Need help improving PCB layout design - Page 1
Forum section: electronics / projects

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 08, 2021, 04:35:19 am
content: I have made an circuit with PCB in EasyEDA and i have allready ordered the first version og the PCB. I know the circuit works, så i need no help regards to that. But if i can remove or add some komponents or maybe change some of the out etc. i will do that if that improves the circuit. My goal is to make a smallest possible board without changing the component packages and sizes. I can change the values and types etc. but not anything else.I'm pretty shure i can do this more efficient, but i just do not know how.Here is my schematic and a screenshot of the PCB layout.

author: james_s
author_rank: Super Contributor
timestamp: November 08, 2021, 04:39:53 am
content: What's with the 51R resistor in parallel to the LED? It's unusual to use a mosfet to control a single LED too, a NPN transistor like a 2N3904 would work just fine. I'm also curious of the reasoning for using a DIP IC when the whole rest of the circuit is SMT.Edit: I guess that resistor is the load needed to keep the power bank on.

author: Psi
author_rank: Super Contributor
timestamp: November 08, 2021, 04:44:02 am
content: This is what I would change.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 08, 2021, 04:44:33 am
content: Quote from: james_s on November 08, 2021, 04:39:53 amThats right. Th 51R resistor is for make Power bank staying onAn SMD version of the 555 timer is not much smaller than an through hole one and it's easyer to solder and it make it also easyer to route my board... i think.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 08, 2021, 04:55:05 am
content: Quote from: Psi on November 08, 2021, 04:44:02 amThanks. I have changed that now

author: mc172
author_rank: Frequent Contributor
timestamp: November 08, 2021, 02:35:36 pm
content: It's a bit unusual to make the copper pour +5V and manually route the ground traces. This may become a problem if you screw it to a grounded metallic object or another board using metallic screws.Also be aware of the screw head diameter if you plan to use screws to mount it.The "GND" terminal on the right hand side doesn't seem to be connected to anything.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 08, 2021, 04:07:29 pm
content: Quote from: mc172 on November 08, 2021, 02:35:36 pmI have both ground and 5V pour.

author: fourfathom
author_rank: Super Contributor
timestamp: November 08, 2021, 04:27:01 pm
content: That 51 Ohm resistor is going to dissipate about 1/2W. What are you using? Even a 1210 size is generally rated for only 1/4W, and you really want a 2x margin for power dissipation (so four big 220R in parallel?).Also, I don't know if you really need that 220 Ohm gate resistor. The '555 has a CMOS totem-pole output, so when there's power there will be defined output (not floating).

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 08, 2021, 04:42:55 pm
content: Quote from: fourfathom on November 08, 2021, 04:27:01 pmActualle a good point there. But the power is just 1 second pulse every 25 second. But i see what you mean. I think my 1206 resistor is 1/2 watt?

author: fourfathom
author_rank: Super Contributor
timestamp: November 08, 2021, 04:55:56 pm
content: Quote from: ElectricPower on November 08, 2021, 04:42:55 pmYes, there are 1/2W 1206 resistors, so that's better. Still. I would like to de-rate them to half that. If you can get rid of that gate resistor then you have room for a second load resistor. I don't know the numbers for pulsed loads, but I suspect that one second is still pretty close to continuous for a 1206 package.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 08, 2021, 05:09:11 pm
content: With one second i meant 1 second load pr 25. second. Not one pulse pr. Second.

author: fourfathom
author_rank: Super Contributor
timestamp: November 08, 2021, 05:37:09 pm
content: Quote from: ElectricPower on November 08, 2021, 05:09:11 pmYes, that's what I thought you meant. I'm just saying that the temperature rise rate of the surface-mount package is probably fast enough that even an isolated one-second pulse takes the peak body temperature pretty close to the continuous level. If you had a 1ms / 25ms pulse train then the temperature would be fairly stable at the effective 1/25 level.But I'm guessing here. I don't know the thermal mass of the resistor, or of the PCB traces, so perhaps one-second pulse at a low duty-cycle is just fine and your single resistor is OK. These resistors usually have pulsed power equations in the spec, but I'm too lazy to check. Also, you might want to have the margin in case your duty-cycle or on-time is accidentally wrong due to component error (since those settings appear to be an option on your board.)

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 08, 2021, 10:34:42 pm
content: Can someone here please help me to find 2 watt 20ohm and 30ohm resistor in 2512 size on AliExpress?I can find 1 watt resistors with these values, but not 2 watt.

author: EPAIII
author_rank: Super Contributor
timestamp: November 08, 2021, 11:12:32 pm
content: Addressing only the physical layout of the board:1. You have four mounting holes for a really small board. Two, one at each end, would provide all the support needed. That would allow you to reduce the width of the board by at least 1/20", probably 1/10" with some adjustments to the spacing of the components. You probably worked with a 1/10" grid but that grid can easily be divided by 2 or even 4 to get a 1/20" or a 1/40 inch grid. And, of course, you can go "off grid".2. I see empty areas on both the front and back sides. The passive components (resistors, capacitors) are arranged on the front in a manner that limits the length of the board. You could squeeze their spacing a bit to get 1/20" or perhaps even 1/10" off the length. But the rear has a large amount of unused space so moving one of the components (resistors) from the front to the rear could easily get over 1/10" overall, perhaps even 2/10".3. By using a 1/2 or 1/4 snap setting while using that 1/10" grid can allow you to space many of the components closer. Your PCB software should have a check built in to warn you when any spacing gets too small. Also, smaller trace widths can be employed in many of the traces in this circuit.4. First, I agree that it is a lot better to use the "copper pour area" for Ground instead of +5. Using it for +5V only invites problems now and in the future. Along with #1 above, I will say that the +5 and Gnd pads on both ends of the PCB are spaced further apart than is actually needed. Bring them to a 0.1", spacing and if you are nervous about that, use oval pads like the ones for the IC. This, along with my suggestion of only one mounting hole on each end, will allow the +5V pad to be spaced further away from any ground traces and the mounting screws will be in far less danger of contacting it. Design it so that the Gnd pads are the ones that are adjacent to the mounting holes.

author: Cerebus
author_rank: Super Contributor
timestamp: November 08, 2021, 11:27:32 pm
content: Quote from: ElectricPower on November 08, 2021, 04:44:33 amThat depends what you mean by SMD. If your goal is to "make a smallest possible board" then it might be worth revisiting your qualification of "without changing the component packages and sizes".A SOIC-8 (at 5x6.2mm) would indeed be smaller than DIP (9.5x7.5mm) - about half the size - and I guess that's what you're thinking of when you say SMD, but there are smaller packages that might persuade you that the step is worth taking in the interests of the "smallest possible board".There's SOT23-5 (at 2.9x2.8mm, 1/8 the size of DIP) or DFN (at 2x2mm), both would be a lot smaller. A version of the 555 is available in both packages (If, and this is always the qualifier for any semiconductor at the moment,ifyou can actually get some). The SOT23 is easy enough to hand solder, DFN really needs hot air but isn't as intimidating to use as it looks at first.One of those "you pays your money and makes you choice" things. If you're prepared to have a slightly bigger board but with THT components you're comfortable with, stick with that. If you're feeling adventurous and want to go really small, investigate the smaller packages.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 08, 2021, 11:39:29 pm
content: WOW! Thanks alot for very good tips guys!You right. I do not need any mounting holes at all actually. And i will deffinitly take a look at all your points here! Thanks alot!But i'm struggle with that load resistor. Do i need a 2W 2512 resistor if i'm going to use lets say an 20 ohm resistor? (If i got an powerbank that requires highet pulse load).I have changed the load resistor to an 2512 size now, for i do not think i can get an 2W resistor in 1206 size?

author: langwadt
author_rank: Super Contributor
timestamp: November 08, 2021, 11:41:02 pm
content: Quote from: fourfathom on November 08, 2021, 04:27:01 pmis the transistor even strictly needed?

author: EPAIII
author_rank: Super Contributor
timestamp: November 08, 2021, 11:46:15 pm
content: The 555 timer IC has a +/-225 mA output current rating. So, unless your LED has a really high current (the 470 Ohm current limiting resistor {~= 10 mA} argues against that) you do not need any transistor or FET to drive it. The LED and it's current limiting resistor can be connected directly to the 555's output terminal and where the other end is connected, +5 V or Gnd will determine when it is on and off. No other connections to the output terminal are needed.If you remove those components from your circuit, the PCB can be significantly smaller. And you do not need that pesky 51 Ohm resistor at all.Spec sheets for almost all electronic components are available on the internet for free:https://www.ti.com/lit/ds/symlink/ne555.pdf?HQS=dis-mous-null-mousermode-dsf-pf-null-wwe&ts=1636377276989&ref_url=https%253A%252F%252Feu.mouser.com%252FSee page 3 for the output ratings.Some published 555 circuits show a load resistor attached to the output. This is ONLY to show that there CAN be a load there. It does not mean that you must have a load resistor. The output of this IC is internally buffered so the value of a load, within the published current limits, or even it's existence or lack thereof, makes NO difference in the operation of the timer.

author: langwadt
author_rank: Super Contributor
timestamp: November 08, 2021, 11:54:09 pm
content: Quote from: EPAIII on November 08, 2021, 11:46:15 pmwill still need the 51R* resistor, the point is to periodically load the powerbank so it doesn't turn off*probably more like 33R to account for the dropout if you want the same current

author: Cerebus
author_rank: Super Contributor
timestamp: November 09, 2021, 12:00:42 am
content: Vishay have quite a lot of application notes on thermal management of SMD resistors. There's probably something in there somewhere that would allow the calculation of the behaviour of the thermal behaviour in face of this 1s every 25s pulsed load, but I don't have the time or the inclination at the moment to go and dig it out. I know that I got something similar from them a while back.

author: james_s
author_rank: Super Contributor
timestamp: November 09, 2021, 12:04:32 am
content: It might be easier to just test it and see what happens. Hook up one of the resistors in question and hit it with greater and greater magnitude pulses until something bad happens and then make sure you are well under that point.

author: fourfathom
author_rank: Super Contributor
timestamp: November 09, 2021, 12:20:24 am
content: Quote from: EPAIII on November 08, 2021, 11:46:15 pmBut the OP is using a CMOS ICM7555, not the original bipolar 555. The output specs are quite different and it looks like it can only sink about 20mA before the VOLrises to one volt.

author: EPAIII
author_rank: Super Contributor
timestamp: November 09, 2021, 02:14:42 am
content: Sorry, I thought the ONLY purpose of the circuit was to light the LED. OK.5V / 51 Ohms = 0.098 A = 98 mA.And the LED seems to be using 10 mA for a total of 108 mA.An LED operating with 110 mA or two LEDs operating at 54 mA will do the same thing.120 to 140 mA rated LEDs:https://www.mouser.com/c/optoelectronics/led-lighting/led-emitters/high-power-leds-white/?q=LED&if%20-%20forward%20current=120%20mA~~150%20mA&mounting%20style=SMD%2FSMT&rp=optoelectronics%2Fled-lighting%2Fled-emitters%2Fhigh-power-leds-white%7C~If%20-%20Forward%20CurrentThere are literally thousands to choose from. I sorted by Forward Voltage <= 3.6 V. I choose the first one which costs $0.59.https://www.mouser.com/ProductDetail/OSRAM-Opto-Semiconductors/KW-DSLP31CC-JXJZ-4O9Q-46A6?qs=81r%252BiQLm7BRQP72vO2qa2A%3D%3DThe Forward Voltage is 3.2 V so the series resistor only needs to drop 1.8 Volts. And it operates at 140 mA. R = E/I = 1.8V / 0.140A = 12.8 Ohms. And P = 1.8V / 0.140A = 0.25W.That resistor should be a lot easier to find. And you still have your current needed to keep the idiotic power supply on line. You can probably raise the value of the resistor a bit and still have 108 mA. You do not need a 2 W load resistor; not even a 1 W. You may need sun glasses when you look at the LED.I would definitely have the ground as a "pour" layer and have it on top. It should extend under the LED and it's series resistor so it can absorb the heat and move it away from those parts. And leave at least 0.1" of space around them so they and that Ground plane can more easily radiate that heat.Quote from: langwadt on November 08, 2021, 11:54:09 pm

author: EPAIII
author_rank: Super Contributor
timestamp: November 09, 2021, 02:20:48 am
content: Oh for pity sake. If he wants to EAT POWER, why on earth use the CMOS version? Use the original 555.Besides, the spec sheet shows that 1 to 2 Volt rise with a 12 V supply. He is using a 5 V supply and I suspect the rise in that Voltage may be less. It is worth a try. And at an available Voltage at the output pin of only 4 V, the series resistor can be even smaller. Or an LED that has a lower forward Voltage can be chosen. I only looked at the first one on the list of hundreds with that current level rating.Quote from: fourfathom on November 09, 2021, 12:20:24 am

author: fourfathom
author_rank: Super Contributor
timestamp: November 09, 2021, 02:35:51 am
content: Quote from: EPAIII on November 09, 2021, 02:20:48 amNot a bad suggestion, I just wanted to point out the difference in specs for the CMOS version he was using.This whole thing would be a great application for one of those 6-pin 25-cent controllers. Six small components (or seven if you double-up the load resistors).

author: EPAIII
author_rank: Super Contributor
timestamp: November 09, 2021, 02:41:05 am
content: Well, I admit I did not notice he was using the CMOS version. I have used that version less than the standard 555.That's what's great about the BB; you get almost instant feedback when you post something. It is great and the OP probably winds up with a much better solution than any one of us can come up with.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 09, 2021, 04:38:40 am
content: Thanks for all good help guysI will check all your sugestions and make som changes. But my biggest question so far is:Where can i find 2W 2512 resistors with value 20ohm and 30 ohm? I prefere to use AliExpress.

author: Niklas
author_rank: Frequent Contributor
timestamp: November 09, 2021, 07:02:45 am
content: One alternative to using a larger package is to split it up on several smaller packages. Instead of 50 ohms dissipating 500 mW, you can connect 2, 3 or 4 resistors packages in parallel and use standard rated 1206 parts. With short duration on the pulses you can use the full power rating of the resistor.100R//100R0.25W per resistor150R//150R//150R0.18W per resistor180R//180R//180R//180R0.14W per resistor

author: Cerebus
author_rank: Super Contributor
timestamp: November 09, 2021, 05:00:02 pm
content: Quote from: Niklas on November 09, 2021, 07:02:45 amWith short duration (and suitable duty cycle) you can use considerablymorethan the full power rating of the resistor. It's all down to quantifying what "short" means for the particular part in question. As long as you don't exceed the pulse power capability of the part, and the pulses are far enough apart for the heat from a pulse to flow to dissipation you're golden.The trick is to look for parts that the manufacturer has characterised for pulse performance. For instance, this is from the data sheet for Vishay CRCW-37 e3 resistors, described on the data sheet as "Standard Thick Film Chip Resistors" i.e. nothing special. Sizes are 1206, 1210 and 2512.The nominal power rating for the 1206 part is 0.25W, but you can see that for a short enough pulse you can put crazy amounts of power through the device (subject to breakdown voltage, ohm's law and pretty much nothing else). It will take ~100W (40,000% of the nominal rating) for 200us. All you have to do then is take the 0.25W rating into account in setting your maximum duty cycle based onaveragepower through the device.

author: fourfathom
author_rank: Super Contributor
timestamp: November 09, 2021, 06:08:47 pm
content: Quote from: Cerebus on November 09, 2021, 05:00:02 pmNice chart! So in this case a 0.5W nominal 1210 resistor can dissipate 1.5W during a 1 second pulse. Given the OP's 1/25 duty-cycle 500mW pulses, there should be adequate margin with his single resistor.So my caution was excessive.

author: Cerebus
author_rank: Super Contributor
timestamp: November 09, 2021, 06:28:11 pm
content: But, my natural caution would be to derate things a bit. Also what's the betting that Vishay's figures are taken for resistors with good heat-sinking on the PCB - something like 80% of SMD resistor heat dissipation is conduction via the board [no pun intended] . There are callouts to various EN and IEC test methods that specify the conditions that the tests are performed under which may specify minimum pad areas for dissipation etc.

author: james_s
author_rank: Super Contributor
timestamp: November 09, 2021, 08:17:01 pm
content: It's worth considering, but in this application I wouldn't be too concerned about it. If this were a design for some sort of life or safety critical device then sure, derating is a good idea, but to keep a power bank turned on? What's the worst that could happen? The resistor fails and the power bank turns off, nobody dies.

author: Cerebus
author_rank: Super Contributor
timestamp: November 09, 2021, 08:28:00 pm
content: What, you think it's not worth making something reliable for its own sake,onlyif it's safety critical? Really?

author: fourfathom
author_rank: Super Contributor
timestamp: November 09, 2021, 08:40:21 pm
content: Quote from: james_s on November 09, 2021, 08:17:01 pmOr the resistor and board catch fire, everybody dies?

author: james_s
author_rank: Super Contributor
timestamp: November 10, 2021, 06:17:45 am
content: Quote from: Cerebus on November 09, 2021, 08:28:00 pmNo, not really. You're not understanding me at all, I just meant I wouldn't obsess over it too much. Look at the data, add a reasonable margin, try it out and confirm it works as expected, it'll probably be fine. No need to analyze a simple hobby project to death. A tiny little chip resistor across 5V is not going to catch fire.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 10, 2021, 07:48:46 am
content: I have tried the circuit on the breadboard with an 1/4watt resistor and the resistor was not very hot after about 30 minutes.

author: Cerebus
author_rank: Super Contributor
timestamp: November 10, 2021, 11:26:55 am
content: Quote from: james_s on November 10, 2021, 06:17:45 amI presume by "add a reasonable margin" you mean "derate it a bit" just like I said. So what was the fuss in the first place? You added all the guff about safety critical. If you didn't think it was really relevant and you didn't want us tounderstandthat, then you'd be better served by not saying it. If you wanted to say that you thought it was overanalyzing then you should have saidthat, not something else. As someone else mentioned recently "we're not mind readers" we're only going to understand what you actually say. Personally I think giving a few seconds thought to derating a component that existsjustto dump energy isn't overanalysing if you want a circuit that keeps working reliably.And talking about what you actually said, I'dunderstand"little hobby project" as a bit unnecessarily dismissive of the OP. Sure, it's not the greatest project in the history of mankind. So what? If it's too insignificant for your consideration, why are you here? If itisworthy of consideration, why not help the OP to make the best job of it he can? I've noticed a tendency recently fromsomeof the professionals on here to sneer at the hobbyists, or people they think are hobbyists, on here - it's not a very attractive attitude.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 10, 2021, 11:33:58 am
content: First off all, thanks for all help!I can say that i have changed the "design" and i have added an 2512 fottprint instead of 1206 for the load resistor. I have also ordered 2512 1W resistors.

author: fourfathom
author_rank: Super Contributor
timestamp: November 10, 2021, 04:42:41 pm
content: Quote from: ElectricPower on November 10, 2021, 07:48:46 amIt sounds like you are OK, but do realize that a leaded 1/4W resistor may have thermal characteristics that are quite different from a small surface-mount resistor. As Cerebus and I have mentioned, with the surface-mount part the circuit board layout (pad and adjacent copper area) will be a big factor.

author: mc172
author_rank: Frequent Contributor
timestamp: November 10, 2021, 05:32:37 pm
content: Does it really need an entire second and that amount of current?

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 10, 2021, 05:34:31 pm
content: Current, yes. It needs at least that current. One second? I really do not know. Maybe not. I trust on Mr Carlsons Lab here... i think...?

author: james_s
author_rank: Super Contributor
timestamp: November 10, 2021, 09:43:28 pm
content: Quote from: Cerebus on November 10, 2021, 11:26:55 amThe "fuss" it seems is entirely from you, ranting for multiple paragraphs over a simple comment from me of the nature of "don't over-think it". I'm sorry, I guess I should have included a trigger warning.

author: Cerebus
author_rank: Super Contributor
timestamp: November 10, 2021, 09:52:34 pm
content: Quote from: james_s on November 10, 2021, 09:43:28 pmStill sneering I see. How do you think that helps?

author: mc172
author_rank: Frequent Contributor
timestamp: November 11, 2021, 02:00:52 pm
content: Quote from: ElectricPower on November 10, 2021, 05:34:31 pmYou mean this one?https://youtu.be/lOL3ic8msas?t=981I scanned through the video to find the schematic and noticed that it says 2.5 seconds on-time. Seems excessive but without that exact power bank to try it on, who knows.It's in your interest to make the pulse as short and as low current as possible, so you will need to experiment with on times, off times and load resistor values anyway. Also, each different model of power bank is going to have different sleep timers, I highly doubt that it's standardised and even if it were, that doesn't seem to stop manufacturers doing whatever they want anyway.Just build one and find out, you can easily modify the on and off times by changing components and you can just put your finger on the resistor to see whether it gets unreasonably hot.

author: fourfathom
author_rank: Super Contributor
timestamp: November 11, 2021, 04:24:51 pm
content: Quote from: mc172 on November 11, 2021, 02:00:52 pmI know this is beating a dead horse, but at least *try* to calculate the power dissipation and temperature rise issue before you build the thing. It's not that hard to get close enough. Later on you can use your finger to verify. I actually have an IR camera attachment for my cellphone that I use to catch any thermal surprises, but if it's going to be at all close, I first do the simple math.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 11, 2021, 04:30:38 pm
content: Yes, i have ordered the pcb now. 120 pcs for just 6 dollar, included chipping to Norway! Amazing

author: mc172
author_rank: Frequent Contributor
timestamp: November 11, 2021, 08:31:27 pm
content: Quote from: fourfathom on November 11, 2021, 04:24:51 pmWhen I said "just try it" I didn't mean to see whether the resistor can dissipate that amount of power, it has already been demonstrated that it is inadequate. I was merely offering up alternate paths to go down. If the duty cycle can be lowered, the power dissipation problem won't even exist. This information is not yet known and it seems like the OP doesn't have any equipment to try it. Just building one provides that equipment.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: November 12, 2021, 10:25:44 pm
content: I find the combination of wanting "the smallest possible" in combination with a DIP NE555 quite weird. Especially since all the other parts are already SMT.

author: fourfathom
author_rank: Super Contributor
timestamp: November 13, 2021, 12:11:06 am
content: Quote from: Doctorandus_P on November 12, 2021, 10:25:44 pmTrue. Even if he already has a bin fill of 555 DIPs, the surface-mount parts are about 10 cents each, and the hassle of mixed through-hole and SMD assembly would certainly drive me in the all-SMD direction.

author: DBecker
author_rank: Frequent Contributor
timestamp: November 13, 2021, 12:49:19 am
content: Quote from: fourfathom on November 13, 2021, 12:11:06 amI expect that he is concerned about soldering the SO-8 package rather than the familiar DIP.Once he tries it, he is likely to find that the SO-8 package is faster and easier to solder than through hole parts. It's only when it gets to smaller pitch that the size becomes a problem.I suggest reducing the number of unique parts. There isn't a good reason to have both a 4.7uF and 10uF capacitors. The power supply bypass is flexible, often optional, and can be whatever value is required elsewhere. Similarly the gate resistor can be the same 470R part used elsewhere. Or use 220R or 270R 0805 resistors everywhere. Five 270R resistors in parallel will handle the current at 5V and nominally be in spec for power. That also gives you an opportunity to experiment with smaller keep-alive current. 10-20mA should be sufficient, it shouldn't take 100mA.I wouldn't flood pour 5V anywhere close to mounting holes or the edge. At minimum there should be a ground ring around the edge. Even if your design rule requires copper clearance to the router line, assume that your hobbyist board will be gripped by a grounded hacksaw blade and bolted down with serrated flange screws.It doesn't apply here, but as a rule put the ground connection closest to the board edge and try to keep the power supply protected by outboard pins.Hand soldering has more constraints than reflow. SMD components should be lined up in even parallel lines. They can be tightly adjacent, but leave a big gutter at the ends for a soldering iron. Think of the angle the tip needs to make, and don't put a tall component at the end of the gutter.On the plus side, hand soldering means that you don't need to be as careful about thermal relief gaps around pads and pins. It only takes an extra second while soldering to add extra heat to a pad that would stubbornly refuse to reflow. No thermal relief gap means that you can better use pin headers and connected wires to carry heat away.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 15, 2021, 04:14:56 am
content: I want to consider all the advice I have received so far and must say I am surprised at how much good feedback I have received so far. I have been testing with 1/4 ohm resistance as load for a whole week now and the resistor does not get hot at all. I have also sat down the pulse on time to just an half second.But I also want to make a device with a USB-C connector at each end. I'm a little unsure how to route the pads on the board, but i only need the power pads. No need for data transfer at all.Here is my schematic. All i have done is adding two USB-C connectors. But i'm unsuure what all the pins on the USB-C footprint are. I also see it is many versions of this connector. Here i have used an 6-pin connector.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 15, 2021, 04:34:22 am
content: Quote from: DBecker on November 13, 2021, 12:49:19 amDo you mean i can remove booth the capacitors?

author: fourfathom
author_rank: Super Contributor
timestamp: November 15, 2021, 07:07:33 am
content: Quote from: ElectricPower on November 15, 2021, 04:34:22 amNo, you need both capacitors. But they can both be the same value. Perhaps use 4.7uF for each of them, and double the value of the timing resistor would work. I believe that 10uF is the maximum value that can be placed on the USB V+ connection, but you should verify that.In general it's a good plan to have as few unique values as is practical. This simplifies production and reduces the chance of assembly errors. "Actually more of a guideline than a rule" as Dr. Peter Venkman once said.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 18, 2021, 03:08:58 am
content: Okay, thanksIs it a good idea to add an little fuse to this regards to user failure when connecting the unit with wron polarity?I know that the power bank probably has reverse polarity protection, but is it a good idea if i want to make it more bullet proof against user failure?If yes... what value fuse do i choose here? I want a small surface mount fuse in 1206 size.Do i choose an resettable automatic fuse, slow blow fuse etc.?

author: fourfathom
author_rank: Super Contributor
timestamp: November 19, 2021, 05:19:10 am
content: Reverse polarity protection? In your case, perhaps just put a series diode in the power connection. Obviously make sure it can handle the current without overheating. This would give you a 0.7V drop in the supply voltage which I don't think will cause any problems, but check the 555 and the FET gate turn-on voltage.If you can't afford the voltage drop, put in a "fuse" device at the power connection, followed by a diode to ground (anode to ground) so it conducts when the power is reversed. I used a "Polyswitch" ( sort of positive temperature coefficient thermistor) resettable part for the "fuse", but that was literally many decades ago, so see what's available now. I think you want a resettable part here.This kind of protection is in general nice to have, but honestly, if you are using USB connectors I don't see how you can get a reverse polarity situation.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 19, 2021, 05:52:16 am
content: Since the end user must solder the wires from the USB cables, it can results in reversed polarity. I know (probable) that power banks has protection against that, but in my head it's no harm to add a little fuse for a penny.But i'm unsure...I just think worst case scenarios here...

author: fourfathom
author_rank: Super Contributor
timestamp: November 19, 2021, 06:07:01 am
content: If the end user is attaching the cable, yes, some protection is in order. I just looked up Polyswitch and similar PTC resettable protection parts and I see you can get a part that will carry 1A, and trip at 1.8A, with an untripped resistance under 0.1 Ohm, in a 1206 package. For this I like the reverse-voltage shunt diode design, just be sure it can carry the trip-current. As I recall the specs on these PTC parts shifts around after a few cycles, so make sure you have margin. In your application shouldn't be too hard to pick a reasonable part.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 19, 2021, 06:10:58 am
content: Why can't i just place a fuse in parallel on the board?

author: fourfathom
author_rank: Super Contributor
timestamp: November 19, 2021, 06:39:14 am
content: Quote from: ElectricPower on November 19, 2021, 06:10:58 amI hope you mean "in series" -- in parallel it's guaranteed to blow whichever way you connect the power, as long as the power source can deliver the current.A fuse in series will protect the power source and wiring, but it's likely that in a reverse-voltage situation the 555 and FET will be destroyed before the fuse blows. That's why I recommend the shunt power diode, so it's the one that is conducting the fault current, not your other components.Or forget the fuse, just use a series diode at the input and accept the small voltage drop. Looking at your circuit, the FET has it's own built-in reverse diode, and it also has those resistors in the drain circuit so it's not going to blow. But the 555 is still in trouble if it sees 5V backwards.

author: DBecker
author_rank: Frequent Contributor
timestamp: November 20, 2021, 02:17:19 am
content: I have a reel of polyfuses, so I might design one into my hobbyist circuit. But they are a bit pricey and specialized for single quantities.A series diode is a simple inexpensive solution for polarity sensitive parts that can handle the voltage drop. The alternative is a shunt diode plus a low value, low wattage series resistor that acts as a fuse. Or just a thin trace on the circuit board. Sure, it functionally kills the circuit and isn't field serviceable, but it's semi-safe and somewhat field repairable.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 20, 2021, 10:47:46 am
content: I see.I have done som changes to my circuit now, and try to add polarity protection with a n-channel. Will that work?Under is an image of my old circuit, and the new with polarity protection:Old circuit without polarity protection:NEW, circuit WITH polarity protection:

author: fourfathom
author_rank: Super Contributor
timestamp: November 20, 2021, 03:30:05 pm
content: Can you describe what this protection circuit is supposed to do? I don't think that new NFET ever turns on, with either +5 or -5V applied to the power connection. With +5 the FET body diode will be forward biased. And that new FET does nothing to protect the 555 from reverse voltage.Why don't you just put a series diode between the +5 input and the rest of your circuit? The ICM7555, LED, and the NFET will all operate at the slightly-reduced voltage.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 20, 2021, 04:15:29 pm
content: Quote from: fourfathom on November 20, 2021, 03:30:05 pmThe reverse polarity protection is only made for protecting this board, and not the powerbank or the units connected to it.Can you please show me where in circuit you wil added an diode?I do not want any drop in voltage to any connected units like an laptop or an mobile phone.

author: fourfathom
author_rank: Super Contributor
timestamp: November 20, 2021, 04:57:37 pm
content: Quote from: ElectricPower on November 20, 2021, 04:15:29 pmLike this?

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 20, 2021, 06:31:13 pm
content: Thanks.But when i do that i think the voltage drop (0.7V) will be to mutch? 5V - 0.7V is only 4.3 volt to charge the units?And will the diode take 2.4A and 4.3V (10 watt)?

author: fourfathom
author_rank: Super Contributor
timestamp: November 20, 2021, 09:29:15 pm
content: Are those two USB jacks charging *outputs*? And the +5V is coming from some external supply that needs a regular "keep alive" load that is provided by the 555, FET, and resistor? I want to be sure I understand the function of your design.Assuming I've got it right, in my modification of your schematic the diode only has to carry the current of the 555 and the FET/load. I believe you were planning on 100mA for this load? If so, a 1A diode should be more than sufficient. The diode is not in the current path that feeds the USB jacks, so they will still get the full 5V.And I thought you weren't concerned about reverse voltage on the USB jacks, but if the user can connect the power supply in reverse then perhaps the jacks do need protection. If so, the series fuse/polyswitch at the power input connection followed by a shunt diode would protect everything.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 21, 2021, 03:22:28 am
content: Quote from: fourfathom on November 20, 2021, 09:29:15 pmThe +5V is 5V from 5V battery power bank. I have added pads manually for the +5V output that goes to the load. (mobile phone, camera etc.) I just want to protect the 555 timer and the component on the board.The finished PCB looks like that:If i connect an diode in series the load is only getting 4.3V. I just want to protect this board, because it is an possibillity for the end user to solder the wires on the wrong way and connect it to a powerbank.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 21, 2021, 03:53:19 am
content: If i do like this and adds an 1N4007 diode in parallel i think my board is protected regards to reverse polarity?I understand that the connected powerbank and the load is not protected, but i assume there is own protection there?I just want to protect the board i posted image om over.

author: fourfathom
author_rank: Super Contributor
timestamp: November 21, 2021, 05:44:03 am
content: That diode would protect the components on your board if diode can handle the powerbank output current. The 1N4000 is a 1A diode, and you really ought not run more than 1/2A through it continuously, just for the sake of margin. I assume that your powerbank can deliver much more than 1A, so in a reverse polarity situation the diode is probably going to fail. If your board normally draws 100mA then you can put a 1/2A fuse in series with the power input, with your diode after that, and the fuse will blow when the voltage is reversed.But I don't see the problem with just using the series diode (as I sketched). Your 555, FET, LED will operate just fine at 4.3V (assuming a 0.7V drop across the series diode). Adjust the load resistor as needed to get the current you want. The 1N4000 will be a good diode for this application. As long as your other loads are connected to the power input, they will still see 5V, not 4.3V.

author: ElectricPower
author_rank: Frequent Contributor
timestamp: November 21, 2021, 06:29:38 am
content: Quote from: fourfathom on November 21, 2021, 05:44:03 amThanksDo you mean something like this?free html hosting

author: fourfathom
author_rank: Super Contributor
timestamp: November 21, 2021, 03:52:02 pm
content: Yes, but in this configuration you don't need the fuse. The diode will block the reverse voltage, passing only a few microamps of reverse current, but this won't be enough to damage your circuits components. With the power source correctly connected your circuit will get about 4.3V (instead of 5V) which is more than sufficient for operation of the 555 and the FET.In this configuration, the only reason for the fuse is if you want to protection should you have a short-circuit or faulty component on the board.