Battery Charger with auto shut-off feature

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yes sir, this is our main design project...

just a few questions sir, to be able to supply 3 cellphone batteries with
3.7V/900mAH, do we need a 3A ac-dc transformer? as for the output voltage, we can only choose a value, greater than or equal to 3.7V, to accommodate 3 cellphone batteries?
 

Ok i get what he wants to do...

He wants to stop the charger from consuming standby power once the phone is charged he trying to make a "greenie product". Mmmm nice idea now to steal it !
 
we can only choose a value, greater than or equal to 3.7V
Click to expand...
Well, actually it would be safest to use the same voltage that the original wall-warts supplied. You might still be in danger unless you know if the wall-wart was current limited.
 
Supposedly, the Vref of the Voltage comparator is 4.2V ( voltage of a fully charged cellphone battery )

The rule of the comparator is to TURN ON when the –V is lower than the Vref (+V)
And TURNS OFF when the –V reached the voltage of the Vref…

On the circuit provided, the –V of the comparator is tapped at the +V of the cellphone battery (this is intended to monitor the battery voltage)

The LED before the transistor (C828/npn was used) indicates the status of the comparator.

Our main problem is this… at NO LOAD condition, the relay tends to turn ON/OFF at rapid motion(repeating process,on/off/on/off/on/off…..)
I’ve understand this because, at no load, the comparator senses the charging voltage instead of the battery voltage which turns OFF the relay (charger voltage is 4.2V and Vref is also 4.2V which turns OFF the comparator). So if the relay is OFF, the comparator senses a ZERO voltage since there is no load… BUT

When we’ve placed a cellphone battery which has 3.8V, again, the relay turns ON/OFF at rapid motion.. this is not expected, because If theres a 3.8V battery load, the comparator will sense a lower voltage than 4.2V Vref, which turns ON the comparator, but why does the relay turns OFF? We also tried to use a variable DC supply as a BATTERY TO BE CHARGE.. we’ve set its voltage to approx 3V, again the relay behaves as is..

What seems to be the problem? On our circuit, we did’nt put a Diode in parallel to the relay. And we did’nt set a hysteysis for the comparator… are there wrong connections on the circuit provided?

Are there any wrong connections on grounding? We’ve also observed that when the ground of the battery charger is not connected at the ground of the Power supply of RELAYS and COMPARATORS, the ON/OFF switching of RELAY is very fast, but when we connect both grounds of the charger and the power supply, the ON/OFF motion of relay was decreased

NOTE: when we pulled out the relay, the comparator behaves as what we have expected.

i hope that all i've said is clear...

thank you and godbless to all.
 

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hi,
Some obvious changes I would suggest.
There maybe others.

Whats the Vref at the junction of 1K/470R, inp to Comp supposed to be?
 
Last edited:
ericgibbs said:
hi,
Some obvious changes I would suggest.
There maybe others.

Whats the Vref at the junction of 1K/470R, inp to Comp supposed to be?
Click to expand...


the Vref should be 2.1V ( 2 x 2.1V = 4.2V )my apologies on the wrong values of resistors, because i did the schematic quickly... on the actual circuit, the ratio of resistors on +V of comparator is 1k/270 ohms, and the Voltage of the power supply is variable, on the circuit, we've came up with a 9.7V making the vref=2.06V (4.12V)

if you don't mind sir, what is the reason why should an 8mA be chosen to flow on the LED?(on our circuit, 9.7 / 2k = 4.85V.... 4.85V/1k = 4.83mA, which drives the LED and the NPN transistor)

on the hysteysis side, my friend suggest me to add a series diode(1n4148) after the RH, for one direction flow of current.is it correct?

what capacitive value should we use in parallel to the power supply on your edited schematic?what is the purpose of that sir?
 

Does that answer you questions?
 
ericgibbs said:
Does that answer you questions?
Click to expand...

i had a different understanding on the concept of Hysterysis, based on what my friend told me, and you can give me sir some clarification about this...

if the resistor ratio is 1k / 270, and the input voltage is 9,7V, the Vref will be 2.06

if i will add an Rh equivalent to 2k ohms, the Vref will change AFTER the comparator turns OFF, which is equal to 1.86V (3.73V)

so if theres an hysterysis, the RH would only parallel to 270 ohms if the comparator shuts down, but in ON state of the comparator, only 1k and 270 ohms are used by the comparator..

i think what you mean by 0.3V hysteryssis is 2.1V (4.2V) - .3V is equal to 1.8V (3.6V) is this it? i'll report back soon, after we tried those suggestions of yours..

so is it the voltage spikes which is the culprit of our problem?

godbless and thank you sir.
 
hi,
Look at this drawing.

It should give you an idea what the hysteresis is doing.

Remember to calculate the Vref for a almost 'discharged' battery and the hysteresis to ensure a fully charged battery before charger shut off.

Is this clear?
 
Last edited:
can't understand it fully (on the Vref=1.5V, this should be 2.1V?)

ok sir, we tried all your suggestions ( shunting a 47uF and 0.1uF capacitor across the supply voltage for RELAY and COMPARATOR, and shunting a Diode across Relay, and finaly, putting a 2.2kohms as hysterysis to the circuit)

we did some measurements on the circuit.

1. Power supply voltage for Relay and comparator ( so as the voltage input for Vref ) is equal to 9.484V and the Vref is equal to 2.033V ( ideally it should be 2.1V )

2. The charging voltage is 4.02V ( 2.01V as a result from 1k/1k ratio for -V of the comparator )

as i've stated on the previous post(s), at no load condition
( if the Charging voltage is 4.2V while the Vref is equal or less than the charging voltage(2.1V), the Relay switches on/off rapidly )

after putting those capacitors and diode as you've suggested on our circuit, the problem didn't removed.....

3. We observed the effects of adding hysterysis on the circuit.

at the +V of the comparator, the ratio of resistors are 1k / 270 ohms.. the supply voltage is 9.484V and it gives a Vref of 2.033V (without RH
but when the 2.2k was connected between the output and 1k/270, the Vref increased from 2.033V to 2.33V and when the comparator is off, the decreased from 2.33V to 1.895V
i don't know the computation for the Vref at ON state of the comparator, but on the OFF state of the comparator, the 2.2k was first paralleled to 270 (which is equal to 240.49ohms) and the ratio was 1k/240.49 then the Vref becomes 1.84, ** the resistors has tolerances, we didn't measure its actual resistances and 1.84 is close to 1.89V(actual Vref at OFF state)

so if the Vref increases due to 2.2k added resistor for hysteresis at ON state(2.33V) therefore the voltage required to turn off the comparator is greater than 2.33V?(4.66V, which should be 4.2V)

so this is another problem? the Vref should not be changed whether there is a hysteresis on the circuit(except for the Vref at OFF state of the comparator)

ok we used the circuit with added components as you've suggested, with a Vref of 2.33V, we tried to increase the charging voltage until it exceeds the Vref, AGAIN the Relay switches back and forth, rapidly, hence there's still a problem.

hope that all i've mentioned here is clear to you sir...

will for the reply... thanks and godbless..
 
hi,

can't understand it fully (on the Vref=1.5V, this should be 2.1V?)

I will study your last post and reply later.

Do you understand the diagram?

EDIT: Adding revised drawing.

I would suggest that you add a variable resistor in the junction of R3/4, this will give you some control while testing.

Look at this drawing, hope it explains OK.
 
Last edited:
i had a wrong understanding about hysterysis, on our actual circuit.

we used 1k/270 ratio of resistors for the +feedback, and to configure the Vref to 2.1V... i was wondering why the Vref increases as the RH is added on the circuit... i realized that, we should include the RH to configure the ratio of resistors for Vref... i'll reply soon as after we revised our circuit.

thank you and godbless sir.
 
hardcore misery said:
i'm having a hard time on setting the Vref(ON state) to 2.05V and Vref(OFF state)=3.6V
Click to expand...

WHY are you trying to set the Vref at these voltages?

Look at the drawing I posted explaining the Vref and hysteresis.

Are you using a variable resistor, to set the Vb/2 or Vref as I have suggested?
 
ericgibbs said:
WHY are you trying to set the Vref at these voltages?

Look at the drawing I posted explaining the Vref and hysteresis.

Are you using a variable resistor, to set the Vb/2 or Vref as I have suggested?
Click to expand...


oops, it's a typoerror it should be
Vref(ON state) to 2.05V and Vref(OFF state)=1.8V

quick question sir, how can we properly discharge a li-ion battery? our problem is we need to discharge the batteries at approx 2.8V or lower than 3.7V.. if we can't achieve that, we have a problem on the Vref at OFF state(1.5V as you've suggested, or 1.8V)

i haven't tried those potentiometer on vb/2, i'm trying to get fixed values of resistors.. thanks sir
 
sir i think theres no problem on the ratio of resistors for vb/2, (1k/1k) the problem is on setting the Vref of ON/OFF states with the correct values of resistors
 
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