Could you look over this schematic please?

[chico]

Hello i had a few threads floating around and I put everything together and made a nice schematic got some software.

I couldnt find the footprints for the proper parts so ignore the pin numbers, but the part numbers are all good.

Any advice is helpful, from making a pretty schematic to issues with my circuit.

The jist of it is that there are 4 faults, if 1 or 2 are enabled an LED turns on and if 3 or 4 are enabled another LED turns on.
These fault signals for 1/2 and 3/4 are combined (OR) and also sent to a NOR gate which wont let me turn on my device if there are faults.
There are 2 LEDs (with thier own enable low switches) which also turn on LEDs if they are enabled and have no faults.
The signals are delayed and go into a schmitt trigger but there is also a signal send out (not delayed) if both LEDs are on.

This is all to learn so please give any tips. Also the resistance/cap values etc are not final.

 

[ronsimpson]

Polarity of 5uF and 4.7uF caps. + is going to ground.
You have a signal that goes thtough a 4072, through a transistor, then a 4002, another transistor, (1G08 and Diode delay thaing).
Change to where the input of the 4002 looks at the 4072 output. The 1G08 should look at the 4002. The diode delay should look at the 4002. This will help you PCB layout and will help the signal quality.

In your delay thing.....the 10k should be removed. Cap is backwards. Remove the 1k.

 

[crutschow]

Typical design practice is to connect the open end of a pot to the wiper if it is being used as a variable resistor. That way if the wiper momentary lifts when you are adjusting the pot, the circuit won't momentarily open circuit.

Normally the ground connections are indicated by a ground symbol.
**broken link removed**

 

[chico]

I had put the 10K there so that when there is no power to the system the cap will still discharge which i desire. Should I do it a better way?

Attaching the logic input to the logic outputs is noted, i thought having it after the transistor would be better since there is more current but i do beleive the signals could get degraded.
Also yes the cap was backwards, whoops :S
Thank you!

A question, if i get a LED with a forward voltage equal to my Transistor Collector voltage do i need to bother with the resistors in series? I imagine I dont.

I have updated the document in the first post

 

[chico]

Normally the ground connections are indicated by a ground symbol.

I couldnt find ground in my program

Regarding the pot, never heard of that but it makes sense

 

[ronsimpson]

The logic looks cleaner.
Your delay does not work!
When the 4002s output is low it will disharge the cap at a rate of 1,000 ohms.
When the 2004s output is high it will charge the cap at a rate of 1,000 ohms.
------------------------------------
You want the cap to charge at a rate set by the 5M pot and discharge very fast via the diode. So remove the 1k and 10k. The 1k is shorting out the 5M pot.

 

[chico]

YES the 1K is bad! and i should have noticed that. Someone mentioned ground bounce which sounded bad so i put that there, maybe i should put an inductor if its a problem.

But is the 10K resistor an issue? I need something to discharge the Cap when the power is off.
If the 4002 is not powered it will not be able to discharge the capacitor by pushing output low, thus i need the current to leak through something. This makes sense to me.
Updated

 

[crutschow]

A question, if i get a LED with a forward voltage equal to my Transistor Collector voltage do i need to bother with the resistors in series? I imagine I dont.

You imagine incorrectly. The LED voltage is the typical voltage that it drops at its rated current, not a voltage that you apply. An LED is a diode and it always needs a resistor in series to limit the current, or alternately it can be driven by a constant-current source.

 

[chico]

You imagine incorrectly. ...An LED is a diode and it always needs a resistor in series to limit the current

Wait but im confused, do i imagine correctly, can i use a 5 V voltage source if my LED forward voltage is 5V. In that situaion it will draw the corresponding current to 5V operation right?

 

[ronsimpson]

If you have a 6 volt light bulb you should give it 6 volts.
A LED is a current device not a voltage device. It converts current to light. Five it 10mA or so.
The voltage for an LED might be 2.5V min, 3V typ and 3.8V max. If you take that LED and forced 3 volts on it lets say it pulled 10mA. At 3.2 volts it will pull 20mA and 3.4 it will pull 40mA. When the diode gets hot it voltage drops. So when hot it pulls 20mA at 3V and 40mA at 3.2V. Now it gets really hot and at 3 volts it now draws 40mA.
This batch of of LEDs might be 3.0 volts but the next could be 2.8 volt parts or 2.5 volt or 3.5V who knows?

 

[ronsimpson]

I see you have not removed the 10k yet.
When the output of the 4002 is high, current of about 1uA flows into the capacitor and the voltage starts up. (current through 5M resistor) When the voltage reaches 0.01 volts you will have 1uA flowing in the 10k to ground. The voltage on the cap will not get above 0.01 volts. You need this voltage to get above 1/2 supply.

 

[Brevor]

You need base resistors for the 4 transistors.

 

[dougy83]

You need base resistors for the 4 transistors.

You don't need any base resistors for the transistors as they're connected as emitter followers.

The 10k definitely cannot be across the 5uF cap if you want it to work.

You also have zener diodes shown on the schematic - are you sure you didn't mean to use schottky or standard diodes?

PS. Please post your amended schematic at the end of the thread rather than just editing the original post attachment and invalidating all subsequent posts.

 

[chico]

OK yes the 10K will be a will be a voltage divider, my mistake... to get the cap to discharge while there is no AC power perhaps I should use a very large resistance instead of 10K like 100M?

That way I can still make the voltage on the delay circuit go above the High Input voltage on my Schmitt in 30s but then over many minutes the cap will discharge (time isnt a big factor in this situation).

 

[dougy83]

to get the cap to discharge while there is no AC power perhaps I should use a very large resistance instead of 10K like 100M?

How about no discharge resistor & let the cap discharge through the schmitt trigger input protection diodes? You can connect a 10k resistor from VCC to GND to speed up the discharge.

 

[chico]

OK if you say that it will discharge through the schmitt while the schmitt isnt powered im ok with that.

 

[dougy83]

Yes it's OK for CMOS logic with diode-clamped inputs: e.g. 4000, 74HC, 74HCT, 74C and some others. If you use LVC (& possibly others), there is no internal diode to VCC but you can place your own from the input to VCC for the desired effect.

 

[Brevor]

You don't need any base resistors for the transistors as they're connected as emitter followers.

Oops I didn't look close enough.