Total Power Consumption

[ZZX]

Guyz, i was interested in calculating the total power consumption of a ranging sensor that consumes per second= 2 amps for 0.4ms and 100ma for the rest of the second. i calculated using integration (area under the curve) and came up with result i wasnt sure is correct or not. So i want some help and if possible also need more tutorials/insight into power consumption calculations of such systems.
Plz help

[Calamar]

2 * 0.004 + 0.1 * 0.996 = 0.1076 Ampere per second
_________________
Best Regards
Daniel H. Sagarra
La Plata (Argentina)

[newguy]

Average I = area under i(t) curve for one cycle/period (length of one cycle)

Using your data, 2A for 0.4 ms, and 100 mA for 0.9996 sec, we get:

Average I = ((2A * 0.0004 sec) + (0.1A * 0.9996 sec))/1 sec

Average I = 100.76 mA.

[Calamar]

sorry 0.4 ms

2 * 0.0004 + 0.1 * 0.996 = 0.1004 Ampere per second
_________________
Best Regards
Daniel H. Sagarra
La Plata (Argentina)

[ckielstra]

[Calamar]

yes, you are right, sorry
_________________
Best Regards
Daniel H. Sagarra
La Plata (Argentina)

[ZZX]

Thank you all for your inputs. Infact i calculated my result correctly and you all verified it. i have a couple more queries here:
1. I assume that 100.76 mA is also my current/hour consumption. Ive assumed since i can calculate :

[ckielstra]

[Guest]

Well its an ultrasonic sensor and im sending the 'ping' once every second. During this time 0.4ms duration consumes 2A and the rest of the second consumes only 100mA. Now the only possible solution to decrease consumption in this way is probably to cut down the sample rate( or ping sending rate) , like probably sending once in 5 second etc.

[ckielstra]

The 2A sounds like a lot but is only active during 0.4ms. From the above calculations you have learned that this peak current has only minimal effect on the total average current. For example reducing the ping rate from once per second to once in 5 seconds will bring down the average current from 100.76mA to 100.15mA, which is a neglectable 0.6% improvement.

Without knowing more details of your circuit it is difficult to give suggestions, but here are some general notes:
- Many of the older lineair regulators have a high quiescent current, the famous LM7805 for example uses 5 to 8mA just for itself. Investigate other parts, for example the S-812C uses only 1uA but is limited to a maximum input voltage of 15V.
- Consider using a switching regulator.
- Which PIC processor are you using? The newer Nanowatt models use less current than the older models.
- Reducing the clock frequency will save power.
- Does the processor need to run always or can you put it periodically to sleep mode?
- Reducing the operating voltage of your circuit will reduce power consumption.
- Do you have LEDs in your circuit? With high-efficiency LEDs you can get the same light output at much lower currents than with cheap LEDs (5mA v.s. 20mA)
- etc.

[Ttlemah]

Yes. It is the 100mA, that 'sounds a lot'. As an example, I did an ultrasonic 'pinger', used in a marine application, which had a similar instantaneous current draw. However the current in the 'idle' times, was less than 10mA. Ckielstra, gives very good pointers to how to achieve this sort of level. In my case, I left the processor running, but used the 'dual clock' option, to switch it to 32KHz. I used a very low quiescent consumption regulator, and actually turned off the power to all but the processor and display, using a SSR. Also remember to be very careful about biasing inputs to a supply rail, when the circuitry is disconnected (a 'floating' input, can create a suprising current draw).
The 100mA, really determines the battery life, and I'd be very suprised if it wasn't possible to reduce this a lot with care. For me, it was the display, that was the 'limiting factor', drawing 8mA on it's own. Another version, used in a buoy, which did not need the display, had a quiescent consumption below 1mA.

Best Wishes

[ZZX]

Thank you both for valuable tips.
However i probably wont be able to reduce the 100mA at idle times since this is what the sensor itself consumes when its idle ! After reading your replies i have the following queries:
1. How can i calculate my PICs total consumption provided the fact im using 2-3 timers and hardware UART and processors running at 4Mhz.
2. Let me know about the dual clock option that you tried

[Gerrit]

Maybe a dombe solution,

why don't switchoff the sonar until it is needed.

save's a lot of power.

Gerrrit

[Ttelmah]

An input that is 'floating', and happens to get to the 'detection level' ofthe chip concerned, can start to make the chip draw significant power as noise on the signal, then results in signal transitions inside the chip. Also in some cases (depends on the internal gate design), there can be a 'crossover', where the internal logic is not biased either off or on, again resulting in extra power being drawn. Hence 'inputs', should always be designed to actually be pulled 'to' a rail. For instance, I have some lines that are used most of the time as output 'drives', but when the device being driven is off, are switched to inputs. To ensure that extra current is not drawn in this state, the lines have 1MR resistors to ground. The extra current when they are driven, is tiny.
What is the time needed for the sonar sensor to 'wake up'?. If it smaller than the gap time between your pings, then turn it off with a MOSFET, or SSR, for at least part of the time. Most have the drive circuit, and receive part seperate, and the receive section on it's own, should not draw 100mA.

Best Wishes

[Guest]

yes, i can turn it off using a fet, since wakeup time is much smaller than the pinging gap. One last thing, is there any difference between not using certain peripheral and turning them off. Cuz if u arnt using them,do they still are consume current. If yes, then how to turn off certain peripherals not in use. Also can i calculate the total power consumed by my pic?