I made a battery charger with some free samples of the maxim max713 chip.

Chip details.

Datasheet.

The 713 supports both NiMH and NiCad while the 712 is for NiMH. Both can fast charge up to 16 cells and support V/T, temperature and time out charge cut off. It can charge the battery while still powering the load. If you don`t need NiCad support get the 712, it`s better for charging NiMH safely because of the way it measures the cut off voltage.

The basic linear circuit is fairly easy to make up, you must use the formulas here to find the values for R1 and Rsense though. Do this with the biggest battery pack you want to use.

1. Choose how many cells to charge. Minimum Input Voltage = Number of cells x 1.9 + 1.5
2. Find out R1. R1 powers the chip. R1 in ohms = (Minimum Input Voltage - 5) / 0.005
3. Decide on a fast charging current. Ifast in mA = Battery capacity in mA / Charge time in hours
4. Find the Rsense resistor. Rsense in ohms = 0.25 / Ifast in A
5. Set PG0 and PG1 to the cell number according to datasheet Table 2.
6. Set PG2 and PG3 to set the cut off time according to datasheet Table 3. Cut off should be slightly higher than charge time.
7. PNP power dissipation. PDpnp =(Maximum Input Voltage - Minimum Battery Voltage) x Charge current in A Check this against the PNP datasheet. This is wasted heat and depending on your cell count range you will need a heatsink and/or fan.

For my charger I chose up to 6 cells (the picture shows jumpers up to 8 cells but it`s not wired up yet). Fast charge current and Rsense aren`t set in stone because they can change if you charge different capacity battery packs.

1. Minimum input voltage = 6 x 1.9 + 1.5 = 12.9v
2. R1 = (12.9 - 5) / 0.005 = 1600. I picked the next lowest resistor at 1.2k.
3. Ifast = 2500mAh / 2 hours = 1250mA.
4. Rsense = 0.25v / 1.25A = 0.2 ohms.
5. PG1 and PG0 both unconnected.
6. PG2 connected to BATT-, PG3 connected to REF pin. With a charge time of 2 hours, the timeout is the next highest at 132 minutes. There will be losses through heat so it`s fine. Also voltage slope cut off is enabled to turn off automagically when the voltage stops rising.
7. PDpnp = (13 - 4) x 1.25A = 11.25W. 2N6109 maximum PD is 40W but it gets lower as it gets hotter. For every degree C above 25 minus 0.32W from 40. If I think it could get up to 60 degrees.. 40W - (60-25) x 0.32 = 28.8W max power dissipation. Well over 11.25W.

Actual charge current above is about 900mA because the fan sucks up a bunch. The jumpers and resistors are really fiddly, I still have to get around to putting the temperature probes on it, changing the jumpers to dip switches or rotary switches, adding 8 cell battery support, and mounting it in a box. The current charge labels on the right are only accurate for a 1000mAh battery but it gives me an idea of what kind of charge rate to expect.

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An Rsense of 0.5 substituted into the calculation in step 4 gives an Ifast of 0.5A. Work backwards into step 3 to find charging time 2.5A/0.5A = 5 hours.

If your batteries are connected in series the voltages add but the capacity doesn't change. In this case your battery pack is 4.8v 2500mAh and your calculations are right as long as you set a cell count of 4 with PG0 and PG1. This is most likely what you have.

If your batteries are connected in parallel the capacity adds up but the voltage doesn't, and in THIS case you have a 1.2v 10000mAh battery and it would take 20 hours at 0.5A.

Can anyone send me a copy of the schematics?

Thanks and regards...

Joseph

Hmm better late than never but the schematic I used is in the datasheet. I just incorporated the LED outputs into it which is also detailed in the datasheet and some jumpers for PGM0/PGM1 adjustment.

CtC made a great looking PCB for it which can be found here. He might let you see his design files if you ask him.

Hello ezekiel181, I was wondering which pins you connected the two led's to and the resistors connected to them.

Thanks

The two LED's and the corresponding current-limit resistors are shown clearly on the schematic(s) in the data sheet. One is power, one is high charge.

I would like to know, if I am applying 12V input to the circuit to charge 6*1.2V @ 800mAh batteries, what voltage should i expect at the output i.e(across the batteries) as am getting 11.5V. Is this right or have i connected something wrong.

Seems high. Double check PG0 and PG1 is set for 6 cells and not 7. Also check R1 allows a minimum of 5mA into the chips supply pin or weird stuff can happen. You should probably get about 10v across all the batteries while they are charging.

Hello Ezekiel,

I would like to thank you very much :)

I was just starting with my project for Battery-Nimh charger where I had to charge a battery of 12V 6Amp battery. I had no clue where I should start with but luckly found your post about Max712. It was really helpfull and it  came to know about lit of things about charging the battery. But still there are some douts like

1. I have 1 Battery which is 12V 3AmpAs and I read voltage of one cell in Nimh is 1.2V. Which means I have 10 cells is this right?

2. But I need to charge 12V 6Amp so I decided to put two battery of 12V 3Amp in Parallel in order to get 12V 6Amp. So now I have a confusion about the no. of cells. How much cells do I have now after putting it into parallel? Must be still 10 right as the voltage remains the same 12V?

3. In order to charge a Battery of 12V 6Amp how much voltage do I need at the output of the Max712 (i mean in the Bat terminal)? The battery I have (12V 3Amp) its datasheet says that it requires a charging voltage of 18V, So do I need to maintain the 18V at the output?which also means I need to have 1.5V more (that is 19.5V) in the input as mentioned in the datasheet of Max712?

4. The last thing you mentioned this (Minimum Input Voltage = Number of cells x 1.9 + 1.5) what is 1.9 refers to? is this the voltage of one cell?

I am really happy to start this project and I am looking forward to hear from everyone regarding my confusions. I hope someone can help me in this :)

Milan

1. Yep 10 cells.

2. Still considered 10 cells for battery voltage purposes but the current capacity is doubled so will take twice as long to charge.

3. See below

4. This is your minimum power supply voltage which in your case will be 20.5v. Pick the closest higher voltage you have available and do the rest of the calculations. I would be careful of the power dissipation in your pass transistor at this high a cell count. Don't worry about your input voltage being higher than 18v, the battery will never see more than about 17v anyway.