Need help improving PCB layout design

[ElectricPower]

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.

[james_s]

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.

[Psi]

This is what I would change.

[ElectricPower]

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.

Thats right. Th 51R resistor is for make Power bank staying on

An 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.

[ElectricPower]

This is what I would change.

Thanks. I have changed that now

[mc172]

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.

[ElectricPower]

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.

I have both ground and 5V pour.

[fourfathom]

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).

[ElectricPower]

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).

Actualle 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?

[fourfathom]

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).

Actualle 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?

Yes, 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.

[ElectricPower]

With one second i meant 1 second load pr 25. second. Not one pulse pr. Second.

[fourfathom]

With one second i meant 1 second load pr 25. second. Not one pulse pr. Second.

Yes, 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.)

[ElectricPower]

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.

[EPAIII]

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.

[Cerebus]

An 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.

That 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, if you 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.

[ElectricPower]

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?

[langwadt]

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).

is the transistor even strictly needed?

[EPAIII]

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%252F

See 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.

[langwadt]

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%252F

See 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.

will 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

[Cerebus]

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.

[james_s]

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.

[fourfathom]

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.

But 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 V_OL rises to one volt.

[EPAIII]

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%20Current

There 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%3D

The 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.

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%252F

See 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.

will 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

[EPAIII]

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.

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.

But 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 V_OL rises to one volt.

[fourfathom]

Oh for pity sake. If he wants to EAT POWER, why on earth use the CMOS version? Use the original 555.

Not 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).