# Voltage input for charging NiMH Batteries

I have 12 AAA NiMH batteries(1000mAH and 1.2V per battery) and I want to know what is the optimum voltage for charging them. I am using a simple Constant Current charger(LM317 and 68 Ohm Resistor(R in the circuit diagram). But i'm unsure on what the input voltage needs to be. My circuit doesn't have the diode.

A constant current source adjusts its output voltage with the load in order to maintain a constant current.

V = IR // holy s*#% it's Ohm's Law


NiMH batteries are fickle fiends to charge, exhibiting temperature-dependent changes in charge and discharge curves. They also don't have a float voltage, so constant voltage charging doesn't work, as you've likely discovered. Energizer has some recommendations regarding charge times:

Typically a moderate rate (2 to 3 hour) smart charger is preferred for NiMH batteries. The batteries are protected from overcharge by the smart charger circuitry. Extremely fast charging (less than 1 hour) can impact battery cycle life and should be limited to an as needed basis. Slow overnight timer based chargers are also acceptable and can be an economical alternative to smart chargers. A charger that applies a 0.1 C rate for 12 to 14 hours is well suited for NiMH batteries. Finally a maintenance (or trickle) charge rate of less than 0.025 C (C/40) is recommended. The use of very small trickle charges is preferred to reduce the negative effects of overcharging.

AAA NiMH batteries have a capacity of 850mAh [varies by manufacturer], so charging with a rate of C/2 to C/3 can be done with a constant current of...

850mAh / (2 to 3 hours) = 283mA to 425mA


An overnight, C/12 trickle charge can be done with a constant current of 71mA. This page mentions that:

Modern cells have an oxygen recycling catalyst which prevents damage to the battery on overcharge, but this recycling cannot keep up if the charge rate is over C/10.

The recommended maintenance charge rate of C/40 can be done with...

850mAh / h / 40 = 21mA


## Smart Chargers

Listed are charging techniques from Energizer, Duracell, and Powerstream:

• ΔV charging: charge at recommended constant current until the cell reaches a peak voltage and decreases (eg. -15mV).

This technique is accurate enough to safely charge at C/2 to C/3 (283mA to 425mA).

• dT/dt charging: monitor cell temperature to both limit maximum temperature and look for characteristic heating rate.

This technique may be used in conjunction with ΔV charge termination to more precisely monitor and terminate the process, allowing the use of higher currents (C/1 to C/2, or 425mA to 850mA).

• Soft start: If the temperature is above 40 degrees C or below zero degrees C start with a C/10 charge. If the discharged battery voltage is less than 1.0 Volts/cell start with a C/10 charge. If the discharged battery voltage is above 1.29 V/cell start with a C/10 charge.

• 1.78V maximum: a single cell must never exceed this.

But what does it all mean!? The input voltage to your LM317 constant current circuit must be enough to support the voltage drop across the regulator and resistor (1.47Ω), drive the required current, and exceed the maximum cell voltage. To source C/1 or 850mA to a AAA NiMH battery, whose internal resistance is at most around 120mΩ, requires (120mΩ + 1.47Ω) * 850mA + 1.2V + 1.78V = 4.3315V. I recommend at least 2V more to reduce the effects of source irregularities like regulation and noise and account for other circuit losses (like that diode you don't have yet). If you're charging 4 cells in series as your diagram indicates, you'll need at least 9.978V (ie: 12V+); 25.034V (27V+) for 12 in series, though I would worry about uneven charging.

• So...short answer = max cell voltage (1.78 V) × number of cells + LM317 dropout at your charging voltage (~2 V) + voltage across resistor (1.25 V, always). – Nick T Jan 8 '11 at 21:52
• @NickT, Aye! [#cells]xI_chargex(120mΩ)+1.2V+1.25V+[#cells]x1.78V. Complication is good for the soul, though (and the snowed-in weekend). – tyblu Jan 8 '11 at 23:34
• Charging an 850mAh battery for 2 hours at 425mA may be arithmetically sound, but starts from the idea that charging is 100% efficient. It's not. I guess depending on your charging method efficiency is limited to about 80%. So if you charge at 425mA, you'll have to charge for 2.5 hours. – stevenvh Jan 9 '11 at 12:18
• @stevenh, One of the listed sources claims 60%-70% efficiency. – tyblu Jan 9 '11 at 12:34
• @tyblu: well, my point was that you can't just divide the capacity by current to find time. – stevenvh Jan 9 '11 at 13:38

Are the batteries arranged in series like your drawing shows? If so your discharged voltage will be around 15V and your charged voltage will be about 16.8V (1.25V/cell discharged, 1.4V/cell charged). LM317 reference is 1.25V across the 68 ohm resistor for charge current of (1.25/68) 18mA charge current. Dropout voltage on the LM317 is for that rate is never more than 2V (page 6 from the datasheet tyblu linked to, very conservative estimate).So minimum input voltage needs to be (conservatively) 2V + 1.25V + 16.8v = 20.05V

How did you choose the value for the resistor? A safe charge rate (if you're watching it to turn it off) is C/10 which should be (100-mAh/10) 100mA charge current. That gives a resistor value of (1.25V/.1A) 12.5 ohms with power rating of (.1^2*12.5) .125W. Are these 12 cells in series or parallel?

• Where did yu get 15V from is that a rounded answer? As if you do 12*1.2 = 14.4V – Dean Jan 8 '11 at 20:34
• I use 1.25V for discharged - mostly because that's what the Roomba charging circuit uses as well. – AngryEE Jan 8 '11 at 20:44

The charging specs on one of the chargers of one of the two most popular battery manufactures is as follows from the charger specs.

Battery Size/ Rated Capacity (mAh) / Estimated Charging Time

AA 1300-2000 8 - 12.5 hr/ AA 2100-2500 13 - 16 hr/ AAA 850 - 900 11 - 12 hr

The specs for the charger current for these three types of batteries is 200ma, with 2 batteries in series at a voltage of 2.8v. Actual measured voltage of the 2 series batteries while charging is 2.885 volts, with a current of 240ma. If you took 10 batteries at 1.2v in series, 12v total battery voltage, with a trickle charge voltage of 13.8v, that is 15% above the battery voltage. With that in mind, then the charge current spec of 200ma, at a per battery rated voltage of 1.2v * 1.15 (Battery voltage plus 15%) would be 1.4 volts per battery (Cell), two in series would be 2.8v, 12 in series would be (1.4v * 12 batteries) at a charge voltage of 16.8 volts across the series battery bank of 12 batteries. From experience, and consulting with a Panasonic battery engineer, I have always used the battery voltage plus 15% charge voltage for trickle charging batteries. In a 24vdc industrial control system, I would run the power supply to 27.6v (24v*1.15) with 2 12v AGM batteries in series for system backup. This has always worked excellently for backup power for control panels on utility power, and most instrumentation, PlCs, and sensors will work fine up to ~30vdc, so I never found this to be a problem for the controls. So, in agreement with the previous post, 16.8v + 2v + 1.25 volts would be the optimum input voltage of 20.05v.

• Can you improve your answer, in terms of readability.. Breakdown into paragraphs and logical approach and explanation. – Umar Mar 12 '17 at 6:27

It's relatively simple. Your battery pack has a maximum wattage. bw.

You source or charger needs to produce at least twice that wattage.

2bw.

The circuit will charge your pack at a rate of efficiency x in time t where t is the time it takes to fully charge the pack @ x.

Now the rest is vague it's been some years since they covered it in school but you can look in a textbook to find out how to set it up properly for your rate of current.

Bw=2Bw times (efficiency) . Usually about 67% with a linear voltage regulator. Or x/t times 100%. Something like that. Your circuit is standard and won't be very efficient since you don't match battery impedence which changes all the time plus constant current chargers bog on modern high capacity batteries you need to add a ocsolator (sp?) at a low frequency rate to produce a sloped high voltage pulse for both charge adsorption and chemistry cooldown along with the constant current.

For example. You charge at a constant current of C/10. Every half a second you pulse a hundred volt positive pulse carrying two hundred millamps into the positive electrode using a separate circuit. Now don't disturb the C/10 constant rate while you do so.

Using the example helps align the dendrites in the cells and produces a much better faster fuller charge that lasts longer.

Lots sa luck.

John.

• Welcome to EE.SE, but your answer is unclear in several respects. What is the "maximum wattage" of a battery pack? Why would the charger need "to produce at least twice that wattage"? "Now the rest is vague it's been some years since they covered it in school ..." You can freshen up by reading the answer at the top of the page which explains the theory. It was accepted as the correct answer eight years ago. – Transistor Jul 7 at 21:54
• Might have been but it's wrong. You questions on my answer make no sense and show little understanding of electricity. – John Musielewicz. Jul 7 at 22:21