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I'm working on transmitting audio through light using amplitude modulation of an LED, received by a solar panel connected directly to an active speaker. Something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

I only have rudimentary understanding of analog electronics (I work mostly with software), so I'm wondering which transformer that should be used. The schematics I find online are all for a 9v or 3v circuit, but since I need a strong light, my LED is 7W 12V. Is an 8 ohm to 1k ohm the correct one? What's the equations for audio transformers and voltage? I only seem to find voltage converter circuits and equations when searching for this online.

The circuit contains parts I already own (LED and solar cell), but I'm happy to order other types if need be.

EDIT: Here are a couple of projects using this technique:
http://www.instructables.com/id/HOW-TO-Send-AudioYour-Voice-Over-a-Beam-of-Light/?ALLSTEPS (photoresitor instead of solar cell)
http://asd.gsfc.nasa.gov/blueshift/index.php/2013/01/22/try-it-at-home-turning-sound-into-light-2/ (No transformer)

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  • \$\begingroup\$ I can't get your link to load but (1) you're running 7/12 = > 0.5 A DC through the transformer. If it doesn't burn it out it will probably drive the core into saturation. (2) A solar cell is unlikely to have a good high-frequency response and will most likely be non-linear and will generate a lot of distortion. \$\endgroup\$ – Transistor Apr 16 '16 at 16:55
  • \$\begingroup\$ (1) Then what is the solution? Add a resistor? If so, which one? (2) Some distortion is not a problem, but will a photoresistor + battery + resistor be a better solution? \$\endgroup\$ – Brean Apr 17 '16 at 14:33
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enter image description here Alright!

So this is the scheme. I omited several circuits because i'm lazy and because i want to concentrate on the system design.

So you have to use a microphone. Maybe an electret, or whatever is available. In this scheme i connected it as an electret. Then you and the first amplifier, where you have to use a potentiometer in the feedback, so you will be able to set your gain.

Then you have a most basic low path filter. 5kHz shoud be enough for your application. Then a comparator.

The negative input should be fed with sawtooth wave of 100kHz. This frequency on on hand is low enough to work with all common components, and is high enough to be order of magnitude higher than your bandwidth (5kHz, remember?).

Next it goes to a gate driver (for some cases you may not even need it) and a MOSFET that turns on the LEDs.

It's pulse width modulation of light. The time of each pulse is proportional to momentary value read from microphone. Why do you need it? It will almost not depend on smoke. Because the detector will only have to detect if the light is on or off, not the power of light.

Pay attention! I drew two white LEDs and one IR. Combination may be different, and for 24V you will have to put more white LEDs (probably 4). But IR is the one that will transmit the data, because it has narrow spectrum and may be nicely matched with IR photodiode.

Now the receiver. A photodiode will read a signal. You will have to set a threshold such that a signal below will be counted dark and above- light.

Then a buffer/comparator (output is 1 if LEDs are on) and a 5kHz filter. If everything worked out well, now you have a recovered audio signal. Now you use an audio amplifier and your speaker.

That's it!

Omited components: power supply, sawtooth waveform generator, IC part numbers.

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  • \$\begingroup\$ Fantastic! I've ordered all the parts now (hopefully), and will give it a shot as soon as they arrive. I hope it's okay to use a 555-timer as a sawtooth wave generator and not opamps? \$\endgroup\$ – Brean Apr 19 '16 at 10:09
  • \$\begingroup\$ You could post full scheme for review. Also simulation has great value. Free simulator: LTSpice \$\endgroup\$ – Gregory Kornblum Apr 19 '16 at 10:37
  • \$\begingroup\$ So, Brean, did you do anything about it? \$\endgroup\$ – Gregory Kornblum Jun 18 '16 at 21:52
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Well, there is such thing "audio transformers". Here, take a look.

Your audio input must be amplified, it will not provide power enough for anything. And LED or laser will not just create PWM.

I would say, the whole idea is a little strange. Although i am not audio design engineer per say, i still wonder, what is it exactly you are trying to achieve?

If you want isolated audio channel with pulse width modulated light, no problem. But why transformer? The receiver must be connected to a comparator, then to a gate driver and a MOSFET. All together would look much like a class D amplifier with optically isolated input.

If you want to transfer power by light, i would suggest separating power channel from data channel.

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  • \$\begingroup\$ Thank you for your quick reply. Yes, I know there is such a thing as an audio transformer, that's why I'm asking for which type I need. The audio input is amplified, that's what an active speaker does. According to the projects posted above in the edit, doing it through a transformer seems to be the most minimal way, and is, even according do wikihow doable. But if you've got a circuit for your solution, I'm happy to take a look. \$\endgroup\$ – Brean Apr 17 '16 at 15:07
  • \$\begingroup\$ Whatever you need, just tell me what is it :) \$\endgroup\$ – Gregory Kornblum Apr 17 '16 at 15:09
  • \$\begingroup\$ Well, to start in the other end: I need a beam of light containing the audio you hear in the room. The audio is a voice talking, and the audio quality need not be good as long as the words are understandable. The electronics driving this thing should be as minimal as possible, and the light needs to be fairly strong so it's perceivable (like a film projector). \$\endgroup\$ – Brean Apr 17 '16 at 16:57
  • \$\begingroup\$ And what are you going to do with the light? It's important. \$\endgroup\$ – Gregory Kornblum Apr 17 '16 at 17:18
  • \$\begingroup\$ It's an art installation in a gallery room. The light needs to be the transmitter of the audio your hear, hence converted back to audio with a solar cell, photoresistor or photodiode. \$\endgroup\$ – Brean Apr 17 '16 at 17:50
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What is your 8 Ohm audio source? 8Ohm is usually a speaker, not a microphone. Do you whish to connect it to the output of a loudspeaker amplifier?

This circuit might work on simulation, but in reality it's hopeless. The constant current through the transformer winding will saturate the transformer core. The 8Ohm source won't be able to drive the any reasonable power into the circuit. The "solar cell" is highly unsuitable.

That you need is an amplifier (a simple operational amplifier circuit will do). You add a constant voltage offset to your audio signal and after you put it to a V-I converter (also an opamp circuit with feedback) to drive the led. That's enough to drive your led.

To receive the signal you need a photodiode with an amplifier (opacity amp as transimpedance amplifier).

First, I need to say that you need to learn the basics of analog electronics before you start thinking about such circuits. Second, even if you make such a circuit, the quality will be very poor. A much better solution would be to use digital transmission with forward error correction coding.

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  • \$\begingroup\$ The audio source will probably be a raspberry pi or an mp3 player of sorts, I think 8 ohm is the standard? According to the projects posted above in the edit, it's highly doable also outside simulation (eg. sci-toys.com/scitoys/scitoys/light/…). The sound quality is not of big importance, but I'd love to do a digital version of it — I just don't have time to write such a thing at the moment. \$\endgroup\$ – Brean Apr 17 '16 at 14:52
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EXECUTIVE SUMMARY: If you want to reproduce that circuit for a 7W power, you will need at least 3.5W of audio, and a transformer with a secondary side rather lower than 1000 ohms (1K) A rough guess would be that you may need something more like 100 up to maybe 300 ohms. Trying to put 7W (LED power) through the secondary side of a small audio transformer can quite possibly burn it out like a fuse because it was never designed to take that kind of current.

BACKGROUND: You have "re-discovered" the MODULATION TRANSFORMER. For more than half a century, this was THE method of modulating AM broadcast transmitters. On YouTube you can see "tours" of large broadcast transmitters where you will see modulation transformers. Some the size of a bread-box for low-power transmitters, and nearly the size of a washing machine for high-power (50KW) transmitters, and up to the size of a mini-van for super-power international short-wave transmitters.

The rule of thumb is that you you need audio power amounting to 1/2 the total power being modulated. So, for a 50KW transmitter, they used a 25KW audio amplifier to drive the modulation transformer. So, for your scheme, you need to calculate the total power you are trying to modulate, and then you need 50% of that power to drive the modulation transformer.

This is an extraordinarily inefficient and costly way to modulate power. In addition to the high audio power to drive the transformer, you need a transformer with the proper primary and secondary winding impedance to efficiently couple the modulation signal, and sufficient wire gauge to modulate the secondary current at minimum DC-loss through the secondary winding. You MAY be able to find a commercial transformer with appropriate windings for your project, but you may spend months looking for it.

Here in the 21st century, there are MUCH more efficient and less expensive ways of modulating power, typically using switch-mode techniques. Even mega-power broadcast transmitters use this kind of high-frequency switching (like Pulse-Width Modulation PWM) for modern solid-state transmitters. The PWM method is also the most common method of creating analog outputs from microcontrollers like Arduino, et.al.

Ref: https://en.wikipedia.org/wiki/Modulation_transformer

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