Pil ömrü nasıl hesaplanır


32

Pille çalışan bir ürünün ne kadar süre çalışacağını nasıl hesaplarım?

İşte elimde ne var:

  1. 2 adet AA, 1.5V, 2700mAH pil
  2. 25 UA Iq ile Voltaj Regülatörü
  3. Voltaj Regülatörü Et =% 80
  4. Aktif Akım = 50mA
  5. Uyku Akımı = 1uA
  6. Görev Döngüsü =% 99,9 (sadece zamanın% 0,1'i aktif)
  7. Aktif Gerilim 3.3V

Şu anki rotayı izledim ve bir cevap aldım. Güç yoluna gittim ve tamamen farklı bir cevap aldım (farklı günler vs günler).

Bunu nasıl yapıyorsun?

Yanıtlar:


22

Hesaplama, muhtemelen bir şey eksik, ama işte yaptığım şey:

1 μA+(50 mA×0.1%)+25 μA=76 μA

76 μA80% efficiency=88 μA

Round up to 100 μA=0.1 mA

2700 mAh0.1 mA3 years

If you're using rechargeable batteries, they'll discharge on their own long before that. Or if any of your other calculations are off (like maybe it's a 98% instead of 99.9% sleep), that will affect it a lot too.


close enough for estimation purposes. The 51 uA is from the 3.3V which would get boosted up from 2.6-3V. The 25uA Iq is from the 2.6-3V battery. You're missing the increase in current (maybe 15-20%) due to step-up effects. But this brings it really close to 100uA, so you're pretty much spot on.
Jason S

My concern is that your calculations don't include the number of batteries. There has to be a reason to put another battery in the product.
Robert

5
With batteries in series, they'll both drain at the same rate and provide the same amount of current. You've got a 3V 2700mAh battery instead of two 1.5V 2700mAh cells.
edebill

1
+1 for adding the point regarding the self-discharge of the batteries.
semaj

Excellent answer. As an improvement i'd say to multiply the battery capacity 2700mAh with 90% or 95% due to the fact that the battery is considered dead after it has discharged about 95% of it's power.
Nikos

13

The output current required from the 3.3V regulator is

1μA×0.999+50mA×0.001+25uA=75.999uA

The output power is

3.3V×75.999μA=250.8μW

The input power to the regulator is

250.8μW0.8=313.5μW

When the batteries are fully charged the input current to the regulator is

313.5μW3V=104μA

If the batteries have a flat discharge curve then you will get a life of

2700mAh104μA = 25837 hours = 2.95 years

Since your batteries are 2700mAh 1.5V AA I am guessing that the discharge curve is not flat. You will need to draw higher currents as the voltage drops.

Also your regulator efficiency probably drops at lower voltages. Again I am guessing since I have not seen the design.

Be careful when calculating using currents. You may be inadvertently assuming that the input and output voltage of the regulator is the same. With an input of 3V and an output of 3.3V it is not a big error. If you do a more accurate estimate of the battery discharge curve it will matter.


we discussed this in the past, and it seems it must be discussed again. Signatures are not allowed on SE sites. I have removed your signature a second time after you rolled back to add it. Your site is interesting, put it in your bio so that all can see it easily.
Kortuk

2
When this message was originally posted signatures were allowed on SE sites so the "Grandfather" principle should be applied.
jluciani

We discussed this months ago in email. Signatures are not allowed. We are not active going through and removing your signatures, but as a question becomes active again or someone flags it we will remove the signature. We are not trying to actively go through and remove your signatures, but we also do not grandfather anything into rules. We removed your signature, you cannot roll it back, we discussed this and you agreed.
Kortuk
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