EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: lumdef on March 15, 2025, 09:01:47 am
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What is the purpose of this single 1N4148 diode? Can current flow from the emitter to the collector in that case?
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The text says it all: "when the voltage on the base of the first transistor sees 0.6V + 0.6V+ 0.6V = 1.8V..."
The first 0.6V above is the VBE of the left-most transistor, the second 0.6V is the VBE of the right-most transistor, and the third 0.6V is the voltage drop on the diode.
Why should there be a problem for the current to flow between the emitter and collector of the right-most transistor? If one were to reverse the diode such that its cathode was connected to the collector, then you would have a problem.
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The diode shifts up or down, all at once, the detected voltages in the "staircase".
If you insert zero, one or more diodes where the D1 is, the input voltage to turn on the first LED will change accordingly (and it will change with the same amount for the following LEDs, too). The "distance" between each voltage step in the "staircase" remains the same, but the beginning voltage of the entire "staircase" moves.
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In very simple terms, think of it as being there to increase the "thresholds" of the transistors. This ensures that LED 1 needs a little more voltage to turn on, preventing it from activating at a very low voltage.
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The diode raises the threshold of the first transistor by 0.6 volts to 1.2 volts. This is often done to increase noise immunity.
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Thank you, it is clear now.