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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

AUTHOR 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

I 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.

A 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.

What a remarkable idea. Thanks for the excellent article!

In 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.