Battery Charger with auto shut-off feature - Page 7 - Electronic Circuits Projects Diagrams Free

hardcore misery · 10th October 2007, 04:31 PM

it has constant current of 312.5maH, and a voltage of 4.18V, is it correct now sir?, and i will use a 1 ohm resistor as limitter? what if the battery voltage is 3.6V, then the Vdrop is 0.6V, now what is the charging current? is it 600mah? or the constant current of 312.5maH?

hardcore misery · 10th October 2007, 04:34 PM

how about on my previous post? is it correct to put the Vref at the -feedback of the comparator? because the sensing voltage tends to decrease as it approaches a fully charged condition?

hardcore misery · 10th October 2007, 04:39 PM

On the actual defense:
1.we will be charging 3 batteries at the same time, to show the multi-charging feature of the charger...

2.we will charge a battery with at least 4V to 4.1V to show the auto shut-off feature of the battery.

i think the panel of judges will not wait to charge a low voltage battery, because we will tell them that it will take a long time to fully charge the battery(ies).

but we have the statistics on the documentation, and we need to know the charging time of the charger

EDITED: it's also important that when the charger shuts off, the battery voltage should be 4.2V.

audioguru · 10th October 2007, 05:04 PM

Quote:

Originally Posted by hardcore misery

is it correct now sir?

You missed many important things.
I am not a SIR because I have never been near The Queen of England.
I have also never been to India.
Call me Audioguru or Hey You.

Quote:

i will use a 1 ohm resistor as limitter? what if the battery voltage is 3.6V, then the Vdrop is 0.6V, now what is the charging current? is it 600mah? or the constant current of 312.5maH?

If the power supply is regulated at 312.5mA then how can it supply more current??
When the battery voltage reaches 4.0V then the 1 ohm resistor limits the current to omly 200mA. At 4.1V then the current is limited to only 100mA.

Quote:

we will be charging 3 batteries at the same time, to show the multi-charging feature of the charger

Your circuit will charge only one Li-Ion battery cell. Each cell will have a different amount of charge remaining. Each cell will need its own amount of charging time and will need its own charger circuit.

hardcore misery · 10th October 2007, 05:13 PM

Quote:

Originally Posted by audioguru

You missed many important things.
I am not a SIR because I have never been near The Queen of England.
I have also never been to India.
Call me Audioguru or Hey You.

If the power supply is regulated at 312.5mA then how can it supply more current??
When the battery voltage reaches 4.0V then the 1 ohm resistor limits the current to omly 200mA. At 4.1V then the current is limited to only 100mA.

Your circuit will charge only one Li-Ion battery cell. Each cell will have a different amount of charge remaining. Each cell will need its own amount of charging time and will need its own charger circuit.

sorry mr.audioguru for calling you SIR, it's because we treat you as an educator, and all of our educator is called b Sir.(insert name here)

btw, our transformer is only 9V/3A, but using LM317, it can go beyond 9.7V. so mr.audioguru, does it take a long time to fully charge a 4V or 4.1V with these set-up?

hardcore misery · 10th October 2007, 05:15 PM

i think its safe for as to set the Icharge at 10% before shutting off?

hardcore misery · 10th October 2007, 05:48 PM

it's hard to compute for resistors for Vref=0.028V

audioguru · 10th October 2007, 06:16 PM

If the unregulated power supply source is not about 10.5V then the LM317 IC might not regulate properly.

Look at the graphs and guess how long your circuit will take to charge.

Look at the graph and guess at what percentage of a full charge will occur if you stop the charging when the current is 10%.

It is easy to calculate a simple voltage divider to reduce 8.8V down to 0.028V. 15k to 47 ohms gives 0.0275V.

hardcore misery · 10th October 2007, 06:24 PM

Quote:

Originally Posted by audioguru

If the unregulated power supply source is not about 10.5V then the LM317 IC might not regulate properly.

Look at the graphs and guess how long your circuit will take to charge.

Look at the graph and guess at what percentage of a full charge will occur if you stop the charging when the current is 10%.

It is easy to calculate a simple voltage divider to reduce 8.8V down to 0.028V. 15k to 47 ohms gives 0.0275V.

its easy to compute without hysterysis..

audioguru · 10th October 2007, 06:25 PM

You just need a tiny amount of hysteresis.

hardcore misery · 10th October 2007, 06:27 PM

Quote:

Originally Posted by audioguru

You just need a tiny amount of hysteresis.

oh, i thought you told me on the previous posts that the resistive values to be use in inputs and RH, should be large?

audioguru · 10th October 2007, 06:36 PM

The resistor that is used in comparator circuits for hysteresis is frequently megohms. 4.7M into 15k to 47 ohms is a small amount of hysteresis. Even if the divider is 150k to 470 ohms the hysteresis is small. Try it.

hardcore misery · 10th October 2007, 06:42 PM

do you have any software or sites for computing hysterysis? we have a program for this, but i'm just trying to input values, and guessing the outputs, i always come out with 110mV at ON state Vref, while the Vref at OFF state is 3.V, or larger, can't output Vreferences with 20mV or higher

are there techniques? sorry for being dumb, it's 1:50am here, and i have final quiz at 8:30am, do i need to sleep?

hardcore misery · 10th October 2007, 06:45 PM

if there's only a calculator which enables you to input the two Vreferences and the resistive values will come out...

hardcore misery · 10th October 2007, 06:57 PM

so mr.audioguru:

the final setup is to regulate the current into 312maH, and 4.2V

and use a 1ohm resistor for limitter, and set the Vref to 950maH * 0.03 at ON state

while at off state the Vref should be 0.2V... i'll be sleeping now, and i hope i can still review my notes in color transmitter of TV and CRT blockdiagram

we have only 2days left, and the other groups were finished on their prototype, while we are behind

i wonder how god can do miracles on our circuit

thank you sir, i'll be back as soon as i wake up.

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10th October 2007, 04:34 PM	(permalink)
hardcore misery	how about on my previous post? is it correct to put the Vref at the -feedback of the comparator? because the sensing voltage tends to decrease as it approaches a fully charged condition?

10th October 2007, 04:39 PM	(permalink)
hardcore misery	On the actual defense: 1.we will be charging 3 batteries at the same time, to show the multi-charging feature of the charger... 2.we will charge a battery with at least 4V to 4.1V to show the auto shut-off feature of the battery. i think the panel of judges will not wait to charge a low voltage battery, because we will tell them that it will take a long time to fully charge the battery(ies). but we have the statistics on the documentation, and we need to know the charging time of the charger EDITED: it's also important that when the charger shuts off, the battery voltage should be 4.2V. Last edited by hardcore misery; 10th October 2007 at 04:43 PM.

10th October 2007, 05:15 PM	(permalink)
hardcore misery	i think its safe for as to set the Icharge at 10% before shutting off?

10th October 2007, 05:48 PM	(permalink)
hardcore misery	it's hard to compute for resistors for Vref=0.028V

10th October 2007, 06:16 PM	(permalink)
audioguru	If the unregulated power supply source is not about 10.5V then the LM317 IC might not regulate properly. Look at the graphs and guess how long your circuit will take to charge. Look at the graph and guess at what percentage of a full charge will occur if you stop the charging when the current is 10%. It is easy to calculate a simple voltage divider to reduce 8.8V down to 0.028V. 15k to 47 ohms gives 0.0275V. __________________ Uncle $crooge

10th October 2007, 06:25 PM	(permalink)
audioguru	You just need a tiny amount of hysteresis. __________________ Uncle $crooge

10th October 2007, 06:36 PM	(permalink)
audioguru	The resistor that is used in comparator circuits for hysteresis is frequently megohms. 4.7M into 15k to 47 ohms is a small amount of hysteresis. Even if the divider is 150k to 470 ohms the hysteresis is small. Try it. __________________ Uncle $crooge

10th October 2007, 06:42 PM	(permalink)
hardcore misery	do you have any software or sites for computing hysterysis? we have a program for this, but i'm just trying to input values, and guessing the outputs, i always come out with 110mV at ON state Vref, while the Vref at OFF state is 3.V, or larger, can't output Vreferences with 20mV or higher are there techniques? sorry for being dumb, it's 1:50am here, and i have final quiz at 8:30am, do i need to sleep?

10th October 2007, 06:45 PM	(permalink)
hardcore misery	if there's only a calculator which enables you to input the two Vreferences and the resistive values will come out...

10th October 2007, 06:57 PM	(permalink)
hardcore misery	so mr.audioguru: the final setup is to regulate the current into 312maH, and 4.2V and use a 1ohm resistor for limitter, and set the Vref to 950maH * 0.03 at ON state while at off state the Vref should be 0.2V... i'll be sleeping now, and i hope i can still review my notes in color transmitter of TV and CRT blockdiagram we have only 2days left, and the other groups were finished on their prototype, while we are behind i wonder how god can do miracles on our circuit thank you sir, i'll be back as soon as i wake up.