Unable to get 3.3V on Proto Board

rathodmanoj

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Hi,
I am new to tinyduino and have recently purchased TinyDuino Processor Board (with Battery Holder), TinyShield USB & ICP and Proto Board 1.

I have stacked up the Processor Board + USB Shield + Proto board and connected the micro USB from computer. My blink  program compiled and executed correctly.

I extracted wires from the Proto Board 1 to connect another device (nRF24L01+) which operates on 3.3v supply. Although, I am unable to get the supply 3.3v on the proto board due to which I am unable to test nRF24L01+ any further. I then removed the micro USB and connected a CR1632 battery. Though still not able to get 3.3v to nRF24L01+.

Is there something I am missing?

Thanks a lot for your help.

Regards,
Manoj.


TinyCircuits

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Hi Manoj,

Sorry for the confusion - the 3.3V rail is there for legacy support and not directly generated from the TinyDuino board.  Instead, Vcc is the main power rail that you can use to power external devices, this is automatically selected between the +5V rail (such as when the USB is present), or VBatt when that is present.  If you are running a coin cell, we'd recommend powering the other device using the Vcc or Vbatt pin.  Note: If you are powering off the Vcc pin, if you plug in USB this pin will go up to 5V, so this may be an issue based on what you are connecting to.  If so, just stick with using Vbatt.

Regards,

Ken
TinyCircuits



rathodmanoj

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Hi Ken,
Many Thanks for the clarification.

It is now clear to me on the usage of these rails. And thanks for suggesting using VBatt. I will use it for connecting nRF24L01+.

Best Regards,
Manoj Rathod.


heliboy

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The problem is that when USB is connected the Vbatt pin only gives 0.6 volts. So you cannot power any proto board device using VBATT since it will not have sufficient voltage to debug while USB is connected. It will only work when the the battery is connected ! You cannot put a 3.3V regulator on the Vcc because though it gives 5V on USB, on battery it only give battery voltage (3V) which is too low for the regulator.

 I think this this is major flaw in the design of this product as is the unconnected 3.3V pin (legacy for what ?).


petulda

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Yeah, I just ran into this too, it's a major flaw in design. I was considering to send the board back as faulty...


TinyCircuits

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I agree that it is confusing, Vcc is the power rail that should be used for powering external peripherals.  The problem is that generating 3.3V from the coin cell is inefficient and will cause it to drain faster.

You can using a low dropout linear regulator (LDO) on the Vcc rail to generate a safe voltage if you are using both the USB and a coin cell battery.  As you mention, a coin cell battery is below 3.3V (typically the are 3.0V nominal), so the output voltage from the LDO will be the input voltage minus the dropout voltage, and you can get LDO's with under 0.1V dropout voltage.  If you external device needs more voltage than this, you'll need to use a higher voltage battery or power supply (like a Li-Ion battery pack).

The power modes are described in more detail at the following link:

https://tiny-circuits.com/tinyduino_overview

I hope that helps explain it better!

Thanks,

Ken
TinyCircuits


 

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