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Any help on a simple logic circuit?

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Junior Member level 1
Jul 29, 2002
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I'm totally new on designing circuit. But I think on this site there should be plenty of gurus. My project is very very simple:

I have three TTL inputs, if they are all HIGH, then the output is HIGH; if either of them is LOW, then the output is LOW. But whenever the output changes from HIGH to LOW, it will not be affected by these three TTL inputs anymore unless you press a reset button.

Can anyone point me a direction on what kind of reference book or docs I should read? If you can, could you mention some chips name (like 74LSXXX) so I can easily locate them?

Many thanks.

find attached the proteus simulation :D press the reset
you need a 4073 :p
**broken link removed**
it will work for you im sure :oops:

you can place diodes also in series with the data line a1 a2 a3

A1 ---->|----
A2 ---->|----
A3 ---->|----

this version is more complete and tested :roll:

you can also place a switch too gnd a pin on the main 3 input and gate via two resistors 1k &10k value

place 10k in series with the pin 5 gate
and place the 1k from the gate via a switch too gnd
to form a not enable pin
so it will never work unless the switch to gnd is not closed

much better colour image


Amazing! How can you draw a circuit so quickly. Anyway, I will find the compoents and test it on a breadboard.

Only two more questions about your figure. What's the use of switches A1 ~ A3? What diodes I should use? Is there any integrated one?

Thanks again.


i cant be bothered too change R9
it should be 1k too gnd
sry about that

corrects the problem at ttl level and 12 volts cmos level

the switches are just too test the bias level was ok

100k too gnd and 10 K buffer zone

the 100K are put there becouse cmos has very sensative gates
so you need too attenuate anyvoltages that will give a miss trip

the diodes use 1n4148 :)

there is no simple solution too a latch

the rest of the switches can go although you need to replace sw1 with a 1K res too use the not enable latch so it still carries the signal for 3 but dosnt latch
or a 1k if you dont want too use this feature or to give you a no latch input by toggle the line low too disable the latch

SW2 can be removed safely
with the enable pin realy is a not disable so when its pulled low the output is turned off or low these are extra feature too what you asked but no matter handy to know

if you download proteus lite
you can view it

this design is realy only half of one chip 7 diodes although you can do without d8 and 6 resistors

although you can do a similar job with a flip flop
or a schmit
this is a nicer set up as it is very flexable too also add other outcomes via just adding a two input and gate type or an or or nor gate

the 74hct4073 is a good choice for TTL levels as it comes in three voltage ranges
the design above wont need altered and will do for any 4073 varient
and still give correct output

too control other logic

quality not componet count makes for reliability {philips moto use too be}

you can use just a simple latch chip there are too many too quote
several cmos will do this also
if you adda counter too the circuit you can have it latch ever six time or seven times etc
but using just a simple latch gives no brain there is more too a latch than just a bolt a clip and a lock

but as i say if you put all your eggs in one basket
you wont know which hen layed layed what one {my old tutor said this }


Thanks for your help. I tried your circuit this week, it works. But it's a little different from what I expect.

Actually, this is for a simple interlock. I expect, in normal condition, the three inputs are all high (which means system is ok) and the output is also high (which will actuate a solenoid). Whenever there is something wrong, one of the input will become low, and the output will also become low (this will shut down the solenoid). But I don't want the system turn back to normal condition automatically, I want turn it on manually when something is wrong.

Simply, 1) when three inputs are high, output is high, 2) if one of the inputs becomes low, the output also becomes low and it will remain low forever regardless the further change of the inputs unless you press a reset button.

The above circuit seems to work like this: if initially the three inputs are high, it will remain high forever; if one or more of inputs is low, the output is low, but when three inputs are all high, the output will become high and stay forever. It just near miss the target.

Best Regards

RS flip flop

What if your three inputs are anded and put to a RS flip flop? Once one of them goes low the flip flop is flipped and stays flipped even if the inputs go high again. you have to manually trigger the other input to get it to flop back.

It would probably be as cheap to use a small Pic. You can get demo compilers to play with. Doing it this way, means you can tweak the circuit to do what you want. Seems rather like overkill for such a trivial task, but just as cheap. Maybe more fun this way?.

Thanks for all the replys, especially !!!MONKEY!!!

I end up with the following circuit and it works just as I want.

The three switches on the left side imitiate the normally open equipments. When these equipments are in normal conditions, the switches are open; when anything wrong, the corresponding swith will become close and fire the interlock. For normally close equipment, just simply connect it to the ground. BTW, the SW1 is for security purpose, cause I don't want to activate the reset button accidentally.

Hi cryyo,

your primary requirement was : if they are all HIGH, then the output is HIGH; if either of them is LOW, then the output is LOW

On your final drawing it's different. Did you change your mind during drawing ?

What it's the purpose of D8 ? If I saw well, !!!MONKEY!!! said you can remove it. If you really simulated in Proteus, what happens with output of U3:A when D8 anode is going down, towards to ground ?

My advice: listen to flatulent's valuable suggestion !



I'm so sorry to confuse you guys. I think I made a terrible typo in my previous post.

Since this project is for an interlock circuit for a vacuum system, there are three equipments to monitor the status of that system. For these three equipments, they all have an alarm relay to indicate abnormal conditions they have sensed. For mormal operation condition, the relay is open. So after modification of !!!Monkey!!!'s original circuit, I find the circuit I posted works just fine.

In my mind, I equalize the OPEN relay with a HIGH signal (partially because one of the manual of an equipment said so) and confuse all of you. I apologize again.

I really have no idea why I use D8, but since !!!Monkey!!! use it, I naturally follow an expert's thought. I have test it with real components, the circuit works the same with or without D8.

As for the RS flip flop that mentioned by flatulent, I have told that I'm almost totally knowledgeless about circuit. I have ever heard the componets mentioned by !!!Monkey!!!, but never a term flip flop ever cross my mind before. After searching the internet, I find numerous flip flops and latches, I really lost. So I decided to keep working on simple diodes, as long as they work for me.



I found one problem with the circuit listed above!

After making a simple circuit board, I tested in my office many times, and the circuit worked perfectly. But when I put it into the system (there are dozens of equipments running all the time), it became just too sensitive: sometimes when I plugged in a nearby equipment, it turned off automatically.

I'm wondering if I need to put some capacitors on both input and output ends to absorb the noise generated by other equipments, so any help on this? If I need put a capacitor, what values it better?



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