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I've seen this following post regarding using the speakers as a microphone: https://skeptics.stackexchange.com/questions/5664/can-a-computer-be-hacked-to-use-a-connected-speaker-as-a-microphone

There are some things I don't understand about it. He says that an amplifier doesn't pass any signal from it's output to it's input. Is this correct? if I have an amplifier such that output=100*input and I force 100*x on the output, won't I get x in the input?

To my understanding, speakers are always plugged through an amplifier to the sound card. The question is whether the amplifier resides in the sound card (where it is also controllable and we can lower it) or in the speaker. In any way, say that when I produce a voltage of x from the sound card I get some sound signal level in the air of gx right on the speakers (g in the appropriate units). Now if I produce a sound in the air that is equal to gx right on the speakers I will get the voltage level x to the sound card. So reducing the amplifier is just a matter of reducing this g. Is this the accurate explanation? and not the explanation inside the link?

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    \$\begingroup\$ Amplifiers generally don't work in reverse. If you manually crank the engine of your car, it doesn't put gasoline back into the tank. \$\endgroup\$ Sep 19 '14 at 13:59
  • \$\begingroup\$ Time goes forward and not backward. This is not an explanation I'm afraid. Can you please explain this? or direct me to something that will? \$\endgroup\$
    – Yaniv
    Sep 19 '14 at 15:41
  • \$\begingroup\$ @Yaniv reading your comments, your fundamental problem is that you don't understand what an amplifier is and how it works. You should do some more reading on that. Base point: amplifiers are made from transistors, and transistors only "work" in one direction. \$\endgroup\$
    – markt
    Sep 19 '14 at 22:15
  • \$\begingroup\$ @markt As I've said in one of these comments, I've tried reading about amplifiers (for example, I've looked at ones in "The Art of Electronics"). What I understood is pretty similar to what Alan said - the voltage on an emitter rises and therefore also on the base. I couldn't understand his comment regarding the micro-amps part and the feedback, mainly because I'm talking about a situation in which I'm forcing a voltage on the output (so the amplifier might resist it, but the equilibrium should still be 100*x -> x as far as I understand). \$\endgroup\$
    – Yaniv
    Sep 20 '14 at 11:07
  • \$\begingroup\$ I'm getting pretty discouraged. Almost all the comments I've gotten in this post are either analogies to something else (light bulb, engine, WTF?) or someone simply stating without any explanations that it can't be done. And that's even for the part of the question I thought was basic. Thanks anyway! \$\endgroup\$
    – Yaniv
    Sep 20 '14 at 11:07
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An amplifier... amplifies. Takes a tiny signal, and beefs it up with more voltage / current / power.
A speaker needs that power, to make a coil move in a magnetic field. Attached to the coil is the cone, this displaces air when the coil moves - and we get sound.

A speaker WILL act as a microphone. It's just not very efficient.

If you get real close, and shout really loud, you will get a mediocre signal. In theory, this get passed back to the input where, since we are reducing signal strength along the way, we would have a microscopic signal.

Then comes the final hurdle: recording this signal.

Computer data, in the form of ones and zeroes, gets chewed through a chip called Digital-Analog-Converter [DAC] before being sent to that amplifier. To record our eensy signal, we need an Analog-Digital-Converter [ADC].

ADC's are weird and wonderful beasts. Your computer has one - it's what the "mic" input goes to. No matter how you quibble about passing signals backwards through an amplifier, I fail to see how you get a digital version without adding your own hardware [ie. planting a bug].

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  • \$\begingroup\$ So you CAN go back through an amplifier? I see there are different answers to this :-) The ADC part can be worked around, by re configuring the sound card if it supports it. \$\endgroup\$
    – Yaniv
    Sep 19 '14 at 13:47
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    \$\begingroup\$ As I noted in another answer, there are 2 paths "back" through an amplifier. If the voltage on an emitter rises, the base voltage will rise too. Trouble is: you're converting milliamps to micro amps here (transistors are current based). The other path is via a feedback resistor, used to stabilise the circuit. BIG trouble is: this feedback signal is used to REDUCE the presence of the signal at the output, so the feedback acts to reduce your external signal even more. The result, sitting at the input of the amplifier, is so laughably small you would typically dismiss it as "noise". \$\endgroup\$ Sep 19 '14 at 14:07
  • \$\begingroup\$ Thank you! Regarding the first path - does it matter that it's micro amps? The voltage difference in the input should just decrease by the gain factor of the amplifier, right? Regarding the second path - I'm not sure I get why it acts to reduce it. I'm keeping the output constant at 100*x! meaning it can't reduce it and the input must go to x (up to the accuracy of the amplifier), doesn't it? Is there somewhere I can read about this? I feel I still haven't gotten a clear answer. Thanks again! \$\endgroup\$
    – Yaniv
    Sep 19 '14 at 15:34
  • \$\begingroup\$ I mean reading especially about forcing an output on an amplifier (or on the subject of speakers as microphone). I tried reading about amplifiers in general. \$\endgroup\$
    – Yaniv
    Sep 19 '14 at 15:36
  • \$\begingroup\$ Regarding the first path - yes, it matters that the signal is down to micro amps. I'm talking drectly through the amplifier [chip] and chances are ther will be 3 or more transistors in a row, to give a total gain of about a million. Thermal noise [background hiss] from the transistors will be bigger than the signal, if you go this route. \$\endgroup\$ Sep 22 '14 at 5:20
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A small lightbulb glows when you put (say) 12V dc on it. However, if you fired a laser beam at the same lightbulb and made it glow exactly the same way you wouldn't get 12V DC out of it.

The lightbulb is representing the speaker and amplifier combined.

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There are some things I don't understand about it. He says that an amplifier doesn't pass any signal from it's output to it's input. Is this correct? if I have an amplifier such that output=100*input and I force 100*x on the output, won't I get x in the input?

By design, and amplifier amplifies its input signal to a signal at the output, and not the other way round. In practice a tiny bit might go in reverse, but that will likely be attenuated by much much more than the amplification factor.

To my understanding, speakers are always plugged through an amplifier to the sound card. The question is whether the amplifier resides in the sound card (where it is also controllable and we can lower it) or in the speaker.

The soundcart contains a (small) amplifier. The old baxes that you plugged into a soundcard were often just a speaker (and did not need any power chord). Nowadays what you plug in is nearly always an amplifier.

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  • \$\begingroup\$ Looking at this amplifier for example: upload.wikimedia.org/wikipedia/commons/6/66/… It seems completely reversible, doesn't it? If I force some Vout, I will get the appropriate Vin, so that the op-amp will continue to operate correctly? And if the soundcard contains an amplifier, how can you ever use speakers as a microphone? \$\endgroup\$
    – Yaniv
    Sep 19 '14 at 13:30
  • \$\begingroup\$ No, it isn't reversible. The output comes from the amplifier. If you push current through the output, you will just burn up the output stage of the amplifier. \$\endgroup\$
    – JRE
    Sep 19 '14 at 13:35
  • \$\begingroup\$ Considering transistors, there are 3 configurations: common emitter, common base, common collector [aka emitter follower]. If you impose a signal on the emitter, there will be a version of it present on the base [just not very big]<br>Also, don't forget - most amps have negative feedback to increase stability, ie. an attenuated version of the output is deliberately fed back to the input. \$\endgroup\$ Sep 19 '14 at 13:48
  • \$\begingroup\$ JRE, can you please give an explanation for this? Looking at some amplifiers at "The Art of Electronics", I can't seem to understand why the amplifier should burn up (and what if I don't "push" current, but actually take away current by decreasing the voltage). Thank you! \$\endgroup\$
    – Yaniv
    Sep 19 '14 at 15:38
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There are two parts to your question:

First, if you force a voltage onto the output of an amplifier, you may just damage the amplifier, which is also trying to faithfully force a voltage onto it's output, as per it's job description.

Second, if you can get to the "raw" connections to the actual loudspeaker inside the box, then it is already working as a microphone right now, whether you connect it to anything or not. Some speakers (like my ancient Creative speakers which came with an ancient Sound Blaster card) have a button to enable or bypass the amplifier in their case, so you can use them without a power source. When the amplifier is bypassed, they will function as a microphone.

Loudspeakers and microphones both convert electrical energy into sound energy (and v.v.) in both directions at the same time, and at all times throughout their lives, EDIT which means you can also use a microphone as a speaker :) don't blow it up though.

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  • \$\begingroup\$ You can only use the speakers as a microphone if they aren't connected to the PC. The sound card of your PC has an amplifier as the last stage, so there's no possibility of getting anything in to the PC that way. Even on sound cards where you need an amplifier in the speaker there is a last amplifier stage in the sound card - it just isn't strong enough to drive a speaker. \$\endgroup\$
    – JRE
    Sep 19 '14 at 13:34
  • \$\begingroup\$ Sure - the speaker (microphone) would need to be plugged into the Mic input on the PC, to use them as a microphone. \$\endgroup\$ Sep 19 '14 at 13:37
  • \$\begingroup\$ Yup. And even then it wouldn't do much good. Most mic inputs have a current source for electret microphones - that would tend to hold the speaker cone still so that you can't get anything out of it asa a mic. \$\endgroup\$
    – JRE
    Sep 19 '14 at 13:44
  • \$\begingroup\$ I've heard that you can use passive speakers as a microphone even if you connect them to the speakers output (by re configuring the pins through the HDA driver for instance), but I still think there should be some amplifier in the way? just maybe you can set it low through the driver as well? \$\endgroup\$
    – Yaniv
    Sep 19 '14 at 13:44
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I'm skeptical that any halfway decent transistor-based amp (which is basically all of them outside of a music or specialty hi-fi shop) will pass anything at all from output to input. They are designed to take an input signal, boost it by some gain value that is usually fixed in the amp itself (the volume control is usually a passive attenuator just before the amp), and force the resulting voltage at the output regardless of what the speaker is trying to do.

In technical terms, the output impedance is pretty low. But even the relatively high output impedance of tube amps and some transistor amps that are designed that way doesn't necessarily mean that there's a path back to the input.

Again in technical terms, the input is usually high impedance, which means that it tries to influence the thing that drives it as little as possible. In a good amp, this high impedance input is usually compared to another high-impedance terminal inside the amp, which is driven from the output through a voltage divider. (that's what sets the gain) The difference is then amplified (a lot) and used to drive the output stage. Basically an opamp with power components.

That being said, it's possible that the scenario in question had a specialized amplifier that may have been designed to do that for a legitimate reason (speakerphone?), or perhaps had some extra monitoring terminals that would show an error signal. Knowing this, the attacker could write his software to take advantage of that particular amp.

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