7943 (SN74HC393N) not working how I expect it to

hp1729

Joined Nov 23, 2015
2,304
To make it more simple for myself I now use a 1hz pulse of the Arduino ;)

By default pin 13 is GND (with a resistor between), but when the signal is high it is 5V (or 3.3V). Is that the correct way? Or is this shortage, and not good?
Yes, 74HC393 should work fine on 3.3 V for power and inputs. Yes, unused inputs should be grounded or to +V, which ever is the disable for that pin. Through a resistor is best but not necessary.
Yes, pull-up resistor on the clock line. Does it count right using the Arduino?
.
 

hp1729

Joined Nov 23, 2015
2,304
Incorrect. If the rise/fall edge duration is longer than i already stated it WON'T trigger ( Clock ).
Yes, you can do at any frequency you want but the above statement must be met!
You are right. That is what that says. I don't think I have ever seen that limitation as a problem. So at 3.3 Volts rise and fall times must be quicker than about 750 ns / Volt, or about 2.5 us?
Thanks for the lesson.
 

Thread Starter

JeroenJK

Joined Dec 9, 2015
7
Yes, with Arduino it counts correct. Unfortunately I don't have a binary to 7segment decoder, but maybe I will be able to build that myself sometime (I have 100 NPN transistors and 100 dual NAND gates ;)).

But for now it works!
 

hp1729

Joined Nov 23, 2015
2,304
Yes, with Arduino it counts correct. Unfortunately I don't have a binary to 7segment decoder, but maybe I will be able to build that myself sometime (I have 100 NPN transistors and 100 dual NAND gates ;)).

But for now it works!
CD4511, cost about $0.50 from Jameco Electronics, is designed to drive a common cathode display. Yes, use resistors. :) The data sheet breaks down the internal circuitry into gates. It would take about 30 gates.
7-segment display is about a dollar.
 
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zenerz

Joined Dec 22, 2014
13
You are right. That is what that says. I don't think I have ever seen that limitation as a problem. So at 3.3 Volts rise and fall times must be quicker than about 750 ns / Volt, or about 2.5 us?
Thanks for the lesson.
Every day is a school day for all of us :)
 

marcf

Joined Dec 29, 2014
289
Are you using 5.0v for Vcc to the 74HC393? According to the data sheet valid logic highs should exceed 0.70 of Vcc. 3.3v/5 is equal to only 0.66 Vcc. Why not use 3.3v for the 74HC303?
 

dl324

Joined Mar 30, 2015
16,916
7493 is the same type as 74HC393, right? That's what I found out on Google, but maybe I'm wrong?
You're wrong.

7493 is a single TTL binary counter that operates from 4.75-5.25V.
upload_2016-1-7_14-7-12.png

74HC393 is a dual CMOS binary counter that operates from 2-6V.
upload_2016-1-7_14-5-39.png
 

marcf

Joined Dec 29, 2014
289
Yes, I realize this, but if you look at the data sheet for the HC393 part and are using a Vcc of 5v, the inputs must go above 70% of Vcc. The TTL version 74LS 393 would work just fine as its inputs only need to go to TTL levels. Also tie unused inputs for TTL Hi , not low and be sure never to try to pull them low with a resistor.
 

cmartinez

Joined Jan 17, 2007
8,252
I didn't see (nor heard being mentioned) a bypass cap on the chip. Practically all digital IC's need a 0.1µF connected between their Vcc and Gnd pins to minimize the possibility of false triggering due to spurious voltage variations.
 

hp1729

Joined Nov 23, 2015
2,304
Yes, I realize this, but if you look at the data sheet for the HC393 part and are using a Vcc of 5v, the inputs must go above 70% of Vcc. The TTL version 74LS 393 would work just fine as its inputs only need to go to TTL levels. Also tie unused inputs for TTL Hi , not low and be sure never to try to pull them low with a resistor.
technically correct. How many times have you built a circuit and not used that capacitor?
It still isn't clear to me if the counter is 7493 or 74HC393...
the pinout he describes is certainly the 74393. He references the 74HC393 specifically.
 
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