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I need to build a circuit where light is converted to sound. The light falls on a CDS cell and the current is passed through the NPN transistor to amplify it.

I know this is a very simple schematic, but what would be the real life model of this? Why is the resistor going to the ground?

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    \$\begingroup\$ Sounds like a laser listener I saw in a magazine once. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 27 '14 at 3:32
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That is a really crappy circuit since it depends on the gain of the transistor being just right. Real transistor gains vary widely, even within the same production lot. Competently designed circuits work from the minimum guaranteed gain of the transistor to at least 10 time that, preferably to infinite gain.

I would start with the transistor stage setting its own DC bias so that the output voltage is near the middle of its range. The signal from the LDR would be AC coupled into that stage. The LDR would have a pulldown at least so that it and the pulldown produce a voltage signal by themselves. This would be the signal fed into the amplifier stage thru a coupling capacitor.

For extra credit, come up with a auto-biasing scheme that keeps the output near the middle of its range regardless of average ambient light level.

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    \$\begingroup\$ The OP wasn't asking for hi-fi and, actually, it's not a bad little circuit at all, seeing as how it's an emitter follower which, with 3V on the LDR and with the LDR sitting at about 10K under ambient light, will put about half of the supply at the top of the emitter resistor. Then, when the LASER modulates the LDR's resistance, the emitter will follow the base voltage undulations, which can be easily cap-coupled to any following gain stages or directly to the headphones. A little sloppy, but perhaps perfect for the OP's purposes, and certainly nowhere near as bad as you put on. \$\endgroup\$ – EM Fields Jul 27 '14 at 15:42
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    \$\begingroup\$ @EMFi: The transistor bias point is very unpredictable, to the point where it could easily be pegged at one extreme or the other. If the transistor gain is high, then the base will be at 3V with no (or very little) change due to LDR resistance. If too low, the output will be very low with little voltage swing. The LDR nominal resistance and the transistor gain have to be matched just right, which can't be done up front due to part variance, even if you could specify the specific ambient level required to get the average LDR resistance. This is lots more than "a little" sloppy. \$\endgroup\$ – Olin Lathrop Jul 27 '14 at 15:52
  • \$\begingroup\$ Because of the fact that the input resistance of an emitter follower is roughly beta times the emitter resistor, a base resistor tied to a fixed supply (like an LDR connected to 3V, for instance) will always elicit an output voltage from the emitter which can be likened to that from a 2 resistor voltage divider, the output voltage range being dependent on the ratio of the resistors. I've worked out a scenario using reasonable resistances and betas, and if you'd like to peruse it, it's at dropbox.com/s/luipdzoaom8i0g2/Light%20to%20sound.png \$\endgroup\$ – EM Fields Jul 27 '14 at 22:30
  • \$\begingroup\$ @EMFi: But that's exactly the point. The apparent base resistance is directly a function of gain, which varies widely, from part to part and with temperature. The datasheet gives you the minimum, but the actual can be sevaral times higher, sometimes many times higher. If the effective base resistance isn't close to the LDR resistance, then you lose gain due to being close to one end or the other. Think of the limiting case where the gain is infinite. The output is always pegged at the supply minus the B-E drop. \$\endgroup\$ – Olin Lathrop Jul 27 '14 at 22:38
  • \$\begingroup\$ Straw man. The argument isn't about what happens in the limit, it's about what happens in the real world with real parts, and you seem unwilling to provide numbers to prove your case, preferring instead to proffer seeming authority-conferring platitudes. \$\endgroup\$ – EM Fields Jul 27 '14 at 23:47
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It is a simple non-inverting transistor amplifier. The resistor to ground is what allows it to be non-inverting. When the transistor is off (the CDS cell not conducting), the Headphone output at the emitter of the Transistor is pulled to ground level through the 47Ω resistor. When the transistor is on, the headphone output gets pulled up to 3V, with the resistor limiting how much current goes through the transistor, and preventing a direct short.

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This is known as an emitter follower. There is no voltage gain in the circuit, this acts as a impedance buffer i.e. it can drive heavier loads that the CDS cell could on it's own.

Light on the CDS cell generates current which goes into the base of the transistor, this current is multiplied by the transistor (this gain is known as Hfe). This current when forced through the resistor generates a voltage which is your output signal. Your Hfe (from some old data sheets) varies but is at least 40 (and can be as high as 100).

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The real world application varies upon the application and the type of the signal. If linearity is not a concern, then the above circuit diagram can be improved to remove the 47(ohm) resistor and the circuit will act as a digital circuit, with almost no sound when there is no light.

On the other hand if linearity is a concern for analog applications, then the above is suitable to provide analog output specially if the amplitude of the light is to be varied on both sides of the normal value (i.e. if the value changes both above and below the dc value). Just check up non-inverting transistor amplifiers for more information for the circuit.

A lot depends upon resistor values of the CDS cell and the resistance of the headphones. Moreover, you might want to connect a capacitor between the output and the headphone to dc couple the output.

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