Calculating Power and Amps for RC Heli Battery Charger and Generator

[cjm181]

Hi

This i suspect is a very silly question. I have a battery charger for rc helis. The charger is supplied with 24Vdc via a pair of computer power supplies. The PSU's are supplied with 240Vac from the wall. I am trying to size up a generator so i can charge at the field.

The charger has two channels, ie it can charge two batteries at once.

The batteries are 6s (22.1V), 3300mA. So for my charger each of them at 4A, the charger will be drawing 24Vdc (thats what the PSU's deliver), and 8A.

So the Power required at the charger input is:

P (W) = I (A) x V

P = 24 x 8 = 192W

So, the supply to the PSU's from the mains is 240Vac, can i use the above numbers to calculate what the amps are coming from the wall? so:

192 = 240 x I

I = 192 / 240 = 0.8A?

So the power at the charger is the same that's coming from the wall?

Back to the generator, if the generator can supply 230Vac @ 1600W, it can supply:

I = 1600 / 230 = 6.9A

But assuming the power is the same anywhere in the setup, if the charger is receiving 1600W at its input, the max amperage the charger can charge at is:

I = 1600 / 24 = 66A?

Are my sumations somewhere near?

Thanks
Craig

 

[Averagesupernova]

You have some misconceptions about voltage and current. Just because a source is able supply a given number of amps does not mean that it will. The number of amps being supplied is dependent on the load.

 

[cjm181]

ok, i understand that. The load will be 8amps will it not? i can set the charger for that! I have an internal amp meter between the charger and PSU's that tell me what the charger is pulling. so if the charger (load) is pulling 8 amps, does the rest of my post work?

I am assuming because i can set the charger, i can technically set the load!

 

[Averagesupernova]

If you figure 192 watts out of your charger and a 100% efficient charger then your math works. The load that your charger puts on the 240 volt source would be .8 amps. However, your charger is not 100% efficient so you will be drawing more than .8 amps.

 

[cjm181]

yes. I see that! There is a lot of heat loss in the system, also in the PSU's. So if i said there was a further 20% loss in these throughout the system, then a 1600W 240Vac generator will have tonnes of capacity?

I will never be charging (a load) more than 10A, so if (ideally) that generator can provide 66A worth of load at the charger, then that generator will have tonnes of capacity.

 

[Averagesupernova]

For the sake argument a horrible efficiency of 20% means that 20% of the power put into the system makes it to the output. So 960 watts in means 192 watts out. Just how bad is the efficiency in your setup? A 1600 watt generator should be able to go along fine with this kind of draw indefinitely I would imagine. Approach the top end of it's rating and I would assume the duty cycle will start to drop off.

 

[CraigHB]

I'm familiar with the equipment you are using, I'm also into model helis.

The maximum power required on the battery side is the charging rate times the maximal battery voltage. Of course this is not the constant requirement, but the peak requirement your PSUs and charger have to cover. That being the case, if you're charging 6S batteries with a maximal voltage of 4.2V per cell peak power demand is 25.2V times your charge rate.

A charge rate used prior to a flight session might might be 2C for about a half hour charge. For your particular battery that would be 2 times 3300mAh for 6.6A. That times 25.2V would yield about 166W required on the batttery side. Ultra high drain LiPos like the type typically used for model helis can have acceptable charge rates up to 10C with 5C more typical. You'd probably want to use a 4C rate at the field for a charge time around 15 minutes. That would be 13.2A times 25.2V for approximately 330W.

The hobby chargers used to charge these types of batteries typically use a synchronous buck-boost converter to step voltages which are quite efficient. It's not unusual to see efficiency in the area of 95%. That means input power only needs to be at the most 10% more than output power with 5% more typical. There's also the efficiency of the power supply which can be higher than 90%. That means worst case would be no less than 80% efficiency from the AC supply to the battery itself. So if you need 330W out, you're going to need worst case 400W in.

In terms of input current on the AC side of the supply, you can arrive at an estimate by dividing out power by voltage. So with a 240V supply and a 400W requirement you'll see about 1.7A drawn by the supply. Of course there's also parallel charging so if you are charging more than one battery at the same time at a particular rate, you can simply multiply out the demand by the number of batteries. That's of course assuming your supply and charger are rated for the increased demand of parallel charging.

 

[cjm181]

Thanks Nova and Craig!

Craig, The charger is an icharger 406duo (2 channel), which i believe can deliver 1400W. The charger is in its own case that i purchased. Inside the case are 2x PSU's, each suppling 12V. So my understanding of this from the guy that manufactured the case, as i am charging 6s lipos its better to supply the charger with 24V. I am not entirely sure of the sizing of the PSU's, but i can assume (i can find out) that they are good for the entire capacity of the charger.

The charger is overkill for my needs, it was a buy good buy once kind of thing.

So as i am poor, i am running with turnigy packs (lol). The heli is a 12s setup, so as you know its two packs per flight. The packs have a max charge rate of 2C, so i will probably aim for 1.5C. After each flight, i intend to charge one pack from one channel, and the other on the other channel.

I was not planning on parallel charging for now, but we will see for the future.

So if i can pick you brains a little further!

I have 6 sets for the heli (12 packs). so if i am discharging the packs during flight to approx 75%, i would need to put back in:

3.3 * 0.75 = 2.5A.

So if i am charging at 1.5C, i can expect a charge time of:

3.3 * 1.5 = 4.95A in an hour

4.95 / 60 = 0.0825A per minuite

2.5 / 0.0825 = 30.3 mins, call it 31 mins to charge a set (one on each channel)

so back to our equations:

P = V * I = 25.2 * 10A = 252W, assuming 80% efficiency, 252 * 1.2 = 302W

So from the generator, i need 302W. So the amperage required from it would be:

302 = 240 * I
I = 302 / 240 = 1.2A

Is all that looking like a good estimate?

So as i progress, let's look at what the max the system can do!

On the face of it if the charger can do 1400W, and the gen can do 1600W then we are well in? But..

The spec of the gen says each 240Vac outlet in a 13 amp socket. Let's assume 80% efficiency again. So the highest current draw from the gen will be:

1600 * 0.8 = 1280W

1280 = 230 * I

I = 1280 / 230 = 5.57A max from gen

so for the charger:

I = 1280 / 25.2 = 50.4A?

So if i am using both channels, that's approx 24A per channel. On a 5.0A pack for example, that would be just shy of 5C

If i am parallel charging as per your coment, rather than the power being spread across both channels, it would just be down one channel? Is that correct?

If it is i can charge 2 sets ( 4 packs, 2 parallel on each channel) i would need a supply to the charger of:

charging each pack at 5A, parallel charging would mean each channel would need 10A, so total for the charger is 20A.

P = 25.2 * 20 = 504W

504 * 1.2 = 605W

At the gen

I = 605 / 230 = 2.6A.

So no matter which way we cut this, the gen will be at max 50% capacity when parallel charging 2 sets of 3.3A packs?

Kr
Craig

 

[CraigHB]

I am not entirely sure of the sizing of the PSU's, but i can assume (i can find out) that they are good for the entire capacity of the charger.

I'd hazard a guess you're using the converted server power supplies that are popular for this application. In that case with two in series each at 750W each you're good for 1500W supply. As the charger is 1400W, it's probably 700W maximum per port. With a 6S batteries at 3300mAh it's entirely possible to use the full capability of your charger and supply when charging four batteries at high rates (2 in parallel on each port).

So as i am poor, i am running with turnigy packs (lol). The heli is a 12s setup, so as you know its two packs per flight. The packs have a max charge rate of 2C, so i will probably aim for 1.5C. After each flight, i intend to charge one pack from one channel, and the other on the other channel.

There's actually a 20C and 40C version of those Turnigy batteries. I believe the 20C ones are rated for 3C charge and the 40C ones are good for 5C charge. Typically the higher the drain rate the higher the maximal charge rate. I'm not positive on those specs so check before using high rates.

So if i can pick you brains a little further!

Discharging the pack 75% means you will be putting back in 75% of the capacity on charge. So with a 3300mAh battery, you'll put back in about 2500mAh. Keep in mind this has nothing to do with the demand on the charger other than the length of time it takes to charge. Peak requirement from the charger and supply is always max battery voltage times charging rate plus losses. Also as batteries wear, capacity falls off and internal resistance goes up. I'm sure you've encountered this with worn batteries lacking the power to do those piro flips.

I'm not going to review all of your calculations as it's too time consuming for me right now, but I'll give you some info on how the batteries charge so you can find it for yourself. Li-Ion batteries charge in essentially two phases, firstly CCM (constant current mode) and finally CVM (constant voltage mode). In CCM capacity is pretty much rate multiplied with time, but this isn't the case for CVM. This is the topping phase of the charge cycle where current falls off as the battery reaches terminal voltage. Current input is non-linear in that phase so you would have to integrate over time to find capacity input. Usually an estimate of around 5% of the overall capacity is good enough.

Typically CVM runs 5 to 15 minutes after CCM completes. The higher the rate the shorter CVM runs. You can also use the fast charge mode of your charger to minimize the time CVM runs with a cost of a percent or two in capacity.

 

[cjm181]

Thats for all that craig. really good help there!

 

[CraigHB]

Welcome.

One other note; the charger can run for additional time in CVM when doing a balance charge. Balancing the cells is a function that typically runs concurrent with CVM, but can add additional time to the charge cycle. It's best to balance charge every time, but you don't have to if you balance every few charge cycles. Typically you can select standard charge or balance charge from within the charger's mode selection menu. If you want to absolutely minimize charge time you can do a standard charge at the field and then balance charge at home.