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In the newest Make magazine, they show how to connect LEDs in line to speakers to have the LEDs light up when the speaker is playing music (pulse with the music). They say you may want to hook up the LEDs to a 9 volt battery in order for the LEDs to come on at lower volumes. They accomplish this via a transistor. I don't get how this works.

The connections are as follows:

NPN transistor:

Base goes to + lead from Amp

Emitter goes to Negative lead of Amp and 9V battery

Collector goes to cathode of LED

Battery:

Positive lead of battery goes to anode of LED (through resistor)

Somehow, when voltage is applied to the + lead of the Amp (music is playing), this turns the LED on?

I would have thought the LEDs would always be on, the connection goes + battery, LED, collector of transistor, emitter of transistor, - of battery. I thought that if a positive voltage was applied to the collector, with the emitter at ground, this would be a closed circuit?

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  • \$\begingroup\$ Please add a schematic next time. You don't need fancy software, a hand-drawn schematic is still better than a description. \$\endgroup\$ – pebbles Sep 7 '13 at 11:17
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The transistor may act as a switch or a variable resistor. If no voltage is applied to the base (more precisely: no current flowing into the base) then the switch is open. As base current is applied it gets amplified by the transistor into an N times larger collector current. The "N" is an important transistor parameter, called \$H_{FE}\$, and it defines the current amplification factor. For general purpose transistors this is often around 100.

So if you apply a base voltage (you need a series resistor!) so that there will flow, say, 1 mA, then there will be 100 mA collector current if the circuit allows it. That means that other components may limit that current to a lower value. Let's assume your LED has a 2 V voltage drop, that will be rather constant for that type of LED. Then assuming the transistor is fully conducting (no voltage drop between collector and emitter) you'll have 9 V battery voltage - 2 V LED voltage = 7 V across the resistor. If we choose a resistor value of 350 Ω then, according to Ohm's Law we have a current of 7 V/ 350 Ω = 20 mA through that resistor and therefore also through the LED. (20 mA is a typical current for an indicator type of LED.)
So, while the transistor would like to draw 100 mA, the resistor will always limit that to the lower 20 mA.

You don't say what the signal from the amplifier is. Is that a line level (500 mV) or a speaker output level (3 V for 1 W)? In the first case the voltage will be too low; a transistor's base has to be at 0.7 V minimum before current starts to flow. If you use the speaker output you can use a 1 kΩ resistor in series with the output to limit the base current.

Also place a diode (1N4148) in anti-parallel with the base: cathode to the base, anode to ground. This prevents too large negative voltages across the base, which would destroy the transistor.

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