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I'm currently trying to build a circuit to transmit audio via laser. It works great and the quality is decent. But I don't get the reason for the diode. If I remove the diode and replace it with a normal wire the music is no longer properly transmitted. I know what a diode does but not how it is used in this special use case. Could anyone try to explain? The "Cont Freq" block is meant to show the music output from a headphone jack of a mobile. The two power sources both supply 5 V each, R4 = 660 Ohm and R2 = 220 Ohm.

The circuit picture drawn by myself. I don't know wether it is correctly drawn but that's how I built it in reality and it works fine.enter image description here

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    \$\begingroup\$ What happened to your previous question? \$\endgroup\$ – Oskar Skog May 30 '17 at 21:00
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    \$\begingroup\$ What's the difference/relation between this and your deleted previous question? \$\endgroup\$ – Oskar Skog May 30 '17 at 21:00
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    \$\begingroup\$ It can't "work fine" (or work at all) with only one wire connected to the section with Q2 and D1. \$\endgroup\$ – Peter Bennett May 30 '17 at 21:07
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    \$\begingroup\$ I think we're supposed to assume that the negative terminal of each of the "batteries" is connected to the negative terminal of "Cont Freq". That's what Trevor did below. In any case, this question isn't any clearer than your first attempt. Reposting the same question over and over again is not going to win you any friends here. \$\endgroup\$ – Dave Tweed May 30 '17 at 22:37
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It helps if you redraw it, though you are missing some grounds.

Note that diode D2 and the diode that is the Base-Emitter junction if Q1 are in parallel. This basically forms a kind of current mirror. Presumably R4 is set to bias them both appropriately.

Any current inserted from the signal source will then be converted into current through the laser diode.

schematic

simulate this circuit – Schematic created using CircuitLab

If you replaced the diode with a wire it would short out the base of Q1 and it won't work.

Still, I'm surprised it works as well as you say it does.

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    \$\begingroup\$ yep, just as surprised (and already was at OP's previous question); I assume the signal actually pulls BT2 up and down, a floating only limited by stray capacitance... what I recommended was to drop the "I tried for hours and this is what kind of works" with a simple Opamp circuit (non-inverting single supply AC coupled opamp voltage amp would do, probably. Maybe one would build a voltage-controlled current source, instead). \$\endgroup\$ – Marcus Müller May 30 '17 at 22:38

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