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There are 3pins need to clear:
Pin 1 (Strobe) when high will push the inputs directly to the outputs, but when it goes low it will hold the previously applied output in the display.
Is it mean 4 bit input pin directly to a~g output pins?
Pin 6 (DIS FEQ IN) when high the output led will be treated as active low however when low the outputs will be active high
If our 7 segment display LED is common anode, then pin 6 should be high?
Pin 7 (VEE) Will be the low value of the output If tied to ground when outputs are low they will be 0V
Is it mean when VEE is 1V, then output Low is 1V?
And for input these pins for High, should we add Pull high resistor?
Thanks
Hi Kang,
Is it mean 4 bit input pin directly to a~g output pins
I may be misunderstanding you question, The decoded binary value will be pushed directly to the outputs as it's decimal number for the seven segments yes.
If our 7 segment display LED is common anode, then pin 6 should be high?
Yes
Is it mean when VEE is 1V, then output Low is 1V?
Yes
And for input these pins for High, should we add Pull high resistor?
It will be fine to simply tie them high
Regards,
Owen
Hi Owen:
Then what function for storbe pin?
Is any rule for pull high resistor? EX: during 1.5k~10k ohm.
Thanks
Hi Kang,
Is any rule for pull high resistor? EX: during 1.5k~10k ohm.
Please see this FAQ on pull-up/down resistors. [FAQ] How do I size pull-up or pull-down resistors?
Then what function for storbe pin?
The function of the strobe pin is to lock the output values. When the Strobe is high the inputs will be converted to the outputs via
When the strobe pin goes low the previous output state will be held and regardless of the input.
For example if strobe is high and your inputs are 0000 the you would see output 1111110. Then when strobe goes low the output is locked to 1111110, so even if the input becomes 1000 the user would still see 1111110.
Let me know if you're still confused.
Regards,
Owen