examination/correction on my MCU application circuit

[black_rider]

what is mcu examination

Hi, I have made a schematics for some MCU based board that I will use as an alarm central. Now I would like you to examin and correct if there are any mistakes, especialy for resistors and capacitors values and for power supply and osilator.
Please do examine/correct optocoupler and darlington NPN array connection.

I also want to ask if I need to add a small capacitor between the optocoupler photo diode input and sensor, switch and lock/unlock signal. Both sensors operate under 12V, but the signal output is a short pulse, the same is for lock/unlock signal from central locking module.

This is forum sized pic, to see a full resolution, please open the attachment.

 

[kender]

spikes damage mcu

At a glance, it looks okay. However, I would add a connector for in-circuit programming of the PIC.
You have optocouplers, but the external sensors and PIC share the same +5V supply. So, there's no galvanic isolation.
Which vibration sensor are you going to use?
Which ultrasonic motion sensor are you planning to use?

 

[CMOS]

mcu correction

ULN2803 is a sinking driver. So all your loads needs to be connected to +12V instead of ground.

 

[pauloynski]

You´ll need to add snubbers or free wheel diodes to the relay´s coils. I don´t think the relays can damage the ULN2803 driver but they can cause erratic behaviour of the mcu.
Regards

 

[CMOS]

You´ll need to add snubbers or free wheel diodes to the relay´s coils. I don´t think the relays can damage the ULN2803 driver but they can cause erratic behaviour of the mcu.
Regards

ULN2803 already has free-wheeling diodes inside. There isn't a need to put external diodes.

 

[pauloynski]

Dear CMOS: it is not that simple. The voltage regulator can´t sink current, only source. When the relays are switched off the free whell diodes carry their current directly to the power supply, but it can´t sink this current. This means a voltage spike will happen that can affect the mcu operation.
Regards

 

[CMOS]

The purpose of free-wheeling diode is not to sink current to ground but to discharge the residual high voltage generated inside the inductive load when current flowing through it is turned off by creating a resistive electrical path between ends of the inductor. Energy is dissipated in the form of heat from diode. This way it does not go into the power rails.
Have a look at the attached schematic showing internal driver stage of ULN2803 with internal and external free wheeling diodes. Can you explain how external free-wheeling diode is different from internal? They both are meant to serve the same purpose.

 

[pauloynski]

Dear CMOS: You´re fully right. But just inspect the circuit and you´ll see that the diodes are not connected the way you have shown (which is right).

Added after 2 minutes:

MIstake of mine. Sorry.

 

[CMOS]

Here is the ULN2803 internal diagram from datasheet. The diodes are connected in the exact same way as I have shown.