Hello,
I have a doubt about the following section in the Ebook.
Volume IV - Digital » LOGIC GATES »The NOT gate
The following excerpt is from the Ebook. (Please refer to the attached picture)
Now there will be current through the left steering diode of Q1 and no current through the right steering diode. This eliminates current through the base of Q2, thus turning it off. With Q2 off, there is no longer a path for Q4 base current, so Q4 goes into cutoff as well. Q3, on the other hand, now has sufficient voltage dropped between its base and ground to forward-bias its base-emitter junction and saturate it, thus raising the output terminal voltage to a "high" state.
If Q4 is cut off, I could not understand how Q3 base emitter junction can be forward biased? With Q4 cut off, D2 and Q3s emitter are in effect cut off from the circuit right? How can the BJT base emitter junction be forward biased with out the emitter being connected to a relatively negative potential compared to the base?
Thank you.
Manulal.
I have a doubt about the following section in the Ebook.
Volume IV - Digital » LOGIC GATES »The NOT gate
The following excerpt is from the Ebook. (Please refer to the attached picture)
Now there will be current through the left steering diode of Q1 and no current through the right steering diode. This eliminates current through the base of Q2, thus turning it off. With Q2 off, there is no longer a path for Q4 base current, so Q4 goes into cutoff as well. Q3, on the other hand, now has sufficient voltage dropped between its base and ground to forward-bias its base-emitter junction and saturate it, thus raising the output terminal voltage to a "high" state.
If Q4 is cut off, I could not understand how Q3 base emitter junction can be forward biased? With Q4 cut off, D2 and Q3s emitter are in effect cut off from the circuit right? How can the BJT base emitter junction be forward biased with out the emitter being connected to a relatively negative potential compared to the base?
Thank you.
Manulal.
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