Feedback on my first PCB design?

Projects
41

Ok - perfect - thanks so much. I think I’ve got enough to send to printing.

One remaining question I have (though maybe non blocking) – I mocked the whole circuit today using 9V from the barrel into a 7805…I seemed to get a very consistent 4.9V out of the 7805, both with and without capacitors.

In other words - the capacitors didn’t seem to do anything. Is there anything I can do to monkey with it and/or use my multimeter with the caps to gauge effects?

I’ve been studying my design in KiCad, and I also noticed my schematic has the symbol for non polarized caps…but the footprint looks polarized; and very odd, it appears as though the pads are reversed on the footprint (e.g. Net-(C1-Pad2) is ground, but it’s circle on the right for one capacitor, and square on the left for another.

  1. Do I need the capacitors in the first place given comment above?
  2. Why would the pads be reversed if they are the same footprint?

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42

Also note - my schematic seems way more complex than it needs to be still, but in addition to my capacitor question above, I updated my design and could use a final check (or hopefully final).

Specifically,

  • if on barrel power I want to power the transmitter off of the 5V coming from the 7805 and the NodeMCU by connecting VIN also to the 7805 output.
  • If on microUSB power direct to the NodeMCU, then I want to power the transmitter by using the VIN to output 5V

…to accomodate for this, I added a diode between the 7805 and the VIN, then I connected the VIN to both the transmitter VCC and to the output of the 7805.

I think I did it all right…but I’m new at this so clearly I could’ve misinterpreted something. Can’t thank the community enough for getting me over the hump on this…I’m excited to send to print as soon as I can verify I’m in good shape here.

One note - the ERC is giving me one warning, but I can’t figure out why.

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43

Based on what Starbird said about nodemu internal regulator: If you know your external power source will be 9…12V you don’t need external regulator for that module.

If you use an external regulator it’s possible and even probable that you don’t need capacitors for it because the model regulates well enough. As I said above, the datasheet of 780x told that the input capacitor is recommended under some circumstances. Note that you don’t have to install capacitors even when you have the footprints on the board. It’s quite common to have footprints “just in case” and if the device resets randomly or does something else which points to power problems the footprints can be populated with extra capacitors. Remember that footprints inside the board outlines come “for free”. You can design as complex systems as you can and then bypass them when you use the board.

But what about the transmitter? It looks like it takes power from the external power source, not from the USB connector of the module. Therefore it may need external power source with the regulator.

I still don’t know what the module pins do. I would guess that VIN doesn’t give power output (because it is power input!) but 3.3 pins give regulated 3.3V out. You should connect the transmitter VCC to 3.3 if it works with 3.3V.

44

In your circuit they probably are not needed.
On the output of 7805 there are probably capacitor which is on the module.
On the input there is probably a capacitor in your power source, but a cable is probably long enough to use one here.

In my opinion it should be changed. You should clearly decide if you use polarized or non polarized.

You can reverse capacitors at schematic with no seen effect (as they are not polarized) and you change the pad assignment to the nets.

45

I don’t understand what you think.
From your previous schematic it looked that transmitter was powered from 3V3.
Now you added diode - so you subtract about 0.7V from 5V reaching 4.3V and use this voltage to power both module and transmitter.

46

I didn’t notice the extra diode. It looks like Waldo thinks that the module offers power from VIN, but as I said, that’s not the case. In this schematic the transmitter doesn’t get any power if the module is powered from USB.

47

To some up: the original design worked (because that’s what you tested, right?) but it’s not safe if the external power source (through the barrel jack) is unpredictable.

If you really know its polarity and voltage there’s no need for the regulator or even for the diode. But if there’s any chance that you or someone else may try to use 24V transformer or wrong polarity, it pays back to have that one diode and the regulator. For simplicity you can forget the capacitors, but it doesn’t hurt to have them in the design because they can be left unpopulated.

48

You might try dropping a ground symbol into the schematic.

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1 Like
49

Thanks folks - replies:

Based on what Starbird said about nodemu internal regulator: If you know your external power source will be 9…12V you don’t need external regulator for that module.

I tested without the regulator, but the 9V going directly to the transmitter there was some definite noise pulsing that I could detect with a radio receiver. I guess it could’ve been that the 9V was too high to feed to the transmitter, but either way, I deduced I need a 5V source on the circuit somewhere. Since I’m using VIN to power the nodemcu in this scenario, I figured I still need the regulator (which can then feed 5V to both transmitter and NodeMCU VIN).

Remember that footprints inside the board outlines come “for free”. You can design as complex systems as you can and then bypass them when you use the board.

I’d bypass them by just jumping a wire between the pads, right?

But what about the transmitter? It looks like it takes power from the external power source, not from the USB connector of the module. Therefore it may need external power source with the regulator.

I still don’t know what the module pins do. I would guess that VIN doesn’t give power output (because it is power input !) but 3.3 pins give regulated 3.3V out. You should connect the transmitter VCC to 3.3 if it works with 3.3V.

Actually, I can’t explain it, but when the nodeMCU is on microusb power, VIN does indeed output 5V. And 5V to the transmitter is giving a stronger signal than 3.3V, so I thought this was better (even if the 5V steps down because of the diode).

I didn’t notice the extra diode. It looks like Waldo thinks that the module offers power from VIN, but as I said, that’s not the case.

@eelik - per above answers, it actually is indeed the case. So given that, did I use the extra diode correctly? I’d prefer not to lose the 0.7V…but without the diode, then on microusb power might I have to worry about power flowing back to the regulator given my current circuit?

50

It’s possible that the USB power is directly connected to VIN. That way it would give power out and to the internal regulator at the same time. In that case the extra diode may be a good idea, although I don’t know if your regulator could handle +5V to its output pin.

But remember that if the voltage in VIN comes directly from USB it’s unregulated. Therefore it may cause noise to the transmitter. 3.3V output, on the other hand, is regulated. In any case you could add a capacitor for the transmitter, too. That could help with the noise.

51

Thanks. Is there a specific metric on the datasheet that describes what a regulator (e.g. 7805) can handle to output pin?

E.g. I’m looking at this one and don’t see anything obvious…(but then again unless it said “ability to handle input V on output pin” it probably wouldn’t be obvious to me…)

52

Suggesting a more comprehensive datasheet.

This one spells out all aspects.

53

Thanks - based on that datasheet and this:

8.1.2 Raising the Output Voltage Above the Input Voltage
Because the output of the device does not sink current, forcing the output high can cause damage to internal low
current paths in a manner similar to that just described in Shorting the Regulator Input.

Seems like the diode to protect output would be needed?

(as an aside the capacitor information in this datasheet was much better than what I previously had, thanks.

54

No worries. That diode would be in order in your case.

55

So D2 is all wrong. It should go from Vout to Vin of the regulator. That being a common practice anyhow in case there are big enough caps at the VReg’s output.
In this case, it actually is necessary.
Voltage-regs often enough are robust enough to withstand a supply on their output. But I don’t suggest this abuse.

Nick

56

Can you explain why? What it does?

EDIT: I hope we are on the same page. I was talking about possible implementation of the module. As a black box it would just have +5V in the VIN pin when the the module is connected to a power source though the USB connector. I don’t see why D2 would be wrong. It would stop current going to the regulator.

57

Yes, it does. But it also prevents the voltage regulator to output the desired voltage.

Nick

58

OK, I understand. The internal regulator may need at least 4.6V (with max 1.3V dropout and 3.3V output). If the external regulator gives 5V and the diode takes 0.7V it’s not enough.

59

Trying to make sure I follow - D2 is wrong because the regulator outputs 5V, and the diode D2 as designed will take 0.7V from that, and 4.3V is not enough to pass to VIN - I think that’s the problem?

If so, I don’t understand what connecting D2 from VOut to VIn of the regulator will do. Wouldn’t that mean that when the NodeMCU is connected to USB power, 5V would still flow from VIn of the NodeMCU to VOut of the regulator? A diode between Regulator VOut and Regulator VIn wouldn’t stop that, would it?

60

Till we don’t have NodeMCU schematic it is divining from coffee grounds.
Name VIN suggests it is to input power supply. It is possible that circuit inside works at 3V3. If I were designing such module to be powered from two sources I would connect two diodes (one from VIN, and second from USB) to protect one power source from another. If diodes are there then when it is powered from USB then no power can go out through VIN.