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I'm trying to use the following IC: SN74S289BJ so that the output is 5v when HIGH and ~0v when LOW, however I'm having problems due to the open collector nature of it, as I've written data to a specific address and the transistor won't sink the pull-up current (meaning that the output LEDs that I use to verify the data are always on) according to the data sheet whenever there is data on a specific output the transistor will be saturated and will cause it to guide the flow of the current to GND.

Currently I have 5v through a 330 Ohms to the collector of the IC 74S289BJ and a LED to show the status.

What would be the correct way to set it up?

Here is my current setup, assume that the address and data are prepared and that the IC is in read mode (WE is HIGH).schematic

Also, the output is inverted because according to the datasheet the output is the data's complement.

Note: IC U2/3/4/5 in the above schematic are SN74H04N HEX inverters

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    \$\begingroup\$ Post a schematic of what you currently have. \$\endgroup\$ – Matt Young Dec 13 '13 at 4:58
  • \$\begingroup\$ 330 ohms is far too low a value for a pull-up resistor. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 13 '13 at 5:18
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    \$\begingroup\$ U2-U5, are they buffers? What voltage level activates the buffers? And what part number? \$\endgroup\$ – Passerby Dec 13 '13 at 6:13
  • \$\begingroup\$ U2-U5 are NOT gates (inversors) \$\endgroup\$ – Tristian Dec 13 '13 at 6:16
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    \$\begingroup\$ Wait. Wait. If the the output LEDs that you use to verify the data are always on, as in Logic High, Then the input to the NOT gate is Logic Low. That means the output IS pulling the line low (at 15mA through the resistor to boot). Did you write 1111 to the data register? Or 0000? 1010? \$\endgroup\$ – Passerby Dec 13 '13 at 7:07
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Typical open collector led setup would be:

schematic

simulate this circuit – Schematic created using CircuitLab

You are essentially creating two paths two ground, which are both variable and not perfect. There is a C-E voltage drop as well. If you measure the output pin voltage compared to ground, when the output pin should be Low (open-collector active), what do you get? The datasheets list that max would be 0.8v. U2-U5, are they buffers? What voltage level activates the buffers? And what part number?

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  • \$\begingroup\$ So, using this setup if I would like to drive another IC how would I connect it, I really have a hard time undertanding this...thanks in advance. \$\endgroup\$ – Tristian Dec 13 '13 at 11:10
  • \$\begingroup\$ @Triztian: Unless you're doing level shifting, OC outputs aren't used to drive other logic inputs. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 13 '13 at 15:10
  • \$\begingroup\$ Ok, but I don't have much of choice because thats the only RAM that I have, I know that the 74S189 is not open collector, but I don't have access to it at the moment. Thanks for the comment anyways. \$\endgroup\$ – Tristian Dec 14 '13 at 20:38
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Open collectors outputs are usually used as low side switches by connecting the emitter to the ground and then feed the ground side of load from the collector. What the open collector can do when used that way is to either sink current (provide ground) to the load when ON or leave the output floating (high resistance across the transistor) when off, the pullup resistor can force a HIGH state while in that high resistance mode.

In your specific circuit the problem may be the LOW level threshold of U2/3/4/5 (74H04) which is about 1.5v with 5v supply.

Another consideration is the max current of the 74HC04 chip which is 50mA total and with the LED resistors missing this can be a big problem.
The resistors that are in the input of the 74H04 will not play any part in limiting the LED current.

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