EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Mugiwara303 on June 19, 2021, 07:33:57 pm
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I have a question about the role of the diode between the base and emitter this is a voltage detector circuit, schematic in attachement
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To protect the base-emitter (BE) junction from significant reverse bias. For NPN transistor, positive base potential relative to emitter cause the junction (which is like a diode) to be forward biased and start conducting. With forward bias, transistor works as a current controlled device. More current between BE means more current between CE. But BE junction may be damaged if reverse bias (base negative relative to emitter) is applied. BE junction of a typical small transistor starts to breakdown at around 5-7V of reverse bias voltage. This breakdown may damage the BE junction, reduce the hfe (basically gain) or cause total failure. Because in this circuit input signal from antenna is unpredictable, the diode is used to clamp any excess negative voltage up to safe limits, so base does not go too much negative in respect to emitter.
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For fun I made this schematic...
[attach=1]
I used Multisim Online -- multisim.com. The "100A/A" is simply the beta of the transistors.
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Not just for protection. Without the diode, the antenna will develop a charge which prevents the first junction from conducting.
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Not just for protection. Without the diode, the antenna will develop a charge which prevents the first junction from conducting.
Fair point, however I don't think that it would cause a problem in this voltage detector circuit, because base has a tendency to float slightly positive. That is because of CB and BE leakage being present. Because the input capacitance is presumably very small I would not expect base DC point to float negative for extended time to disrupt the work of circuit.
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Thank you, one more thing to add is when I put a shotky diode instead of the 4007 the circuit doesn't work at all and without any diode the range of detection decreases
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A Schottky diode has a higher reverse leakage current than a normal silicon diode, this will most likely be biasing the transistor hard off.
You may want to try substituting a normal signal diode for the 4007, something like a 1N4148 which will have lower junction capacitance. The 4007(1kV) might have been chosen for very low leakage at very low voltage though.