Simple NOT gate problem! HEELPPP

[fouadalnoor]

Hello guys, I am simply trying to create a potential divider with a NOT gate. As one of the resistors in the potential divider increases with resistance the the voltage going out decreases and thus the voltage going into the NOT gate increases and the output should be low.

But for some reason I can't seem to get my circuit to work as it should. Looking at the attached image of the circuit (using multisim) the light probe is always on.

The problem seems to lie with my input but I just don't know why.

Hope you guys can help.

Thanks.

 

[crutschow]

Have you looked at the data sheet for the 74LS04? The input has 1mA flowing out of it to generate a logic 0 at 0.8V. Your input impedance is too high to ever bring the input below 0.8V (check the input voltage during your simulation).

You either need to reduce the input resistance or go to a CMOS circuit (such as the CD4096 or 74HC04) which has negligible input current.

 

[fouadalnoor]

Well I now tried to swap it with a CMOS 40106BP_5V but to no avail... still wont work.

I couldn't find the chips that your talking about on multisim so I couldnt swap them. Also I dont know how to change the input impedance of the chips so im stuck now...

 

[crutschow]

Well, a 40106 is a synchronous binary counter, not a gate, so I wouldn't expect it to work, why would you? Do you ever look at the characteristics of the IC you are trying to use before putting it in the circuit? Random electronic design does not work well. If you can't be bothered to look at the data sheet for the IC you are using, then any attempt to help you is rather wasted.

Look for 4069 in the CMOS digital parts list.

I was referring to the impedance of the resistors at the input to the gate, not the gate impedance, which of course you can't control.

 

[ericgibbs]

Well, a 40106 is a synchronous binary counter, not a gate, so I wouldn't expect it to work, why would you? Do you ever look at the characteristics of the IC you are trying to use before putting it in the circuit? Random electronic design does not work well. If you can't be bothered to look at the data sheet for the IC you are using, then any attempt to help you is rather wasted.

Look for 4069 in the CMOS digital parts list.

I was referring to the impedance of the resistors at the input to the gate, not the gate impedance, which of course you can't control.

hi Carl,
I think you have a typo ref the HEF40106
Eric

 

[crutschow]

Please accept my apology, I misread the Google search. So with foot-in-mouth I retract my peevish comments.

That is a HEX inverter which should work in the circuit I believe. Show us the new circuit and your simulation results.

 

[fouadalnoor]

Ok, the images are attached below. Note that when the input is high or low seems to make no difference...

 

[shaneshane1]

Ok, the images are attached below. Note that when the input is high or low seems to make no difference...

does it work if you remove your voltage divider and put a "pull down" (10K to ground) resistor on the input and use your 5V supply to activate your input? the 10K ensures that the input WILL go low without the applied 5V input...

 

[crutschow]

I believe the simulator needs a ground connection on the circuit for proper simulation and I don't see one.

 

[fouadalnoor]

Ok guys, now I have added a normal ground to it and two simple pull down resistors. It seems to work as expected.

The problem is, as soon as I want to create a potential divider with it instead (by simply taking the input connection and placing it between the two resistors) it wont work as it should?!

I keep getting 5v on the input even though im changing the second resistor...

 

[fouadalnoor]

sorry, the circuit I tried with the two pull down resistors looked like the images attached and they both worked as expected...

Its only when I try this stuff with a potential divider that it seems to fail...

 

[fouadalnoor]

FINALLY:

Ok, it worked just now. I am actually pretty sure it was the ground that had to be proper.

umm, in the real circuit do I just connect my chip to the negative terminal of the battery when connecting it to ground?

 

[ericgibbs]

FINALLY:

Ok, it worked just now. I am actually pretty sure it was the ground that had to be proper.

umm, in the real circuit do I just connect my chip to the negative terminal of the battery when connecting it to ground?

hi,
On most project circuits, -V of the battery is 0V/Gnd and the +V is the supply to the IC's
Always have a capacitor across the +V and 0V, about 100nF and a 10uF, this will reduce electrical interference within the circuit components.

 

[fouadalnoor]

Ah, thanks you very much for your help, my whole project seems to be working now. Though I have (as always) another little problem that I can't figure out...

I have made another little circuit and connected it to an AND gate. Its very straight forward really, and it seems to work, though the probe that is supposed to light when its ON does not. Even though everything else works and there is a 5v going across as shown on the picture.

I have edited out all the extra bits to show what I am talking about.

The first picture shows it working correctly when I just use the AND gate symbol instead of the actual chip.

In the second picture its still working, though the probe is not on for some reason...

 

[crutschow]

What is the switch level of the probe? The 4081 has a suffix of 5V so it may only output 5V, even though the supply is 9V.

 

[fouadalnoor]

Thank YOU SOOO MUCH, I have been looking at that thing for like 2 hours and couldn't figure out what I did wrong lol.

The simulation now fully works, I will be making my PCB soon. I will probs post more problems as I go along.

 

[crutschow]

Thank YOU SOOO MUCH, I have been looking at that thing for like 2 hours and couldn't figure out what I did wrong lol.

The simulation now fully works, I will be making my PCB soon. I will probs post more problems as I go along.

The moral is, if the circuit is not working, always use a voltmeter or oscilloscope to check voltage levels and verify that the levels are really what you expect.