# MCP73844 Helping with charging current formula on datasheet

I am trying to design the needed circuit to use this part.

I don't quite understand the formula to figure out your charging rate. The example shows 100m ohm for 1A charge.

I can't tell if I can charge at a higher current than that.

Thanks for the help. I have been reading the datasheet over and over and need a little help.

I download the pdf file. I quickly read the datasheet and try to figure out the section you have a problem with. In the datasheet, it show two example ( 1 A @ 100 m-ohms ) and ( 500 mA @ 220 m-ohms ) . The formula is simply V = R I. V is Vsense , R is R sense and I is the charging current. The internal circuit show this section a op-amp, measure the V sense. When I calculated V sense, it is both 0.1 V so depend of the current you want going toward the battery, and V sense is 0.1, calculated R sense.

Example : 2 A charging = I, V sense = 0.1 so R sense is 50 m-ohms. P is 0.2 Watt so you need a 0.05 ohms @ 1/4 Watt ← I pick a 1/2 watt - safer. You may have a hard time to find that resistor…

Thanks!

So the input voltage doesn't matter? I will be using 12v 5A regulated power supply.

So what limits the charge current? heat?

How did you calculate V sense? is it because of the 1k resistor in the internal circuit?

so .1/2 which is .05 oh okay...

Will be using this battery

http://www.all-battery.com/li-ion1865074v6600mahrechargeablebatterymodulewithpcb.aspx

Capacity 6600mAh Voltage 7.4V, peak at 8.4V Max. charge current 3A Max. discharge current 6.6A

Found this

http://www.mouser.com/Passive-Components/Resistors/_/N-5g9n?P=1z0vod6Z1z0x7b2Z1z0z7ym&Keyword=0.05+ohms+resistor&FS=True

First one would be find?

Then with this P-channel MOSFET?

http://www.mouser.com/Search/ProductDetail.aspx?R=IRF7404TRPBFvirtualkey57370000virtualkey942-IRF7404TRPBF

Plus .1uf Cap and .10uf cap per the example

looks like i will be set for 2A charging.

Let see... MOSFET 6 A handle -> OK R sense 0.05 1/2 W at 1% ---> excellent... 2A setting --> OK Supply is 12 V @ 5A OK but the chip is Vdd max = 13.5 V ---> caution --> yellow Battery to be charge : a 7.4 V peaking at 8.4 V The chip MCP73844 is design for a 8.4 V charging voltage ----> Caution --> red

my opinion...

Techone: Let see... MOSFET 6 A handle -> OK R sense 0.05 1/2 W at 1% ---> excellent... 2A setting --> OK Supply is 12 V @ 5A OK but the chip is Vdd max = 13.5 V ---> caution --> yellow Battery to be charge : a 7.4 V peaking at 8.4 V The chip MCP73844 is design for a 8.4 V charging voltage ----> Caution --> red

my opinion...

The PS is a solid 12v so should never go over that. This is what I found... unless you can find a better one. (I have the 5v 5A one it is great) http://www.gadgettown.com/Switching-Power-Supply-12V-5A.html

Well from what I understand on lithium 's (i have 2 on my current robot, rebuilding it) is that a 3.7v is fully charged at 4.2v so it charges at 4.2v. The 2S one is 7.4 but fully charged at 8.4. which is why the charger is at 8.4v, is that right?

I didn't post this, the battery would protect it self.... Cut off voltage Over-Charge Protection: 8.5V Over-Discharge Protection: 5V

I would be charging 2 batteries possibly with this power supply.

1 battery for logic and one for motors with the charge embedded. Depends on how much space I will have.

Well from what I understand on lithium 's (i have 2 on my current robot, rebuilding it) is that a 3.7v is fully charged at 4.2v so it charges at 4.2v. The 2S one is 7.4 but fully charged at 8.4. which is why the charger is at 8.4v, is that right?

Well. the 8.4 V can from the datasheet of the chip. the 844 model is setup at 8.4 V. The other chip number 841, 842, 843 are setup differently. ← To my understanding of the datasheet. Depend of the chip number you have access ← I thinking you trying to “hack it up” ← my guess , you are stuck for the charging voltage, but you can modify the charging rate ( 1 A, 2A 3 A ← what ever )

My opinion is : caution… you don’t want to “burn”, “blow up” or “damage” the battery being charge. I recommend to experiments and see what happen. I am just an hobbiest who went to DeVry in 1990’s , work as a tech for 3 years in the 1990’s and I try to re-kindle the passion I have with electronics. I am not familiar with this chip, but commun sense tell me : caution and experiments.

Well worse case I guess the protection PCB would kick in.

Hard to test with the tiny IC. I have the 840 coming in which is the 8.4v one. I got a few samples.

How did you calculate the Vsense?

What is their example 100mOhm? .1/1=.1 Hows does that equal 100mOhm?

Hard to test with the tiny IC. I have the 840 coming in which is the 8.4v one. I got a few samples.

Yes that is the correct one to use. You match the chip to the datasheet of the battery you are using and according to the website you would want to use the 8.4v setting.

looks like i will be set for 2A charging.

Just because you CAN charge the battery at this rate doesn't mean you have to. Unless you need to rapidly charge the battery for some reason you may want to scale back the charge rate to some thing lower. Keep in mind the higher the current the larger wattage components you'll need to use. Plus the IC itself uses a ground plane under the IC to help dissipate the heat, so the higher the charge current the larger the are will need to be.

Thanks, I think I am leaning towards the 1A, even though from experience a 6600mAH battery takes forever to charge at around that current rate. I could do something like 1.2 or so, like the Adafruit charger.

I still like to understand the formula. I don't see how it comes out to 100Mega Ohm. And how Vsense is calculated to .1

It is milli ohms. Not Mega. Vsense is 0.1 V The chip has to see that voltage to operated. Check the internal circuit of the chip. It is using a op-amp. So the chip is consern, Vsense is a constant of 0.1 Volt. You simply use V = R I to find the rest. In this case, the I is know, so find R. I see two explample in the datasheet.

Thanks, I thought so, makes more sense milli... Should of followed instinct ;)

Does it show the .1 in the datasheet or is the given being a op-amp.

I will try it again at home...

Thanks.