Voltage drop in battery under load - Page 1
Forum section: electronics / beginners

author: jwbrooks
author_rank: Contributor
timestamp: July 23, 2013, 03:06:33 pm
content: The problem:I've built a relatively simple circuit involving two 3V CR1216 coin batteries in series to power a simple circuit. I measure 5.5 volts across the two batteries when they are unloaded, but when I plug them into the system, it drops to 3 volts. I need at least 3.5 volts to ensure the 2.5V regulator correctly powers the other circuits. If I use a 5 volt DC power supply instead of a battery, I don't have this problem.My limited understanding:I imagine my problem is that I'm drawing too much current, and the batteries are having trouble maintaining a high voltage. Something to do with the internal resistance of the batteries maybe?The questions:What is going on, and how can I solve this? My application needs to stay as small and lightweight as possible (thus my use of two CR1216 coin batteries). Thanks for your help!The circuitTwo CR1216 batteries (ideally 6V) is regulated by a 2.5V regulator. The 2.5V line now powers a quater bridge strain gauge, an amplifier, a pulse width modulator, and an IR LED. I can provide more info on the circuit if need be, but I imagine that this is a pretty simple question.CR1216http://www.amazon.com/dp/B000RB1GDM/ref=pe_385040_30332190_pe_175190_21431760_M3T1_ST1_dp_1Strain gaugehttp://www.vishaypg.com/docs/11224/125un.pdfPulse Width Modulatorhttp://www.linear.com/product/LTC6992-1Amplifierhttp://www.ti.com/lit/ds/sbos069/sbos069.pdfIR Emitterhttps://www.sparkfun.com/products/2412.5V regulatorhttp://www.st.com/web/en/resource/technical/document/datasheet/CD00000544.pdfMisc. resistorsBackgroundI'm relatively new to the electronic and DIY community, and I find that I'm always running into what must be very basic questions. I'm hoping EEVblog will be a good place where I can ask these. If there is a better place that I should be asking these questions, please let me know. Thanks!

author: c4757p
author_rank: Super Contributor
timestamp: July 23, 2013, 03:12:23 pm
content: Power it with a stronger supply so it comes up to full working voltage and measure the current draw. Use current limit and be ready to shut it down - it must be taking quite a bit to drop two lithium batteries to half voltage...

author: Neganur
author_rank: Unknown
timestamp: July 23, 2013, 03:29:16 pm
content: Those coin cells are not up for the task, the datasheet specs them at 46 mAh and they're supposedly lasting ~740 hours under continuous load of 46 µA and I'm guessing your circuit needs more than that. The regulator alone needs some quiescent current.What's the current consumption when you're using the 5V supply?

author: Galaxyrise
author_rank: Frequent Contributor
timestamp: July 23, 2013, 03:32:51 pm
content: Internal resistance is correct for the reason the voltage sags. Looking ata datasheetfor an energizer CR1216, the internal resistance starts at 2 Ohms. Using Ohms law, I compute that dropping 2.5V to internal resistance would take 625mA. That's quite a lot for the circuit you've described, so the next step is figuring out what's using so much power.Looking at the devices you've listed, my guess would be that you omitted a current limit resistor for the LED. Otherwise, nothing leaps out as consuming that much power.

author: IanB
author_rank: Super Contributor
timestamp: July 23, 2013, 03:53:18 pm
content: It definitely seems to be a case of drawing too much current from the battery. The voltage dip is approximately equal to the current drawn times the internal resistance of the battery.When you connect your circuit up to the 5 V power supply, use your multimeter to measure the current used (put the meter in series with the device being powered). Compare the measured current with the expected current. You can get the expected current by calculating how much current each element of your circuit ought to be using and adding up all the contributions.If your expected current is too large you will need a bigger battery.If your measured current is larger then your expected current then you need to trace back and find out what is not working as expected.

author: free_electron
author_rank: Super Contributor
timestamp: July 23, 2013, 04:31:45 pm
content: you are using a linear regulator ? do you know you are wasting 60% of your batteries energy in that regulator ?

author: jwbrooks
author_rank: Contributor
timestamp: July 23, 2013, 06:10:31 pm
content: Wow. Thanks for all the feedback!Is my regulator considered "linear"? I also didn't know it consumed so much power. What are the alternatives?I'm using a 330ohm resistor in series with the LED.On the 5V power supply, my multimeter says it's pulling 16 to 18mA depending on the duty cycle of the LED. That's not much at all. I'll check the data sheets and lookup expected power consumption. I'll post the results back here in a bit.I put the two batteries back in and observed the same drop in voltage. I was a bit concerned that maybe the mid point of the two batteries was shorting (my board is a bit cramped), but I measured 1.1 V at the mid point and 2.2 across both. The batteries die very quickly (thus 2.2 instead of 2.5).

author: IanB
author_rank: Super Contributor
timestamp: July 23, 2013, 06:23:46 pm
content: If you look at thedata sheetthe CR1216 is designed for a continuous drain of about 0.05 mA into a 60 kilohm load, or an intermittent pulsed load averaging to about the same thing.So if you are trying to draw 18 mA you are exceeding the design load by about 360 times.I think you need a much bigger battery. You are effectively taking a watch battery designed for microamps and loading it down with a near short circuit.

author: free_electron
author_rank: Super Contributor
timestamp: July 23, 2013, 06:32:07 pm
content: well. you start with 6 volts. send it through the regulator to make 2.5. that means 6-2.5 = 3.5 volt gets 'eaten ' by the regulator.if you draw 10 milliampere your circuit consumes 10 ma x 2.5 volt = 25 mW.the regulator 'burns' 10mA * the 3.5 volts across it or 35mW.in other words : you are burning off more than you use ... a very big waste of energy.use a step-down regulator.

author: mariush
author_rank: Super Contributor
timestamp: July 23, 2013, 07:57:16 pm
content: Yes, LD1117 is a linear voltage regulator. It is a nice one, as it can do 0.8A with only about 1v of voltage drop, but it will still regulate by dissipating the voltage difference as heat, as free_electron says.Putting the fact that batteries may not be able to give you the current amount you want...If you want the highest efficiency, you'll have to go with a step-down (buck) regulator, which will be a bit more efficient.If you want to go with the lowest number of parts, you could still put the batteries in parallel and use a linear regulator that doesn't require such a high voltage drop.For example, you have these :http://uk.farnell.com/texas-instruments/tps73125dbvt/ldo-reg-5-5vin-2-5v-0-15a-5sot23/dp/211262530mV voltage drop, so you could use batteries as low as 2.6v, up to 150mA if the batteries are up to it.http://uk.farnell.com/micrel-semiconductor/mic5320-jgyd6-tr/ldo-dual-2-5-1-8v-0-15a-6tsot23/dp/229342935mV drop , same otherwisehttp://uk.farnell.com/texas-instruments/lp3985im5-2-5-nopb/ic-v-reg-linear-2-5v-smd/dp/1469024RL60mV drophttp://uk.farnell.com/nxp/ld6806td-25p/ldo-2-5v-200ma-5sot753/dp/209550580mV drop, 200mA etcIf you want through hole, Microchip has some very nice linear regulators :http://uk.farnell.com/microchip/mcp1700-2502e-to/ic-ldo-250ma-2-5v-to-92-3/dp/1851939About 60mV voltage drop at 100mA, 180mV at about 250mAAll of the above are very easy to use... just a couple of ceramic 1-10uF capacitors so they don't take much room on the board.But you still have the current consumption left... 16-20mA is a lot for those batteries.. even the cr2032 is designed for 0.2mA which is 100 times less than your needs.You might be able to get a small lithium battery that would be able to give you that current capability, something like this comes to mind:http://www.hobbyking.com/hobbyking/store/__23310__Turnigy_nano_tech_90mah_1S_15C_Round_Lipo_.htmlhttp://www.hobbyking.com/hobbyking/store/__7565__ZIPPY_50mAh_20C_single_cell.htmlhttp://www.hobbyking.com/hobbyking/store/__7568__ZIPPY_240mAh_20C_single_cell.html

author: SeanB
author_rank: Super Contributor
timestamp: July 23, 2013, 08:03:55 pm
content: Can you not use a larger coin cell, like a CR2450? That will both have lower internal resistance and higher capacity. A single cell and a low dropout regulator may work well.

author: IanB
author_rank: Super Contributor
timestamp: July 23, 2013, 08:12:53 pm
content: If you need more power you might consider a CR2 lithium cell:http://data.energizer.com/PDFs/1cr2.pdf

author: Bloch
author_rank: Unknown
timestamp: July 23, 2013, 08:18:04 pm
content: Quote from: jwbrooks on July 23, 2013, 03:06:33 pmHow much current are you using for the IR diode ? My guess is that is the biggest current user.

author: jwbrooks
author_rank: Contributor
timestamp: July 24, 2013, 04:29:10 pm
content: Hmmm. This is all good information. Thank you, everyone.The CR2450 is only rated for 200microamps. The Zippy batteries are larger than I want, but I may not have a choice. The CR2 just seems too big. Thanks for thoughts.I don't have an easy way to measure the current draw in the IR diode. Based on my playing with the duty cycle in the diode, it appears to only be around 2-4 mA. I'll likely be redesigning the circuit based on the ideas here and will keep this in mind.Also thank you everyone for the explanation and feedback of the linear regulators. I'll incorporate one of these into the next design.Everyone, thank you very much for your help with this problem! I feel like I've already learned quite a bit with this post alone, and if no one minds, I'll be posting here in the future with other basic questions.- John

author: IanB
author_rank: Super Contributor
timestamp: July 24, 2013, 05:29:14 pm
content: It's also worth mentioning perhaps that the key chain sized and pen sized laser pointers that you find everywhere are typically powered by a couple of LR44 button cells. Given that laser diodes must need a few mA to run, perhaps that will give you an avenue to explore?

author: hlavac
author_rank: Frequent Contributor
timestamp: July 30, 2013, 01:51:35 am
content: Did you put a capacitor across the battery to cover any spikes in power consumption?

author: jwbrooks
author_rank: Contributor
timestamp: July 30, 2013, 01:55:46 am
content: No, I did not. What capacitance do you recommend? Are these type of batteries bad about spiking?

author: hlavac
author_rank: Frequent Contributor
timestamp: August 04, 2013, 04:14:31 pm
content: Yes they are bad, they have high internal resistance so when they get loaded too much most of the power is lost on the internal resistance and voltage drops a lot, enough to brown out your circuit.Try 10uF or more.

author: SeanB
author_rank: Super Contributor
timestamp: August 04, 2013, 04:30:31 pm
content: I get a very good increase on life of these cells in intermittent loading ( remote control fobs) of about 3 times just by adding a small 10-22uF 16/25V tantalum capacitor across the battery terminals. Helps a lot, just the battery eventually goes so high in resistance that it no longer works for 2 presses in quick order.