74LS14 testing circuit pls - Page 1
Forum section: electronics / beginners

author: Chriss
author_rank: Frequent Contributor
timestamp: April 12, 2018, 08:59:15 pm
content: Hi!Can somebody share with me a simplest testing circuit for a 74LS14 Schmitt Trigger pls.I have some strange situation with this ic's and I cant figure it out.I give power to the ic as in the datasheet described.I put a LED through a 330Ohm resistor to the Y1 andset a +5v to A1 and the led is all the time on.No matter if I set a +5v or GND to the A1.Pin. Also to other gates too..So, does I mess something or are my ic's bad?Some of them with the same setup didn't lite up the LED.I just swap the ic in the same circuit...

author: Zero999
author_rank: Super Contributor
timestamp: April 13, 2018, 09:52:57 am
content: Are the LED and resistor connected from the output to 0V?If so, your existing circuit sounds like the correct way to test it and barring any mistakes, such as: the IC's 0V in not being connected or confusion over which input goes to which gate, the IC is bad.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 13, 2018, 11:19:25 am
content: Yes.The LED is connected exactly as you described.Some gates are didn't recognize anything, some gates are non-stop on or off. ..Sent from my GT-I8260 using Tapatalk

author: ebastler
author_rank: Super Contributor
timestamp: April 13, 2018, 01:29:26 pm
content: If you have tried multiple ICs which do not work, it seems likely that your test environment has a problem. Do you have a clean supply voltage? (How do you generate the +5V?) It is advisable to have one electrolytic capcitor on your breadboard, e.g. 470µF, and one small 100nF capacitor at each IC. The capacitors are connected between +5V and GND to filter the supply voltage.If you want to test the behavior of the Schmitt trigger in more detail, you can connect a potentiometer between +5V and GND. (Use any value between 470 Ohm and 10 kOhm.) Then connect the middle pin of the potentiometer, which will have an adjustable voltage, to the Schmitt trigger's input. As you turn the potentiometer up and down, you should notice the hysteresis of the Schmitt trigger: The voltage where it turns the LED ON is higher than the voltage where (coming down) it turns the LED OFF again.

author: TomS_
author_rank: Frequent Contributor
timestamp: April 13, 2018, 02:12:34 pm
content: Quote from: Chriss on April 12, 2018, 08:59:15 pmAs I understand it, TTL inputs effectively "float high", so applying a logic high and removing it does not alter the state of the input, which is perhaps why you see no change.Try adding a pull down resistor to the input and then apply a logic high to it and see if that helps.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 13, 2018, 02:42:04 pm
content: Ok, here is the circuit I use...I have two of these ic's.1. IC Have 3 gates which are switching nice but 3 are didn't even try to switch.They are only on low state no matter what I apply on the input A1 pin.2. IC have only one gate "working" but full of noise and very long switching time.Here is my circuit:

author: PA0PBZ
author_rank: Super Contributor
timestamp: April 13, 2018, 02:50:46 pm
content: Connect the LED not to ground but to 5VConnect the 10K resistor not to ground but to 5V, and switch the input to ground.LS14HIGH Level Output Current -0.4 mALOW Level Output Current 8 mA

author: Chriss
author_rank: Frequent Contributor
timestamp: April 13, 2018, 03:05:06 pm
content: I tried it also...No changes...I disconnected the LED and just put the scope to the pin to check the signal etc. etc...no lucky...Actually this connection without the LED should work on the scope as far as I know, or?

author: ebastler
author_rank: Super Contributor
timestamp: April 13, 2018, 03:09:31 pm
content: Quote from: Chriss on April 13, 2018, 03:05:06 pmYou have a scope?! That's new information, isn't it?So, what do you see on the scope?Also, could you please comment on my question regarding your 5V power source and the suggestion regarding capacitors? If you have a scope, you can of course check the quality of the power supply too.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 13, 2018, 03:42:38 pm
content: QuoteYea, I have a scope. Sorry I didn't mentioned before, I'm so pissed off because I ordered 2 new IC's.Yesterday I got them.Put on breatboard just to test them and I saw I have a problem like I described.Checked several times does I done something wrong but not really found the problem.So I decide to ask more people for watching what the hell did I don't do right.It's a simple input - output stuff, nothing complicated...Power supplay:I use the power from my bench power supply which is nice and clean without any noise.I have a two channel scope.CH1 is on the output pin of the IC, CH2 is on the VCC pin on the IC.CH2 is clear and nice.Here is a picture from my scope on the 1. gate of the IC:On the same IC but on gate 3 or gate 4 there are no Hi-Lo with the same input setup.On the same IC on gate 6 I have non-stop Hi with the same input setup.I assume, this IC (one of that two I ordered) is bad, because only one gate is working.On the 2. IC there on the 3. gate I have the same signal as on the 1. IC but the rising and falling edges aretake around 10mS to reach the Hi and Low state.The other gates are not reacting or some of them sometime have some strange noise but withoutany Hi-LOW stage.I use always the same input setup as on the 1. IC where you can see the scoped signal...Pls. send me a circuit if mine is not correct.Thank you.

author: KrudyZ
author_rank: Frequent Contributor
timestamp: April 13, 2018, 04:38:10 pm
content: How about looking at both the output and the INPUT?LS gates need as much as 400 uA of input current when driven low.In addition this is a Schmitt trigger input with hysteresis, so your pull-pown might not be strong enough to reach a proper low state on the input.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 13, 2018, 04:43:06 pm
content: Let's say If I set a pull-up resistor to the input pin and switch the input to low direct to ground that should deliver enough current or?But no deference, pull-up/down resistor, input signal Low or High no matter, just several of gates are working.Could it be I killed the IC's because I tested them in my breatboard but I didn't grounded all the unused inputs?I deal with such of IC's and never had this kind of problem like now.

author: StillTrying
author_rank: Super Contributor
timestamp: April 13, 2018, 04:55:13 pm
content: 10k is too high to pull a LSTTL input low.Check for bad connections to 0V (pin 7) and well as 5V.As KrudyZ said, scope both the input and output of one gate at the same time, to see if you have at least one gate working.The 100u to 470u cap is to protect the IC from any voltage surges during switch on and off.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 13, 2018, 05:41:31 pm
content: Ok, here is the working one gate of one of the two ic's.If I try all the other input to feed with the same signal and setup like this working onenothing is happening or the outputs are only HI or LOW.I assume my IC's are bad and I have one IC with only one working gate.This is what I'm think and I cant get believe this because the IC's are brand new!

author: PA0PBZ
author_rank: Super Contributor
timestamp: April 13, 2018, 06:06:59 pm
content: It wouldn't surprise me if you killed the ICs pulling too much current from the HI output.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 13, 2018, 06:16:20 pm
content: Hmmm...Thanks for this valuable micro schematic.It could be that I killed one of the IC's where I tried the LED with a HI level output.But what is with the other one IC?I didn't connected the LED to the other one IC, just the scope...And on that just scoped IC I have only one gate working.So, the best testing way would be to test it with a LED on a LOW level output setup.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 13, 2018, 06:20:33 pm
content: Could it be does the IC's was killed from ESD?because they was just stabbed into foam board and putinto a PVC small bag when it was delivered to me.

author: PA0PBZ
author_rank: Super Contributor
timestamp: April 13, 2018, 06:26:12 pm
content: Quote from: Chriss on April 13, 2018, 06:20:33 pmI don't think so, LSTTL is very hard to kill with ESD. Since you never used the other IC with a LED we can forget about killing it that way, so where did you buy them, what did you pay, how do they look and what current do they consume on VCC? I wonder if you got fake ones but I can't imagine why anyone would fake an LS14...

author: TomS_
author_rank: Frequent Contributor
timestamp: April 13, 2018, 06:26:58 pm
content: Quote from: Chriss on April 13, 2018, 05:41:31 pmWhere did you buy them?I dont think you would blow them by not pulling unused inputs up/down. There must be some other explanation.Personally, in future I would look at AHCT family chips. They have TTL compatible inputs, are super quick (nanoseconds switching time), use very little power because they are CMOS, and can source/sink more current (20mA each way, give or take). Unless its for legacy, I dont think theres much point to using older TTL families any more these days?

author: TomS_
author_rank: Frequent Contributor
timestamp: April 13, 2018, 06:28:13 pm
content: Quote from: PA0PBZ on April 13, 2018, 06:26:12 pmsnap

author: StillTrying
author_rank: Super Contributor
timestamp: April 13, 2018, 06:34:04 pm
content: On the scope pic the input signal isonly justgoing LOW enough to reach the 14's lower hysteresis level, which can be quite low ~0.6V, it could do with going a bit lower."Could it be does the IC's was killed from ESD?"In theory yes, in practice no.Edit: While I'm closing down old windowz, I post this here.

author: Ian.M
author_rank: Super Contributor
timestamp: April 13, 2018, 07:49:30 pm
content: I just dug out a crusty old Motorola 1981 date code CERDIP 74LS14 and breadboarded up a simple test jig. I used longer jumpers for the gate input and output for ease of moving them around the chip. Its currently blinking away at about 6Hz.LED steady ON - input not connected or stuck high or output stuck lowLED off - input stuck low or output stuck highLED blinking - gate O.K.You need such a low value of feedback resistor R1 to guarantee the output low can pull the input down to the logic '0' threshold. Its a little bit abusive as when the output goes high, its pulling about 8mA to 10mA from it initially to charge the capacitor, but as that's only for a brief part of the circuit, and the IC can withstand a single output direct short to ground for up to one second, its unlikely to cause any issues unless the IC is really crappy. I wouldn't however use this circuit long-term in a permanent build.The same circuit will do for testing any inverting Schmitt trigger input bipolar TTL gate - just tie extra inputs high (via 1K) or low to reduce the gate's logic function to a single inverter, and apply the test jig to one input at a time. If you want to test a CMOS gate, decrease C1 to 4.7uF and increase R1 to 33K. It is *NOT* suitable for testing non-inverting or non-Schmitt input gates.If your PSU wires are longer than a few inches, put 100uF across them at the breadboard.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 13, 2018, 08:08:17 pm
content: Ian.MThank you, I will try your method later.Here is the IC how they look like:Is this maybe a fake one?

author: ebastler
author_rank: Super Contributor
timestamp: April 13, 2018, 08:21:22 pm
content: Quote from: Chriss on April 13, 2018, 08:08:17 pmCertainly a vintage one - from 1985?! Where did you get these?The EL manufacturer prefix does not ring a bell for me; but I certainly don't claim to know them all.The body material looks unusual (color, sharp edges). Maybe from the former Eastern block countries?Anbody?In any case, this seems like a chip someone dug up from the bottom of a long-forgotten box; so it might well be fishy...

author: Chriss
author_rank: Frequent Contributor
timestamp: April 13, 2018, 08:27:40 pm
content: QuoteI found this one as equivalent in AHCT version: SN74AHC14 but the main caracteristics are the same as the LS.Can you pls give me a model of AHCT? which would be nice for me.I have to use the IC to convert the photo interrupter signal to a logic Hi-Lo signal...I got that ic from a local electronic component store in Hungary.

author: Ian.M
author_rank: Super Contributor
timestamp: April 13, 2018, 08:45:53 pm
content: 74ACH..../74AHCT.... are excessively fast, which causes problems if you have long wires or don't have good grounding and decoupling - which is a typical problem when breadboarding.74HC..../74HCT.... are a bit slower (similar to 'classic' or LS TTL) and more forgiving. They are far more breadboard friendly.Only get the 74xxCT... TTL logic level compatible input version if you *NEED* a low threshold voltage, either for compatibility with bipolar TTL outputs, or 3.3V CMOS logic outputs etc. Otherwise the 74xxC... families have more or less symmetrical input thresholds and significantly better noise immunity. All 74xxC... families drive rail-to-rail (except open drain parts or if excessively loaded), but for many, they are still better at sinking than sourcing, so higher current LEDs are best wired active low (between the output and Vcc, with a series resistor).

author: KrudyZ
author_rank: Frequent Contributor
timestamp: April 13, 2018, 08:50:34 pm
content: Quote from: Chriss on April 13, 2018, 05:41:31 pmThe rise and fall times of that purple trace are VERY slow.Look at the time base of the scope...

author: Chriss
author_rank: Frequent Contributor
timestamp: April 13, 2018, 08:53:17 pm
content: Yes. The purple trace is the output from the photo interrupter which is connected to the input of the gate.Sent from my GT-I8260 using Tapatalk

author: Ian.M
author_rank: Super Contributor
timestamp: April 13, 2018, 09:12:27 pm
content: So you've only got about 3V input amplitude, which means that if you are running the logic from 5V, you *NEED* real bipolar TTL or LS TTL or 74HCT not 74HC logic.It may be worth using a comparator (e.gLM339), with resistors to set the threshold and hysterisis. As that comparator has an open collector output, don't forget the output pullup resistors.

author: ebastler
author_rank: Super Contributor
timestamp: April 13, 2018, 09:25:48 pm
content: Quote from: Chriss on April 13, 2018, 08:53:17 pmHuh? I thought you had a pull-down resistor and a switch to +5V connected to the input?!(Your schematic in reply #5?)It's hard to do remote troubleshooting when we don't have complete and consistent information about the boundary conditions. Your photo interrupter's output might just be too weak for the LS14 input.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 13, 2018, 09:35:44 pm
content: In the first version I had that with the switch, after I found that good gate I connected also my photo interrupter to the74LS14 good gate and made the measurements...I know it is hard to remote troubleshooting...I was thinking also about the LM339 to use, but I found more simple to use a 74LS14 in my circuit than the LM339.I tested in my circuit that one IC 74LS14 which have a good gate and it is doing the job well.I also connected it to my uC and started to count the triggered signals and works great.I don't know, maybe I will give a try to another IC but now from the 74HCT version and something not from the year '85.I just was wondering about my situation with this IC's because I was playing before also whit such of IC's and I never had a burnt oneor situations like now, except if I repaired some electronics with a bad Schmitt trigger...So, back to the situation, I assume more or less my schematic shouldn't kill the IC's.Maybe that version where I connected the LED to the IC.But I have also a bad IC on which was the LED not connected.I will try to get a new ic but now one from the 74HCT version and make a new test..I don't know, maybe I messed up something.Just one more question:If I don't connect the unused pins to gnd, can that kill the IC?

author: rstofer
author_rank: Super Contributor
timestamp: April 13, 2018, 09:35:59 pm
content: Some of MY rules for logic (regardless of family)Apply Vcc and Gnd to the proper pins with short leads from the PSAdd 0.1 ufd capacitor between Vcc and Gnd at EVERY chip and keep the leads as short as possible. There are sockets with built-in capacitors - that is short!Always pull inputs high and switch them low. The size of the pull-up resistor depends on the logic high threshold and the pin input current. If I don't care about power consumption, 1k is the right answer.Never pull a load high, always pull it to ground. This is important for TTL and less so for CMOS but I do it that way anyway.

author: rstofer
author_rank: Super Contributor
timestamp: April 13, 2018, 09:41:14 pm
content: Quote from: Chriss on April 13, 2018, 09:35:44 pmIt can be for CMOS.You need to treat the inputs and outputs properly. You can pull inputs to a solid ground but I would never pull them to a solid Vcc. I ALWAYS use a resistor to limit the logic high current (others will have a different opinion). ALL inputs on ALL devices need to have some kind of connection. Up or down, your choice but they can not be allowed to float.Outputs: Never pull an LED up, always pull it down, read the datasheet for maximum output current - both high and low. Connect the resistor between Vcc and the LED anode, connect the LED cathode to the output pin.ETA: You can tie multiple inputs to the same pull-up resistor but you have to make sure the resulting pin voltage is higher than the logic 1 threshold with all inputs sinking their maximum logic high input current.

author: TomS_
author_rank: Frequent Contributor
timestamp: April 14, 2018, 08:47:29 am
content: Quote from: Ian.M on April 13, 2018, 08:45:53 pmSorry for the hijack, but this is burning a hole in my understanding - which was that the AHC/AHCT versions "simply have a lower propagation delay" (thats the way TI marketing material positioned it anyway). If the HC/HCT versions are a bit slower propagation wise, how does an input transitioning an output a bit quicker cause such significant issues? Or perhaps I have missed/do not understand another detail...Re using xHCT parts, I have been looking to stock these specifically because so many microcontrollers and other logic devices seem to be 3V3 parts these days, particularly at the mid-high end, and the xHCT parts are more compatible due to lower logic high requirement. So its more or less about stocking parts which can be as universal as possible.Thanks!

author: Ian.M
author_rank: Super Contributor
timestamp: April 14, 2018, 09:05:37 am
content: Its all down to risetimes and load capacitance. The faster the risetime, the bigger the current spike drawn from Vcc or Gnd to charge or discharge the load capacitance when an output switches and the greater the resulting threshold shift experienced by other gates on the same chip caused by the disturbance at its supply pins. Then its a matter of noise levels on the inputs and the probability that the threshold shift will momentarily erode the logic input threshold margins enough to cause a glitch. Its a square law effect - the current spike amplitude increases in proportion to the speed of the transition, but its duration also reduces proportionately, so a 20% reduction in risetime makes it 44% harder to decouple the supplies adequately and provide a reliable ground (stiff enough at DC and low enough impedance up to an equivalent frequency to double the rise time)With very fast logic you get into the situation that you need to source terminate longer lines to control the risetime to tame the spikes.

author: Zero999
author_rank: Super Contributor
timestamp: April 14, 2018, 01:37:25 pm
content: Quote from: Ian.M on April 13, 2018, 09:12:27 pmOne way round that is to add a diode to the pull-up resistor. It's a bit of a crude hack and only works at fairly low frequencies.

author: Ian.M
author_rank: Super Contributor
timestamp: April 14, 2018, 03:02:37 pm
content: The other dirty trick is to lift the ground of *ALL* the 3.3V logic by one diode drop, which can work reasonably well with faster signals. However it barely makes the typical 70% guaranteed logic '1' threshold, so isn't great for noise immunity. Also, beware of 5V logic with 80%/20% thresholds, as even with a diode drop boosting it, the 3.3V logic wont reliably cross the 80% threshold unless you overvolt the 3.3V logic and/or undervolt the 5V logic.

author: joeqsmith
author_rank: Super Contributor
timestamp: April 14, 2018, 04:33:47 pm
content: Quote from: Chriss on April 12, 2018, 08:59:15 pmPicture of your setup may have helped. Could be wired wrong. Some time ago, I played with a bunch of old DIP inverters. I started out trying to replicate someones setup using some whiteboard.https://www.eevblog.com/forum/projects/waveforms-in-a-74ls04-ring-oscillator/

author: Chriss
author_rank: Frequent Contributor
timestamp: April 14, 2018, 07:12:03 pm
content: Ok.I have some questions:1. Is my ic fake or not?2. Is it worth to buy the HCT version and use that new technology for upcoming projects or can I use also the old 74LS14 series?I digged through tons of data sheets, resources on the net but found millions of variations of circuit made with the 74ls14 which are actually crossing between the theory and real world hardware circuits.So I found also a blog where somebodi I think from Porugal wrote about a light detector circuit where actually my circuit was used. I mean the guy who wrote the text on that blog didn't copy - paste my circuit but it is the same meaning how they made his electronic light switch.If my circuit could kill the ic an/or are wrong then millions of other circuits are still wrong in the world, what are not impossible.No one datasheet show any so to say "tipical" schematic how to use the ic.Can we figure out a tipical wireup for this ic what could work in real world?What a kind of crappie ic is this when it can't handle a LED diode.Because of 20mA max current they will be killed....Than it is a "must have" to put a transistor to the output. Without the teansistor a such of ic can be really easy killed.Alk these is to shitty.At the end I come to a mind not to use this crappie ic's and interchange them wit an LM339 and finishing the job.So, what is your opinion about a circuit how to use this crap inverter to switch the input from a photo interrupter and the output should send the signal to a uC.Sent from my GT-I8260 using Tapatalk

author: ebastler
author_rank: Super Contributor
timestamp: April 14, 2018, 07:53:20 pm
content: Quote from: Chriss on April 14, 2018, 07:12:03 pmAd 1.It seems unlikely that it is a fake, but given its age and not-so-mainstream origin and package, maybe it has not aged well and has degraded over time.Have you tested it with (a) clean power supply, decoupled with suitabe 100nF capacitor, (b) strong, well-defined input test signal from pullup resistor and switch to GND, and (c) checking the output on an oscilloscope, without a load connected? It is not clear to me from the thread if you have done tests where all three conditions were met. If the anser is "yes", ad the chip still did not behave, then it is probably bad.Ad 2.A key benefit of an HCT chip would be that you can get one from current production, ruling out the possible degradation of your Hungarian LS chip. Also, the higher input impedance of the HCT may be helpful if you feed it a wimpy signal from a photosensor. So yes, I would give it a try.

author: rstofer
author_rank: Super Contributor
timestamp: April 14, 2018, 08:11:46 pm
content: There is nothing wrong with 74LSxx devices. It is up to you to read and understand the capabilities and limitations as given in the datasheet. Every single number printed in the datasheet means something. Sometimes it is a factor in your design, sometimes it isn't.All TTL logic is asymmetric in terms of pull-up and pull-down current. That's why loads are always pulled down. This isn't true for CMOS...For the 74LS family, IOL(pull down capability) is 8 mA - hardly enough to light an LED but it will work. OTOH, IOH(pull up capability) is only 400 uA and is incapable of driving an LED.If you look at IOLand IILyou will see where the device output can drive 20 inputs. Same for IOHdriving IIH. I wouldn't push the 20 gate fanout but I would certainly do 10 or 15 gates. The point is, the 74LSxx devices are perfectly capable of doing performing their intended function in their intended environment. It's up to you to design around the capabilities and limitations.http://ee-classes.usc.edu/ee459/library/datasheets/DM74LS14.pdf

author: Chriss
author_rank: Frequent Contributor
timestamp: April 14, 2018, 08:59:27 pm
content: ebastler:QuoteYes, I done these tests several times gate by gate and only one gate had a good signal of the two IC's.That's made me crazy because I bought 2 IC's and I tried them and they are looking bad.I tested before too in the past this type of IC's with my scope without loads and all they was ok, with a nice inverted signals,but these two ic's didnt do anything. Or non stop low, or non-stop high, or non-stop nothing. Just one gate is acting well.Here is two circuits what I was used also before for testing and button debouncing.When I say "before" I mean years back...Is the 1. circuit good enough to just test the gates on this 74LS14?I ask because I done this type of connections in my projects and they still working even after 7 years.Also the 2. circuit, is that good enough because that is also working years back till now...This circuit I used for button hardware debouncing for uC's...When I tested the 74LS14 in the past, I never pulled up the not used inputs, and I never faced this kind of problem like now.I should maybe avoid this type of wiring, to let the inputs floating...

author: Ian.M
author_rank: Super Contributor
timestamp: April 15, 2018, 12:44:17 am
content: For the second circuit, it works with a switch, but if you attempt to drive it from a transistor, unless you reduce the 1K resistor, you may not meet the guaranteed logic '0' threshold.e.g. theOnSemi 74LS14 datasheetsays it can take up to 0.4mA to pull the input down. Therefore there can be up to 0.4V across the 1K resistor and to get the input under the 0.6V guaranteed logic '0' threshold (parameter: 'Negative-Going Threshold Voltage') the left end of the resistor has to be below 0.2V. However the typical output '0' voltage is 0.25V so you are already S.O.L. if you want to reliably drive a low pass filter like that from another gate on the same chip. Keep that resistor under 500R and you'll get away with it.RC debouncing for 74HC/HCT is much easier as their input current is negligible so you can use much lower value capacitors and higher series resistors.

author: ebastler
author_rank: Super Contributor
timestamp: April 15, 2018, 04:33:36 am
content: Chriss - when you say „this type of IC“ has worked for you in the past, do you mean the exact batch of Hungarian ICs you are using now, or some orher brand? If you were successfully using something else before, it seems pretty clear that the Hungarian chips are bad - toss them!

author: amyk
author_rank: Super Contributor
timestamp: April 15, 2018, 05:18:15 am
content: Not fake, but certainly very old and possibly defective.Does the bottom look similar to this?http://www.silirium.ru/pictures/ussr-Electronpribor-KR531IP3-brown-back.jpgA quick Google suggests these parts were found in Commodore C64shttp://cbm-hackers.2304266.n4.nabble.com/CBM-s-CP-M-Z-80-carts-for-the-64-td4656117.htmlQuotehttps://www.forum64.de/index.php?thread/45601-west-trifft-ost-russ-ramchips-im-c64/Not sure what the EL manufacturer prefix is either, but Elektronpribor is a good guess.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 15, 2018, 05:14:07 pm
content: Ian.M:Thanks for this nice explanation, I really appreciate it.I will definitively switch to the 74HC/HCT series IC's.ebastler:When I say „this type of IC“ has worked for me in the past, I mean this hexagon Schmitt trigger IC's but from other manufacturer.I never faced with this brow casing IC.I have experience with 74xx series IC's but from other manufacturer.amyk:No, the bottom didn't looks like on the link.The bottom is simple brown plastic like the upper side.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 16, 2018, 11:54:20 am
content: Ok, today I got new IC's but now SN74S14N in black package...Please give me suggestion how to correct breadboard this IC.In the past I used the 1. circuit from my schematic above.I read somewhere on the net does all the not used input pins should be pull-up or down.I never did that before and I never killed any IC this type of Schmitt trigger.But now I'm a bit afraid and wish to change things if I need to make my circuits better.What would be your simplest setup for a quick test on breadboard the SN74SL14N?Thank you.

author: iMo
author_rank: Super Contributor
timestamp: April 16, 2018, 12:02:15 pm
content: Based on the package and type of font used it is an old Soviet chip (most probably an export marking)..The former eastern countries produced 74xxx series as well, they used black packages (with different form and fonts used), however.This is the simplest test you may try:http://www.learningaboutelectronics.com/Articles/7414-schmitt-trigger-oscillator-circuit.phpYou may use any resistor 1k-100k and capacitor 10nF-10uF with your test. You may wire the inputs of the other free inverters to the oscillator's output and observe their outputs - you have to see pulses as well..

author: Ian.M
author_rank: Super Contributor
timestamp: April 16, 2018, 12:23:33 pm
content: Are you sure about that part number? Google doesn't return *ANY* datasheets for SN74S14N or even 74S14. If that's what's on them, odds are they are fakes. IMHO its more likely you've got SN74LS14N chips.Your circuit n.o. 1 or my circuit from reply#21will do to test them, one gate at a time. Don't leave them running for a long time on my circuit - it pushes them too close to their limits.Bipolar TTL (all types) inputs float high well enough that you can leave unused gates' inputs floating when breadboarding. However the recommendation for the final build is to pull up to Vcc via a 1K resistor. You can often group several unused inputs to one resistor. Strictly speaking its only original 7400 series TTL that you *had* to use a pullup resistor for to avoid latchup if the input experienced a spike on the Vcc rail. One generally avoids grounding unused inputs unless its required for the logic function, as that consumes extra current. Don't leave spare inputs of used gates floating - that can result in big trouble with glitches when another pin on the same chip changes state.CMOS is *MUCH* easier - just tie the input direct to either rail AT THE CHIP - it wont waste current and it wont latchup.

author: PA0PBZ
author_rank: Super Contributor
timestamp: April 16, 2018, 12:30:27 pm
content: In the same post he calls it a SN74SL14N so my bet is a keyboard problem.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 16, 2018, 12:31:51 pm
content: Uff!Misspelled. .. it is an LSSent from my GT-I8260 using Tapatalk

author: rstofer
author_rank: Super Contributor
timestamp: April 16, 2018, 01:25:08 pm
content: I already gave you the 'rules' for digital circuits: all inputs are terminated to appropriate logic levels, they are nexer left floating. You can tie pins to solid ground but I don't tie them to a solid Vcc, I always use a pull-up resistor. If I do the math, I can tie several unused inputs up through a single resistor. I can also tie unused inputs up to some legitimate output signal. This affects fan-out but it might not matter.For a breadboard circuit, I will probably use a pull-up between 1k and 4.7k. If I'm going to actually use the circuit, I might make certain that I always meet the VIHrequirement. And then I will still use a lower value - probably between 1k and 4.7k.FWIW, it is better to use a lower value resistor when debouncing switches. Mechanical switches take a certain voltage to break through the oxide layer and a certain current to maintain continuity. 1k seems to work pretty well.Logic high input current is lower than logic low input current. In terms of power consumption, it is probably better to tie unused inputs to logic high through a pull-up resistor.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 18, 2018, 02:00:44 pm
content: What is wrong with this circuit?I use the P832 Photo interrupter and the SN74LS14 to transform the signal from the P832 to a square signal.When I interrupt the photo signal with a card board the interrupter actually goes low aroun 0.25V but the 7414 won'tinvert the input.This SN74LS14 is a good one now.Here is my circuit what I use:I also tried the R3 to change from 1K up to 10K but no big difference...What does I miss?Thank you...

author: PA0PBZ
author_rank: Super Contributor
timestamp: April 18, 2018, 02:16:27 pm
content: What happens if you remove the P832, does the 74ls14 output go high now?Did you try to just short the 74ls14 input to ground with the P832, without using any resistors?

author: Chriss
author_rank: Frequent Contributor
timestamp: April 18, 2018, 02:20:08 pm
content: If I remove P832 the 7414 doesn't go high, but If I short the 7414 input to GND then the signal goes high.I also tried with a simple transistor BC337 as a switch wired up to the input of 7414 and worked fine.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 18, 2018, 02:23:20 pm
content: I think I have to put another value for the R3 but I can't figure out what value could it be.

author: PA0PBZ
author_rank: Super Contributor
timestamp: April 18, 2018, 02:26:05 pm
content: Quote from: Chriss on April 18, 2018, 02:20:08 pmDid you try my other suggestion?

author: Chriss
author_rank: Frequent Contributor
timestamp: April 18, 2018, 02:37:33 pm
content: QuoteYes, I tried, nothing happens. Signal on my scope still high.But here is something I found in the datasheet of the P832:I set the R3 to 1K and connected the Colector to +5v.With this setup the 7414 works great on the scope.I'm a bit afraid does I do something wrong?Or can I use the above schematic to make my project working?I have to setup another 3 photo interrupter to the 7414.What do you think?

author: Ian.M
author_rank: Super Contributor
timestamp: April 18, 2018, 02:52:30 pm
content: You cant pull a TTL input low reliably with a resistor over a few hundred ohms so putting the phototransistor between In and Gnd is essential as it probably cant handle driving a 470R resistor with its emitter. Once you've moved the phototransistor, I'd put a pullup resistor on the input as well if the phototransistor has enough drive to handle the extra current and still reach a valid logic '0' level with plenty of margin.CMOS 74HC/HCT here would be a lot easier - it removes most of the constraints on resistor values etc. caused by the input current that has to be sunk to pull the input low and lets you use the Toshiba datasheet circuit with a much higher resistor value.If you don't want a low going pulse when the light path is interrupted, use another gate of the 74xx14 to invert the output. If you *MUST* have more than 3 inputs on the same chip and *MUST* use a 74LS14, you'll need to add extra transistors to turn the phototransistors into Darlingtons so they can handle the approx 10mA to drive high with a 470R pullup. Any jellybean small NPN will do, but put 1K B-E to sharpen up the turnoff. Personally, I wouldn't bother, unless its driving a lot of other gates or is in a high risk environment for ESD I'd just drop in a 74HCT14.

author: StillTrying
author_rank: Super Contributor
timestamp: April 18, 2018, 03:39:57 pm
content: How do you expect the opto to pull the signal high with R1 there.This should work with any version of '14 inverters, - it's as in Ian's 1st paragraph.

author: rstofer
author_rank: Super Contributor
timestamp: April 18, 2018, 03:46:39 pm
content: Quote from: Chriss on April 18, 2018, 02:00:44 pmThat circuit just isn't going to work.Assume the transistor is in saturation, call it 0V drop between collector and emitter (it'll never really be that low) then CALCULATE the output voltage going to the LS14. You'll only get about 5V / 11 or around .46V. That 30k/3k voltage divider is killing you.On the sensor, put a 10k resistor between the Vcc pin and 5V - connect this collector junction to the LS14Remove the resistor in the emitter and ground it.The datasheet uses 0.5 mA when they specify VCEand, for 5V, this requires 10k Ohms.The 10k resistor will drop, at most, 10k * 20 uA = 0.2V when driving the LS14 input high so the gate will see about 4.8V and it only needs to see 1.9V to get above the threshold. In other words, 10k is fine.Use the circuit in Reply 59. I would still go for the 10k resistor...

author: Chriss
author_rank: Frequent Contributor
timestamp: April 18, 2018, 09:16:41 pm
content: Thank you guys!I really appreciate your help.Actually, the P832 I salvaged from an old printer pcb.The schematic above in my post #52 what is not working is taken from the printer pcb.But the on the junction where I connected my 7414 is connected to a pin froma uC on the original pcb.So, however, that is in my case not working with the 7414.Please check this if I understand in the right way what is standing in the datasheet for the 7414:IOHdoes it mean when the input is feed withlowsignal, the Imaxon the output is the max current of -0.4mA what the 7414 can delivery.IOLdoes it means when the input is feed with ahighsignal, the Imaxon the output is the max current of 8 mA what the 7414 can delivery.Does I understand it correct?I ask just to clear the meaning because of my English language understanding...I trying to understand the datasheet and wish to find the mayor parameters to set up the 7414...

author: rfengg
author_rank: Regular Contributor
timestamp: April 18, 2018, 11:07:59 pm
content: IOH does it mean when the input is feed with low signal, the Imax on the output is the max current of -0.4mA what the 7414 can delivery.IOL does it means when the input is feed with a high signal, the Imax on the output is the max current of 8 mA what the 7414 can delivery.You are correct.When the input is low , as its an inverter, the outout goes high.At this time , the maximum current that the TTL gate can deliver is 400uA to reliably hold the output at the high state.The 400uA also determines the fanout of the logic family i.e., the number of other devices you can connect to the output.When the input is high , as its an inverter, the outout goes low.At this time , the maximum current that the TTL gate cansinkis 8mA to reliably hold the output at the low state.Please also remember the IOH and IOL are dependent on the exact logic family (74XX vs 74SXX vs 74LSXX etc etc.....) and also for the military versions starting with 54, so you need to be diligent to look up these on the data sheet. For eg, on the TI SN7414, the recommended IOL is 16mA while for the SN74LS14 its 8mA and 4mA for the SN54LS14.

author: rstofer
author_rank: Super Contributor
timestamp: April 19, 2018, 01:27:03 am
content: I => currentO => outputH => highThis is the maximum output current when the pin is high. It is shown as (-) because the current is coming out of the pin. Source current, the pin is sourcing the current.I => currentO => outputL => lowThis is the maximum output current when the pin is low. It is shown as (+) because the current is going into the pin. Sink current, the pin is sinking the current.You can now see why we don't try to pull LEDs up! 400 uA just won't work. Even 8 mA of sink current won't provide the 20 mA that a normal LED would like to see. The LED will light up but not to full brightness.

author: rstofer
author_rank: Super Contributor
timestamp: April 19, 2018, 01:52:35 pm
content: OOPS!The transistor in the opto device has to sink the collector resistor current PLUS the IILof the 74LS14 and this total current should be about 0.5 mA. The LS14 has a max IILof 0.4 mA so that only leaves 0.1 mA for the resistor current.5V / 0.1 mA = 50k Ohms, not 10k - 52k should work.The 52k resistor times 20 uA (IIH) will drop 1.04V so the LS14 will see 3.96V on the pin. Far above the threshold voltage of max of 1.9V so everything should work fine.

author: StillTrying
author_rank: Super Contributor
timestamp: April 19, 2018, 04:40:27 pm
content: Quote from: rstofer on April 19, 2018, 01:52:35 pmAfter much DS reading, I think the 0.5mA is just a 'test condition', it's not saying that it comes out of saturation if it's a bit higher. With 20mA through the TX, I make the RX sinking up to about 4mA before it starts to come out of saturation.http://pdf.datasheetcatalog.com/datasheets/toshiba/TLP832(F).pdf

author: rstofer
author_rank: Super Contributor
timestamp: April 19, 2018, 07:46:03 pm
content: That's correct, the 0.5 mA number is used for the rise/fall time test. The datasheet also gets involved with the current transfer ration which relates the LED current to the transistor collector current. I didn't see where they gave any advice on VCESat at any particular LED or collector current.It is true that the resistor could be much smaller but it doesn't need to be. The LS14 input high current is quite low and the required sink current is well within the test circuit range.I would try it and see. One disadvantage of high value resistors is possible noise.In this schematic, we see 22k resistors used plus a 0.1 ufd capacitor between the collector and ground:http://www.junun.org/MarkIII/datasheets/Controller_SCH.pdfThese optos are not slot sensors and the data may not apply directly but it does show one way of dealing with any noise that may occur on the collector.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 19, 2018, 09:01:51 pm
content: Ok, this is going a bit over my head, this calculations...Long time ago when I was in school and do the math's, but I shouldpractice again...So, it theInput LOW Currentor theInput HIGH Currentwhat is in the DS of the LS14 actually thecurrent what is needed to feed into the input of the gate so the gate can hold the output LOW or HIGH?Maybe tonight if I don't drop to sleep will try this circuit with a 52K resistor between the VCC and collector.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 19, 2018, 10:47:13 pm
content: I tested already the circuit like rstofer mentioned in post no #64.It is working great.Thank's all of you with such of help and info sharing.

author: rstofer
author_rank: Super Contributor
timestamp: April 21, 2018, 03:16:54 am
content: It's good that it works but you should really measure some voltages to prove that it works. The LS14 input is a good candidate. What is the voltage when the input is '0' and again when it is '1'. How does that line up with the threshold voltages in the datasheet?You could try to reduce the LED current with increasingly large resistors until the logic '0' voltage starts to rise. Just to see what the Current Transfer Ration (CTR) actually is. This is not really important and may be difficult to test.

author: Chriss
author_rank: Frequent Contributor
timestamp: April 21, 2018, 09:51:02 pm
content: When the input is "0" the voltage is 190mV.When the input is "1" the voltage is 5v.

author: rstofer
author_rank: Super Contributor
timestamp: April 22, 2018, 12:19:29 am
content: That seems about right. VCESat for most transistors is about 0.2V so 190 mV is pretty close. The transistor is saturated - it is fully ON.The 5V reading is also about right. There's very little current flow through the resistor so there is very little voltage drop.Glad it works!