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04-04-2016

Circuit combines power supply and audio amplifier

Susanne Nell

EDN

The circuit in Figure 1 can help if you must transfer dc power and audio over a pair of copper wires. One application for such a circuit is a low-cost door-opening system with speech input. The circuit uses only one IC, the well-known LM317, a low-cost power-supply regulator. Using this chip, you can modulate the adjustment-pin input with the audio signal from an electret condenser microphone, connected between the output and the adjustment terminals of the IC. The LM317 regulates the output in such a way that the voltage on the microphone is always 1.25 V dc. This application uses a WM34 electret microphone, which comes in a standard 10-mm capsule from Panasonic and is common in low-cost equipment. You can use nearly any electret capsule, because the well-regulated voltage on the microphone never exceeds 1.25 V. Every electret capsule contains an integrated JFET-based impedance converter that translates speech into a current flowing from the source to the drain terminal. This current through the microphone modulates the voltage on the variable resistor, RP. Because the output of the LM317 must follow the voltage on RP, you obtain a low-impedance audio signal riding on the output dc voltage.

Circuit combines power supply and audio amplifier
Figure 1. A novel circuit uses the adjustment pin of a regulator IC to provide audio amplification.

The microphone directly modulates the adjustment pin, so a smoothing capacitor, such as C1, for noise and hum does not influence the level of the audio signal. C1 shunts some of the audio signal to ground, but the LM317 compensates for the loss with internal gain. To avoid excessive losses in the LM317, use a capacitor with as low a value as possible. The circuit works well without a capacitor, but values as high as 47 µF do not present a problem. Using RP, you can adjust the dc output voltage and the gain for the microphone signal. For proper operation, the LM317 needs to deliver a minimum current of 4 mA from its output terminal. If your design uses no loudspeaker, you can connect a load resistor to sink this 4 mA. Designs using low-impedance loudspeakers must also have load resistors. You must add the ac current in the audio signal to the minimum current requirement of 4 mA. For an 8 Ω loudspeaker, you need a minimum resistive load of 470 Ω to avoid distortion.

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