Power Boost

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douglove
 
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Power Boost

Post by douglove »

I am building a location datalogger using these parts...
* Ultimate GPS - https://www.adafruit.com/product/746
* 10 DOF - https://www.adafruit.com/products/1604
* Arduino Micro - https://www.adafruit.com/product/1315
* 2 or 3 FRAM boards - https://www.adafruit.com/products/1895

I would like to use the power boost charger (https://www.adafruit.com/products/1895) and a battery (lipo or lion).

My questions are...

1. Which battery should I use to power this using the power boost, so that it will operate for around 16 hours?
2. Does the power boost charger pass the USB lines through to the Arduino? I would like to be able to communicate with the Arduino to read the data but only expose 1 usb port (power boost port) from the enclosure.

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adafruit_support_mike
 
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Re: Power Boost

Post by adafruit_support_mike »

1) It's hard to estimate current consumption for that many parts off the top of my head, but I'd guess you're looking at a load of about 75-100mA. The charger boosts voltage by about 140%, and is about 90% efficient, so based on that guess you'd pull about 120-160mA from the battery. To do that for 16 hours, you'll need 1920-2560mAh of energy.

Those numbers suggest a 2500mAh cell would come close to what you want, but wouldn't leave much margin for error. Your best bet will be to measure the actual current consumption of the circuit and multiply up from there.

2) Nope.. only the VCC and GND lines go through the Power Boost.

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douglove
 
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Re: Power Boost

Post by douglove »

Thanks for the reply Mike.

Based on your numbers, I am thinking about using the 4400mAh (https://www.adafruit.com/products/354).

I plan on wiring the 5V pin on the power boost charger to the VI pin on the Arduino Micro (5V model) and ground to ground. If I have the battery connected to the power boost charger can I plug in the USB to the Arduino Micro in order to transfer data or upload the sketch? Or do I need to unplug the battery from the power boost before connecting the USB to the Micro?

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adafruit_support_mike
 
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Re: Power Boost

Post by adafruit_support_mike »

Connecting 5v to the Vin pin probably won't work.

Power from the Vin pin goes through a P-mosfet (T2) to the input side of the NCP1117-5 voltage regulator. The output of the regulator is tied directly to the Arduino's +5v rail, and the gate of T2 is tied to the +5v rail.

The NCP1117-5 needs at least 5.5v to create a true 5v output for the +5v rail. T2 needs the voltage at Vin to be at least 1v-2v higher than the voltage on the +5v rail before it will allow current to flow from Vin to the NCP1117-5.

That works beautifully if the voltage at Vin is at least 6v. The NCP1117-5 has enough headroom to operate, the voltage at +5v is really 5v, and T2 sees enough gate-source voltage to open and keep everything working.

If you connect less than 6v to Vin, the NCP1117-5 won't have enough headroom, so the voltage on the +5v rail will be somewhere below 5-actual-v. It will probably rise to about 3.5v, because T2 needs that much to open and send power to the NCP1117-5, but there's no telling how much current will flow under those conditions.

The USB-5v pin connects to the +5v rail through another P-mosfet (T1). If the voltage at Vin is more than about 3.5v, T1 will shut off, isolating USB-5v from the +5v rail.

The worst possible condition would be to have equal 5v supplies connected to Vin and USB-5v. Under those conditions, the NCP1117-5 would only produce about 3.5v, but the true voltage at the +5v rail would have to be low enough for power to flow through T2. That would also be low enough for power to flow through T1, so the power reaching the +5v rail would be a mix of weak connections from both Vin and the USB cable.


To power the Arduino, connect the 5v output from the boost converter to the Arduino's 5v pin (aka: the +5v rail). That will bypass the NCP1117-5 voltage regulator and keep T2 shut off, isolating the Vin pin from the +5v rail.

You need a voltage at the Vin rail to keep USB-5v from causing problems though, so connect a diode from +5v to Vin. Having 4.4v at Vin will keep T1 shut off, isolating USB-5v from the +5v rail. The diode will also prevent any back-connection from Vin to the +5v rail.

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douglove
 
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Re: Power Boost

Post by douglove »

Mike, you are awesome. I finally got time to work on my project again. I connected everything up the way you suggested and it works perfectly. Thanks again.

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adafruit_support_mike
 
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Re: Power Boost

Post by adafruit_support_mike »

Whew! It's always a relief to hear it worked. ;-)

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douglove
 
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Re: Power Boost

Post by douglove »

I picked up a charger doctor and measured my project power usage.

1. The charger doctor shows 0.06a. Does that mean that it is using 60 mah?
2. Above it was stated that the power booster is about 90% efficient. Does that mean that if I had a 1000 mah battery that equates to a 900 mah accounting for the efficiency loss?
3. To calculate the approximate number of hours that my project should run, is this equation correct?
time = (battery capacity * .90) / (charger doctor reading in mah)?

Thanks for all the help.

Doug

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adafruit_support_mike
 
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Re: Power Boost

Post by adafruit_support_mike »

douglove wrote:1. The charger doctor shows 0.06a. Does that mean that it is using 60 mah?
Close.

The Charger Doctor measures current, which is an instantaneous value like speed. A battery's capacity is measured as energy, which is a cumulative value like distance.

The value of 0.06 means your circuit is drawing 60mA of current at any moment. If you run 60mA of current through the circuit for an hour, you'll have used 60mAh of energy.
douglove wrote:2. Above it was stated that the power booster is about 90% efficient. Does that mean that if I had a 1000 mah battery that equates to a 900 mah accounting for the efficiency loss?
Pretty much, but that's a power calculation (voltage times current) rather than a single value.

If the PowerBoost's output is 60mA @ 5v, your circuit is using 300mW of power. That means it needs to draw at least 300mW of power from the battery. If the battery voltage is 3.7v, the current will be 300mW/3.7v=81mA.

That value doesn't take the charger efficiency into account though. Only 90% of the power coming out of the battery makes it to the load, so to get 300mW out of the PowerBoost you need to draw 300mW/0.9=333mW out of the battery. Assuming 3.7v from the battery again, the battery current will be 333mW/3.7v=90mA.
douglove wrote:3. To calculate the approximate number of hours that my project should run, is this equation correct?
Not quite..

The base equation is: P.out = 0.9 * P.in

We break that into pieces as: V.out * I.out = 0.9 * V.in * I.in

Shuffling things around to get the ratio we want produces: I.in / I.out = V.out /( V.in * 0.9 )

We know all the values on the right hand side, so: I.in / I.out = 5v /( 3.7v * 0.9 )

Plugging and chugging we get: I.in / I.out = 1.5

The time estimate will be: time = stored-energy / current

And the current is I.in, which we know: time = stored-energy /( I.out * 1.5 )


You had the right general idea, but the power-scaling thing isn't obvious.

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douglove
 
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Re: Power Boost

Post by douglove »

Thanks for breaking that down.

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