Applying voltage to output of a boost-converter

Hi,
I have a project with this schematic:

As power sources, I designed the board to receive power from either a 5V DC wall adapter (which goes to CN6: 5V-IN) or from 18650 batteries. Initially I planned to use 2x18650 with a 5V buck-converter but later on I decided that I could also use 1x18650 with a 5V boost-converter. This is why I left them both on the schematic, so I decide later on.
Currently, the project is running fine with 1x8650 and this boost: https://www.aliexpress.com/item/4001054709058.html

I am happy with the outcome, but I was thinking for some new features on the SW part which requires more power and I thought: what if I power the board from both DC adapter and 18650 battery?
And, I did.
I placed the multimeter set on milli-amps on the CN4 connector and started to measure the power consumption from the battery.
When powered from battery, the current drawn from the battery shows ~1mA during deep sleep and when I connect the DC adapter, the current drawn fell down to 0.04mA. Even when I wake-up my WemosD1, the current remains at 0.04mA. Which means, that when the DC adapter is connected to VCC (which is also the output of the boost-converter) there is almost no current consumption from the battery.
This makes me think that I could actually power the board from the DC adapter and also put a battery, as a backup for example.

And, my question is: can I do this safely? Is there a risk? For example, the boost might fry, or something, if there is voltage applied at its output? Even if the voltage is the same as its rating (being a 5V boost)?

I’d use a boost/buck converter and add a manual jumper on the output for Vcc.

You might consider two Schottky diodes too, one for 5v, one for the converter :thinking: .

We don't have a schematic of the boost module you're using, but typically they have a schottky diode in the output line, the output of which charges the capacitor. So if you apply 5V or higher to the module output, any backflow should be blocked by the diode. I think this should work fine as far as the battery and boost regulator are concerned. But if the external 5V source might be connected, but powered down, then there is no protection from the boosted battery output flowing back into that 5V brick, or whatever it is. Of course it may have its own blocking diode built in, but you need to measure to see if any current flows in that situation, and if it does, I think you should add a schottky diode in that line too.

What you have is an odd version of a load sharing circuit - odd because the external power source feeds in after the boost converter, and there is no charger circuit. If you're still in a position to consider modifications to the circuit, you could add the battery charging function, and configure a normal load sharing circuit. The the external source, when plugged in, would power the load (through the boost converter) and also charge the battery independently if needed. The battery would power the load only when the external source is unplugged. You would need to add a charging module, a schottky diode, a P-channel mosfet, and a resistor. I can post a circuit if you're interested. Basically, you would have the 5V brick plugged in all the time and powering the load, and the battery would function as UPS backup.

I think the 0.04mA is just the quiescent current of the boost converter.

Edit: I should have said - I am not an engineer. So if you want a definitive answer, you should wait for one of them to reply.

Hi,
Can you please post a schematic of everything in your project connected.

The schematic you posted is good for PCB manufacture with net names, but very hard to use for troubleshooting.

A full diagram with wires would be very helpful.
Show what you have connected to the connectors to.
You can still use net names for Vcc and gnd, but interconnections of signals need to be shown.

Thanks.. Tom... :smiley: :+1: :coffee: :australia:

Hi,

Sorry if I didn't actually post the schematic, but I can't find all the components I use and it was easy for me just to enumerate them.

So, here they are:

Connections are:

  • One button + the led are connected to CN1 (since they share the GND)
  • The other button is connected to CN4 and its led to CN2
  • The 18650 battery output comes, of course, to CN5
  • The MG996R is connected to CN3
  • DS1307 is connected to H2
  • 5V DC plug is connected to CN6 (optional, only if battery+boost is missing)

Rest of the connectors are left unused (for current project), since I designed this PCB as a multi-purpose, for various projects using analog sensors or i2c devices.
Now, the initial plan was to: either run the board on a 5DC plug, but don't use a boost or buck, or use a boost or buck with 18650 batteries. So, this is why I didn't even add some switch or jumper to the input of CN6.
So, right now I have a PCB without a boost/buck added to it, but it runs directly on 5V from the DC adapter.
And I have another one with the boost, running from battery, and my question from this topic comes from an idea that came up with: what if I used both sources of power: the DC plug and the battery?

The thing is that, if the project is running on battery, the controller will be mostly in deep-sleep (remotely unreachable). So, I wanted to add a mode, when I leave home, to be able to control it remotely. But, when I do this, I'll have to power it from a DC plug instead of the battery (since it will be always connected to WiFi and listening to HTTP clients), so the battery will just be used as backup when the city power goes off.

Thanks!

True. But initially I didn't even though I would use both power sources in the same time. More, I am scared of working with rechargeable batteries and this is why I didn't even plan to add a charging circuit on the PCB, but plan to charge the battery outside of it (scared not to make it explode, even though I only use protected batteries) .

Yes, that was my though also.

Hi,
SORRY but really need a connected diagram, including the servo connections.

Why do you need to buffer the servo signal with two BJTs?

serv

Tom... :smiley: :+1: :coffee: :australia:

Because my WemosD1 IO logic is 3.3V and the servo PWM signal works only with 5V logic.
What you see in CN3, is VCC going to servo VCC and the output of the BJTs going to PWM pin of the servo. Also, I used a current sense resistor in there (R8) which is used to detect servo stalls, which is connected in series with the GND pin of the servo.

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