Volume IV - Digital » LOGIC GATES »The NOT gate doubt

Discussion in 'General Electronics Chat' started by manulal, Jul 25, 2012.

  1. manulal

    Thread Starter New Member

    Dec 6, 2010
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    3
    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 Q3’s 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.
     
  2. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    It's true there is no DC current flowing in the output if there is no output load, but during the transition from output low to output high there is momentary current flow from Q3 to charge the output capacitance and change the output state. The change in output state is due to Q3's base-emitter junction being forward biased. During steady-state conditions the Q3's base-emitter junction remains forward biased even though no current flows because the base remains a few tenths of a volt above the emitter to supply Q4's leakage current.
     
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  3. manulal

    Thread Starter New Member

    Dec 6, 2010
    17
    3
    Thank you very much for your reply.

    “During steady-state conditions the Q3's base-emitter junction remains forward biased even though no current flows because the base remains a few tenths of a volt above the emitter to supply Q4's leakage current.”

    But if there is no load (no load is shown in the diagram) and Q4 is cut off, how will this leakage current flow in the first place?

    Manulal.
     
  4. manulal

    Thread Starter New Member

    Dec 6, 2010
    17
    3
    Did I understand it wrong?

    “a few tenths of a volt above the emitter to supply Q4's leakage current.”

    Did you mean Q3’s leakage current?
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,805
    Think of it this way.

    Imagine putting a variable resistor between the output and ground. Now do you see the current path between Vcc, through R2, through Q3, through D2, through the load to ground? Now imagine that we make the resistor of moderate size, say 10kΩ. The voltage at the output will be about two diode drops below Vcc, or roughly 3.6V (a tiny amount less because of the drop across R2, but that should be pretty small). Now what happens as the resistor size is increased? The amount of current flowing decreases, which means that not as much voltage appears across the base-emitter junction of Q3 or across D2, so Vout goes up. The bigger we make the load resistor, the smaller the current and the less the voltage drops across those two junctions and, therefore, the closer Vout gets to Vcc. In the limit as the load resistance becomes infinite (or if we remove it entirely), the voltage drop across both will be zero and we will have 5V at the output. Now, any current that the output has to deliver (including current that leaks through Q4) has to come through Q3 and D2 and so there will be a slight drop in voltage, but to get it down low enough that it no longer qualifies as a "high" output, you've got to load it pretty aggressively.
     
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  6. manulal

    Thread Starter New Member

    Dec 6, 2010
    17
    3
    @WBahn

    Thank you for your kind explanation.

    “In the limit as the load resistance becomes infinite (or if we remove it entirely), the voltage drop across both will be zero and we will have 5V at the output.”

    What I understood from the above is that even if the load is removed entirely there will be a base current in Q3. This is flowing through Q4 as a leakage current even though Q4 is in cut off stage. Am I right in this?

    Manulal.
     
  7. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    Yes, there is a small leakage current through Q4 when it is off.
     
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  8. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,805
    Yes, although that current will probably be so small that you will still see something very close to 5V at the output. In fact, if you measure it, the meter you measure it will likely present about a 10MΩ load and require a little bit of current (about 50nA) which may well be more than the leakage current in question.

    Also, as rule it is best to use the default font when you reply. It is the one people are used to reading and having all the font tags can make quoting portions of your response difficult.

    Cheers!
     
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