I was just trying to learn how a transistor actually works by tinkering around myself. I have an Arduino running a flashing light program. I hooked up an NPN transistor to test out how I can make it into a switch by limiting the base pin with a potentiometer. So, I have the emitter hooked to ground, Collector hooked to the LED which goes to positive. Now I put a resistor in the base pin spot and when I touch the resistor, it turns on the circuit. From the little research I did, I learned that humans can emit a voltage? How true is this or am I just messing up a circuit somehow?
Circuit Activation – How Touching It Turned It On
lednpnpowertransistors
Related Solutions
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.
Do you have enough headroom to increase the current sensing resistor slightly an drive the BJT from a potmeter in parallel?
Of course you could optimize R3 to a series of resistors with ratios 270:100:180, where 270 is near ground, 100 is your pot and 180 at source.
Best Answer
So I gather that you have something like the schematic below, except that I added R1 because I can't bear to draw a device for burning out LEDs and transistors.
You're in a room that has a strong electric field at 50 or 60Hz, and you're a fairly large conductor compared to the bare base lead of the transistor. You've got some small resistance to ground, and the transistor base leaks to ground anyway.
So what the base of the transistor sees when you touch it is a fairly strong line-frequency source in series with a small-valued capacitor, and maybe a large-valued resistor. This combination is enough to pump current into the base of the transistor on the peaks of the electrical cycle. Apparently in your case, that causes enough collector current to make the LED visible.
Most likely, it's as simple as that. If you have an oscilloscope you can see this just by touching the probe (it's a nice quick test that the probe is connected, in fact). Also, if you have an oscilloscope, you could put it on the collector of the transistor and see the voltage vary at the line rate when you touch the transistor base.
simulate this circuit – Schematic created using CircuitLab