I'm trying to make a DIY Arduino Uno board.
The power supply was designed as follows.
Will there be a problem if I implement the circuit as it is?
What will happen if I connect a zener diode (5.1V) to the output of the regulator (7805)?
Q1) Is this the correct circuit? Isn't that right?
Q2) Can you explain the reason in detail?
No.
The LM7805 requires capacitors directly on the input and output pins, to ground. While the actual values depend on the manufacturer of the specific LM7805 then using a capacitor equal or greater than 10uF aith a 0.1uF ceramic capacitor in parallel, On both input and output as close to the pins as you can get.
C1 and C2 are a bit far away. C2 should be 100nF. The zener diode is a bit superfluous.
The reason I designed it that way is
The reason for placing the zener diode is,
When the lm7805 element dies, a zener diode is designed to prevent overvoltage.
I'd love to hear your reasons for thinking it's unnecessary.
Thank you very much for your interest and reply.
If the 7805 dies short circuit it will take the zener out and you'll get 12v on your Arduino. If it does open circuit the zener does nothing.
Because when / if a LM7805 dies it normally dies open circuit not closed circuit.
If it did indeed die short circuit then your Zener diode would simply fry due to the the current down it, unless it had a current rating above that of what ever is supplying the input to the regulator. This would take a big stud mounted Zener on a large heat sink to be effective.
I understand what you mean. Thank you for answer.
Finally, I have a question for you.
Can you briefly introduce the 5VDC, overvoltage protection circuit? The intended use is to use as much power as the Arduino Uno.
Why are you concerned? No one else bothers and generally speaking there isn't a problem.
You could use a big fat 5.1V zenner, but then you need to be absolutly sure there is a fuse of suitable rating in the incoming supply to prevent the fire.
That is negligible, about 40mA or so. But your power supply will not be limited to supplying just that much current but much more. So you need to consider the maximum current it can supply, not what you want it to supply.
The minimum current from a mains AC adaptor ( sometimes called a wall wart ) would be 0.5 Amp with 1 to 3A being typical. Perhaps if you told us what that was we could calculate how much heat you would dissipate if the regulator failed short.
@jhaine Thank you very much for your reply.
@Grumpy_Mike Thank you very much for your reply.
@srnet Thank you very much for your reply.
By any chance, what fuse is applicable here?
I would appreciate it if you could let me know the product name, etc.
It seems to be of great help to me.
Sorry but in electronics a fuse is too slow to protect damage to electronics.
An old joke is that electronics are a device to stop a fuse from blowing.
Oldie but goodie. Seriously, fuses are generally to protect wiring so it doesn't cause a fire with a short circuit.
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