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A Matter of Efficiency

Thread starter #1
Hello,

I am currently working on a project that consists of both Master and Slave radio stations, in which the Slave operates on battery power remotely from the house where the Master resides. Powering the slave is my primary concern now, and I wondered which two options below would be more efficient (if you can think of more, share!). Most of the hardware I have cannot operate under 2.2V minimum, and so either 3.0V or 3.3V is necessary.

Which would be more efficient?

1. Use two AA alkaline battery cells in series to achieve 3.0V (typical alkaline battery life for a single AA cell is 2400 mAH)

2. Use two AA alkaline battery cells in parallel (typical alkaline battery life would be around 4800 mAH total with two cells in parallel) with a boost regulator that converts the 1.5V battery supply up to 3.3V using the MCP1624

The power design, depending on which parameter is more efficient, will power a ambient light sensor that consumes around 658nA of current. When a notable change in ambient light is perceived by the light sensor, it will wake the PIC up from sleep (while sleeping, it consumes 20nA) trigger an IR pair to see if an obstruction is present, and also wake up the slave RFM12B transceiver, do some housekeeping, report to the Master, and then go back to sleep. So most of the time, the PIC, TRX, and other hardware will be sleeping, with a negligible consumption current.

According to the MCP1624 datasheet, its quiescent current is typically 19uA (can I depend on that value and what about my load?), so I was thinking that having the 2 AA battery cells in parallel yielding 4800 mAH with the MCP1624 boosting everything up to 3.3V.

What direction should I go?

Thanks.
 
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Thread starter #3
That does seem like it would be more efficient. Do you think I should have a regulator with the two AA batteries in series or just the standalone series AA battery cells?

Would there be any other power designs that might be more efficient than two AA battery cells in series? Otherwise I will just go with that.

Thanks for the prompt reply.
 

audioguru

Well-Known Member
Most Helpful Member
#4
Energizer has datasheets for their batteries on their website.
Alkaline cells have different capacities at different load currents.
The capacity is only 1200mAh when the voltage for two cells in series has dropped to 2.2V in about 6 hours at 200mA.
The capacity is 2400mAh when the voltage for two cells in series has dropped to 2.2V in about 100 hours at only 24mA.
 

Mr RB

Well-Known Member
#5
Are there wires between the master and slave?

Another thought for the light sensor and sleep would be to use a timed sleep cycle for the PIC, and the PIC turns the light sensor off (so they both sleep). Assuming what the light sensor detects happens slowly it only needs to be tested a few times a second? Or less?
 

MrDEB

Well-Known Member
#6
a schematic would be nice.
boosting the 1.5v to 3.3 is out of the equation if battery life is a concern as ronsimson pointed out. It takes battery power to boost from 1.5 to 3.3 You can't get a free ride.
 

bountyhunter

Well-Known Member
#7
Hello,

I am currently working on a project that consists of both a Master and Slave stations, in which the Slave operates on battery power remotely from the house where the Master resides. Powering the slave is my primary concern now, and I wondered which two options below would be more efficient (if you can think of more, share!). Most of the hardware I have cannot operate under 2.2V minimum, and so either 3.0V or 3.3V is necessary.

Which would be more efficient?

1. Use two AA alkaline battery cells in series to achieve 3.0V (typical alkaline battery life for a single AA cell is 2400 mAH)

2. Use two AA alkaline battery cells in parallel (typical alkaline battery life would be around 4800 mAH total with two cells in parallel) with a boost regulator that converts the 1.5V battery supply up to 3.3V
Boost will be pretty inefficient at low VIN. Also, paralleling batteries is a bad idea.
 

MrAl

Well-Known Member
Most Helpful Member
#8
Hello,

I am currently working on a project that consists of both a Master and Slave stations, in which the Slave operates on battery power remotely from the house where the Master resides. Powering the slave is my primary concern now, and I wondered which two options below would be more efficient (if you can think of more, share!). Most of the hardware I have cannot operate under 2.2V minimum, and so either 3.0V or 3.3V is necessary.

Which would be more efficient?

1. Use two AA alkaline battery cells in series to achieve 3.0V (typical alkaline battery life for a single AA cell is 2400 mAH)

2. Use two AA alkaline battery cells in parallel (typical alkaline battery life would be around 4800 mAH total with two cells in parallel) with a boost regulator that converts the 1.5V battery supply up to 3.3V using the MCP1624

The power design, depending on which parameter is more efficient, will power a ambient light sensor that consumes around 658nA of current. When a notable change in ambient light is perceived by the light sensor, it will wake the PIC up from sleep (while sleeping, it consumes 20nA) trigger an IR pair to see if an obstruction is present, and also wake up the slave RFM12B transceiver, do some housekeeping, report to the Master, and then go back to sleep. So most of the time, the PIC, TRX, and other hardware will be sleeping, with a negligible consumption current.

According to the MCP1624 datasheet, its quiescent current is typically 19uA (can I depend on that value and what about my load?), so I was thinking that having the 2 AA battery cells in parallel yielding 4800 mAH with the MCP1624 boosting everything up to 3.3V.

What direction should I go?

Thanks.

Hi,

Theoretically there is no difference, but in practice other things enter the picture which changes the outcome quite significantly.

Theoretically if you start with two cells 1.5v each and each has 0.1 ohm series resistance and you connect them each to their own 1.4 ohm resistor (two complete circuits) the power dissipation is 2.8 watts in the load resistors and 3 watts delivered from the batteries, for an efficiency of 93.3 percent. If you instead connect them in series then to keep the same power load you have to connect the two load resistors in series too so then we would have a 3v supply powering a 2.8 ohm resistor and with the series resistances of both cells now in series that means again we have 2.8 watts in the resistors and 3 watts from the batteries for an efficiency of 93.3 percent again.
So there is no difference at least theoretically.

However, once you start connecting the circuit to other things, even just plain wires, everything changes. And if there is a device such as a switching transistor that has to handle the power then there is going to be some loss there too. There are several reasons why the switching transistor might cause a loss of power, mainly voltage drop and switching losses. These losses add to the total required input power in addition to the theoretical above, so we end up with a less efficient circuit.

So the best bet is to use the cells directly whenever possible, but we do have to determine if this is really possible first. The cells voltage drops over time so we have to determine if the cell can provide the right voltage for the circuit for the time we intend to run it for. Some applications will not pass this little test, so we are then forced to use a switching converter.

If a switching converter is really necessary, then the best bet here is to keep the input/output voltage differential as low as possible. In short this means rather than going from 1.5v to 3.3v it would be better to go from 3.0v to 3.3v as the efficiency will be better.

So to recap, we first decide if we really need a converter, and if not, use the batteries directly any way possible. If we do need a converter, then we try to keep the input output voltage differential as low as possible.
 
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Thread starter #9
Energizer has datasheets for their batteries on their website.
Alkaline cells have different capacities at different load currents.
The capacity is only 1200mAh when the voltage for two cells in series has dropped to 2.2V in about 6 hours at 200mA.
The capacity is 2400mAh when the voltage for two cells in series has dropped to 2.2V in about 100 hours at only 24mA.

Hi AG,

I found the following link online: http://www.powerstream.com/AA-tests.htm

The Slave is going to consume a negligible current (certainly less than 100mA), but I did find the 100mA dishcharce curve below:

View attachment 60470

Where "DC AA" specifies a Duracell coppertop and "RS" is a RadioShack Enercell Plus. Unfortunately they did not compare an Energizer.

When the PIC wakes up, it begins its cycle of communication which is only for a short moment - perhaps on the order of milli-seconds. When the RFM12B transceiver is waked up, it will start transmitting for that short time, and while it does so, it will consume about 25mA. Because of this, the circuit should last years on two AA batteries in series, shouldn't it?
 
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Thread starter #10
Are there wires between the master and slave?

Another thought for the light sensor and sleep would be to use a timed sleep cycle for the PIC, and the PIC turns the light sensor off (so they both sleep). Assuming what the light sensor detects happens slowly it only needs to be tested a few times a second? Or less?
The Master and Slave devices both communicate through the RFM12B transceiver.
 
Thread starter #11
a schematic would be nice.
boosting the 1.5v to 3.3 is out of the equation if battery life is a concern as ronsimson pointed out. It takes battery power to boost from 1.5 to 3.3 You can't get a free ride.
Hi MrDEB,

I do have a block diagram, so you can get a rough idea of how the Slave will work. But I have decided to program the Slave's settings through the Master now, so ignore the keypad.

View attachment 60471
 

ronsimpson

Well-Known Member
Most Helpful Member
#12
Many PICs have a brown out detector or can monitor the battery voltage. It is a good idea to shut down or reset the micro when the battery starts to collapse.

I once built a power supply that tried to use up all the power in the batteries. It sucked the batteries down to 0.5 volts/cell while keeping the micro at 3.3 volts. The problem is that the batteries, when discharged, will leak in a couple of months. I now stop using power when the voltage reaches about 1 volt. There is not much power in the 1.0 to 0.5 volts area.
 
Thread starter #13
Hi Ron,

The brown-out detector is an excellent idea, thanks. I noticed that with discharge curves I posted earlier, they determined the amp-hour (AH) value when the batteries reached approximately 1V.
 

ronsimpson

Well-Known Member
Most Helpful Member
#14
The older PICs do not have a voltage reference. Analog voltages are measured against VCC. When living off a battery you can not really measure the battery voltage. I have measured battery voltage in a indirect method. Connect a resistor from an unused I/O pin to a diode pointing at ground. When this I/O pin is low or 3-state no power is lost. When this pin is high there will be 0.6 volts at the diode. This 0.6 V will be almost constant over the batter life. The battery will start at 3.3V and drop to 2.0V. By measuring the 0.6V against the VCC you can see the ratio of VCC compared to 0.6 volts. This gives a crude measure of battery voltages. This needs to be done every minute or less so the power lost in the measurement is little.
 
Thread starter #15
Hi,

Theoretically there is no difference, but in practice other things enter the picture which changes the outcome quite significantly.

Theoretically if you start with two cells 1.5v each and each has 0.1 ohm series resistance and you connect them each to their own 1.4 ohm resistor (two complete circuits) the power dissipation is 2.8 watts in the load resistors and 3 watts delivered from the batteries, for an efficiency of 93.3 percent. If you instead connect them in series then to keep the same power load you have to connect the two load resistors in series too so then we would have a 3v supply powering a 2.8 ohm resistor and with the series resistances of both cells now in series that means again we have 2.8 watts in the resistors and 3 watts from the batteries for an efficiency of 93.3 percent again.
So there is no difference at least theoretically.

However, once you start connecting the circuit to other things, even just plain wires, everything changes. And if there is a device such as a switching transistor that has to handle the power then there is going to be some loss there too. There are several reasons why the switching transistor might cause a loss of power, mainly voltage drop and switching losses. These losses add to the total required input power in addition to the theoretical above, so we end up with a less efficient circuit.

So the best bet is to use the cells directly whenever possible, but we do have to determine if this is really possible first. The cells voltage drops over time so we have to determine if the cell can provide the right voltage for the circuit for the time we intend to run it for. Some applications will not pass this little test, so we are then forced to use a switching converter.

If a switching converter is really necessary, then the best bet here is to keep the input/output voltage differential as low as possible. In short this means rather than going from 1.5v to 3.3v it would be better to go from 3.0v to 3.3v as the efficiency will be better.

So to recap, we first decide if we really need a converter, and if not, use the batteries directly any way possible. If we do need a converter, then we try to keep the input output voltage differential as low as possible.
Most of the hardware that the Slave utilizes is only on rare occurrences. It would be nice if I could get away with just two AA alkaline battery cells in series, so hopefully a converter isn't necessary. Below you can find the pseudo code for how the Slave and Master will work. I also have provided a summarized list of all the hardware the Slave utilizes along with all of the device's consumption current:

Code:
Pseudo Code for Slave
1. At first, the following Slave hardware is asleep/shut off except for the ambient light sensor:
     a.	PIC18F25K80 - 20nA sleep current
     b.	RFM12B - 300nA standby current
     c.	        IR LED Emitter (turned off)
     d.	IR Receiver (turned off)

2. The ambient light sensor, which is ON while the rest of the hardware is asleep, consumes a mere 650nA. When it detects a noticeable change in ambient light, it begins its cycle:
     a.	Wake up the PIC
     b.	Quickly use IR LED pair to determine if obstruction is present and turn off after detection/no-detection is completed.
     c.	        Wake up the transceiver and communicate to the Master:
                       1.	If an obstruction was detected by the IR pair, report to the Master.
                       2.	Check for Slave configuration updates and save configuration parameters in flash.
                       3.	Report battery status using brown-out detector or through the RFM12B transceiver module (it has its own internal battery monitor).
                       4.	Communicate any other pertinent information, etc.
                       d.	Configure all of the hardware except for ambient light sensor to go back to sleep and repeat cycle once a noticeable change in ambient light is perceived.
That is basically how the Slave will operate, and you can see that most of the time the Slave will be sleeping except for the light sensor. Estimates will vary, but usually each cycle will occur every 24 hours. Thus, I am trying to make this circuit extremely low power so that it will last years on two AA alkaline cells.

What do you guys think?
 
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#16
The older PICs do not have a voltage reference. Analog voltages are measured against VCC. When living off a battery you can not really measure the battery voltage. I have measured battery voltage in a indirect method. Connect a resistor from an unused I/O pin to a diode pointing at ground. When this I/O pin is low or 3-state no power is lost. When this pin is high there will be 0.6 volts at the diode. This 0.6 V will be almost constant over the batter life. The battery will start at 3.3V and drop to 2.0V. By measuring the 0.6V against the VCC you can see the ratio of VCC compared to 0.6 volts. This gives a crude measure of battery voltages. This needs to be done every minute or less so the power lost in the measurement is little.
Actually, most all PIC's that I know off that have an ADC, have an internal 0.6v diode reference. It usually can't be used as an ADC reference but it can be sampled by the ADC.
Basically doing the same thing you just mentioned but the hardware already exists internally.
 
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3v0

Coop Build Coordinator
Forum Supporter
#17
The chip EN0 is using is a modern on with a HLVD, HIGH/LOW-VOLTAGE DETECT, it uses 4 bits to set the trigger point and another to set the direction.

For general battery applications ... the device voltage
decreases. When the device voltage reaches voltage,
VA, the HLVD logic generates an interrupt.
From the PIC18F66K80 FAMILY datasheet section 26.

If you have not looked into the newer chips you may want to give it a go. Lots of fun stuff like the CTMU. It is used for touch switches but also do other useful tasks.
 
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