Help: Circuit Design 4in NOR Gate

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eminence said:
How would you layout the schematic to achieve the desired result?
Click to expand...
Previous post and figure were updated. I assume the transistor used in your circuit is PNP, right?

Pls refer to the figure below. I think there is a problem. No matter the output of NOR gate is '0' or '1', the transistor will still be triggerred or ON. When the output of NOR gate is '0', the voltage at point B1 will be 0V, and therefore Q1 should be ON. When the output of NOR gate is '1' (HIGH or 5V), the voltage at point B1 will be near to 0V because 100k resistor is much larger than the 4.7k resistor. As the result, Q1 should be ON. Pls correct if if I'm wrong.
 


I am not sure why he hooked that 4.7k resistor to ground. I thought exactly the same thing as your thinking when I saw it. So I tried the circuit with and without the resister there. But it made no difference as I don't think the NOR gate was changing.

I too have edited my previous post.

Simon
 

eminence said:
What are pull up resistors?

From what I can tell the gate is not going to change because the inputs are not +V, but I am not sure? Can the 4002 gate be changed with a GND signal fed to the inputs?
Click to expand...
You need to add a 10k pull-up resistor for each input to the NOR gate. See figure.


Well, it could be the NOR gate did not work, or the transistor part did not work.

4002 is a CMOS 4-input NOR gate. What is the supply to the 4002? Is Pin-14 connected to +5V or +12V?
 

nicleo said:
4002 is a CMOS 4-input NOR gate. What is the supply to the 4002? Is Pin-14 connected to +5V or +12V?
Click to expand...

+12v

Simon
 

XNOX_Rambo said:
C'mon guys - don't make it harder than it is...
Click to expand...
This could be a good option.
 

XNOX_Rambo said:
C'mon guys - don't make it harder than it is...
Click to expand...

That is the first thing I thought of, but you have forgotten one thing that would only work if all doors were open. Otherwise it would still be getting 12v GND from the other doors. So it would not detect 1 or 2 or 3 or 4, but just all or none.

That is why I used 4 relays in my circuit but it used too much power.

Unless you have done something I have not noticed.

Thanks,
Simon
 

eminence said:
That is the first thing I thought of, but you have forgotten one thing that would only work if all doors were open. Otherwise it would still be getting 12v GND from the other doors. So it would not detect 1 or 2 or 3 or 4, but just all or none.
Click to expand...
Simon, try to add those pull-up resistors and see whether the problem solved or not.
 


Thanks. I will have to go and buy some 10k resistors tomorrow as I don't have any. Apart from the pull-up resistors can you see any other problems with the diagram?

I tried shorting the base leg of Q1 to GND and to +12v but it did not switch the relay, I was trying to simulate the output of the NOR gate. What is going wrong there?

Thanks,
Simon
 

Here is a bad thing: Unless the person behind the counter has a degree in electrical engineering, he is then a saleperson.

eminence they sell Alfa Romeo cars in Australia? I know they are famous carabinieri cars in Italy but that far in Australia?
Do not tell me they even sell Fiat UNO there.
 

The purpouse of the diodes is to block the +12V from the other door switches.
Or has the function of a diode changed somehow lately...? :wink:
 

Besides resistors, probably you should get few extra PNP transistors. May be the one you have damaged already.

By the try to measure the voltage of NOR's output when ALL inputs are '1' (HIGH). Also, measure the voltage of NOR's output when ALL inputs are '0' (LOW).
 


Alfa Romeos are available in most countries. FIATs have not been available here since the early 1970s. Anyway I could tell you a lot about the Italian car industry in Australia but that would get me no further towards a solution to this problem.

Even worse would be to get a FIAT Panda here, thankfully not.

The guy in the store is an electronics hobbyist and a has a degree of knowledge. That is why I have come here for further assistance.
 

XNOX_Rambo said:
The purpouse of the diodes is to block the +12V from the other door switches.
Or has the function of a diode changed somehow lately...? :wink:
Click to expand...

Well how does the signal from the door you want get through? Also the door switches do not have +12v running through them at any time. Just 0v GND (ground) and 12v GND. Perhaps I just don't understand your diagram properly. Could you explain it to me further?

Thanks for your ideas,
Simon
 

Ah sorry - I had missed you switch drawing in the middle of all the others.
The "12V GND" was puzzling, but now I see what you mean.
I will modify my drawing and post it shortly.

/Rambo
 

This is in principle a DTL gate.
When all doors are shut (switches closed) the transistor will not be sourced
with any current to its base - and therefore not conduct.

If any switch is opened the corresponding resistor will source the transistor base
with current (about 1,2 mA when using 10 kOhm resistors) and the transistor will
conduct energizing the relay.

The diodes prevent the other switches from grounding the base of the transistor -
i.e. acting as an OR gate.

Please get back to me if you have any questions.

/Rambo
 


Thanks for you ideas. How can I tell which transistor to use? I will peobably have to use an equivalent anyway as we don't get the same range as the USA.

I have attached my version of your diagram and I am wondering if you think it will still work. I have added the extra diodes to protect the switches in the doors. I have already burnt two of them out when their wires were accidently shorted to +12v by someone who was helping me. So they need to be protected from +12v go down the line to them.

They are a magnetic reed style witch so if +12v goes down the line they fuse together and then burn up.

Let me know your thoughts,
Simon
 

You should get a transistor based on the current rating of the relay coil.

In my opinion, after adding D1 ~ D4, the circuit will not work anymore because, for example, when door1 is close, D1 is not short and therefore the cathode of D1 is still at 12V. What do you think?
 

I have been offered this as a solution on another forum. What do you think to it?

Ignore the cabin lamp.

Simon
 

nicleo said:
You should get a transistor based on the current rating of the relay coil.
Click to expand...

Well that makes sense.

nicleo said:
In my opinion, after adding D1 ~ D4, the circuit will not work anymore because, for example, when door1 is close, D1 is not short and therefore the cathode of D1 is still at 12V. What do you think?
Click to expand...

Do you think it will destroy switches without them (D1 ~ D4)?

Thanks,
Simon
 

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