Is my pull-down resistor value correct?

Question

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 V_il is about 0.8V while the I_il 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}

Answer 1

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.

Answer 2

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.

Answer 3

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.