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I'd like to create a simple circuit that makes an LED brighter depending on the amplitude of an audio signal. My idea is to use the following schematic ;

schematic

simulate this circuit – Schematic created using CircuitLab

My question is this - what happens when the voltage of the AC signal goes negative? Does any current flow out of the base of the transistor? And if so, what happens to the current through the transistor? Also, I appreciate that this isn't a great circuit to indicate the amplitude of the AC audio signal, but it should more or less work right?

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2 Answers 2

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The led will not even light up until the signal amplitude is higher than the transistors Vbe (typically 0.65v for a bjt) unless it is DC biased by this amount. I would advise AC coupling the audio signal into the pre biased transistor, in order to get a better proportial relationship between signal amplitude and led brightness. Also reverse biasing the b-e junction is ok until you exceed the
V(br)EB at which point R2 will limit the Ieb (reverse base current) and the Ice current will be 0, Icb will be very small until V(br)CB is reached.

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  • \$\begingroup\$ This is a good answer. A few other notes: The circuit has a strong dependence on beta of the transistor, which is a variable characteristic. It may be necessary to tweak the resistor values on an case-by-case basis. Depending on the output impedance of the signal, the base resistor may load it a bit, and could attenuate it or add distortion. You need to take the trouble to estimate the peak base current, the desired peak current for the LED, the transistor beta, and use appropriate resistor values as a starting point. You don't want the transistor in saturation, but in linear region. \$\endgroup\$
    – mkeith
    Nov 18, 2014 at 18:02
  • \$\begingroup\$ That's a very good point - which in fact means I can use the circuit instead as a more-or-less-right clipping indicator for line level output. Which is actually what I really want anyway (typical... not asking the proper question). And if I divide down the AC signal a bit too I can get the clipping level to more-or-less exactly 1v. \$\endgroup\$
    – Dave Branton
    Nov 18, 2014 at 22:10
  • \$\begingroup\$ It may work. In essence you are making use of the base emitter Voltage as a reference. When your divider output is over around 0.6V, the base current will start to flow. Note that base emitter voltage has a strong temperature coefficient. So this circuit won't be great in that respect. It may tend to drift as temperature changes. \$\endgroup\$
    – mkeith
    Nov 21, 2014 at 6:36
  • \$\begingroup\$ I would probably use a comparator with open collector output. The positive terminal would be Vin, the negative terminal would be held at 1.0V (somehow) and the output would turn on whenever Vin > 1.0V. \$\endgroup\$
    – mkeith
    Nov 21, 2014 at 6:37
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Your circuit will basically work. But when the base voltage is lower than the emitter, then the PN junction will be reverse biased, and if

$$ V_{EB} > V_{(BR)EBO} = 6V $$

Your base-emitter may breakdown, and there will be a current flow out from base, the current will be limited by \$R_{2}\$.

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