What are the rules to be followed to implement a IC(...like 74 series) on Genral purpose PCB,
i need use 74193 in my circuit, and i have done with the design.
But my problem is i dont know how to implement IC's Practically (Connections the i need to make, to exclude High impedance state etc.,).
What are the rules to be followed to implement a IC(...like 74 series) on Genral purpose PCB,
i need use 74193 in my circuit, and i have done with the design.
But my problem is i dont know how to implement IC's Practically (Connections the i need to make, to exclude High impedance state etc.,).
Here it is...also tell me where all i should use pull-down resistors,
i tried the design in a breadboard, but there was something wrong,
so i've posted this thread
Here it is...also tell me where all i should use pull-down resistors,
i tried the design in a breadboard, but there was something wrong,
so i've posted this thread
hi,
Gate outputs are normally pulled low or high.
You have to devise a logic gate combination that keeps the unused clock input held high while the other clock input is being clocked on a rising edge.
Do you follow OK.
BTW: do you plan to use the parallel load pin.?
Also have you got the CLEAR pin held low.?
EDIT:
Hint. Consider using NAND logic gates. Lets know how it goes.
hi,
Gate outputs are normally pulled low or high.
You have to devise a logic gate combination that keeps the unused clock input held high while the other clock input is being clocked on a rising edge.
I may have to use Pull-down resistors for this, if i'm correct...
may i know should i use pull-down resistors in all inputs of IC, so that i will eliminate High input state?
74xx ICs are extremely old. Nobody uses them today except school kids. They need a regulated 5V supply, use a high supply current and a high input current.
Maybe you should use CD4xxx Cmos logic ICs. They work from an unregulated 3V to 18V supply and their supply current is almost nothing. Their input current is nothing.
Does this mean 'i have to use pull up/ pull down resistors at output'? No.
I may have to use Pull-down resistors for this, if i'm correct...
may i know should i use pull-down resistors in all inputs of IC, so that i will eliminate High input state? No
I'm not using "LOAD" Use a 1K resistor to +5v to hold this pin high, if not being used
Yes, i'l use pull down resistor for "Clear" Connect directly to 0V, if you dont plan to use Clear/Reset
i may have to use it because i need the
"previous state as output" when there is no input...
(Because there will be no input at some times...
and 74279 does'nt have pull-down resistor embedded in it...)
May i know why u told so?
You are living in a very different world than most of us (in India?). Most of your parts are extremely old. We strain to remember them.
Only school kids from India call us SIR. A SIR is somebody who was Knighted by the Queen Of England.
A very old fashioned 74xxxx IC needs a very high input current to ground for its input to be seen as a logic low. The input must be forced to 0V with 1.6mA to make it low.
You are living in a very different world than most of us (in India?). Most of your parts are extremely old. We strain to remember them.
*********** India call us SIR. A SIR is somebody who was Knighted by the Queen Of England *******
Ya from India! It's something like giving respect to an elderly person...
I'm sorry if it hurt you, i'l never be calling so to anybody here, from now on!
A very old fashioned 74xxxx IC needs a very high input current to ground for its input to be seen as a logic low. The input must be forced to 0V with 1.6mA to make it low.
Ya from India! It's something like giving respect to an elderly person...
I'm sorry if it hurt you, i'l never be calling so to anybody here, from now on!
Can you please elaborate,... i dont know why "1.6mA" is required to Ground an Input
hi,
Please call me whatever you wish, Eric would get the job done.
If the S/R is TTL, the open circuit inputs will assume a logic high, as 'agu' points out to pull the TTL input to a logic low the PullDown resistor must draw 1.3mA from the TTL input.
I would not normally recommend using PD resistors on TTL inputs.
For Reference:
In a high speed logic circuit I would recommend you do not leave TTL inputs, whose outputs are connected to a following device input, open circuit. [Floating]
This could cause mis-operation of the circuit due to electrical 'noise'
Why not just build the circuit from the details you have been given.
Just imagine that all TTL inputs have built-in pull-up resistors so in order to pull an input low you need a pull-down resistor which will draw the 1.3mA Mr Gibbs sir was talking about, in order to make the voltage low enough for the input to be low.
Ouch.! that smarts, does this mean I am no longer on your Christmas gift list.
BTW: did you hear the latest Govt fudge word, the unemployed are categorised as 'economically inactive', it means they can juggle with the unemployment figures even more.!!