You need to use EasyEDA editor to create some projects before publishing
How to drive 5V relay with SMD componant?
2794 18
mannewman 6 years ago
Hello With a PIC microcontroler i want to drive 5V/10A SONGLE relay (LCSC ref C35449) to command rolling shutter motor : R1(4.7K) + R2(47K connected to ground) + NPN transistor and with a free-wheeling diode on coil ==> classic ! But someone could tell me what SMD ref used in my circuit ? is it necessary to use an optocoupler (instead of transistor) to isolate microcontroler Is it easy to solder SMD ? Thanks
Comments
andyfierman 6 years ago
@mannewman, Welcome to EasyEDA. Please show us a schematic of what you mean by: `R1(4.7K) + R2(47K connected to ground) + NPN transistor and with a free-wheeling diode on coil` Your questions: 1) `...someone could tell me what SMD ref used in my circuit ?` No because you have not specified what voltages you are using to drive the circuit you wish to build with. Just saying: `With a PIC microcontroler i want to drive 5V/10A SONGLE relay` does not give enough information about which PIC and what the high and low output voltages - and at what currents they are quoted - are for that microcontroller. It also does not specify what actual supply voltage you are using for the relay. We assume 5V because it is a 5V relay but what is the tolerance on this and can it supply enough (extra) current to drive the relay coil? 2) `is it necessary to use an optocoupler (instead of transistor) to isolate microcontroler` Probably not because the relay provides isolation anyway but you have not stated what voltage you may require isolation at. 110V mains? 230V mains? 2kV? You still need to specify the isolation voltage in order to check that the relay you have chosen will provide it. 3) `Is it easy to solder SMD ?` Do you mean by hand? If yes then you will need a good quality pair of tweezers, a fine point (1mm or less) soldering iron preferably with a short reach (distance) from hand grip to tip and a steady hand. You will also need to use nothing smaller than about 0603 parts for resistors and caps because smaller than that they are very hard to pick up and hold. Preferably 0805. If you have any 0.05 inch pitch parts such as the PIC then unless you have had a lot of practice or training, you probably will not be able to solder them without bridging the pins. You will need some good quality desoldering braid and preferably some solder flux. Buy spare parts to make up for those that ping out of the tweezers and are never seen again. * If you are new to soldering make yourself a training PCB project with lots of SMD parts on to practice on *before* you try to build a real project. Design it so you can easily see the joints and can check them for continuity and shorts using a DMM. See: https://electronicsclub.info/soldering.htm
Reply
mannewman 6 years ago
![Shematic][1] [1]: /editor/20171214/5a328929a9815.PNG Output current PIC 16F88 = 20mA Relay drive ~230 v / 10A (rolling shutter motor)
Reply
mannewman 6 years ago
Hello all Someone could help me ? my schematic is it correct concerning values SMD components ? and i need to use optocoupler or no ? Thank you
Reply
andyfierman 6 years ago
1. Increase R4 & R7 to 100k. This increases the base current available into the relay driver transistors. Ensure that you select transistors with an hfe at about Ic = 50mA of at least 100. 2. Increase R5 & R8 (Rseries) to 1k to reduce LED current to approx 3mA (Iled = (5V-Vfwd_led)/Rseries) where Vfwd_led is between approx 1.8V and 2.2V for ordinary red, green and yellow leds or about 3.5V for high intensity leds and blue or white leds. 3. Add a series R for LED1 (see 2). 4. Check to see if you need a pullup resistor for the LEARN switch like on the RESET switch. 5. Add power supply decoupling according to the manufacturers advice in their datasheets and apps notes. 6. Learn more about electronics design (our simulation tool are great for this. See (3) in: https://easyeda.com/andyfierman/Welcome_to_EasyEDA-31e1288f882e49e582699b8eb7fe9b1f). :)
Reply
mannewman 6 years ago
Ok andyfierman I follow all your advice and come back to you with the whole schematic computed with SMD component For 1. : i would like delete R4 and R7 to save space on PCB For 2. and 3 : ok for 3 mA / 1K For 4 : pullup resistor provide by program in PIC 16F88 For 5 : visible in my next shematic For 6 : perfect this link to learn thank you for answering me quickly (it's not so easy for me to translate in English from French !!!)
Reply
andyfierman 6 years ago
### Removing R4 and R7. Make sure the PIC outputs driving the relay driver transistors are configured so that they can pull to ground when sinking 0mA as well as that they pull up to very near 5V when sourcing 1mA. Check the PIC datasheet. ### Decoupling About 100nF ceramic across the PIC 5V to ground. You will also need about 10uF across 5V to ground at each relay driver plus about 100uF where the 5V and 0V comes into the PCB. Please also see: https://easyeda.com/andyfierman/Power_supply_decoupling_and_why_it_matters_-451e18a0d36b4f208394b2a2ff7642c9
Reply
mannewman 6 years ago
Hello andyfierman Here are the rest of the schematic and your remarks ! ![Version 2][1] [1]: /editor/20171217/5a36599920525.PNG
Reply
mannewman 6 years ago
and the PCB : ![PCB2][1] [1]: /editor/20171217/5a3665433beb8.PNG
Reply
andyfierman 6 years ago
Next steps: see also (1), (2), (4), (5) and (6) in: https://easyeda.com/andyfierman/Welcome_to_EasyEDA-31e1288f882e49e582699b8eb7fe9b1f
Reply
mannewman 6 years ago
Ok i will check your document you did not see any big mistakes in the schematic ? So i really thank you andyfierman for your great help with this design and your availability !!!
Reply
andyfierman 6 years ago
@mannewman, `you did not see any big mistakes in the schematic ?` You have placed 10uF C3 and C4 in the wrong place: `You will also need about 10uF across 5V to ground at each relay driver` *Not* from the PIC outputs to ground. * One comment: before you proceed any further you **must** check that you have sufficient high voltage creepage and clearance distances around **all** your traces carrying 230Vrms. * You must also check that the traces that are carrying 10Arms without overheating. **You are dealing with lethal voltages and high powers. If you do not understand what I am discussing here then stop your design until you have done enough research to find out and understand.** Sorry but I can not check your schematic - or your PCB - in detail because you have not provided enough detail to be able to do that without spending hours trying to work out what your various devices are doing and what the nets are between them. Without that information it is takes far too long and is not possible to offer any meaningful comments. If you would like a detailed design review then I can do that but that is not a free service and you would have to offer far more information than you have so far. To give you an idea of what you should have available when you ask anyone to review a design, here is an (incomplete) document about what needs to be supplied for a professional design review: https://docs.google.com/document/d/e/2PACX-1vRhMMBzCZ_orIFr56QyqYI8HcWsFA00evgh1cQ_069MJRU2YUJXGWzyX3mMVnxT-SNz1mydEicu6Bfh/pub
Reply
mannewman 6 years ago
This is the right place for C5 now ? ![enter image description here][1] [1]: /editor/20171218/5a3783a548515.PNG
Reply
andyfierman 6 years ago
Yes, that's right now.
Reply
mannewman 6 years ago
Thank you but i don't understand the interest to add others capacitors between 5V and GND, C1 and C2 already present on Power supply ? ![enter image description here][1] [1]: /editor/20171218/5a3791356ef0d.PNG
Reply
andyfierman 6 years ago
Every time you then a relay on our off, you step change the current being drawn through the traces to them by about 90mA. This will cause a voltage step in the 5V rail. Decoupling the 5V real at the relay driver will reduce the size of this voltage step by supplying the step current change from the capacitor instead of the 5V suppl. This will improve your power integrity. This is good electronics design practice.
Reply
andyfierman 6 years ago
@mannewman, Let's try that again without Autocorrupt: Every time you turn a relay on or off, you step change the current being drawn through the PCB traces to them by about 90mA. This will cause a voltage step in the 5V rail. Decoupling the 5V relays at the relay drivers will reduce the size of this voltage step by supplying the step current change from the capacitor instead of the 5V supply. This is good electronics design practice because it will improve your power integrity.
Reply
andyfierman 6 years ago
To help you understand why you need to decouple the relay drivers individually, I have now added: **More examples of power supply decoupling in action** to: https://easyeda.com/andyfierman/Power_supply_decoupling_and_why_it_matters_-451e18a0d36b4f208394b2a2ff7642c9
Reply
mannewman 6 years ago
Ok andyfierman i check the PCB carefuly and i launch production i give you a feedback after test thank you again
Reply
Login or Register to add a comment
goToTop
你现在访问的是EasyEDA海外版,使用建立访问速度更快的国内版 https://lceda.cn(需要重新注册)
如果需要转移工程请在个人中心 - 工程 - 工程高级设置 - 下载工程,下载后在https://lceda.cn/editor 打开保存即可。
有问题联系QQ 3001956291 不再提醒
svg-battery svg-battery-wifi svg-books svg-more svg-paste svg-pencil svg-plant svg-ruler svg-share svg-user svg-logo-cn svg-double-arrow -mockplus- -mockplus- -mockplus- -mockplus- -mockplus- -mockplus- -mockplus- -mockplus-@1x -mockplus-

Cookie Notice

Our website uses essential cookies to help us ensure that it is working as expected, and uses optional analytics cookies to offer you a better browsing experience. To find out more, read our Cookie Notice