You need to use EasyEDA editor to create some projects before publishing
Where to find one for scheme revision?
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Sergey Romanov 5 years ago
I am new in electronics. Just created my first scheme. Please suggest me a resource where I can hire a professional to revise my scheme and make sure that I've made all smart choices?
Comments
andyfierman 5 years ago
If you are a beginner in electronics then you probably do not want a professional Design Review as that normally costs money. What you want is experienced people on this forum to look at your design but to do that you will need to make it public and post the url here. Is this it? [https://easyeda.com/sergey.romanov](https://easyeda.com/sergey.romanov) If you really need a paid-for professional Design Review, you can message me. * But before you do anything else make sure that you have read all the advice given in (2) of: [https://easyeda.com/forum/topic/How-to-ask-for-help-and-get-an-answer-71b17a40d15442349eaecbfae083e46a](https://easyeda.com/forum/topic/How-to-ask-for-help-and-get-an-answer-71b17a40d15442349eaecbfae083e46a) _especially_ sections (2), (4) and (6). It will save you both time and money. :)
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andyfierman 5 years ago
Sorry, I meant to point here: [https://easyeda.com/sergey.romanov/dmx-splitter](https://easyeda.com/sergey.romanov/dmx-splitter) One thing I noticed straight away is that the resistor in series with your mains side LED is too low. The voltage drop across the led will depend on the colour and type of LED but will be between about 2V and 4V. For an indicator LED, the typical current though it will probably be around 20mA (50mA max in the form of positive or negative half sinewave pulses) so with up to 240V rms you need a higher resistor value of around R = (240V-4V)/20mA approx 12k Ohms. However the power dissipation in this resistor will be about (240V-4V)*20mA/2 (because it's every other half cycle) approx 2.4W so you probably want to drop the LED current to no more than 5mA (and put up with a dimmer LED) to get the resistor dissipation down below about 0.6W. Then, if you're using SMD resistors, split the resistors into two of half the value to (i) halve the power dissipation in each again and (ii) double the creepage and clearance distances by splitting it across the gap between the pads of each resistor. An alternative is to use a series RC voltage dropper which may end up with a lower dissipation but is a bit more complicated to design. To help with fault finding in an installation you might consider putting an LED both sides of the fuse so a quick look up at the unit on a lighting gantry shows if there's power on both or only the input side of the fuse. I have also seen some circuits with a LED, anti-parallel diode and dropper connected across the fuse which lights up if the fuse has blown but that has some safety implications since there is a leakage path across an open fuse. You might also want to look at this project for some ideas on reverse polarity and overvoltage protections if that is important to your application: [https://easyeda\.com/example/Uberclamp\_Schematic\_PCB\_and\_BoM\-r4YgysK2k](https://easyeda.com/example/Uberclamp_Schematic_PCB_and_BoM-r4YgysK2k) :)
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