# How can I effectively reduce the voltage needed to activate a transistor?

I have built a circuit which basically connects the line out (audio output) of a music playing device to a set of LEDs (actually a huge strip of around 200 LEDs), so they flash in time with the music (from internet tutorials - I'm a bit of beginner).

simulate this circuit – Schematic created using CircuitLab

My circuit works very well using my laptop as the audio device (connecting my circuit to the headphone jack on it). But when I use something smaller such as an iPod, the lights barely turn on at all.

I've tried using a Darlington Pair (below), but that makes the issue worse. This is why I think the issue is that the audio line out is not reaching the 0.7 volts across the base and emitter that the TIP31C transistor needs to activate (the Darlington Pair means it now needs 1.4 volts to activate).

simulate this circuit

From my research, it looks like using an op amp might be the way forward, to amplify the audio line out signal before the TIP31C transistor. Would somebody be able to suggest one, and which inputs I should connect to?

I've also read that Germanium transistors only need 0.3v across the base and emitter to activate, would that be useful?

• I removed the explicit 'thanks' from your post. You may express your thanks by upvoting any good answers that come in, and via comments to those answers. (Upvotes are the preferred channel.) Jun 6, 2013 at 11:57

In short: you can't. The 0.6V threshold for a BJT is a consequence of the physics of silicon P-N junctions.

A germanium transistor would work, but you will have to mail-order it, and it will be expensive.

A rail-to-rail op-amp indeed may be an option.

However, another solution is to make the voltage of your audio signal higher, rather than making the transistor threshold lower. You could do this two ways:

Make the emitter voltage lower

simulate this circuit – Schematic created using CircuitLab

Now, the audio signal is 0.6V higher than the emitter. Of course, you'd have to come up with a way to get a 0.6V power supply, and probably adjust it to get just the action you want. There's another way...

Add a DC bias to the signal

simulate this circuit

Here you can adjust the pot to add some amount of DC bias to the signal to get the sensitivity you desire. The capacitor serves to isolate this DC from your audio source while allowing the AC signal to pass. This is called capacitive coupling.

R4 exists to limit the base current in case R1 is adjusted too far. There's no point in biasing the signal above 0.7V since that would mean the transistor is always on, so R4 also makes the useful adjustment range of R1 wider.

Also, notice in both cases I've added a resistor to the transistor base. You don't want to make this mistake.

• I tried adding a DC bias to the signal, one problem though! My circuit is 12V, 4A (it comes from a transformer which is needed to power the 300 LEDs). So when the TIP31C transistor is not on (so current is not flowing through LED strip), the pot was having to take 48 Watts, which blew it. I cannot find any pots that have a power rating nearly that big. Any suggestions? Jun 7, 2013 at 14:53
• @CraigWalton 48 watts? How do you figure? 12V on a $1k\Omega$ pot is $(12V)^2/1000\Omega = 0.144W$. This may be too much for a tiny trim-pot, but any panel mount pot will handle it fine. You can also use fixed resistors instead of a pot. Jun 7, 2013 at 15:48
• @CraigWalton also, I just happened to find this answer on capacitive coupling and DC biasing: electronics.stackexchange.com/questions/60694/… Jun 7, 2013 at 15:59
• I misunderstood the power, I thought it would be 12V*4A = 48W. I read through that "Capacitive Coupling/DC biasing" question and answer, it makes a lot more sense now. I'm struggling to figure out what value of capacitance to use. I know that I have to use F = 1 / (2 π R C) where F is the lowest frequency (20Hz), R is the impedance it will be driving in Ω, C is the capacitance. Will the impedance in your DC biasing circuit above just be whatever the "bottom half" of the pot resistance is, that is, as if there were 2 fixed resistors, it will just be the bottom resistor? Jun 8, 2013 at 11:29
• @CraigWalton no, it will be some combination of the two of them. If you read through Olin's answer I linked above, he goes through calculating the impedance in detail. The pot is equivalent to R3 and R4 in his circuit, although he has them connected slightly differently. You could connect the pot similarly. Or, use $1\mu F$ and probably it will be fine. If you find it doesn't respond to bass well enough, make it bigger. If you are just blinking an LED, you don't need super high fidelity. If you want to understand more, google "input impedance common emitter" or ask a new question. Jun 8, 2013 at 11:38

You can use an op-amp that accepts input to the negative rail, e.g. LM158, to drive the main switching transistor (BJT or MOSFET), thus:

simulate this circuit – Schematic created using CircuitLab

The above arrangement will cause the LEDs to light up at less than 150 mV peak to peak input signal.

• For higher gain, reduce R2.
• If the LEDs remain on all the time, reduce the gain by increasing R2.
• To increase the maximum current through the LEDs, reduce the value of R4 (and vice versa)

The BAR28 Schottky Diode is added to shunt the negative part of the input signal to ground, to prevent exposing the op amp input to too low a voltage below the ground rail.

I too would recommend an op-amp circuit, like the LM158 already suggested. Its a good way to ensure that the circuit can be easily altered to accommodate several different audio sources. My only caution is that if you use a diode to block negative signal as shown, be sure to add a resistor to the input, or you'll risk clipping the audio, and cause audible distortion. I've found typical ear-bud impedance to be in the neighborhood of 32 ohms, so a resistor around 1K or higher should prevent this problem. (Sorry-- I'd have added this suggestion as a comment, but I don't have enough "reputation" yet)