies low frequencies, feeding on “all-bass” sound to the instrument amplifier. Naturally, the bigger the speaker used with the amp, the more powerful the bass: use 15-inchers with the Bass Booster and you can rattle the windows. Bass Booster is powered by an ordinary 9 volt transistor radio battery. It can be assembled on a small printed board or on a veroboard using point to point wiring. The booster connects between your instrument and its amplifier through two standard RCA Jacks

Parts List:
P1 = 50K
P2 = 100K
R1 = 22K
R2 = 470K
R3 = 47K
R4 = 10K
R5 = 470R
R6 = 1K
Q1 = 2N2222
C1 = 2.2uF-25v
C2 = 100nF-63v
C31 = 00nF-63V
C4 = 3.3uF-25v
C5 = 470uF-25v
D1 = 5mm. Red Led
Q1 = 2N2222
B1 = 9v Battery
J1 = RCA Audio Input Socket
J2 = RCA Audio Output Socket
S1 = On-Off Switch
Bass Booster Circuits.
Connect your electronic guitar or other electronic instrument to input jack J1; Connect output jack J2 to your instruments amplifier’s normally-used input. With power switch S1 off, key S2 so the instrument feeds directly to the instrument amplifier. With P2 set full counter-clockwise (Off), turn power switch S1 on, key S2 once, and advance P2 for the desired Bass Boost level. To cut back to natural sound just stomp down on S2 and key the Bass Booster out. Don’t worry about leaving power switch S1 on for several hours of a gig. The circuit pulls less than 1mA from the battery, so battery will last many, many months.Bass Booster Circuit .

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a simple DC voltage booster

Joule Thief: a simple DC voltage booster

A joule thief circuit lit from a 0.5V AAA battery. Click to enlarge.

A Joule thief is a simple circuit that acts as a DC to DC booster, raising a supply voltage by several volts. In this iteration it uses the exhausted voltage of a alkaline battery and boosts it enough to light a blue LED that requires 2.8V to light. With an otherwise dead 0.5V AAA battery, it will light a blue LED and run for days, using (at the moment) just under 2mA of current. Its much much dimmer than using a fresh battery or running the LED with a proper current through it. Giving the Joule thief circuit 3V from two fresh batteries pulls 75mA through the LED, making it very, very bright and probably short-lived.

A schematic is below. The circuit works like this: When first turned on, current flows into the inductor and produces a magnetic field in the toroid. While this is happening, no voltage appears at the base of the transistor, so the transistor remains off. The LED sees at first no voltage and while the inductor fills up, it only sees a maximum voltage of the battery, which is not enough to pass the diode. Once the inductor is charged, the battery voltage appears at the base of the transistor, turning it on. This allows the right side of the inductor to want to dump the energy it has stored in its magnetic field as quickly as possible, and this gives us a high-voltage that appears across the inductor. When that voltage exceeds 2.8V the LED turns on and lights up until the voltage drops below, triggering the sequence to begin again. I measured the frequency of the on/off oscillation and it seems to run at about 34kHz; the multimeter said between 68 and 72kHz but a radio showed there was 34kHz signal as well, which I assume was the fundamental (and the 68kHz one a harmonic). It did change in frequency a bit while on.

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