3.5 mm trs jack power carrying points

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venky srinivas

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Apr 22, 2014
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gyuz i wanted to know which point of the 3.5 mm trs jack carry the negative and positive current that is the TIP,RING,SLEEVE out of these which carries current....
 
Solution
nukemaster
Honestly my the way that Li-ion chargers work, I do not think you will ever get power out of the usb port unless it was designed for it(on the go or similar has been designed from the ground up to not only give you power, but make sure it is 5 volts.).

I have looked quite deep into getting power out of such devices and it simply does not work. Looking at MANY different usb chargers they are all built to prevent any power from moving in the reverse direction. This protects the IC's(these ic's can do many things and are not that easy to identify from a tear down video) dealing with the charging as well as the device providing the power.
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Are you dealing with Audio(AC, so the + and - are flipping all the time) or are you dealing with something else(charging maybe?).

I mean current has to flow from tip to speaker(or headphone) to sleeve and ring to speaker(or headphone) to sleeve. If they are powered speakers, the current flow is VERY low and even headphones tend to be under 10-15ma.

Maybe an idea of what you are trying to connect will help.
 
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thank u nuke your were so helpful for the question u asked what i m trying to connect that i cannot tell u thank u can u tell me that from micro usb 2.0 can the current from the battery be extracted even the mobile doesnt support otg compatibility
 
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While some devices will allow power from the USB ports(even with the design intent to be allow power in). This does not work on all devices(many devices have a small part called a diode that only allows the power to flow IN and not out) while USB OTG devices to this by design.

I am not seeing what USB has to do with the 3.5mm jack unless your device charges from it's 3.5mm jack.

If this is the case, It is best to just use a multi meter to find out how the power flows. Some chargers will list it as well.

For instance a Cobra 2-way radio(walkie-talkie) charges from its headset port and uses sleeve as negative and ring as positive. Tip goes unused in this system. So If one wanted to, they could charge it by applying power on the sleeve and ring(that device need 12 volts making USB no good for it without other hardware).
 
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What exactly are you trying to power off the phone? I do not have that phone to say if it will or will not put out power on its usb connector.

the OTG cable just allows supported devices to use usb hardware like flash drives and keyboards. Having power is to run these devices.

I can not recommend using a phone as a power source when you can get usb power packs for a low price these days.\

I still do not understand exactly what you are tying to do here. You want to power the phone or power something else off the phone?
 
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Ok well I have high doubts the 3.5mm jack will do it, but a OTG to USB cable may give you 5 volts to work with. If your phone has the built in diode, you will not get any power out of it.

This is still not always the recommended path to take, but it may work.

The outer pins on usb provide the power. Pin 1 is +5 volts and pin 4 is negative(called ground most of the time). Please remember to select the appropriate size resistor or your LED will burn.

Another small option would be a cheap flashlight that would not require the phone to run.

Image from www.apcc.tk
256y6xi.gif
 
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You can not tell if it has a built in diode without a meter testing the usb plug. I do not think I have ever tried to take power out of an input on a device like that. I know that some users with battery powered usb changers have had issues of power flowing back into the battery when it dies.

If you are building a flashlight with an LED, you need a resistor to limit the current flowing to the LED, without it the led can burn out rather fast.
 
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k thank u nuke but last question to u if i use 3.5mm slot then it will also require a resistor??? i have searched out about resistors and how to connect but one thing i did not got it that how to select the resistor they are many different sizes can u suggest me for a normal camera flash the resistor ...
 
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The resistor depends on the current the LED can handle(and the forward voltage of the LED in use. These are not like camera flashes that use much more power.), now you can not just go plug in an led to a 3.5mm jack and expect it to light up. Headphones operate at very low voltage levels. That said you would also have to play a tone to even get it to light.

Some microphones operate with phantom voltage, but chances are this is not going to be how the one on a cell phone works.

You are far better off with something that is just battery operated. If you find you are not getting enough light for image taking, you may be able to build a holder to mount a flash light on the phone(or the phone on the light). to give you faster images(more light means the shutter stays open for less time).

Please note that an actual camera flash is MUCH brighter, but the cell phone sensors are optimized for lower light anyway(and the small size helps with this at a cost to overall image quality).

Edit,

Here is a calculator that allows you to determine the resistor needed.
http://ledcalc.com/

You still need to know info on the power source(voltage) and led(current and forward voltage).

Again, due to inexperience, I recommend against messing with the phone. Get a battery pack for your light instead.
 
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I still do not see HOW you will get the power out of 3.5mm. Unless you can write an app to tell the phone to either listen(you may have a very small bit of current to bias a microphone) or play a VERY loud sound(since audio is technically voltage.).

The input voltage of the battery does not even guarantee the output voltage at any point in the system. Buck converters will lower it as needed and boost converters or charge pumps can increase the voltage beyond what the battery provides.

As an example I have a volt to 24 dc-dc converter(-12 , 0 , +12 technically). Now does this mean running something like a headphone amp off or active combiner of it will produce 24 volts? not likely in fact I would not be surprised if I even needed 0.5-1 volt to make my headphones quite loud. This does not cover the fact that music is AC(diodes are DC).

EDIT. I did some tests for you.

I am going to take the guess work out of it for you.

The lowest forward voltage diode I have will light at about 1.2 volts(it just lights very dim). Now that is VERY loud for audio on many pairs of headphones. The AC nature of audio means you will get a blinking light(and only with the LOUDEST parts of the song) unless you play a sine wave into it(60 hz does work, but I am betting it is not ideal and half the time the led should be out, but it flashes very fast). Adding a full wave rectifier would turn this AC into pulsing DC, but would requite about 0.7 or more volts more due to losses in the rectifier. This is quickly becoming a pain in the but to do with audio

and YES, I just I just set this up to test for you(I used my computer as the sound source since my MP3 player lacks the needed voltage to run it. I also ran everything into an opamp voltage follower to make sure that if I killed anything it would be a 58 cent opamp. It was power from +/-12 volts. I used NO resistor because I did not care if I killed the LED and I know the current from the opamp was going to be low at the required voltage.). This LED has a much lower forward voltage then anything a Cell uses for a flash so you will need MORE voltage to even light it(2-3 volts would be a guess). That is just simply not going to work with most headphone driving devices.

I took this a step further and tried to get power out of a usb device(Creative Zen). No such luck.

Slight light at the end of the tunnel.

If your device uses TRRS(left + right + mic) then you may have mic power. Now how much current this can provide as well as its voltage and if it is on all the time are questions I can not answer.

Would would have to take a meter to see if you have mic power and how much it is.

In my tests, using USB OR audio is just not going to work out well. I did these tests so you do NOT have to. A 4 cell AA or AAA batter holder would give you a nice power source with low resistor values all without potential damage to your phone.
 
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I do not use Facebook to be honest. Do not even have a login for it.

But I did get some of the testing out of the way for you :)

For finding just the right resistor if you happen to have a viable resistor(if you want to start very safe 10k will only let 1miliamp into a 2.7 forward led. This is a very safe start.) you can start out dim and slowly brighten the light to the level you get when the flash was on a phone. At that point you can measure the resistance and voltage drop to get all the info you need for that LED.
 
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k nuke my mobile also connects TRRS 3.5 mm jack so can u tell now how to connect the mic wire to get power , u were telling that it can be possible using the mic power so tell me the connection to the led positive and negative terminal......
 
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On most phones it would be the right right next to the sleeve. Again, you you pull too much power(mic's do not tend to need much power) you may damage something.

Using a meter and a TRRS to TRRS cable you can find out for sure. Sleeve is almost always ground so you would just meter for DC under 12 volts to find the location of power.

I am on the other side of the world in Canada.

If you can get your led working on 5 volts from USB, you can always connect it to an external usb charger instead of the phone.

5 volts on a 3 volt drop would need something in the 120 - 150 ohm resistor. to get into the 15-20ma range. Lower value resistors will allow more power, but if you exceed the led power rating it will burn(sometimes it even changes color before it fails.).
 
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so now nuke tell what should i go for usb or audio jack pls tell me in simple step for connecting these to the led and in micro usb should i connect the sense to the black wire as in otg cable then only it will pass the current?? and the big problem for me is i do not have a multimeter
 
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