Hey all
Complete noob here - so can someone explain the concept of 'pullup resistors'?
Im looking at the datasheet for the 74LS07 and it says 'the open collector outputs require external pullup resistors for proper logical operation'.
As Pauline Hanson said, 'please explain'?
Thanks
"Simon Scott"
collector
operation'.
The output of the buffer is not capable of supplying a voltage, it can only pull the pin to ground. In order to provide a high level output, a resistor must be used to "pull up" the pin to Vcc (supply voltage).
The output pin of the device is connected to the collector of a transistor. When the transistor is on, the output is shorted to ground through the transistor, giving 0 volts output. When the transistor is off, the output is essentially disconnected from the device. So if you want that output to be high, you connect it through a pull-up to the positive supply. That way, another device input can look at that pin and it will see either 0 or V+.
Open collector outputs like this allow you to connect multiple device outputs together for a "wired OR" If any one of the outputs is grounded, the overall conection is grounded. Only if all connected outputs are floating will the pull-up bring the output up to V+. You can't do this with the usual active outputs, because they are either connected to ground or V+, so if you connect 2 of these together and one is on and the other is off, you get a short across the power supply and that lets the smoke out of the chips.
Best regards,
Bob Masta dqatechATdaqartaDOTcom D A Q A R T A Data AcQuisition And Real-Time Analysis
I am a novice also and appreciate this lucid explanation. Thanks, Bob.
The "open collector output" is about like a switch, with one side connected internally to ground and the other end left open-ended as the "output." So it looks like:
/ / ,------o o---------> out | gnd
(View the above in a fixed-spaced font.)
Note that if the switch is open, that "out" isn't connected to anything at all. If it is closed, everything is fine, as it is connected directly to ground. But if the switch is open, a resistor to the +V rail will "tie" the output line to something when the switch isn't closed. When closed, the switch with dominate and the resistor to the +V won't interfere much, except to start sourcing current. But that's acceptable in these cases.
One useful reason for open-collector (or open-drain) outputs is that they can be simply tied together with other outputs for a "wire or" (or "wire and" depending on point of view) effect. But another reason may be that the +V rail needs to be something other than the normal voltage rail used elsewhere by the logic.
Jon
Got it, thanks guys :)
So by using a resistor to connect the output pin to a supply, if the output is low the supply is grounded (and therefore reads 0 to anything downstream) but if the output is 'high' (ie, disconnected) then the downstream logic gets volts from the supply.
Yes?
No. The *supply* is not grounded when the output is low, the end of the resistor that connects to the downstream logic is. The output is low (0V or close to it) when the transistor is turned on. When it is turned off, ground is removed from the end of the resistor that is connected to downstream logic. Since the other end of the resistor is connected to the supply, the downstream stuff "sees" the supply voltage (instead of 0 volts) through the resistor.
Ed
Thanks Ed, yeh, thats what I meant - the supply is grounded through the resistor when the pin is low......
While it is true that there is a current path from the supply, through the resistor to ground when the open collector output is low, we wouldn't normally say "the supply is grounded", because it isn't. We would usually say "the signal is grounded", or "the open collector output is grounded", to make it clear that we are talking about a signal, not about the power supply itself.
Saying "the supply is grounded" implies a short circuit between the power supply and ground - a serious problem.
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If suffered from this before - I think it comes from a highschool 'electricity flows like water' education which does more harm than good.
I do know what you mean though. The pin is ground so downstream (there's that water again!) logic sees ground. When the pin is disconnected, the downstream logic seems 5V (or whatever) from the resistor connected to a supply.
Without the resistor you would indeed be connecting the supply to ground which would be a real problem :)
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