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I'll be using a momentary push button, SW1 as shown in the figure. With that, I also decided to make a D flip-flop act like a latch by connecting the switch to the clock.

My question is if my pull-down resistor value is correct. I'll be using 74LS DIP IC gate specifically HD74LS74A. According to the data sheet, the Vil is about 0.8V while the Iil for the clock input is 0.4mA. Based on my readings in forums, I divided the two resulting to 2KΩ.

Should I aim for 2KΩ or should I use a higher or lower value for the pull-down. Sorry I'm not very familiar with this topic.


simulate this circuit – Schematic created using CircuitLab


marked as duplicate by SamGibson, ThreePhaseEel, Daniel Grillo, Dmitry Grigoryev, Voltage Spike Dec 15 '16 at 22:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ Very well put for one of your first tries. Well done. And welcome. \$\endgroup\$ – Asmyldof Dec 13 '16 at 22:58
  • 4
    \$\begingroup\$ Yes, as I explained in a comment on your original question, the 10kΩ pull-downs on LS TTL inputs which you showed there are too high. Your new question is effectively a duplicate of the questions I linked to in that comment, especially the answers to this one, where the maximum of 2kΩ pull-down value is explained. \$\endgroup\$ – SamGibson Dec 13 '16 at 23:01
  • \$\begingroup\$ @SamGibson Ahg. You had to rain on my parade just before I go to bed? \$\endgroup\$ – Asmyldof Dec 13 '16 at 23:08
  • \$\begingroup\$ @Asmyldof - Sorry, no raining intended! ;-) OP's previous question wasn't about the pull-downs, hence my highlighting of the pull-downs was tangential to his original topic. This new question (which uses information from the questions I linked in that comment) is a duplicate IMHO, but your answer nicely highlights why pull-ups were (are) much more commonly used with TTL. \$\endgroup\$ – SamGibson Dec 13 '16 at 23:11

You are making a small mistake.

Your V(il) reading is a maximum, so anything below it is okay, but above might not be seen as a low input.

You can also see that the I(il) input current is negative, which means current comes out.

So your pull-down resistor will need to be lower to guarantee operation.

However, you can also put the switch in the bottom part, since a switch has a very low resistance when pushed, and then put a 1k to 10k pull-up resistor. Because you can see that the minimum input high voltage is only 2V and that the current it draws at a high input is ten times as small.

That would look like this:


simulate this circuit – Schematic created using CircuitLab

With 40μA sourced sunk by the clock through your pull-up, the 10k will only lose 0.4V, which gives 4.6V. Very well above the minimum input high voltage.

When you push the switch, that's likely less than 10 Ohms to ground, so with 0.4mA coming out of the clock the voltage will nearly not be above ground. 4mV is very negligible in digital circuits running on 5V. Even in 3V3 and 1V8 circuits 4mV is assumed 0.

Last Edit

Of course, if your data depends on the clock (such as with a flip-flop driven toggle circuit), you will soon find you also want a capacitor in there.

If you put in a capacitor next to where ever your resistor ends up being, you will smooth out accidental multiple pulses. We call those accidental pulses when we talk about buttons "contact bounce". It's a very common thing and should be very easy to find on this site and Google.

With 10k, you would need between 100nF and 1μF most likely. Also depending on the button type and such.

  • \$\begingroup\$ Sorry, this kinda confused me. So technically, my pull down resistor would work if it is less than 2k Ohms;, say 1500 k Ohms;. However, pull-up resistors make the circuit normally closed because as the button is pushed, current stops flowing; am I right? \$\endgroup\$ – Orlando Lewis Dec 13 '16 at 23:16
  • \$\begingroup\$ And also, what do you mean by 'sunk' here? \$\endgroup\$ – Orlando Lewis Dec 13 '16 at 23:24
  • \$\begingroup\$ @OrlandoLewis Yes, with a lower resistor it will probably work. It may even work with 2k, since the datasheet gives limits, but 2k is considered risky (may work some times, may not others). Generally in digital systems you want big margins and low currents. A 1k (the number I'd pick, or lower) would drain 5mA when the button is pushed. Pros don't like that. If the button is to ground, you will make the signal low of course when you push, but when you release it will go high, so with a pulse button.... "sunk" is an English word: dictionary.com/browse/sunk \$\endgroup\$ – Asmyldof Dec 14 '16 at 7:15

Since bipolar TTL (74nn, 74LSnn, 74ALSnn) inputs source current, it is recommended practice to put the switch between input and Ground, with a pull-up resistor of 5 - 10K. This way, when the switch is closed, the input is very definitely a Low, with no need to hope that the pull-down resistor value is low enough resistance to guarantee that the input will be seen as Low. Having the switch to Ground also reduces the current consumption when the switch is open.

  • \$\begingroup\$ Why is the resistor 5-10K... Where did those value come from? \$\endgroup\$ – Orlando Lewis Dec 14 '16 at 0:01
  • \$\begingroup\$ The value of the pull-up resistor is not at all critical - you want a low enough value to allow enough current to ensure the input is held high with the switch open, but not so low that it draws excessive current when the switch is closed. \$\endgroup\$ – Peter Bennett Dec 14 '16 at 0:41
  • \$\begingroup\$ So how will I know if it is low enough? What values from the data sheet are used? Is it a rule of thumb to use the said range for 74LS DIPs? \$\endgroup\$ – Orlando Lewis Dec 14 '16 at 1:00
  • \$\begingroup\$ I suppose that the pull-up resistor value is determined by rule-of-thumb/common practice/what parts I have handy/what I've seen others use. In other word, there is nothing scientific about choosing the value. The 2K suggested by Jack Creasey in his answer "feels" lower than necessary to me, but isn't wrong. I know this is somewhat of a non-answer, but that's the way things are, sometimes... \$\endgroup\$ – Peter Bennett Dec 14 '16 at 1:17

Your 2k Ohm value is only correct to set an absolute maximum V(IL) value, you should make it less than 2k to ensure you generate a lower voltage than the
V(IL) spec for the device.

However I'd suggest you should not implement the switch this way. Your switch may be some distance (inches or more perhaps) from your logic PCB, and it is poor policy to run a supply voltage (+5 VCC in this case) to any panel mounted switches. The reason is that any short to Gnd will potentially short out your supply voltage. This might not damage your logic or power supply, but it will cause a reset or malfunction of your logic.

I'd suggest the following if you want to use an LS7474 device:


simulate this circuit – Schematic created using CircuitLab

Notice in the schematic above the *SET input would be pulled low by the switch, and would actually set the 'Q' high (if it was not already so) immediately. The circuit is immune to switch bounce, and the Q remains set until cleared by your other logic or microprocessor pulling the *CLR low. Notice from the datasheet that if both *CLR and *SET are pulled low at the same time, the Q output remains high until either *CLR or *SET go back to one, where the state of Q is defined by the last one to go high.
Your circuit as shown shows no way to set 'Q' to an initial state, or to clear 'Q' after you have read/used the value. This one above uses the *CLR to initialize or reset Q.

The 2k used as a pullup is a rather arbitrary selection. It's best to have some current flow in the switch (it's called a wetting current) and in this case you will have 2.5 mA from the 2k and 0.8 mA from the *SET for a total of 3.3 mA in the switch when depressed.

  • \$\begingroup\$ I find this complicated. I'm not using any PCB. I also have a different momentary push button for the clear and I was thinking to connect the preset to my +5 V side \$\endgroup\$ – Orlando Lewis Dec 14 '16 at 0:58
  • \$\begingroup\$ @Orlando. Then the circuit still suits......you simply use another push button to the *CLR side of the LS7474. How is it more complicated than using the CLK pin? \$\endgroup\$ – Jack Creasey Dec 14 '16 at 1:14

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