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radium98

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i would like to know is it possible to replace a 7403 by a 7400 IC
 

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First of all, the 7403 is an open-collector output, the 7400 is totem pole. Secondly, your circuit doesn't make sense to me. Aside from the fact that you've got no component values, you've got an LED (D1) and R11 connected to, what I'm guessing, is your VCC. The other side of the LED goes to the gate inputs, and is connected to another resistor, R8, that also goes to VCC. What are you trying to accomplish here? There's no current path for the LED.
 

barry forget what is the circuit do , my question is simple and clear , replacement of the ic 03 by 00 without any mess , or have to remove the pullup resistor
 

Your question is neither simple nor clear. Without knowing how you're using this in a circuit, no one can give you an answer.
 

Depends on size of PU R. If its rated such that when output is low
the Iout meets chip < Imax output rating and its Vout when low
does not look like a Vih into follow on gate.



Regards, Dana.
 

Should be fine for this, just some wasted pullup current.
 

Because if they wanted to work for a living and be
forced to stay current, they'd be out in industry. Their
real world experience, if any, may be of that vintage.

I had one professor who, every year, made a
whole semester out of designing an EKG signal
RC 60Hz filter. Can't get a whole lot lazier than that.
I took the class as a freshman and saw the same
junk on the same board and heard the same rap
while passing by as a senior.
 

Despite of all unclear points: Two 7403 gates are performing wired-or function in this circuit, it can't be achieved by 7400 without additional components or restructured logic.
 

My point exactly, but OP says “forget what is the circuit do”.
 

sorry for the delay reply , i have read all the answers , the 2 pullup resistors are the 11k
i have fitted 7400 because at the moment i did not find 7403 .
pictures show the entire pcb ,it is a 19khz pilot lock for an RDS coder ,i have seen that there me a slight little difference or a missed error in the schematic .if anyone need it i will post it , but i still need confirmation about if i can use it and what is the role of the 7403 here , thepll part it working okay .
 

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i would like to know is it possible to replace a 7403 by a 7400 IC
No.
Not without modifications.

But "modifications" means one has to look for the circuit's function, which you don't want us to do....

Kaus
 

KlaustST it is not a secret , here it is , it is a simple circuit of a pilot lock , i need to synchro the st with the rds signal , but i see if you look to the wiring and pinout in the schematic it is different from the pcb , so why i think guys up say that , my only question is i do not have 7403 i only have 7400 is it okay only by removing or keeping the 11k resistors ,picture are for the rds and for the pilot , if that may help .
 

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Hi,
my only question is i do not have 7403 i only have 7400 is it okay only by removing or keeping the 11k resistors
In my post#12 there is a clear "No".
and in post#2 and post#9 there is the explanation "why".

Klaus
 
Despite of all unclear points: Two 7403 gates are performing wired-or function in this circuit, it can't be achieved by 7400 without additional components or restructured logic.
Not cleanly, but true TTL Has a very asymmetric output drive (spec @ 4mA low, 400uA high) and you could still get functioning wired-AND nodes if you are willing to eat the excess supply current and heat. What the real VOL is, as lashed up, you could inspect.
 

presuming a No , so the original 7403 must be inserted :)
 

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