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I want to trigger a transistor using iPhone's audio output jack. My goal is to turn off an LED(which is fed by an external +5V battery) by sending an audio signal from iphone's output jack. At the beginning i thought it is easy but i encounter many problems. I tried to do it with a transistor but it seems the iphone audio output is too weak even for a transistor.

I am confused every time i find different ideas. Let me be more precise, here are the fatcs:

1-) The output voltage of iphone is very low, something like 300 mV

2-) The output is an audio signal

And here are the problems:

1-) How to amplify this voltage? I found some different ideas after some googling such as:

a-Using transformers. This guys says one needs micro-transformer

http://web.eecs.umich.edu/~prabal/pubs/papers/kuo10hijack-islped.pdf

b- Using capacitor. Here one guy is suggesting using a very simple circuit. I tried it didn't work:

Power LED from headphone jack

c- Using op-amp

Can you assist me a little bit about this?

2-) Do we need to rectify the audio signal?

Here is the circuit i was trying at the beginning. I thought it is simple. But the transistor is not triggering with the audio signal:

enter image description here

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  • \$\begingroup\$ Hi.. Did you get that to work for you? I need something similar \$\endgroup\$
    – Piotr Kula
    Commented Apr 2, 2012 at 16:30

2 Answers 2

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A transistor will need about 0.65V to start conducting, so the iPhone's output is indeed too low. The lowest part solution to get a higher voltage will be the transformer. You'll need a winding ratio of 1:10 to get a decent output level.
You have to rectify the amplified signal. If you were just to average it using a capacitor you'd get 0V. You don't need full rectification, however. You can do without the negative part of the signal; there will always be enough positive signal to work with. Feed the signal via a diode to a capacitor. This will get you a peak detector: the capacitor will load to the positive peaks in the signal. Of course when the music stops the charge should also disappear. You can discharge the capacitor via the transistor's base resistor (missing in your drawing), and if you want to discharge faster have another resistor parallel to the capacitor.

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  • \$\begingroup\$ hello stevenvh, thanks for your answer. i am quite new in the field. what do you think of this one: img687.imageshack.us/img687/7417/62752803.jpg if you think that this works i would like to have your opinion about the values of the elements. many thanks \$\endgroup\$
    – user16307
    Commented Mar 30, 2012 at 9:22
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    \$\begingroup\$ @cmd1024 - the rectifier should come before the capacitor, otherwise, like I said, the cap's voltage will be zero. Put the diode where you now have R1. R1 can be left out. \$\endgroup\$
    – stevenvh
    Commented Mar 30, 2012 at 9:27
  • \$\begingroup\$ million thanks man, i have nobody to ask about these.. so this is the latest version imageshack.us/photo/my-images/580/20027568.jpg what about these values: C1 19 uF - 100 uF. D1 = 1N4148 or anything. R1 = 10k R2 = 220 ohms. should i use a microtransformer? \$\endgroup\$
    – user16307
    Commented Mar 30, 2012 at 10:02
  • \$\begingroup\$ @cmd1024 - you're welcome! That's what StackExchange is for. \$\endgroup\$
    – stevenvh
    Commented Mar 30, 2012 at 10:04
  • \$\begingroup\$ sorry,, so this is the latest version imageshack.us/photomy-images/580/20027568.jpg what about these values: C1 19 uF - 100 uF. D1 = 1N4148 or anything. R1 = 10k R2 = 220 ohms. should i use a microtransformer? – \$\endgroup\$
    – user16307
    Commented Mar 30, 2012 at 10:06
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Consider also this different solution: since you have an external supply, if you don't need a lot of current on the LED you can also use an Op-Amp to provide a high-impedance input and a lot of gain. You could even use it in open-loop mode for just on/off switching, just biasing the inverting input with a voltage divider (you can decide the ratios to obtain the right sensitivity).

In that case, just pay attention to pick a rail-to-rail Opamp that works with the 0/+5 V supply and can drive enough current for the LED. If you want to use a more powerful LED, you can still use the Opamp to drive a transistor.

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