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I'm new to electronics. I'm trying to figure out how this circuit works to fade the LED light in and out:

http://www.555-timer-circuits.com/images/circuit-fading-led.gif.

This is how I think it works (of course, I'm wrong):

  1. When you initially connect the circuit, power flows from the 9V power source (I know electrons actually flow in the opposite direction, but I'm sticking with convention) to pin 8 and 4 on the 555 timer. Both the "set" and "reset" comparators inside the 555 timer output 0, causing the flip-flop inside the 555 timer to keep its initial state (i.e. 0) and output 0, which gets passed to the output stage inside the 555 timer, causing the output stage of the 555 timer to allow pin 3 of the 555 timer to act as a source of power. The 33K resistor (i.e. R1) dramatically reduces the voltage across pins 6 and 2 to some value less than 3V, such that both the "set" and "reset" comparators within the 555 timer continue to output 0. However, the current flowing through pin 3 begins to charge the 100uF capacitor (C1), as well as flow through the base of the NPN transistor (Q1), which also allows the much greater current from the 9V power source to flow through the LED, illuminating the LED (although maybe not very brightly, as the current flowing through the base of the transistor is very low and the amount of current that gets through the transistor from the battery is in proportion to that current).
  2. As C1 becomes more charged, less current flows into C1 and more current flows through the base of the transistor, making the LED light shine brighter. Eventually, the charge across C1 results in a voltage across pin 6 and pin 2 greater than 6V, causing the "set" comparator to output 1, which sets the state of the flip-flop to 1, which causes the output stage to change pin 3 from a power source to a power sink. C1 starts to discharge across R1, as well as through the base of the NPN transistor (Q1) - first quickly, causing the LED to continue to shine brightly at first and gradually become less bright as the charge across C1 diminishes.
  3. Eventually, the charge across C1 results in a voltage across pin 6 and pin 2 that is less than 3V, causing the "reset" comparator to output 1 (the "set" comparator already started output-ing 0 when the voltage dropped below 6V), causing the flip-flop's state to change to 0, causing the output stage to allow pin 3 to act as a power source again.
  4. The process repeats.

Is my thought process correct? Did I miss anything?

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2 Answers 2

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Q1 is an emitter follower. The voltage on the emitter is the capacitor voltage minus 0.7V or so.

If the LED Vf is 2V then the current through the LED is about (Vc1-2.7)/470. So at 3V on the capacitor, the LED is almost off.

The base current is about 12-25uA (depending on transistor beta) when the capacitor is at 6V (peak) so it will lengthen the cycle a bit, depending on the type of 555. A bipolar type will output about 7.8V with light current, so the current through R1 will be only about 55uA, so it might fail with a BC547A.

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  • \$\begingroup\$ Thanks for your response! Why would the voltage on the emitter only be affected by the capacitor voltage? Wouldn't it also be affected by the voltage coming from pin 3, while pin 3 is in power source mode? And how about the voltage from the battery? Doesn't that figure into the voltage at the emitter? Also, why might it fail with a BC547A if the current through R1 is around 55uA? \$\endgroup\$
    – lostinthecloud
    Dec 27, 2017 at 23:21
  • \$\begingroup\$ 1. When the transistor is in the active mode (which it is here) Vbe is 0.7V. If we ignore the base current the emitter voltage is the capacitor voltage minus the 0.7V. The current through the resistor from pin 3 will eventually change the capacitor voltage, just as you say. 2. The minimum gain of a BC547A is only 110, meaning the peak (with Vcap = 6V) base current could be as high as 63uA (at room temperature, more at low temperature), so the capacitor will never get to 5V, it will stall at maybe 5.6 or 5.8V, so pin 3 will stay high, the LED will stay on fairly bright and that's that. \$\endgroup\$
    – Spehro Pefhany
    Dec 28, 2017 at 4:18
  • \$\begingroup\$ @lostinthecloud The current coming from pin 3... flows through the capacitor which affects the capacitor voltage. \$\endgroup\$
    – user253751
    Jun 30, 2020 at 10:16
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In the emitter follower configuration, Q1 acts as a current amplifier. Within limits, the emitter voltage always is 0.6 V or so less than the base voltage. As the base moves up and down, the emitter tracks it minus the offset cause by the base-emitter junction forward voltage. The reason for the transistor is that it can pass around 100 mA of current from the collector to the emitter for every 1 mA of current into the base.

In your circuit, if the LED were attached directly to the R1-C1 node it would try to draw so much current through the 33 K resistor that the voltage drop across the resistor would be too much, and the current would be so low that the LED would not light. With the transistor in there, only about 1% of the LED current is coming from pin 3 through R1. The rest comes directly from the 9 V source. So your statement in #2 about "less current flows into C1 and more current flows through the base of the transistor" is incorrect the way you mean it. The base current into the transistor does increase slightly as the R1-C1 voltage increases, but only because the the current requirement of the LED increases as the voltage across it and R3 increases. The load connected to the emitter is a form a negative feedback called emitter degeneration. This is what stabilizes the circuit at each instant as the capacitor voltage ramps up and down.

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  • \$\begingroup\$ Thanks! Are you saying that the current through the base of the transistor increases as the charge across the capacitor increases because the voltage at the base of the transistor increases as the charge across the capacitor increases (as opposed to the way I described it in my original post, which made it sound as though the rate of the flow of electrons into the capacitor gradually slowed as the capacitor became more charged)? Does the flow of electrons into a capacitor slow down as a capacitor charges? \$\endgroup\$
    – lostinthecloud
    Dec 28, 2017 at 0:08
  • \$\begingroup\$ Yes, electron flow decreases as the capacitor voltage increases toward the voltage at pin 3 (either charging or discharging). But the base current does not increase strictly because the base voltage above ground is increasing. If the emitter had a constant current load instead of an LED and resistor, the base current would not change as the capacitor voltage ramped up and down. Not enough characters to cover all of it in one message. \$\endgroup\$
    – AnalogKid
    Dec 28, 2017 at 1:58
  • \$\begingroup\$ The current through the emitter is the sum of the base and emitter currents, roughly 1:100 in a small signal transistor. Base voltage rises, this increases the base current because one circuit is from the base to the emitter through the base-emitter junction, basically a diode, through the resistor and through the LED to GND. So base current increases, causing an increase in collector current of 100 times more. \$\endgroup\$
    – AnalogKid
    Dec 28, 2017 at 2:06
  • \$\begingroup\$ As the emitter current increases, so does the voltage across R3, and as the voltage across R3 increases, the base current decreases. Eventually a point of stability is reached, until the base voltage changes again. With a voltage ramp on the base you get a current ramp through the LED. \$\endgroup\$
    – AnalogKid
    Dec 28, 2017 at 2:06

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