HEX Inverter Alarm Circuit with Reed Switch

Thread Starter

Tiger7

Joined Mar 9, 2018
1
Hi, I am a complete beginner in electronics but I urgently require help for a GCSE (UK secondary school qualifiaction) project: an alarm system which uses a reed switch and HEX inverter to detect the opening of a door, triggering the buzzer and red LED. For some reason, the HEX inverter doesn't seem to be working: when I take out the HEX inverter, nothing happens - the three outputs just continuosly flare up regardless of the HEX inverter. I have connected the ground pin and power supply pin as well as all the un-used input pins to ground.

I have used a 555 astable to try and create oscillations for the buzzer and LED output, but when the circuit is switched on, all the outputs (2 LEDs and a buzzer) continuously flare up.

Below is my PCB design which replicates the 'Yenka' circuit design at the bottom.

Capture.PNG Yenka - Alarm circuit.PNGThis is the 'Yenks' circuit design which does not include the volatage regulator (the power supply has been reduced to 6V to replicate the effect of a L7805 voltage regualtor) or the reed switch (replaced by an SPST switch which is connected to the NOT logic gate.
 

AnalogKid

Joined Aug 1, 2013
10,173
There are a couple of changes that would make the circuit less complex, but your problem probably is that there is no pull-down resistor at the IC3a input to GND. CMOS inputs are extremely high impedance, and can remember their last input for a long time when floating.

Also, add a 0.1 uF cap across the power and ground pins of each IC.

To protect the 555, increase both timing resistors 10x, aand decrease the timing cap 10x. This reduces the peak current into the Discharge pin.

ak
 
Last edited:

Dodgydave

Joined Jun 22, 2012
10,590
If pin 3 of the 555 is permanently high, then pin 2 is not getting a charge up voltage from the resistors 5k6, 1k, or the 100uF capacitor is shorted out .
 

jbeng

Joined Sep 10, 2006
84
This may or may not affect the overall function of your circuit, but looking at your PC board layout, the green "power" indicator LED appears to be shorted out by two square pads of unknown function...
 

sghioto

Joined Dec 31, 2017
3,664
You definitely need a pull down resistor on pin1 of IC3A. On the PCB it shows a unmarked component where the reed switch connects to pin1 of IC3A. Is that a jumper?
For the circuit to work properly the reed switch or SPST needs to be normally closed. The door opening is detected when the switch opens.
SG
 

takao21203

Joined Apr 28, 2012
3,702
put a led on the inverter output for testing
maybe the ic is bad. if it is socketed you can bend the pins and use flywires for another gate which maybe works, if you only have one IC.

transistors and LEDs all can be bad, recommend to get a component tester and DMM, as well you should not use 6V for 74XX IC they may work for some time, but its not good.

Test the output stage too, tap 1K resistor after the 10K to vcc and see if anything happens.

The NE555 could be bad. But i dont think it will be damaged by 1k ohms and 5.6k ohms thats not happening in my opinion

the output stage is somehow strange i dont understand why two transistors are needed and why they are in series, what is the transistors you use? they could be bad or wrongly inserted, and the arrangement of them, I dont understand it fully.

The LED could be directly from the NE555 and 1 transistor for the buzzeres, could use 1K with modern LED with 330R it will be irritating too bright.
 

sghioto

Joined Dec 31, 2017
3,664
he output stage is somehow strange i don't understand why two transistors are needed and why they are in series, what is the transistors you use? they could be bad or wrongly inserted, and the arrangement of them, I dont understand it fully..
The output circuit is a standard darlington configuration, possibly not needed in this case but works.
SG
 

MisterBill2

Joined Jan 23, 2018
13,722
The output circuit is a standard darlington configuration, possibly not needed in this case but works.
SG
Every input to the hex inverter IC needs to be tird to either V+ or ground, if it is not in use.AND the input with the switch needs to have apull-down resistor as well, so that it is either at V+ or at zero volts. CMOS gates will drift into a linear mode if the inputs are left floating.
 

sghioto

Joined Dec 31, 2017
3,664
Why are you quoting me? What you are suggesting has already been discussed. If you are going to participate please read all prior post.
SG
 

MisterBill2

Joined Jan 23, 2018
13,722
Why are you quoting me? What you are suggesting has already been discussed. If you are going to participate please read all prior post.
SG
I quoted you because I am not an expert at participating in this discussion format just yet. I did read the other parts, but it did not seem that the comment about CMOS inputs was well understood. In addition,b the comment about a pull down resistor only addressed the one section of the hex inverter IC, but ALL inputs need to be biased either high or low, not just the one being used.
 

sghioto

Joined Dec 31, 2017
3,664
The TS mentioned the unused inputs in post 1 when describing his PCB layout. He hasn't replied since his original post anyway so I guess he got a "F":D.
I apologize if I sounded harsh.
SG
 
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