I would like to know more detail on how all the transistors in the ttl turned on.

Discussion in 'General Electronics Chat' started by lynnfaiz, Jun 29, 2015.

  1. lynnfaiz

    Thread Starter New Member

    Dec 16, 2012
    29
    1
    This thread was separated from TTL circuits Help. (NOT, AND, OR, ETC.)

    Hi Wendy and all.
    i have read through the e book on the ttl chapter. it a good explaination there.
    specifically on nand gate or inverter, I would like to know more detail on how all the transistors in the ttl turned on.
    Is it caused by the Base Voltage or the Base Current?
    It would be very helpful if someone can explain how the transistors in the TTL turn ON technically. The explaination using the diode does help but doesnt give a clear picture (technically) how Q1 that is OFF state can turn on the next transistor.
    Really need help on this. and i thanks u all in advance!

    Mod edit: please don't hijack, now you have your own thread.
    Scott Wang.
     
  2. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Original thread: TTL circuits Help. (NOT, AND, OR, ETC.)

    It is current on TTL, but this is reflected as voltage. A DVM will give a good picture of a 0 or a 1 on a TTL Gate. .7V and less is a 0, 3.2 V and above is a 1.

    If you need help understanding TTL just and quote this post, I'll get notified. :)
     
    lynnfaiz likes this.
  3. dl324

    Distinguished Member

    Mar 30, 2015
    3,250
    626
    Transistors are current controlled devices, so injecting current into the base is how they're turned on.
    In digital circuits (e.g. TTL), transistors are either on or off. The driver provides sufficient base current to the transistor it's driving to cause it to saturate.

    If you give a specific example, you can get a more specific answer.
     
    lynnfaiz likes this.
  4. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    Is this the TTL NAND circuit your are referring to?
    [​IMG]

    For the purposes of this discussion, the transistors can be modeled as current-controlled devices.
    But you need to understand that Q1 here is not always acting as a normal transistor. The base-emitter and base-emitter junctions do indeed just act as diodes when both inputs are high.

    If both inputs are high, then the only path for the current from R1 is through Q1's forward-biased base-collector junction (the reverse of normal transistor operation) and into the base emitter junction of Q2. (at this point Q1 is not being used as a normal transistor)
    Q2 is then turned ON and its emitter current goes into the base of Q3, also turning Q3 on and generating a logic-low at the Output.

    Now if one (or two) inputs goes low then R1's current starts flowing through Q1's base emitter junction. Q1 now acts like a normal transistor, turning on and saturating, which pulls the base of Q2 to near the input level (ground).
    The current into the base of Q2 now stops and there is a momentary current flow from Q1's collector to emitter which pulls any residual charge out of the base junction of Q2.
    Q2 (and thus also Q3) thus turn off and Output goes high.

    Make sense?
     
    Last edited: Jun 29, 2015
    lynnfaiz likes this.
  5. lynnfaiz

    Thread Starter New Member

    Dec 16, 2012
    29
    1

    Thanks Crutschow
    Yes i am referring to that diagram.
    If i am to refer to the inverter diagram (in the e textbook), is there a way to explain how Q1 works with using the diode?
    Doesnt the diode use to make an easier understanding on the multiple emitter of the BJT?
    So, Q3 turn ON because of the emitter current from Q2 and not because of the voltage at base of Q3?
    Please help me to clarify.
    Thanks in advance!
     
  6. lynnfaiz

    Thread Starter New Member

    Dec 16, 2012
    29
    1
    ttl-nand-gate.gif

    Hi All, referring to this circuit, Can anyone help to explain, during LOW state (output LOW), what makes Q3 turn OFF and how do you calculate the Q3 base voltage.
    And one more, during HIGH state (output HIGH), Q3 is ON, is there any current flowing at R3 and R2?

    ***should this be a new thread?
     
  7. DickCappels

    Moderator

    Aug 21, 2008
    2,664
    634
    Thank you for asking.

    It doesn't need to be in a new thread because it is the same topic as your other posts in the thread.
     
    lynnfaiz likes this.
  8. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    You can look at the diode equivalent if it helps you understand.

    [​IMG]
    When one or both inputs are low, the current from R1 go through the input diode(s) to ground and Q2 and Q3 are off.
    When both inputs are high the current from R1 goes into the base of Q2, turning on Q2 and Q3.

    As to what turns on Q3, there's a running argument about that on these forums. From a solid-state physics point-of-view a BJT is modeled as a voltage-controlled current-output device, but that is only really useful for small-signal AC type circuit design.
    For this type of circuit (digital switching, large-signal) a black-box current-controlled current-output model is generally more useful, even if it's not supported by the solid-state physics model.
    So it does take a base emitter voltage of about 0.7V to turn on Q3, but at the same time a current flows through Q3's base-emitter junction (in this case from Q2's emitter) From this a larger collector current flows that is Beta (transistor current gain) times the base current.
    Thus it takes both base voltage and base current to turn on a BJT.
     
    lynnfaiz likes this.
  9. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    [​IMG]
    When the output is low (inputs high), Q2 is ON. This pulls the collector of Q2 (base of Q3) low.
    Thus there is no current to turn on Q3 and it is off.
    When Q2 is ON it's emitter voltage is at about 0.7V from the base-emitter voltage of Q4.
    Since Q2 is ON (saturated) its collector voltage is very close to its emitter voltage or also about 0.7V.

    When the output is high, and Q2 is OFF and Q3 is ON there is no steady-state current through R2 and R3 if there is no load on the output.
    But there is momentary current through them when the output goes from ZERO to ONE as required to charge the various circuit stray capacitances.
     
  10. lynnfaiz

    Thread Starter New Member

    Dec 16, 2012
    29
    1
    thanks for the reply.
    if the output is high and i have a load connected to it. The current sourcing action will happen at R2 or R3. and what causing Q3 to turn on is the current pass through R2 to the Q3 Base and its Base Voltage?
     
  11. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    With no load, Q3's base voltage equals Vcc (no current through R2) and Q3's emitter is about 0.6V below Vcc due to the base-emitter diode drop.

    A load added to the output will cause current to start to flow out of Q3's emitter.
    This will lower Q3's emitter (and thus also the base) voltage sufficiently so that there is enough current through R2 to provide the necessary base current to turn Q3 on.
    This base current equals Q3's emitter (the load) current divided by the transistor Beta current gain.
    Beta typically is somewhere around a hundred.
     
  12. lynnfaiz

    Thread Starter New Member

    Dec 16, 2012
    29
    1
    when Q3 is ON due to base current flowing from Vcc to R2, then we will also have current flowing in R3. Am i right?
    how much would the current roughly be in R2 and R3?
    The current would be different if there's a load connected to and the gate is in HIGH state. Am i right?
     
  13. crutschow

    Expert

    Mar 14, 2008
    13,056
    3,245
    I answered your question in Post #11. Why are you re-asking the question?
    The current in R3 equals the output load current minus Q3's base current.
    Typically the base current (from R2) is equal to the Q3's emitter current divided by Beta and as I already stated, Beta is typically around 100.
     
  14. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Some thing to realize is that Q1 in that configuration has gain, it is not just diodes back to back (though that model also works).

    Are you having trouble visualizing the current paths Lynn?
     
Loading...