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I have an issue with my binary up-counter using a 74LS74AN D-Flip Flop
The issue is the following:
I am trying to toggle the flipflop state by sending it a 1Hz Square wave (5V Peak) to make it toggle every clock cycle (Rising Edge triggered)
This is the schematic I came up:
enter image description here

And this is the datasheet:
https://pdf1.alldatasheet.com/datasheet-pdf/view/12660/ONSEMI/SN74LS74AN.html

For some reason the flipFlop will not toggle when the LED is connected. I have measured at the D-Pin with the LED connected and it reads 3.3V roughly. Which is expected when the 74LS74AN Outputs at 3.5V. According to the datasheet High Level is 2V.
I already tried out another chip. I got the same issue there.

My idea is:
!Q Starts out HIGH, so when the CLK goes HIGH the FlipFlop will latch HIGH, so !Q will go LOW. On the next clock cycle the FlipFlop will latch LOW so !Q goes HIGH again.

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Connect the LED from Vcc through the resistor to the Q output (with the proper LED polarity, of course).

schematic

simulate this circuit – Schematic created using CircuitLab

The 74LS outputs can sink 8mA so the resistor should be no less than about 400 ohms. Try 510 ohms.

The sourcing capability is much less, as shown in this datasheet. The currents (-0.4mA and 8mA) represent the maximum loading with which the output is guaranteed to have valid logic output voltages including noise margin, if you abuse the output and don't need it to have a valid logic level you can get much more current but I don't advise that.

enter image description here

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  • \$\begingroup\$ Ty I hooked up a 1k Resistor and it works now. \$\endgroup\$ – Jan Krüger Jul 13 '19 at 21:31
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The LED seems to be loading Qn too much.

A couple (well, 3) ways to fix this.

  • turn the LED and resistor around, tie the resistor to VDD. TTL sink current is much better than sourcing.
  • Drive the LED from Q.
  • add a buffer to the LED.
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In general, you can't drive an LED with a TTL output and still use that output as a logic signal. Your flip-flop can only provide 400\$\mu\$A when the output is a valid logic 1. Your LED circuit is sinking so much current that the flip-flop can't provide a good logic 1 on the !Q output.

You need to use a separate buffer to drive the LED. Something like a 7407 might be a good choice. Connect the input of the buffer to !Q; use the buffer output to drive the LED circuit.

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  • \$\begingroup\$ Okay I thought about that too, but in a video where I got the idea from I saw it working without an buffer. Ty for you help. \$\endgroup\$ – Jan Krüger Jul 13 '19 at 21:25
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    \$\begingroup\$ @JanKrüger: The CMOS equivalent (74HC74, or other parts with a "C" in the middle) can source much more current than the LS family, so would probably work in your original circuit. \$\endgroup\$ – Peter Bennett Jul 13 '19 at 21:57
  • \$\begingroup\$ That's not true 'in general' at all. The LS family has a particularly weak output drive for logic high. HC/HCT and later families have much stronger and balanced outputs. Downvoting, I'm afraid. \$\endgroup\$ – TonyM Jul 14 '19 at 7:16
  • \$\begingroup\$ @TonyM But HC and HCT are not TTL, they are CMOS. The HC series isn't even TTL compatible. Notice that I didn't say "7400 series", I specifically referred to TTL. \$\endgroup\$ – Elliot Alderson Jul 14 '19 at 14:23
  • \$\begingroup\$ You're maintaining that HCT and all later TTL families 'generally' (mostly) can't drive an LED load (<5 mA) while outputting a logic high? All the TTL families over those 30 years? (The OP's question was LS so, by 'TTL', no-one's talking about a 60's/70's 7474, nor did my comment.) \$\endgroup\$ – TonyM Jul 14 '19 at 19:01

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