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How does an open collector work?

  1. Nov 22, 2016 #1
    I'm slightly confused as to how an open collector works.

    When a logic low input, the NOT gate is high, turning on the transistor and allows the current to flow from collector, through the load to the emitter, which is connected to ground. In this condition, does the load operate?

    When a logic low high input, the NOT gate is low, turning off the transistor. So in this condition, the load doesn't operate?

    What is the need for the NOT gate?

     

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  2. jcsd
  3. Nov 22, 2016 #2

    vk6kro

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    This is effectively two inverters, one driving the other inside the integrated circuit. So, the effect is that, logically, they cancel each other.

    However, the benefit of using the IC is that a very small amount of power at the input of the IC is used to control a much larger power in the load.
     
  4. Nov 22, 2016 #3
    How is there two inverters, it's only one. Maybe it's the way I've drawn it.

    6794892030_0b1f2919bf.jpg

    Here’s a simplified model of what is inside each buffer channel.

    The buffer input goes into a logical NOT gate. The output of that NOT gate goes to the base of an NPN bipolar transistor. The emitter of the transistor is connected to ground and the collector of the transistor is connected to the output. This is the “open collector.”

    When a logical input to the SN7407N is low, the output of the NOT gate is high, so the base of the transistor is held at a voltage above the emitter. This “turns on” the transistor, which means that if there is any voltage (above about 1.5 V) connected to the collector– that is, connected to the output of the SN7407N channel –current will flow from the collector, through the transistor to ground.

    When a logical input to the SN7407N is high, the output of the NOT gate is low, so the base of the transistor is held low, at the same voltage as the emitter. The transistor is off, and does not conduct current. That is to say, no current flows to or from the output. It’s as though the output is simply not connected to anything.

    That in bold is copied from a webpage, but I'm having trouble understanding how it works.

    When there's a logic low, the output of the NOT gate is high, so the transistor is turned on and current will flow from the collector through the transistor to ground. In this condition, will the load operate or not?

    I don't understand why there is a NOT gate, can you explain?
     
    Last edited by a moderator: Nov 22, 2016
  5. Nov 22, 2016 #4

    vk6kro

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    Yes, the open collector transistor is also an inverter.

    A voltage on its base great enough to turn the transistor on will cause the output voltage to drop as the collector conducts current.

    So, the transistor acts as an inverter. Current flows through the load causing it to operate. If it is a lamp, this may be enough current to make the lamp filament glow, for example.

    The input to the NOT gate may be less than a milliamp, while the current in the load may be many mA. So less power is drawn from whatever is driving the IC than if it tried to drive the load on its own.
     
  6. Nov 22, 2016 #5
    Okay, the transistor is also an inverter. What is the need for the other inverter?
     
  7. Nov 22, 2016 #6

    NascentOxygen

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    The first stage gives isolation. It provides gain, and squares-up the input so that the O/C stage is driven by a healthy binary signal. Otherwise, with poor quality input the O/C stage could easily be only half turned on and operating in its linear region.
     
  8. Nov 23, 2016 #7

    Svein

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    Here is a schematic of one of the gates in a 7407:
    7407circuit.GIF
    The input transistor (and the next) is there to present a standard TTL load to the other IC's. The two output transistors are there to fulfill the specifications for the 7407.

    Comment: The 7407 is really old technology, it burns a lot of power in order to do a simple thing.
     
  9. Nov 23, 2016 #8

    NascentOxygen

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    It's strange to have the input shown internally solidly grounded.
     
  10. Nov 23, 2016 #9

    Svein

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    Agree. I picked the schematic from the website of one of the original suppliers. There should be a reverse-biased diode from the input to ground. Here is the correct schematic:
    YOW1R.png
     
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