The 7490 Decade Counter is an integrated circuit, or IC, that sequences or "counts" through ten numbers. The numbers range from zero to nine, and each is represented by four digits of Binary-Coded Decimal, or BCD. Each digit of the BCD number is produced at an output of the 7490. For example, a decimal "2" is 0010 in BCD and the combined 7490 output is: QD=0, QC=0, QB=1, QA=0. That output has electrical significance because each "1" stands for 5V and each "0" stands for 0V. The term "MOD-6" means that the 7490 will have a Modulus of six and will only sequence through six numbers, rather than ten. Wiring the 7490 as a MOD-6 counter requires two specific connections to the 7490 reset pins.
- Skill level:
Other People Are Reading
Things you need
- 7490 decade counter
- Electronics breadboard
- 7490 datasheet
- Jumper wires
Insert the 7490 into the breadboard. The notch on the chip should be at the top and the chip should straddle the blank line that divides the rows. Pin 1 of the 7490 is the first pin to the left of the notch.
Locate the 7490 pins labelled "VCC" and "GND." Refer to the 7490 datasheet for the pin description. Connect VCC to breadboard power and connect GND to breadboard ground. Use jumper wires for the connections.
Locate the 7490 pins labelled "R9(1)" and "R9(2)." Use jumpers to connect both pins to breadboard ground.
Locate the 7490 pin labelled "CKA" or "Input A." Leave the CKA pin open for now. A frequency generator, clock pulse or another counter will be connected to this pin. An alternating (+5V to 0V) signal at the CKA pin causes the 7490 to sequence through the numbers. The speed of that signal determines how fast the numbers will be appear at the 7490 output pins.
Locate the four output pins of the 7490. They are labelled "QD," "QC," "QB" and "QA." Connect a jumper to each output and leave the other end free.
Turn the breadboard counter-clockwise so that the 7490 is horizontal. Insert the free end of each output jumper into adjacent breadboard rows. Follow the left-to-right order of the outputs, which is: QD, QC, QB, QA. The jumper ends will form a horizontal line which will be parallel to the 7490. This is where output indicators, such as LEDs, will be connected.
Locate the 7490 pin labelled "CKB" or "Input B." Connect QA to CKB with a jumper. This connection sets the counter to BCD-output mode.
Locate the 7490 pins labelled "RO(1)" and "RO(2)." Connect QB to RO(1) with a jumper and connect QC to RO(2) with a jumper. This sets the Modulus to six. In BCD, six takes the following form: QD=0 QC=1 QB=1 QA=0. The 7490 will reset to BCD 0000 when both RO(1) and RO(2) receive a "1" from QC and QB. The MOD-6 output sequence is as follows, from zero to five: 0000, 0001, 0010, 0011, 0100, 0101.
Tips and warnings
- A working knowledge of binary and BCD is helpful for understanding this tutorial. In digital electronics the number zero is always used. That is why a MOD-6 counter goes from zero to five rather than from one to six. In BCD, digits are called bits. The most significant bit (QD) and least significant bit (QA) are extremely important for proper interpretation of BCD numbers. That is why the outputs were placed in order. You may notice that the output will be BCD 0110 for a fraction of a second. This is called a "glitch" and it is a normal part of 7490 functioning. Both RO(1) and RO(2) have to momentarily receive a "1" from BCD 0110 to reset the count to 0000. Additional electronic components may be used to bypass that glitch.
- Take precautions for static electricity when handling integrated circuits, especially the CMOS type.
- 20 of the funniest online reviews ever
- 14 Biggest lies people tell in online dating sites
- Hilarious things Google thinks you're trying to search for