Electronic – SN74LS74AN trouble

digital-logicintegrated-circuit

I purchased an SN74LS74AN ic and the pinout is (unless I am mistaken) like this:

I did the following:

pin 14 to power, pin 7 to ground, pins 1 & 4 via 10k ohm resistors to power, pin 6 to pin 2 (Q' to Data input), pin 3 to power with a simple push button

also, pins 5 and 6 go to ground via 100 ohm resistors and blue LEDs

my intention is to make a circuit so that when I push my push button (which goes to pin 3), the LED which is 1 goes to 0, and the LED which is 0 goes to 1 and so on.

here it is in falstad

i believe the clock input for this chip is edge triggered

Anyhow, it is not working. can someone help me figure out why? bear in mind I am a beginner.

Edit1: by not working I mean either one of the LEDs is turned on. i don't have any mechanism of predicting which one will be on. pressing my push button to pin3 does not change the current state.

Kindly,

Best Answer

There are quite a number of changes you need to make.

1) Your LED current limit resistors are too small. A 74LS74 has a nominal rating of 16 mA when low. Assuming 2 volts Vf on your LEDs, producing 0 volts at a low output will draw about 30 mA. Try increasing the resistors to 300 ohms or so.

2) You have not shown any treatment of your unused inputs. Logic ICs should never have floating inputs. Connect unused inputs together and tie them to +5 with a 1k pullup resistor, since they are active low, and you don't want them active. In general, for TTL/LSTTL, the standard termination for unused inputs is via pullup, even when the input can be tied either way. The reason is that the input current is much less when high than when low, so less power is wasted on the unused function. By the same token, TTL/LSTTL outputs can sink much more current (when low) than they can source (when high), which is why your LED connections are correct - except for the resistor values.

3) As st2000 has answered, you need to condition your input. As it happens, you can do this cheaply by using the unused half of the chip as a SR flip-flop.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

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You are right about the NPN transistor, that is what you need in order to achieve that behaviour, however, since the µC would have to control the cathode of one of the displays, that would be a problem since most µC can only source / sink a few mA. Assuming each segment needs 10mA that would require the pin to sink 70 mA.

The most simple solution is to have two NPN transistors capable of sinking at least 100 mA (such as the common BC547).

Diagram bellow:

Diagram

Being D1 and D2 the 7 segment displays (please ignore the reference). The anodes on the LEDs will connect to the µC pins.

In this configuration only Q1 conducts when the µC outputs 1 and when it outputs 0 only the Q2 conducts.

Note
It might be necessary to add a high value pull-up resistor (for example 100 kΩ) to the base Q2.

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Based on comments it seems that limiting resistors will be necessary: one each between the P1-P8 LED state outputs and the LED arrays, for a total of 8 resistors.

If 3.2V 20mA LEDs are used, then 100 Ohm resistors will be appropriate.

(5.0V-3.2V) ÷ 0.020A = 80 Ohms 
(rounded up to the next appropriate standard resistance value) = 100 Ohms

It could also be beneficial to hook up the reset pins on the CD4017's to the Arduino for control. The P1-P8 LED state control pins can have their duty cycles controlled for LED brightness control. Strategic use of the clock signal is necessary to cycle through each column of LEDs within each of the eight layers.

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