Need help improving PCB layout design

[ElectricPower]

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.

[ElectricPower]

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.

Do you mean i can remove booth the capacitors?

[fourfathom]

Do you mean i can remove booth the capacitors?

No, 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.

[ElectricPower]

Okay, thanks

Is 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.?

[fourfathom]

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.

[ElectricPower]

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...

[fourfathom]

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.

[ElectricPower]

Why can't i just place a fuse in parallel on the board?

[fourfathom]

Why can't i just place a fuse in parallel on the board?

I 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.

[edit] 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.

[DBecker]

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.

[ElectricPower]

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:

[fourfathom]

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.

[ElectricPower]

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.

The 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.

[fourfathom]

The 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.

Like this?

[ElectricPower]

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)?

[fourfathom]

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.

[ElectricPower]

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.

The +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.

[ElectricPower]

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.

[fourfathom]

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.

[ElectricPower]

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.

Thanks

Do you mean something like this?


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[fourfathom]

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.