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More about Chargers

By Stan Yeo  

Introduction

This article will hopefully reduce the number of burnt out chargers and mains regulated power supplies (PSU). Lower modeller's expectations as to how quickly batteries, Lithium Polymer (LoPo) in particular, can be charged with the equipment they have and assist in the selection of a new charger or electric flight packs. The article is biased towards charging LiPo batteries as most chargers are capable of fast charging the majority of receiver / transmitter batteries and very few of us are still using Nickel Metal Hydride (NiMh) electric flight packs.

LiPo Charging

The charging regime for the Lithium family of batteries is different from other types of battery that we use in that the charge is terminated on battery (cell) voltage. Also it is a two stage charge, i.e. it starts as a constant current charge where the charging voltage rises in line with the battery voltage and finishes as a constant voltage charge where the charge current drops off to almost zero as the battery voltage gets closer to the charge voltage (Ohms Law). This can be observed if you watch the digital display on the charger in the later stages of the charge. The changeover generally occurs when the battery voltage is within about 0.1 volts of the cut-off voltage. When the combined voltage of the cells on charge equals 4.2v (Lithium Polymer) per cell then the charger terminates the charge. For a 3 cell LiPo pack this will be 12.6v. The cell cut-off voltage varies with cell type i.e. Lithium Ion is 4.1v. But please remember most chargers will also terminate the charge and timeout for safety reasons after 2 to 4hrs. Sometimes it is useful to disable this facility particularly when cycling NiCad / NiMh packs otherwise a false capacity result will be returned.

Expected Charge Times

With NiCad / NiMh battery packs the formula for calculating the charge time of a fully discharged pack is:

Battery Capacity / Charge Current x 1.4 = Charge Time (hrs)

i.e. For a 1000mA Pack being charged at 100mA (standard 10% charge)

Charge Time = 1000/100x1.4 = 1.4hrs i.e. 1hr 24mins.

Unfortunately this is not the case for Lithium packs due to the final stage of the charge being at constant voltage with a declining charge current. I would surmise this adds another 10 - 15 minutes to the charge time resulting in a typical full charge for a LiPo pack of around 1hr 45mins instead of 1hr 24mins for a 1C charge. C = Pack Capacity.

Cell Charging Voltage

To 'push electricity' back into the battery to charge the battery the charge voltage must be greater than the battery voltage to overcome the resistance of the battery. The battery resistance will vary from battery to battery, type to type, as a consequence so will the voltage differential.

The full charge cut-off voltage for a single LiPo cell is, as stated, 4.2v so for the purposes of this article and simplicity we have opted for a charge voltage of 5v per cell. For a 2 cell pack this would be 10v (2 x 5v) and a 3 cell pack 15v etc. etc. I am aware there are a number of variables that affect the charge voltage per cell but those of you that understand this relationship will also understand the relationship between charge voltage, charge current and the rated output of the charger so please accept this generalisation as I am trying to help those who do not!

Chargers

Based on our experience the most popular chargers are the ones with specification maximums of 50 watts / 6amps / 6 cells (LiPo). This for the non-electrically minded modeller is where the confusion arises.

Aide Memoir: Watts = Volts x Amps.

This can be rearranged as Volts = Watts/Amps or Amps = Watts/Volts

If the charge current for a 50w charger is set at the 6 Amps maximum the out put voltage would be 50W / 6A = 8.33v. Hardly enough to fully charge a 2 cell LiPo to the required 8.4v.

Caution . We do not recommend running chargers at their rated capacity. We would recommend a maximum 90% loading for charger longevity. In this case 50w/0.9 = 45 watts.

Practical Examples

Taking a typical 50w / 6A / 6S charger operating at 90% capability i.e. restricting the maximum output to 45w as suggested above results in the following maximum charging currents when the 5v per cell charge voltage mentioned above is used.

Single Cell = 45w/5v = 9 Amps Greater than 6A max. current limit.

2 Cell (2S) = 45w/(2x5)v =4.5 Amps. Max. Pack capacity 4500mA @ 1C

3 cell (3S) = 45w/(3x5)v = 3 Amps. Max. Pack Capacity 3000mA @ 1C

4 Cell (4S) = 45w/(4x5)v = 2.25 Amps. Max Pack Capacity 2250mA @ 1C

5S = 1.8 Amps and 6S = 1.5 Amps.

If for example you are using a 2200mA 6S LiPo pack popular with a particular size of electric helicopter this results in a charge time of just over 2hrs instead of the normal 1 ¾ hrs @ 1C due to the charge current only being 1.5A.

2.2A/1.5A x 1.4 = 2hrs 3minutes.

For most modellers it is not a problem as they will have more than one flight pack but it does raise of issues. One being do you use 2 x 3 cell packs in series and have a second charger to hand. Another being using two 3 cell packs instead one 6S does allow you to use the packs in other models only requiring a 3S pack. In addition should one pack fail in a pack you have only lost 3 cells as opposed to 6 cells!

As can be seen from the above calculations the issue also starts at 3S with cells of more than 3000mA capacity. Some cells can be charged at higher than the 1C rate. This in turn reduces the size of the battery that can be safely charged. In this situation our advice would be to look at buying a higher spec. charger such as a 200w 8S 10A with a 200w mains regulated power supply (probably a cheaper option than buying an all in one AC/DC one). A charger with this specification is capable of charging a 6S pack @ 6A

i.e. (200w x 90%) / (6 x 5v) = 180w / 30v = 6A

Regulated Mains Power Supply (PSU)

What applies to the charger also applies to a 13.8v Regulated Mains Power Supply (PSU) re power output. If the PSU and charger both have digital displays then you will notice a difference in the current reading between the two devices. There are two reasons for this. Firstly the output voltage of the charger is likely to be different from that of the PSU, remember its about power i.e. Watts . Secondly even if the output voltages are the same the current will be higher on the PSU due to losses in the charger. To compensate for this add 15-20% as a safety margin between the maximum output being used on the charger to that expected of the PSU.

To Sum Up

If you are new to electric flight and not of an electrical mindset please study this article and try to understand the concepts. They are no more complicated than A = B x C or B = A / C. For those who take issue with my simplistic approach I do apologise. I am just trying to lift the mist for those who do not. Also please read the other articles on electric flight on our website www.phoenixmp.com.

Happy Landings

Stan

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