Arduino pro mini v5 on-board MIC5205 150mA LDO Regulator

I am currently working on a hand prosthetic and am developing a wearable wristband to read emg signals. The wristband contains the following:

  1. Arduino Pro mini 5V ( Typical supply voltage: +5.0 V, Typical supply current: 10-25 mA)

  2. Transmitter Module ( Typical supply voltage: +5.0 V, Typical supply current: 03-10 mA)

  3. Myoware muscle sensor ( Typical supply voltage: +5.0 V, Typical supply current: 09-14 mA)

I am currently looking for a battery to power these components. The battery needs to be as small and as light as possible. It also needs to be rechargeable. There is no minimum operation time due to the project being a prototype, however, the longer the duration the better. I am currently looking at the s 200mAh 2S 7.4v 20C LiPo Battery ( Search results for: 'batteries lipo batteries eflite umx beast type lipo battery' )

QUESTION 1) Will this battery work fine with the components mentioned above?

QUESTION 2) will it work for roughly hours as I calculated?

0.2 Ah /(25 mA+10 mA+14 mA)= 4 h

QUESTION 3) looking at the Arduino on-board voltage regulator (http://www.micrel.com/_PDF/mic5205.pdf) I calculate that there will be 0.379 W power dissipation is the value correct?

Pd=(7.4 V-5 V)*150 mA + 7.4 V *2.5 mA = 0.379 W

QUESTION 4) I also calculate that the voltage regulator temperature rise will be 83.83 C is the vale correct and is it normal or do I need a heat sink?

TempRise= 0.379 W*220 = 83.83 C

4.8V batteries may be smaller and more suitable for your project. Just a thought.

Hey,
So you reckon powering the mini with 4.8V is better than using a 7.4V?

QUESTION 3) looking at the Arduino on-board voltage regulator (http://www.micrel.com/_PDF/mic5205.pdf) I calculate that there will be 0.379 W power dissipation is the value correct?

Pd=(7.4 V-5 V)*150 mA + 7.4 V *2.5 mA = 0.379 W

No, because you're not actually using 150mA. And the 7.4V is "nominal" for a 2s LiPo pack (more like 8.4 initially.)
so, more like (worst case): Pd=(8.4 V-5 V)*(25 mA+10 mA+14 mA) + 8.4 V *2.5 mA

You might be a lot better off using a 1S LiPo and a boost regulator to get 5V, since a boost regulator will give you 80+% efficiency, while the linear step down is worse than 50%...

Thanks for the insight. So a 240mAh 1S 3.7v 25C LiPo Battery with a 5V Step-Up/Step-Down Voltage Regulator would the job?

http://www.overlander.co.uk/lipo-batteries-240mah-1s-3-7v-25c-sport-hubsan-x4-mini-quadcopter-type.html

https://www.coolcomponents.co.uk/5v-step-up-step-down-voltage-regulator-s7v8f5.html

Yes, that should last a few hours per charge. Maybe 4 to 5.

Would I need a PCB for the LiPo battery or is that unnecessary?

Not needed, just the connector.
For example, this remote control use a 3.3V/8MHz Promini running from a 1000mAH LiPo that is tiewrapped to the bottom of the card. A Max1811 (not plugged in here) is used to charge the battery from a 5V source (such as USB).

Here's the circuit

But the battery datasheet says "Never discharge Li‐Po battery below the Lowest Discharge Voltage 3v per cell"

and the voltage regulator data says "takes an input voltage from 2.7 V to 11.8 V and increases or decreases the voltage to a fixed 5 V output'

Thus wouldn't the Voltage regulator discharge the battery to 2.7V causing irreversible damage which will deteriorate the battery performance and cycle life ?

http://www.overlander.co.uk/fullymax_warning_sheet_li_poly.pdf

If you look at the schematic, you will notice I am powering the promini via the VCC pin, not Raw, so the onboard regulator is bypassed. The battery never sees more than 4.2V from the max1811 charge control chip.
The promini goes into power down sleep mode after waking up from a button press and transmitting the character via virtualWire library.
The battery lasts about a month between charges, when I notice that it is not sending anymore I plug it in to charge it.
A battery disconnect transistor would probably be good to keep the battery from overdischarging.
At the same time, it's been running like that since ~Jan 2011. Maybe battery life has been degraded some, I don't monitor it all that closely.

I also intend to power through the VCC pin bypassing the linear voltage regulator.
So Ill have the LiPo battery connected to an external switching voltage regulator connected to the VCC pin. The external voltage regulator I have been referring to will discharge a battery to 2.7V which is smaller than the limit of >3V LiPo cell discharge. Perhaps I am misunderstanding something.

You'll have the battery connected to the input of a switching regulator, with nothing to stop the regulator from overdischarging the battery? Yes, that could lead to battery issues.
You need a part like this to disconnect the battery from the load

Values here may need adjusting to meet 3V instead of something lower

You need a battery charge control circuito to charge the battery still.

The two components seem rather complex. Is there no simpler solution to breaking a circuit when battery input voltage is lower than a set limit?

For example will this PCB disconnect the circuit at 3V ?
Over Discharge Protection Over Discharge Detection Voltage 2.5±0.063V
Over Discharge Release Voltage 3.0±0.075V

For charging the Battery can't I use something as simple as this ?http://www.overlander.co.uk/chargers/lipo-battery-balance-chargers/overlander-rc3-lipo-charger.html

Wasferd:
The two components seem rather complex. Is there no simpler solution to breaking a circuit when battery input voltage is lower than a set limit?

Probably not.

Disconnecting a battery below a certain voltage is a particularly complex problem, since the circuit which does this must draw no current once disconnected, but draw no current when connected either as clearly that would actually discharge the battery - the very thing you need to prevent.

This rules out for example Zener diodes as threshold references; the entire circuit must be CMOS.

To make efficient use of battery, step-up or step-down converter is needed (depends of battery configuration). Good one have efficiency above 90%, 80% is a bad one :slight_smile:
Don't use on-board LDO for this.

Most such regulators have 'shutdown' pin. You can get regulator with very low shutdown current (microamps) and implement soft shutdown.

This is my quite old project powered from li-ion batteries with this feature:

On the top-left part of schematic you have power section.
There are 2 components:

  • voltage measuring circuit (D1, R7)
  • shutdown circuit (D2, R9, C7, D3)

How it works:
Start:

  • you press power switch
  • C7 is already discharged, so it sets shutdown pin of converter to high
  • converter is enabled and passes power to the rest of circuit
  • cpu starts
  • first thing cpu does is putting PB2 high
  • PB2 is connected via diode D2 to shutdown input and it sets high voltage there before C7 is discharged

Stop:

  • cpu use ADC to get battery voltage
  • if voltage is too low, it finishes all operations and sets PB2 to low
  • C7 discharges through R9
  • when C7 is discharged, converter stops
  • whole circuit draws only converter standby current

Of course it's only one option. If you don't trust your code, it's better to use just battery protection circuit. But soft shutdown is good choice when you need clean shutdown (for example to finish write to sd card etc.).

So this PCB will not work?
Over Discharge Protection Over Discharge Detection Voltage 2.5±0.063V
Over Discharge Release Voltage 3.0±0.075V

majek:
To make efficient use of battery, step-up or step-down converter is needed (depends of battery configuration). Good one have efficiency above 90%, 80% is a bad one :slight_smile:
Don't use on-board LDO for this.

Most such regulators have 'shutdown' pin. You can get regulator with very low shutdown current (microamps) and implement soft shutdown.

This is my quite old project powered from li-ion batteries with this feature:

On the top-left part of schematic you have power section.
There are 2 components:

  • voltage measuring circuit (D1, R7)
  • shutdown circuit (D2, R9, C7, D3)

How it works:
Start:

  • you press power switch
  • C7 is already discharged, so it sets shutdown pin of converter to high
  • converter is enabled and passes power to the rest of circuit
  • cpu starts
  • first thing cpu does is putting PB2 high
  • PB2 is connected via diode D2 to shutdown input and it sets high voltage there before C7 is discharged

Stop:

  • cpu use ADC to get battery voltage
  • if voltage is too low, it finishes all operations and sets PB2 to low
  • C7 discharges through R9
  • when C7 is discharged, converter stops
  • whole circuit draws only converter standby current

Of course it's only one option. If you don't trust your code, it's better to use just battery protection circuit. But soft shutdown is good choice when you need clean shutdown (for example to finish write to sd card etc.).

Would I be able to use the SparkFun LiPo Fuel Gauge and program the Arduino to enter sleep mode at 3V?
If I disable the power LED and since I am bypassing the Linear regulator the current consumption will be 0.0058 mA is that good enough?

http://www.home-automation-community.com/arduino-low-power-how-to-run-atmega328p-for-a-year-on-coin-cell-battery/

Wasferd:
Hey,
So you reckon powering the mini with 4.8V is better than using a 7.4V?

I was merely suggesting it as a possible option, since the voltage is so close to operating voltage and it doesn't need to be dropped as much, possibly giving you better efficiency.

most efficient solution is running an m328 directly off 1s lipo. avr can monitor voltage and shut down to virtually zero current at any point. promini is not capable of this and rube goldberg regulator schemes a huge waste too. kiss.