# Effects of Pulsed vs Direct Current on Battery Depletion

I'm doing some work with a portable, cell-battery-based electronics and someone on the design team mentioned that pulsed current will more quickly deplete a battery than constant current, even if both have the same total energy consumed. Is this true? What is the reasoning behind it if it is true?

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–  davidcary Oct 1 '11 at 12:04

P = I^2R.

So the power loss in the resistor goes up with the square of the current. Multiply this by time and you get energy (joules) lost to the resistor. Example: Vbatt = 10 V and Rbatt = 1 ohm. 100 Joules required.

• Ipulse = 2A,
• Time to deliver energy = 6.25 seconds (total "on" time),
• Pbatt = 4 W, and
• Power lost in battery = 25 Joules

continous method Rload = 40 ohm:

• Icont = 0.249 A,
• time to deliver energy = 42 s,
• Pbatt = 0.06 W, and
• Power lost in battery = 2.5 J.

The difference is the i^2 term!

Most importantly is the 0.06 W vs 4 W internal power dissipation. The 4 W will generate signiciantly more heat than the 0.06 W due to the thermal resistance of the cell to ambient. This heat will elevate the batteries temperature, and this elevated temperature will also reduce the batteries capacity.

So this is the secondary effect of using pulsed power, not only is the battery dissipating more energy due to the I^2 term, it is "losing" available capacity due to elevated temperature.

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When you discharge battery, pulse current means that during these peaks current has to be higher, proportionally to your duty cycle.

Higher current means higher looses in battery itself due to internal resistance. This is much more pronounced in usual prime batteries, and less in Lithium & NiMh due to their lower internal resistance.

I.e. if internal resistance is 0.5 Ohm, and your peak current is 1A, you are loosing 0.5V right inside battery. But with continuous discharge at 0.2A, you are loosing just 0.1V.

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But for the same energy drawn from the battery, energy loss in the internal resistance would be the same as well! (unless this resistance increases with increasing current) –  stevenvh Jun 20 '11 at 14:42
No, the lower voltage drop - the lower energy loses with the same energy drawn. Compare: 1sec*0.2A*0.1V = 0.02J looses for continious mode, 0.2sec*1A*0.5V = 0.1J looses per second for pulse mode. –  BarsMonster Jun 20 '11 at 14:44
What I mean is: if you draw 1A instead of 0.2A you consume the same energy in 1/5 of the time. You lose 5 times as much per unit of time, but since the time is 1/5 the total loss is the same. –  stevenvh Jun 20 '11 at 14:49
Yes, if you look at my calculations, you would see that in first case I multiply 1sec * 0.2A, and in the second case I multiply 0.2sec * 1A - the difference in energy loss occurs due to different voltage drop on parasitic resistance inside battery (0.1V in first case and 0.5V in second) –  BarsMonster Jun 20 '11 at 15:01
I now see that my comment is the same as stevenh's first comment. –  Federico Russo Jun 20 '11 at 15:45

Both energy losses and negative effects on the battery will usually be higher for pulse discharge than for steady DC discharge. The degree depends on battery chemistry, construction, past history and more but overall pulsing is almost sure to be worse.

If the same energy is taken by either steady or pulsing discharge then the pulses must be at higher current. Even if the energy out is what the battery sees and not what the load sees, then higher discharge rates will decrease the available amp hours for equal delivered energy. If the energy out refers to what the load or user sees then the situation is even worse as the battery will have higher internal losses at higher current so lose more internally to deliver the equal energy externally.

Take as an example a 2000 mAh = 2Ah battery discharged at 1A continuous or at 2A with a 50% on/off duty cycle.

(1) Cell internal impedance will almost certainly be higher at higher current but even if it is the same.

Wattage loss DC = I^2 R = 1^2 x R = 1R
Wattage loss pulsed = I^2R x 50% = 2^2.R x 1/2 = 2R.

ie doubling current will double losses if Rinternal does not rise.

(2) Worse - if this is feeding a switching power supply, the requirement is NOT constant mean current but constant energy rate delivered to the converter. The higher current will lower the terminal voltage so at 50% duty cycle the current will not be double but will be 2 x V_1A/V_2A
where V_1A is the terminal voltage when 1A is drwan etc.

(3) Cells last longer when discharged at lower rates relative to their amp hor rating. eg a 2Ah cell discharged at 2A will provide fewer full discharge cycles than the same cell discharged at 1A. In both cases the cell is fully discharged but the effect on the system is exacerbated by higher rates. The reasons vary with chemistry. LiIon cells are mechanically pumped to death mechanically as Li metai is added and removed (hence the success of liFePo4 which maintains an Olivine solid cell structure when 'empty"). NimH undergoes secondary reactions which are affected by temperature and internal potentials. Lead acid cells tend to mechanically destroy their plates with current being a factor in rate.

So - overall, multiple factors combine to reduce effective battery capacity and lifetime under pulse discharge.

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There is another AMAZING benefit from pulse discharge.

It can actually improve ESR of old batteries if not damaged by shorting or heat.

It that was invented by an old neighbour of mine , and manufactured by my employer EMS firm "was Nortel then C-MAC ...sold to Solectron" in Winnipeg. The product was a little pulse circuit in parallel with LEAD ACID or ni-cad batteries and called "Solartech" . Company has long since sold its patent rights and owner made millions. We made many and I test one on a large motive power tractor at Air Canada for hauling planes. Not only did ESR improve but specific gravity improved after full charge. proving to me that the battery health was improved. The theory was the 25Khz or so 10 nS pulses (Tr) of small current resonate the sulphate crystals that coat the electrodes and cause battery life reduction. So in a sense it rejuvenated the electrodes. Now there are many failure modes in batteries such as metallic particles in acid causing leakage and self heating or boiling of electrolyte.. But for many rechargeable batteries. It worked. He was able to sell them in large quantities to truck companies and tractor companies. For cars, people tended to abuse batteries and it was cheaper to replace the battery than add on $50 or$75 rejuvenator. But it was also found effective other chemical battery technologies. That's all I can say.

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That's interesting. I wonder if any research has been done as to what the best frequency is for various chemistries. I will tune my buck converter to just that next time. –  Joel B Apr 11 '12 at 19:50

Barsmonster suggests that your internal loss is larger when pulsing. That's not quite correct. Say the internal resistance is 1/10 of the load resistance. Then the power dissipated in it will be 1/10 of the load power. Over time this will cause 1/10 of the energy loss, no matter what the load pattern is. So if the battery is drained 1/10 of its capacity will be lost in the internal resistance.
This leads to the assumption that the battery's capacity is the same for whatever use if the internal resistance is a constant. If the internal resistance would increase with increasing current then pulsing would cause a higher internal loss than loading it with a constant current, and there would be more energy available if the load is constant.

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You say "the internal resistance is 1/10 of the load resistance". But the load resistance is different for the pulsed case and for the non-pulsed case. If he draws 1A in the pulsed case (as opposed to 0.2A in the non-pulsed case), that means lower load resistance and therefore the influence of the internal resistance is greater, isn't it? –  AndreKR Jun 20 '11 at 17:02
I mentioned that my reasoning presumed a constant internal resistance (I even wrote it in italics), and I also said that pulsing would cause a higher loss if it wasn't. Barsmonster doesn't mention anything about a variable internal resistance. On the contrary, he uses the same value for his calculation of both cases. –  Federico Russo Jun 20 '11 at 17:20
P = I^2R: The power loss inside the battery is proportional to the square of the current being drawn. So the load pattern makes all the difference. In your example you have missed the fact that the load resistance changes for a different load pattern, and so the proportion of losses in the battery varies and is not static. –  smashtastic Jun 20 '11 at 19:13
Yes, we assume the internal resistance is constant, but the load resistance is not. –  AndreKR Jun 20 '11 at 20:58
I think we're using 2 contradictory definitions of "constant". Russo is right: if I assume the battery resistance is some "constant" fraction of the load resistance (say 1/10 the load resistance), then the load pattern doesn't matter. However, BarsMonster and AndreKR are also right: if I assume the battery resistance is some "constant" absolute quantity (say 1 Ohm), then the load pattern does matter. My understanding is the BarsMonster assumption, while not perfect, is closer to reality -- it better explains why measured battery voltage appears bounce up after disconnecting a heavy load. –  davidcary Jun 22 '11 at 23:30

Rather than add to a lo-o-o-ong line of comments to barsmonter's and Federico's posts. I will add a post:

Barsmonster is correct. This may not be the only reason pulsed batteries last longer, but Barsmonter's reasoning and math bear re-reading for those who are not yet convinced.

The key to this is understanding that when current is pulsed, the external load resistance which the battery sees is much lower, which makes it's own internal resistance a higher percentage of the total load. Let's look at two examples of equal energy being drawn from a battery.

Take a 10V battery with 100mOhm internal resistance, providing 1 Joule to two different loads. For the first load the draw is continuous for 100 seconds. For the second load the draw is a sharp pulse once every 100 seconds.

Case A (continuous load): load = 10K Ohms. I = 10V/10KOhms = 1mA. Power to the load = (I^2)(R) = (0.001)(0.001)(10K) = 10mW. 1J/10mW = 100 seconds to get 1J into the load. During this time the battery internally dissipates (0.001A)(0.001A)(0.1Ohm)(100s) = 0.01mJ. Efficiency = 1J/(1.00001)J = 99.999%

Case B (Pulsed load): load = 1 Ohm. I = 10V/1Ohm = 10A. Power to the load = (I^2)(R) = (10A)(10A)(1 Ohm) = 100W. 1J/100W = 0.01 seconds to get 1J into the load. During this time the battery internally dissipates (10A)(10A)(0.1Ohm)(0.01s) = 0.1J = 100mJ. Efficiency = 1J/1.1J = 90.91%.

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in case B, dont you have 100% efficiency for the other 99.99 seconds based on your math? –  Kortuk Oct 11 '11 at 3:46