NIMH Charger LM317T - Problem

[Suraj143]

Here is the charger circuit for my NIMH batteries. The problem is every time I see the batteries are charging with a very limited current some thing like 50mA and decreasing current. It’s not charging with the R3 calculated value 250mA.

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But when I charge with this circuit it’s charging with the calculated value (250 mA) and decreasing the current.But when I measure the voltage while charging its showing 3.2V.

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I want to use the first circuit.I adjusted the R3 but no progress.Please help me.

 

[ericgibbs]

hi suraj,

Recheck the value of the 3R3 for 250mA.
You require approx +0.6V across it at 250mA

The divider chain resistors seem a little low?

The first circuit is voltage controlled, with a current limit, so the current will fall off as the battery charges.

The second is not voltage limited, its current limited, so it will keep driving current into the battery, not a good idea.


Regards

 

[Suraj143]

Earlier I adjusted the divider resistors something like R1 - 240ohms & R2 - 374Ohms.But it didn't give me a result.I have been testing this circuit for many days please help me.

 

[ericgibbs]

hi,

What is the voltage across the battery when you are using the first circuit. Are you sure the battery is not almost fully charged ??

Whats the voltage at the junction of the 120R and 159R??

The 3R3 should be 2R2, you have already 10mA from the divider chain flowing in the Rs resistor [the 3R3].

The voltage across the current sense resistor should be 0.6V at the charge current you are designing for,
the 3R3 will drop 0.6 at 182mA, take away the 10 mA from the divider and you are left with only 170mA.

Dont use the second circuit, it could over charge the battery.

EDIT: Allow for the small voltage drop across the meter, when checking the circuit.
If the 'top' resistor is 240R then the 'bottom' resistor should be about 300R. [use a 500R pot]

 

[Suraj143]

Hi Eric thanks for your great input.

There was a problem occurred when I was measuring with my multimeter. Now it seems ok I believe.
I adjusted the current sense resistor to 2.2 ohms as you suggested.

These results I got when measuring. (Voltages across the resistors when charging the batteries)

Ex – 1

Battery voltage = 2.86V (batteries almost full charged condition)
R Sense = 0.1V
R1 = 1.22V
R2 = 1.24V
Charging Current = 5mA

Ex – 2

Battery voltage = 2.71V
R Sense = 0.62V
R1 = 1.22V
R2 = 1.48V
Charging Current = 240mA

I have some doubts about charging please help me. Sorry to trouble you.

Questions

1) Is my values ok?

2) When my batteries dead it’s showing 1.1V to 1.2V not 0V.what is the correct voltage after discharge the batteries?

3) I want to discharge my batteries how can I discharge my batteries otherwise it will take long time to check with my circuit. Is there a separate circuit for discharge as well?

4) If I charge half charged batteries with this circuit it’s charging with the calculated current only few minutes, when the voltage rise the current limits. So all the other hours it’s decreasing the charging rate. (Charging with the calculated current only with few minutes)

When the battery voltage equals it’s charging with the voltage. No current sending to the batteries but meter reading is (5mA-10mA).

5) How did you calculate the 10mA from the divider chain? & the other values 182mA?
Thank a lot
Suraj

 

[ericgibbs]

>> Questions

1) Is my values ok?
They look nominal to me.

2) When my batteries dead it’s showing 1.1V to 1.2V not 0V.what is the correct voltage after discharge the batteries?
A battery should NEVER be discharged to 0V, else it will be damaged.
On the web look for charging/discharging NiMH batteries, a graph showing the range of voltages which the battery must be operated.

3) I want to discharge my batteries how can I discharge my batteries otherwise it will take long time to check with my circuit. Is there a separate circuit for discharge as well?
You can discharge your batteries using a suitable resistor, I would suggest that you discharge at the C5 rate, look at the web graphs.

4) If I charge half charged batteries with this circuit it’s charging with the calculated current only few minutes, when the voltage rise the current limits. So all the other hours it’s decreasing the charging rate. (Charging with the calculated current only with few minutes)
I would expect this to happen, you cannot be sure where the battery is on its charge/discharge curve.
Some users recommend that you fully discharge the battery, before recharging. Its important that you use the recommended charge/discharge rates if you want to maximise the life of the battery.

When the battery voltage equals it’s charging with the voltage. No current sending to the batteries but meter reading is (5mA-10mA).
This current is a residual current and should not harm the battery.

5) How did you calculate the 10mA from the divider chain? & the other values 182mA?
The 10mA is from ohms law, you know the voltage across the divider and the sum of the dividers resistors and I=V/R

The 182mA, is calculated from the +0.6Vbe transitor 'turn on voltage' and and the 3R3, ohms law again.

EDIT: added 2 *.gif from web. Remember you have two cells in series.

 

[ericgibbs]

hi suraj,

>> But when I charge with this circuit it’s charging with the calculated value (250 mA) and decreasing the current.But when I measure the voltage while charging its showing 3.2V.

So that you can easily see that you are overcharging using the second circuit, look at the attached gif.

I have rescaled thegraphs to suit two cells in series.

 

[Suraj143]

Hi Eric thanks for your great help. Now I really understood about this circuit.
My problems solved.

Thanks a lot

Eric is best.

 

[Suraj143]

Current Sensing

Hi Eric I have a small question about this circuit .Can you please take a look.

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Here is a bulb added for the charger circuit. When charging the bulb turns ON.

D1 / 22R = 27mA .When R Sense passing 27mA the bulb turns OFF.
When I add a 47R then sense current to the bulb will be 12mA.So when R Sense passing 12mA the bulb turns OFF.

My question is how this Q2 transistor detects the current or voltage to turn OFF the bulb.

 

[ericgibbs]

hi suraj,

At the bottom left of your drawing , showing 0V, that is not 0V, its the negative supply input.

All the charger circuit current is flowing thru the D1 diode and 22R resistor combination.
So Q2 will be biased ON if sufficient current flows thru the 22R, the Lamp will be switched ON.

The 470R [Q2 base] is connected to the 'high' end of the charge current sense resistor, the 2R2.

So as the battery becomes fully charged the Rsense current falls, so the voltage across the 2R2 falls. also
the voltage drop across the 22R also falls and the Lamp is quickly switched OFF.

The D1 diode is used to clamp the voltage drop across the 22R. to 0.7V.