# Pull down resistor value in debouncer circuit  I am from computer background and have made debouncer circuit (SR) using NOR gate IC HD74LS02P as a part of digital electronics project. I have used pull down resistor at all inputs of the gates used. But when R (Reset) pin is provided input high, the output Q becomes low, and now when as soon as R pin is provided input low the output Q becomes low. But after 15 minutes the output Q becomes active even if R and S are both low and previous output Q was also low. Is there a problem with pull-up resistor value ? I have used 100 ohm resistors. I have used old mobile phone charger which gives 5.13 V and 0.48A current. I have attached my readings as well as electrical characteristics of IC

• There is an easy-to-use schematic tool button on the editor toolbar. Add one in so we can see what you are talking about. It also has a simulator built-in so you can test your circuit. Why are you using 100 Ω resistors? On a 5 V supply they will draw $I = \frac {V}{R} = \frac {5}{100} = 50 \ \mathrm {mA}$ which is very high for a logic circuit. – Transistor Oct 8 '17 at 14:52
• And a tip - TTL does not normally use pulldown resistors. Typical input resistors are 1k to +5, with switch inputs connecting to ground. – WhatRoughBeast Oct 8 '17 at 16:35
• @Transistor Thanks. I have attached photo of my readings and result. Can you explain the exact problem area in circuit and suggested solution. – Riddhi Akbari Oct 8 '17 at 18:38
• @WhatRoughBeast Thanks. I have attached photo of my readings and result. Can you suggest exact resistor value. – Riddhi Akbari Oct 8 '17 at 18:40
• @RiddhiAkbari try increasing all resistances to 1k and see if it works then. 100 Ω is way too small. – Harry Svensson Oct 8 '17 at 18:43 Figure 1. The relevant details from the 74LS datasheet.

Understanding the inputs:

• We can see from (1) that to guarantee a low reading on the input we need to pull the input below 0.8 V.
• From (2) we can see that the $I_{IL}$ - the current into the input when held at 0.4 V is -0.4 mA. The negative sign tells us that current will flow out of the input into whatever is pulling the input low.
• From the line above we can calculate the required pull-down resistance from Ohm's Law as $R = \frac {V}{I} = \frac {0.4}{0.4m} = 1k \ \Omega$. Any higher resistance may not pull the input down adequately. Any lower resistance means unnecessary loading of the previous stage.
• From (3) we can see a characteristic of TTL which is that they pull high by themselves - although it's best not to rely on this - and that only 20 µA is required to hold it high.

Understanding the outputs:

• From (5) we can see that the $V_{OH}$ - the high level output voltage - can supply only 400 µA - not enough to power your LED reliably.
• Below (5) we can see that the $V_{OL}$ - the low level output voltage - can sink 4 to 8 mA and still stay low enough to drive another TTL chip reliably. This determines the maximum fan-out of the TTL chip - the number of TTL inputs one TTL output can drive. Figure 2. 7400 NAND gate. Source: Wikipedia's Transistor-transistor logic which gives a detailed explanation of the workings.

To understand why all of the above is so it helps to look at the 74xx internals. (I coulnd't find a 74LSxx output stage diagram so this standard TTL output will have to suffice.)

The inputs:

• The inputs are formed by a dual emitter transistor. When disconnected no current flows through the emitters. Pull up requires little or no effort - just enough to make sure that noise or EMF doesn't turn them on.
• To pull low the base current has to be "sunk" by the external circuit.

The outputs:

The output is named a "totem-pole" structure after the native American Indian carved wooden totem-pole.

• The output pull-up is weak for a few reasons. At (1) we can see that there is a 130 Ω pull-up internally. This is the first cause of voltage drop on the output. We can also see that the high-side transistor switch is NPN so the emitter can only go as high as 0.7 V below the base voltage which itself will be less than 5 V due to R2.
• Meanwhile, at (2) we can see a strong pull-down switch formed by the lower NPN transistor.

This explains why the TTL pulls down better than it pulls up.

But after 15 minutes the output Q becomes active even if R and S are both low and previous output Q was also low.

I suspect the output voltage is dropping with time due to heat. You are probably overloading the output and the output voltage and input threshold voltages are changing over time.

Try increasing your resistors to 1k and disconnect your LEDs. Monitor the output with a voltmeter and see if they remain stable.

To power the LED you can try reversing it to see if pulling low works better. Limit the current to about 5 mA and you should be OK. With 2 V across the LED and 2.5 across the resistor this would be $R = \frac {V}{I} = \frac {2.5}{5m} = 0.5k = 500 \Omega$. 470 or 560 Ω would be fine.

Given the relative pullup/pulldown characteristics of TTL, I suggest you rebuild your circuit as follows: simulate this circuit – Schematic created using CircuitLab

Your original circuit has various problems:

1) Your inputs are either floating or grounded. Floating TTL inputs will ordinarily pull up to a HIGH, but not strongly so, and pickup of noise is a potential problem.

2) More importantly, connecting an output to ground through 100 ohms is just asking for trouble, since a high output must provide much more current than is specified. Don't do it.

3) As you may have noticed, I've connected the LED to +5, rather than ground, and reduced the current limit resistor. TTL can sink much more current than it can source, so always pull down external loads.

As @Transistor has suggested, overheating is probably the source of your problem.