74HC14 stability problems.  - Page 1
Forum section: electronics / projects

author: deth502
author_rank: Frequent Contributor
timestamp: October 08, 2013, 01:06:17 am
content: working on a project, this is the delay timer. push start, pot r1 (50k) sets the delay time from 0 to about 5 sec. at that time, the buzzer sounds for about .5-.75 seconds. the "reset" pin goes low simultaneously with the sounding of the buzzer, and stays low for the duration. the stop button sets everything back to zero to start again.the reset isint hooked up to anything, i just added that to the drawing for my reference as to where it will go. the only thing on my breadboard not shown is a bypass cap on the power pin of the 74HC14.problem is, it dosent work reliably. some times, nothing happens with the start button. probing it shows me that when pushing the start button, the +v only goes to ~1-2vdc, and not to the 5vdc supply voltage. (only when the problem is present) i dont know why. other times it works perfectly.is there something here im forgetting/leaving out? like i said, i have the bypass cap on the ic's +v input. other than that, is there something i should be doing to help with the stability of this circuit?

author: deth502
author_rank: Frequent Contributor
timestamp: October 08, 2013, 01:09:16 am
content: and as you can see, all 6 gates of the hc14 are used so there are no floating gates on the ic that could be causing the problem.

author: dannyf
author_rank: Super Contributor
timestamp: October 08, 2013, 01:25:08 am
content: QuoteC1 is too large: with a short push of the start button, C1 isn't sufficiently charged so when you release the start button, the output of the 2nd gate gets pulled down by C1.

author: Bertho
author_rank: Regular Contributor
timestamp: October 08, 2013, 01:32:00 am
content: The 100k pull up on the start may be not enough to pull the input through the hysteresis. The input consumes current while pulling it through the hysteresis area, which can be quite substantial (1mA order). The 100k does not allow this current to flow, leaving the input under the V+ threshold. Try a 1..10k resistor.The 1M positive feedback may also be too large. The leakage current would drop 1V. It should be fine though, but I suggest halving it.

author: deth502
author_rank: Frequent Contributor
timestamp: October 08, 2013, 01:33:56 am
content: Quote from: dannyf on October 08, 2013, 01:25:08 ami actually bumped it up to 100uf to get the timing i wanted with the largest trim pot i could find (the 50k) im looking for a delay of between 0 and about 5-6 sec.ill do a little more digging and try to find a bigger trim pot so i can drop the value of c1 back down again and see what happens. thanks.

author: dannyf
author_rank: Super Contributor
timestamp: October 08, 2013, 01:56:56 am
content: A higher pot doesn't work.Think of a relaxation oscillator with a pin tied to the start / reset buttons. Think a NAND gate like HC132.

author: w2aew
author_rank: Super Contributor
timestamp: October 08, 2013, 01:58:24 am
content: Please don't forget to include a power supply decoupling capacitor. Noise on the supply from the switching can easily disrupt circuit operation.

author: deth502
author_rank: Frequent Contributor
timestamp: October 08, 2013, 07:36:52 pm
content: Quote from: w2aew on October 08, 2013, 01:58:24 amalways, the first thing i do when i come across a problem is, if at all possible, power the circuit by battery to eliminate noise. ive already tried that here.

author: edavid
author_rank: Super Contributor
timestamp: October 08, 2013, 07:52:55 pm
content: Quote from: Bertho on October 08, 2013, 01:32:00 amThis is not correct, the 74HC14 never draws input current. It draws more supply current in the hysteresis area.

author: edavid
author_rank: Super Contributor
timestamp: October 08, 2013, 07:54:01 pm
content: Quote from: dannyf on October 08, 2013, 01:25:08 amEasy to fix this by adding a diode from the start button to C1 (move the resistor to the other side of the start button).

author: deth502
author_rank: Frequent Contributor
timestamp: October 09, 2013, 11:34:12 pm
content: ok, update, and a walk through my thought process.gates a and b form the switch. when start is momentarily pressed, it pulls gate a outputs low, which pulls b's high, which feeds back through the 1M resistor. if i did the diode to c1, id have to hold down the start button for the entire delay while it charges. not an option.i dropped the 100k pull-up on the start button to 47k, dropped the 100uF cap to a 47uF, and put a 100k trim pot in. with the smaller cap drawing less power to charge, and the smaller resistor putting more voltage to gate a to insure pull-up seems to have fixed the problem. and the 100k trim pot is getting me pretty close to my desired time delay range.thanks for the input!

author: free_electron
author_rank: Super Contributor
timestamp: October 10, 2013, 12:04:19 am
content: Quote from: dannyf on October 08, 2013, 01:25:08 amyou don't understand hos this works. the first two inverters form a flipflop. the switches themselves have no impact on c1. c1 is charged by the second inverter.

author: free_electron
author_rank: Super Contributor
timestamp: October 10, 2013, 12:14:14 am
content: Quote from: edavid on October 08, 2013, 07:54:01 pmand another one that does not understand how this works. read my previous post.The first two inverters combined with te 1meg form a flipflop.basically what happens is the following ( forget all other parts attached to the first two inverters )if you press stop you set a hard logic zero on the input of the first inverter. this inverter goes high, the second inverter makes this back low.this LOw is fed back via the 1 meg to the input. release the stop button. and the inverters stay in their current state. the low from the second inverter goes through the 1 meg to the input of the first , closing the loop. it's like a snake biting its own tail.press the START button and you apply a logic high via the 100K. Technically you have a voltage divider vrom vcc thru 1k , thru 1 meg to the Vol (voltage output low) of the seocnd inverter. 100k 1meg .. meh . 4.5 volt. that s aogic high allright.So now the snake bites its tail but the other direction. you essentially 'latch the last known state into the inverter ring.This is exactly how semiconductor memories are made. We don't use real flipflops with gates. An inverter is a pmos and an n-mos stacked. so two n-mosses and two p-mosses to the trick. instead of a resistor we make the channels of the p-mosses high ohmic. and the mosses can share the same well. bingo. instant memory cell that retains what yo throw in it.if you press BOTH buttons, STOp will win as it is the 'hardest' signal. If you want start to win : put the 100 from the stop witch to ground. and wire Start directly to vcc.Now, i see one potential problem. discharging that 100uf , thru the diode and the cmos output... that will damage the chip over time. shunting 100uf to ground gives a tremendous peak current... put a current limiter in there. 470 ohm on 5 volts limits you to roughly 10mA ( a liitel bit more) , well within spec of what the gates can handle (look it up , probably 25 mA ) so you can get away with 220 ohm.so simply put that resistor in series with the diode.it is possible that peak current during discharge is already upsetting the inverters and your circuit stops working for a while ...

author: deth502
author_rank: Frequent Contributor
timestamp: October 10, 2013, 03:05:16 am
content: yes, exactly. then once the output of gate 2 goes high, it charges the r/c timing circuit of r1 and c1 for "x" time delay. after the delay, a buzzer sounds and simultaneously the timer starts. the timer starts from taking the reset pin from high to low, which is the output of gate c. gate d in an inverter. at the same time as the reset pin goes low, the buzzer sounds for about 1/2 to 3/4 second determined by the 2nd r/c circuit of c2 and r2.Quote from: free_electron on October 10, 2013, 12:14:14 ami see what your saying here, makes sense. i can easily do this to add some protection to the ic.