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- Miron Cristea
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- Wednesday, 29 January 2020
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When the input current is null, the transistor is open (conducts current) presumably into the next gate, which opens the next gate Zener diode and turn-off the next transistor. In this way, the base voltage of the first transistor is low (its Zener diode off) and the collector voltage is V_Z (Zener of the next gate is on) i.e. high. Results that base-collector junction of the first transistor is forward biased, i.e. the transistor is saturated. This is not good for the switching speed.
Any comments?

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