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[Design Briefs] Analog Pink-Noise Source Drives Earphones Or Speakers Dick Cappels ED Online ID #5310 July 21, 2003
Acoustic white noise is useful as a masking sound to improve privacy in offices, or as a soothing sound field to help concentration and sleep. It has also been said to aid in therapy for people suffering from tinnitus (ringing in the ears). White noise is a signal containing components at all frequencies with equal amplitude. The circuit shown in the figure generates pink noise (which is white noise rolled off) sufficient to drive a pair of small speakers or earphones. It's well known that an avalanched emitter-base junction of a junction transistor is a good noise source. In this circuit, Q1 is biased to its emitter-base junction avalanche voltage by inverter Q2, and in doing so, generates a millivolt or two of white noise. A2 and A3 provide 50× gain each, for a total of 2500×, producing volts of noise from the junction. A4 serves as a unity-gain inverting amplifier. When the headphones or speakers are connected between the output of A3 and A4 to form a full bridge output, voltage gain is doubled and the maximum drive voltage approaches 14 V p-p. A1 buffers a 4.5-V reference made from the two 470-Ω resistors. The base of Q1 isn't connected. The full-bridge op-amp circuit can provide plenty of drive voltage for an efficient pair of speakers, but not much current without additional buffers. So if it's used to drive a pair of speakers or earphones, you'll get the greatest power output by driving them in series rather than parallel. The noise spectrum can easily be shaped to compensate for the characteristics of a particular set of speakers or earphones by controlling the bandwidth of the amplifiers following the noise source. For example, placing a small capacitor (0.001 µF for a start) across one or both of the 20-kΩ resistors will result in a high-frequency boost of the amplitude of high-frequency components. Placing a small capacitor across the 1-MΩ resistor (47 pf for a start) will cause a decrease in the amplitude of high-frequency components. |
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Reader Comments
Found the article interesting. Have you looked at a case where the "tone" of the signal can be modified (probably positive feedback) to have a bass (waterfall) tone and a treble (rain) sound? AUTHOR RESPONSE: Steve, Thank you for your kind comment. As for more advanced effects, no, I hadn't looked at those techniques. If you run across some simple ways of getting those effects, I'd appreciate if if you would pass them along. Regards, Dick Steve Corley -August 07, 2003AUTHOR RESPONSE: Q1 is connected correctly in the schematic (base is open). This particular connection reverse biases the emitter-base junction at a little over 4 volts and adds one diode drop from the forward-biased base-collector junction. The only significant difference between connecting the base of Q2 to the base of Q1 and connecting the base of Q2 to the emitter of Q1 is one diode drop across Q1. Mr. Melnick is correct about the effects of adding capacitors across the 1 Meg Ohm feedback resistors. He might also want to take into account the -6 db/octave roll off for the LM324's open loop response that intercepts the two 50X gain stage's closed loop gain a little over 10 kHz, for a total of -12db per octave, into account as well. Though I am not sure about the commonly accepted definition of "brown noise", Mr Melnick statement concerning the way to make "pink" from white" does appear to be in agreement with what I have heard recently from other readers. Regards, Dick Cappels Dick Cappels -August 05, 2003I believe that it is the collector, not the base, of Q1 that should be open. Use of a capacitor across the 1-Meg resistor will result in a 6-dB/oct rolloff and produce 'brown' noise. A 3-dB/oct filter is required for producing pink noise from a white-noise source. Nick Melnick -July 28, 2003A reader, one Mr. Shapiro, found that A1 in the circuit is not working properly. Looking at the circuit, I see an inexcusable mistake in the circuit. A1 should be set up as a voltage-follower. The inverting and non-inverting pins (pins 2 and 3) are swapped. If you connect A1 pin 3 to the junction of the two 470-k resistors and connect A1 pin 2 to A1 pin 1. After making that change (swapping pins 2 and 3), the output of A1 should be 50% of the power supply voltage -- which is about 4.5 volts if you are powering it from a 9-V battery. I am very sorry for the inconvenience this error must have caused some readers. Thanks to Mr. Shapiro for bringing it to my attention. Dick Cappels -July 27, 2003What a remarkable idea. Thanks for the excellent article! Tom -July 22, 2003In the schematic, Q1's base connects to Q2's base, and the emitter lead is the one that is not connected. It is the emitter-base junction that will break down in the vicinity of 8 volts... the emitter-collector voltage is likely an order of magnitude higher and will not function as shown. Also, the emitter-base breakdown is in the close neighborhood of 9V, and more reliable operation can be had using 12-18V for the noise generator's source. Ed Ganger -July 22, 2003 | |||||||||||||||||||||||||
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