Li-ion BQ2057C charger, help!

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Hi All

I have made a Li-ion charger BQ2057C 4.2V 1A.

I have made my design just like Figure4 on page 10.

The only components i have changed are RSNS =0.1R, so i can get 1A output and MOSFET is IRF7416
http://www.irf.com/product-info/datasheets/data/irf7416.pdf

Other then that nothing is changed.

Ok my problems

Vin can be from 5V to a max of 18V.

But Vout should be no more then 4.2.

So when Vin is 5V, Vout it about 4.6V, and a Vin increase, Vout stablize to about 5V.

To me this seems its not right, at first i though it could be the IC, so i changed them. It made no differnce. But as the Vout is not linear to Vin, i knew the Mosfet is working.

And another , if i plugg in a battery even its full, the Vout, does become 4.2V. But if the battery is low and its charging, the voltage will be about 3.45V, but as increase Vin from 5V to higher, i see battery starts to take more current then 1A.

I now think the problem is with R2. But what effect does this have on the switching. If i am right in saying this value is the problem, how would i calculate to get a correct value?

Regards

DJ

Thanks

Regards

DJ

R2 is just a pull-up for the OC output on CC, the circuit isn't sensitive to its value.

It sounds like it's operating correctly to me. For Li+ power supply designs I assume 5.0V for the maximum supply line voltage for exactly that reason - charging voltage goes well above Vbat. If the internal battery voltage is 4.2V and the internal resistance is 300-400 mOhms, then at 1A charging current you're already at 4.5-4.6V. If you have a connector, clip leads, and such adding a few tenths of an Ohm you can get up to 5V pretty quickly.

How much more than 1A does the charging current go? If you measure the voltage across the VIN and VSNS pins it shouldn't exceed the Vsns spec. If you're at 1.1A or 1.2A don't worry about it. You are probably be within tolerance of the reference and the sense resistor.

In regards to 5V being the max? Why? The Li-ion IC can have 18VMax. I thought this will be like an LDO ?

Now as i increase voltage my current also increase to about 1.65A, and even more.

When i was soldering, i add bit more flux, so when current starts to increase the temp also increase, causing the flux to react.

But as i increase Vin, it starts getting closer 5.2V output, but once Vin gets above 14V, its starts to reduce Vout to about 4.9V.

Using ohms law it understandable why current increase with voltage, but the limiting resistor with Mosfet, should compenstate.

When you say voltage across the VIN and VSNS pins do you mean after the sense res, becuase the reading i get is 4.9V, which because expected as value of resitor is very low.

Dj

Thanks

Regards

DJ

Sorry, my 5V comment was confusing. You typically route the battery voltage around to other voltage regulators, ICs, etc. I assumed 3.0V to 5.0V as the voltage range for that main supply line so I could make sure all the other regulators and ICs hooked up to it were happy. So it wouldn't shock me to see up to 5V on that line.

The charger does act like an LDO. You've got something like 50 C/W thermal resistance on that pass transistor, and 1A of charging current. If you put in 12V and get 5V out at 1A that's (12V - 5V) x 1A = 7W dissipated power. Trise = 7W x 50C/W = 350C rise. You'll toast that pass transistor. Depending on how high your run your input voltage you may need to reduce the maximum charging current for these thermal considerations rather than battery charging considerations. Keep the max Trise below 100C, 150C worst case for a hobby project.

The charging IC monitors the voltage drop across Rsns (this is measured across VCC and SNS) and compares that to an internal reference, 105mV typical . With a 0.1 Ohm resistor, if you exceed 1.05A then you get too high of a voltage across VCC and SNS and so IC detects an over-current situation and starts to shut down that big pass transistor, Q1. As the IC increases voltage to the gate of Q1 it starts to turn off, limiting the current. It's a very simple control loop.

Make sure you have a good connection through R2. You need that for Q1 to start to shut down and reduce current.

Measure the voltage across the VCC and SNS pins. As your charge current increases your voltage drop across this resistor should increase by Ohm's law. When the charge current is way too high, say 1.5A, what is the voltage difference across VCC and SNS pins? Measure this by putting one robe on the VCC pin and one on the SNS pin (or the traces connecting to them, close to the pins)

How are you measuring current?

Hi Ray

Quote:
Sorry, my 5V comment was confusing. You typically route the battery voltage around to other voltage regulators

Well I will be having 3.3V LDO.

Quote:

The charger does act like an LDO. You've got something like 50 C/W thermal resistance on that pass transistor, and 1A of charging current. If you put in 12V and get 5V out at 1A that's (12V - 5V) x 1A = 7W dissipated power. Trise = 7W x 50C/W = 350C rise

By the way at 12V current increase to about 1.6A.

So for Vin 5V 1A that is (5-4.63)*1A=0.37W. So 0.37*50C/W=18.5C, would this be right, when you say rise that is above room temp?

Ok I have measured voltage between the VCC and SNS and at 5V, itâ€™s about 105mV, as I increase voltage in this also increase.

Quote:
With a 0.1 Ohm resistor, if you exceed 1.05A then you get too high of a voltage across VCC and SNS and so IC detects an over-current situation and starts to shut down that big pass transistor, Q1.

This is not happening, what happens is that the current keep increase and IC start to heat up.

By the way I am not use a Transistor but Mosftet.

R2 has got a good connection, but as I increase voltage the gate voltage keep increasing as well. But current does not decrease.

I am sure there is no short circuit because when battery is disconnected them there is not current being drawn.

Yes I have measured voltage be placing probes between the two pins.

I measure current two way,

1 . Power supply shows how much current is drawn

2. Use an Amp meter between the Li-ion Ic and battery

It seems that the Mosfet is not shutting down when Vin increase, could this be due to my resistor value.

DJ
-Update-
I just tired increasing value of R2 to 10K, and it did not make much differnce, i also tried decreasing it to 470R, and noticed that the current reduced as as Vin increased.

I use to use Lm3658 previosuly, i was expecting this IC to work in similar way, where by when output remains stack at 4.2V 1A, but input current increase as Vin decrease and decrease when Vin increase.

DJ

Thanks

Regards

DJ

Anyone got ideas?

Thanks

Regards

DJ

Hi All

I am not sure if this is correct. I had a look at the datasheet of the MOSFET and noticed that the Vgs when set at about 3V, will give let more then 1A pass through.

So at 5V, the sense resistor is keep current below 1A but as Vin increase, current increase rather then staying static. So i guess if i add a series reistor on the gate of mosfet to the pull up reistor, that should improve things.

DJ

Thanks

Regards

DJ

Try the TI forums, there is one for those chips.

Leon

Leon Heller G1HSM

Thanks

Thanks

Regards

DJ

Just back from a trip to Rome on business without internet access. Now I"m getting my fix!

djoshi wrote:

So for Vin 5V 1A that is (5-4.63)*1A=0.37W. So 0.37*50C/W=18.5C, would this be right, when you say rise that is above room temp?

Yes, at 5V you'll be fine. You mentioned flux bubbling around the part so I assume you were in a high dissipated power state which I think would have to be from a high input voltage.

Quote:

This is not happening, what happens is that the current keep increase and IC start to heat up.

By the way I am not use a Transistor but Mosftet.

R2 has got a good connection, but as I increase voltage the gate voltage keep increasing as well. But current does not decrease.

A MOSFET is a type of transistor, and a P-Channel MOSFET is the type you are using and what you need.

When you increase supply voltage the gate voltage should increase roughly equal to it. So if you go up by 1V on Vin then gate voltage should go up roughly 1V. This keeps a constant Vgs and therefore constant current through the FET.

Quote:

Yes I have measured voltage be placing probes between the two pins.

I measure current two way,

1 . Power supply shows how much current is drawn

2. Use an Amp meter between the Li-ion Ic and battery

It seems that the Mosfet is not shutting down when Vin increase, could this be due to my resistor value.
DJ

-Update-
I just tired increasing value of R2 to 10K, and it did not make much differnce, i also tried decreasing it to 470R, and noticed that the current reduced as as Vin increased.

Since you are seeing gate voltage increase with Vin it sounds like R2 is doing its job.

I would probably try something like this if I had the circuit here in front of me. Increase Rsns to 100 Ohm. Then put a potentiometer on the output in place of the battery. 100 Ohm Rsns should give 10 mA current limit, and a 100 Ohm load would allow up to 50mA to be drawn (5V / 100 Ohms). So use something like a 5k pot. If you only have a very low power pot then you might need to put a 1k fixed 1/4W resistor in parallel with a 5k pot or something like that.

Start with the pot all the way open. At 1k Ohms you'd see 5mA (5V / 1k). Slowly reduce Rpot and watch the current increase and see where, if ever, the current gets limited by the IC.

This will let you work with a low power circuit without worrying about damaging a Li+ battery. And you'll have a circuit where you can calculate and control exactly what should happen.