Greetings everyone
I have been in the process of designing a high power relay board
Yes I know it's already a commercial product exists but honestly I don't trust it as the traces seem too narrow for the claimed current it should be able to handle + Where is the fun in that
Here is my Schematic (Any suggestion is much appreciated)

Now according to the Trace width calculator I used
For 25 Amps I need a trace width of 25.5 mm (I am using 26 mm here)
Which is quite impossible as even when I enlarge the PCB size
The traces would still interfere with the Ralay and the terminal block

Is my calculator wrong?
Any Idea is welcome

 

You need to swap the base and emitter of the transistor also the power indicator LED will be running at 100mA.

For the trace width the result is going to depend on the inputs to the calculator - copper thickness, allowed temperature rise...
The values below give 6.5mm.
https://www.7pcb.com/trace-width-calculator.php

 

A 4N25 has a nominal current transfer ratio of 0.2. The relay current is nominally 77 mA from the data I can find. There is not the slightest chance that that optocoupler can directly drive that relay. In fact, it is very unlikely that any optocoupler will directly drive that relay. You will require an additional transistor to handle the relay current. Any common type such 2N3904, 2N2222A, 2N4401, MPS-A06 would do. A darlington such as an MPS-A13 or a small MOSFET might be better if you want to keep the input current to the optocoupler low.

EDIT there are darlington output optocouplers rated for 100 mA output current, so it should be possible to drive the relay directly with the coupler - e.g. 4N33

 

Using standard pcb copper thick 0,035 mm then 25mm is needed @ 10degree temp rise.

Picbuster

 

Your opto-coupler and relay will never work the way the schematic is drawn.

Also, why not move the relay off the board and use a relay with screw terminals?
You will need 10AWG wire for 25A switched by the relay.

 

if you leave PCB mask open along the trace, you can put some solder on trace to fatten it up. larger cross section can carry more current.

R1 need to be about 1k.

what is the purpose of R2,R3 and U1?

 

if you leave PCB mask open along the trace, you can put some solder on trace to fatten it up. larger cross section can carry more current.

R1 need to be about 1k.

what is the purpose of R2,R3 and U1?

Once or twice I've seen manufacturers overlay heavy power tracks with tinned copper wire.

 

You can use thinner traces if you solder tinned copper wire over the trace.
In the past I have used the outer braid from coaxial cable directly soldered to the print when I made a 50 volt 25 amp psu for a high power linear amplifier for my H.F transceiver.
Below is a photo of tinned copper wire soldered to thin traces of an experimental power supply I was making.

 

You can use thinner traces if you solder tinned copper wire over the trace.
In the past I have used the outer braid from coaxial cable directly soldered to the print when I made a 50 volt 25 amp psu for a high power linear amplifier for my H.F transceiver.
Below is a photo of tinned copper wire soldered to thin traces of an experimental power supply I was making.

Doing that with ANY kind of stranded wire is a seriously bad idea!

 

Doing that with ANY kind of stranded wire is a seriously bad idea!

WHY? when it is flattened and soldered to the trace it is almost as good as a strip of copper soldered to it, AND it was done to the servo's in a military control system designed by one of the top military contractors, please qualify your statement.

 

Soldering wire to a trace is a labor intensive/old amateur trick and wouldn't be done in a "real product" build in the past 25+ years or anything where cost is a concern (which is everything outside of military applications for the most part)..
Now one can easily just go heavy/thicker copper laminates (10+oz) or other methods to achieve an increased current rating.. laminated bus bars,etc....

As stated the calculator is giving you the numbers based off the entered temperature rise which you chose as 10deg C..
And those calculators from my experience are quite conservative..
Most typical FR4 circuit boards can be allowed to run up to 105-150deg C temperatures (UL MOT rating) ..
Enter 50degC for temp rise and watch that width DROP...
Even better is to enter the correct value for your application.. (UL MOT of PCB laminate minus max ambient minus a safety factor temp or something like that)

Then look into a thermal relief for those pins to avoid excessive/unequal soldering time between that and other "normal" traces..

 

WHY? when it is flattened and soldered to the trace it is almost as good as a strip of copper soldered to it, AND it was done to the servo's in a military control system designed by one of the top military contractors, please qualify your statement.

Sooner or later you'll get a stray strand.

 

Sooner or later you'll get a stray strand.

Ah yes, if not properly soldered down in the first place, but as far as a cheap and quick solution when experimenting with a circuit, and also saving the cost of the vastly more expensive heavier gauge copper clad pcb, it certainly is a perfectly valid solution.
The high power supply for my linear has worked for over 25 years with no signs of fatigue.
For D.C I don't see a problem if done properly, but would not recommend it for R.F or large scale production.

 

Thank you so much everybody
Sorry for my late reply as I almost just gave up there
I had been researching this for the past week and even watched Dave's EEVBlog episodes which were helpful in their own but not for my case
Needless to say they are soooooo long and boooooring
Honestly I didn't think my circuit was that terrible
I thought I would do much better for my first attempt (Gonna double check that Opto-Couplers specs once more)

 

if you leave PCB mask open along the trace, you can put some solder on trace to fatten it up. larger cross section can carry more current.

R1 need to be about 1k.

what is the purpose of R2,R3 and U1?

R2 and R3 are to make a voltage divider to achieve the Opto-Couplers operating voltage
U1 is the Opto-Coupler itself

EDIT:
Also why R1 should be 1k?
The LED forward voltage is 3.3 Volts (Blue LED) and the Supply voltage is 5 Volts
with a 100 Ohm resistor I get a current of 17 mA which I say is just perfect to compensate for other components accuracy

 

Okay I actually didn't notice it until now
the Opto-Coupler in the Schematic seems to be a way way less powerful one than i intend to buy
The one here is the "4N25M" while the One I have is just the "4N25"
Data Sheet linked below for your verification
And Thanks again

https://www.vishay.com/docs/83725/4n25.pdf

 

I have just seen an "indestructables" article where they actually use de-soldering braid to repair tracks. I guess that as the braid is coated with some sort of flux makes it easier to solder. Never tried it, but might be worth a try.

 

EDIT:
Also why R1 should be 1k?
The LED forward voltage is 3.3 Volts (Blue LED) and the Supply voltage is 5 Volts
with a 100 Ohm resistor I get a current of 17 mA which I say is just perfect to compensate for other components accuracy

The relay is 12V and is fed from the same VCC as the LED. How did you get 5V for the LED?

 

The relay is 12V and is fed from the same VCC as the LED. How did you get 5V for the LED?

I totally forgot that, as the relays I am prototyping with now operate on 5 Volts I calculated the resistor for the LED based on them and forgot that I need 12 Volts here
Thanks for pointing this out
(I feel really stupid now)

 

The first reasonable compromise is to accept 30°C temperature rise in the track. The second, which is often not an option with low-cost fast-turn prototype board houses is to use 2 ounce (0.0028", 0.07 mm) copper. Even 4 ounce copper begins to impose limitations on trace & space width. Anything over four ounce is going to cost substantially more because it ties up the board fab's plating line for a long long time.

Again, a 4N25 has a specified current transfer ratio (CTR) of nominally 0.2. That means for each mA through the input emitter, you can expect only 200 µA through the transistor. For 77 mA for the relay, you would have to put 385 milliamps - over a third of an ampere - through the input emitter. It would be instantly destroyed.