Wired-or Int Lines
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Here is a sample circuit with open collector “wired-OR” interrupt lines... Either Chip A or Chip B can interrupt the microprocessor. Both interrupt lines come in on the same IRQ line. Because both are open-collector and each has a pull-up resistor, no “bus fight” will occur. Pullup resistor
Chip A INT/
Chip B INT/ uP
C
C
B
B
Chip A
Chip B INT/ E
E
These 4 scenarios can happen...
No one interrupts the uP Chip A interrupts the uP Chip B interrupts the uP Both Chip A and B interrupt the uP
Chip A INT/
Chip B INT/
INT/
HIGH LOW HIGH LOW
HIGH HIGH LOW LOW
HIGH LOW LOW LOW
When no interrupts are asserted, all INT/ lines are HIGH. This is because neither BJT is conducting – no current flows through the BJT or the pullup resistor. That means the voltages at the collectors are HIGH. Consider the second row, when Chip A asserts it’s interrupt. The BJT turns ON and conducts, so Chip A INT/ goes LOW. This pin is also connected to the uP INT/ and Chip B INT/. That means the voltage at all those places is LOW. So Why isn’t there a short circuit at Chip B, because Chip B’s INT/ line is high? The critical point is that Chip B is NOT CONDUCTING. It looks like a high impedance open circuit to the rest of the world. So there is NO short circuit to ground through the Chip B BJT. That means point C can be pulled low EXTERNALLY by Chip A, and the current flows through both resistors and then through the Chip A BJT only.
Chip A INT/
Chip B INT/ uP
C
C
B
B
Chip A
Chip B INT/ E
E
These 4 scenarios can happen...
No one interrupts the uP Chip A interrupts the uP Chip B interrupts the uP Both Chip A and B interrupt the uP
Chip A INT/
Chip B INT/
INT/
HIGH LOW HIGH LOW
HIGH HIGH LOW LOW
HIGH LOW LOW LOW
When no interrupts are asserted, all INT/ lines are HIGH. This is because neither BJT is conducting – no current flows through the BJT or the pullup resistor. That means the voltages at the collectors are HIGH. Consider the second row, when Chip A asserts it’s interrupt. The BJT turns ON and conducts, so Chip A INT/ goes LOW. This pin is also connected to the uP INT/ and Chip B INT/. That means the voltage at all those places is LOW. So Why isn’t there a short circuit at Chip B, because Chip B’s INT/ line is high? The critical point is that Chip B is NOT CONDUCTING. It looks like a high impedance open circuit to the rest of the world. So there is NO short circuit to ground through the Chip B BJT. That means point C can be pulled low EXTERNALLY by Chip A, and the current flows through both resistors and then through the Chip A BJT only.
