# How do I amplify a dc voltage that is too weak to trigger the base in a transistor?

I'm trying to take the voltage output from a 3.5mm audio jack and have it light up an LED according to it's voltage. I was going to attach the left audio cable to the base of a 2222n transistor which would act as my amplifier to trigger the LED to light. Unfortunately, the voltage reading off the wire is like .2 or .02 and isn't high enough to trigger the flow. Is there any way to increase the voltage to trigger the transistor or would I have to use a transformer of some sort (or an op amp??)

• Is it an audio signal that you want to trigger the LED? May 19, 2013 at 20:48
• You will need an external power source like a battery... you can't amplify a voltage unless it has a lot of current to back it up, which 3.5mm audio does not. Take a look at this question: electronics.stackexchange.com/questions/54980/… May 19, 2013 at 20:58

The problem which you have correctly identified (good for you!) is that the incoming signal has small voltage swings around 0V, but the transistor requires around 0.6V to begin to turn on. The solution is to bias the transistor to a given operating point: providing a base voltage which causes a degree of turn-on that exists when there is no signal. Then the signal is shifted to that same voltage level, so that is swings around that voltage rather than zero (or whatever is its original DC offset).

For AC signals, this level shifting is easily accomplished. We simply connect the AC signal source to the transistor's base not via a wire, but through a capacitor. The capacitor blocks DC current, put passes through the AC voltage swings, superimposing them onto the voltage of the target node.

simulate this circuit – Schematic created using CircuitLab

In this circuit, the sine wave generator is configured to generate a sine wave with an amplitude of only 0.1 mV. Yet, the positive peaks of this wave cause peaks of 1.5mA of current to flow through the LED. This is because of the bias set up by R1 and D1 creates a voltage which, conveyed via R2 to the base of Q1, sets up sets up that transistor's operating point so that it is slightly turned on, just on the verge of being much more turned on. In fact, according to the simulator, about 50 microamperes already flow through the LED when there is no incoming signal. So from this operating point, any voltage swings in the signal cause a response in the transistor. (Why a diode? Because a diode has a similar voltage drop to the base-emitter junction of a transistor of the same type, e.g. silicon.)

Negative swings in the input are ignored, and the flashes of peak current dumped through the LED are proportional to the amplitude of the positive swings, so it should vary in brightness with the signal level. The resistor values have to be adjusted based on your LED.

The responsiveness to small signals is quite sensitive to R1, R2 and R3. If R1 is too small, for instance, then VBIAS will turn on the transistor more; small signal swings cause more LED current than before. R4 needs to be adjusted based on the supply voltage, the type of LED, and the desired maximum current that is to be supplied to it.

R3 is only 0.22 $\Omega$ which is deliberate: that is to provide only small feedback to stabilize the operating point of the transistor against thermal runaway, without sacrificing a lot of current gain that would make the circuit less sensitive. In a pure on/off LED switching circuit, we would not have R3, but here we are keeping the transistor slightly turned on all the time, with a slight quiescent current, which brings in the risk of thermal runaway.

R5 protects the base-emitter diode of Q1 against current driven by the input, because the emitter resistor R3 is insufficient.

One problem with the above circuit is the small input impedance, essentially dictated by R3. This is fine for being driven by a speaker or headphone output, but it's too small for line level outputs which expect something in the neighborhood of 10K. The behavior is also very nonlinear. A doubling of the input peak from 0.1V to 0.2V much more than doubles the LED current. The behavior follows the nonlinear VBE versus collector current curve of the transistor. We can address both issues with these changes:

simulate this circuit

The first notable difference is a much larger emitter resistor R3 which provides a lot more negative feedback to stabilize the bias point. By itself, that costs us a lot of gain, but we can recover some of that by bypassing R3 with a capacitor to ground. R5 is no longer necessary. R2 increases because the original 2.7K would detract from the newly increased impedance of the base. R1 is slightly decreased to boost VBIAS a little bit to compensate for some loss of sensitivity.

According to the simulation, the input impedance is about 8.3K at 1 kHz, which is reasonable for line level, when we are not actually trying to preserve the frequency response of the audio but only lighting a LED. It drops to around 6.4K at 10 kHz.

Calculating the input impedance from simulation: obtain a plot of the current flowing into C1. Check that this is in phase with the input voltage. Then divide the peak-to-peak input voltage by the peak-to-peak current.)

• Thank you, this was perfect! I need to study the circuit some more to understand every detail but I know what the end goal is which is to provide a voltage that will basically be sitting just below 0.7v so that the signal will bump it up over the "on" point. Kind of like an offset voltage so to speak. May 20, 2013 at 1:29
• @SonicD007 It's not that perfect! The circuit has a low impedance: basically that of R5. It should be okay with an output that can drive headphones, but it's low for a line out output that expects a "line level" type input. The way to improve this is a larger R3, where we recover lost gain by bypassing R3 with a capacitor to ground.
– Kaz
May 20, 2013 at 1:57

Looks like Kurt sees it the same as me. AC / audio trigger circuit: -

V2 is the audio input.

It can be made to run off a smaller supply but I think about 3V3 will be the limit and also it'll have to be a normal LED that drops about 1.8V.

Higher supply voltage (no more than 12V) means you can trigger a higher power LED.

• I don't quite understand the circuit you posted. I'm still a novice so I don't fully understand what the capacitor and resistor do to the audio signal there. This is the circuit I had built which doesn't trigger the npn transistor. i.imgur.com/ivmsWAR.png?1 Any advice on how to change the circuit I have now to get this working would be very much appreciated. May 19, 2013 at 23:14
• Your circuit should be posted in your question, and it is useless unless you show some component values. Possibly all you have to do is reduce the emitter resistor, but it's anyone's guess at the moment. May 20, 2013 at 0:08

You are thinking of the transistor as a digital switch. You think that .2 volts isn't enough to turn on a transistor that needs .6 volts to do so. But a transistor was also meant to operate as an analog device. Such circuits "bias" the transistor by setting the base voltage up into the "turning on" region, and they do this through a high resistance. Then your "small" signal can coupled in and it will cause the transistor output to vary. The output will be amplified, and if that's not enough, you can do this again with another stage until the signal is large enough to do the job you set out to do.

An op is just a (large) bunch of transistors again, doing things much like I just described.

The circuit posted by Andy uses a somewhat fancier current source, which in turn is like the voltage and high resistance I described, only it works better (higher resistance and self-adjusting).