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Question concerning increasing voltage for phone auxiliary speaker

Pharaday

Jan 18, 2016
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I have taken guts of a speaker that you plug into your phone with an audio cable and replaced the speaker with LEDs so that any signal going into the audio cable causes the LEDs to light up, brighter if loud and dimmer if quiet. Simple.

It uses a little 5 volt rechargeable battery. My problem is that even one LED does not get very bright from the input that I'm giving it. Increasing volume or amplifying the signal would solve this. Unforch, I cannot do that.

My question is what would blow up if I gave it a battery with a little bit more voltage so that it will power a few LEDs at decent brightness.

Would that work? It feels even more silly after having typed it out, if I'm honest....

Here's a pic, you can see the blue battery on the right, the circuit board on the left, and one of the diffused LEDs I'm using in the middle.

20191020_131012.jpg
 

Audioguru

Sep 24, 2016
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You do not say the color or voltage required by the LEDs. You also do not say the voltage of the battery that you are using.
If the LEDs do not have a current-limiting resistor or circuit then if you increase the battery voltage the LEDs will be too bright and burn out in a moment.

I think if you have a white LED and it needs about 3.2V and your battery looks like a Li-PO that is 4.2V when fully charged but it is almost dead at 3V. Then the LED is very dim without the phone signal and the signal adds a little voltage that makes LED a little brighter. To make the LED a lot brighter then you must simply amplify the signal.
 

Pharaday

Jan 18, 2016
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My bad, The battery seems to max out at 4.7 volts. It's charged by USB so I know it's cool with 5 volts. I harvested the LED so I'm not sure of it's specs, but it's one of those surface mounted tiny ones. Not the smallest of the surface mounted LEDs but about middle-sized surface mount LED. It gets a good brightness without risking burnout from a 9 volt battery and a 10,000 Ohm resistor.

The LEDs are a dime a dozen I'm just concerned that I might put too many amps through that circuit and burnout some part of it. I'd like to try giving it 7 to 10 volts to see if it could run more than one LED but not at the risk of the circuit.
 

Audioguru

Sep 24, 2016
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If your battery is a Lithium type and is at 4.7V then it is over-charged and is ready to catch on fire or explode. You need a lithium battery charger circuit that can be fed USB 5V. A Lithium battery charger output is never higher than 4.20V.

You are biasing an LED with a DC voltage that is high enough to make the LED glow dimly then you are modulating the LED voltage with the phone signal to make it brighter when the phone signal is loud. You must limit the maximum amount of LED current to avoid destroying the LED. The datasheet for the LED will list its maximum allowed continuous current, but yours will fluctuate with the phone signal.
 

BobK

Jan 5, 2010
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You are biasing an LED with a DC voltage that is high enough to make the LED glow dimly then you are modulating the LED voltage with the phone signal to make it brighter when the phone signal is loud. You must limit the maximum amount of LED current to avoid destroying the LED. The datasheet for the LED will list its maximum allowed continuous current, but yours will fluctuate with the phone signal.
What makes you think that? The OP says he replaced the speaker in an amplified speaker with the LED. The amplifier is being run by the battery, which might well be a 4 cell NiMH which would be 4.8V.

The amplifier is either bridged (likely) or not. If it is a bridged class D amp it can output nearly 4.8V when driven with a high enough signal. If it is an ordinary class AB the max peak amplitude is probably around 2V at best.

Bob
 

Audioguru

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So then the circuit board is not "the phone", instead it is the amplifier from an amplified speaker.
Driving an LED from the output of an amplifier will result in no light from the LED most of the time then the LED brightly flashes for peak levels from the amplifier.

I think it will look better if the battery biases the LED so that it almost lights with no signal then its brightness will follow the level of the amplifier output, and the LED will be dim for low levels, medium brightness for medium levels and bright for high levels.

The battery is the wrong shape for four AAA or AA cells but instead it looks like a single-cell Li-PO.
 

Pharaday

Jan 18, 2016
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You guys are like titans of electrical knowledge and I am but a man gazing up at you in awe. I really appreciate your input.

Yes this is seperate from the phone. Its just a little amp and speaker that you can plug into a phone or anything. I haven't measures the max output but I can.

Please excuse this novice, but could you explain how I could bias the LED with the battery have it change the LED intensity according to volume? Bias means to put positive battery wire to positive leg of led and negative wire to negative leg? Bias = to power?

To me, if I connect that battery to the LED right now (and the 4.7 volts didn't instantly burn out the LED), the LED would just be steadily on.

Anyways, I think I could add a simple transistor amp between the speaker output and the LED with it's own power source. I could amplify my signal that way I guess but we're getting kinda..... convoluted here hahaha!
 

Audioguru

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The LEDs need DC but the amplifier output is AC. Without biasing the LED then it does not light until the signal AC peak voltage exceeds the forward voltage required to light the LED which might be 3V. Then the LED is turned off most of the time.

I suggested to bias the LEDs through a series resistor so that they are almost lighting and use the audio signal from the amplifier to increase the LED voltage for brightness control. Of course you must limit the maximum current with the value of this series resistor so that the LEDs do not blow up.

If I was building this circuit then I would use a peak detector circuit to hold the peaks for a moment so that the LEDs do not simply blur on and off at the audio frequencies.
 

BobK

Jan 5, 2010
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What @Audioguru means by biasing the LED is that the audio output should be put in series with the battery and a resistor. This means the battery voltage is added to the audio voltage, and that voltage is applied through the resistor to the LED.

But you cannot use the same battery for this as the one that is powering the amplifier board.

If you want to try this, I would use a separate 3V battery for the bias.

Bob
 

Audioguru

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You do not need a separate battery to bias the LED. simply use a voltage divider to set the bias voltage then capacitor-couple the audio to the LED and its current-limiting series resistor.
 
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