LM317 Ni-Cd charger

I have constructed one LM317T based Ni-Cd charger for 4 series cells. Charging voltage I applied is 6.30V. But in the battery pack, there is one 4007 diode is connected for safety. So there might be some voltage drop. The problem is as the charging is going on the output voltage also increased. I wondered y the current flow indicator doesn't stop. After some hours, I checked the output voltage it was 6.55 V! y this happens! Also what is the maximum charged voltage of a fully charged Ni-Cd AA cell.what is the voltage drop for 4007 I have to add for setting LM317?

transistor495 said:

I have constructed one LM317T based Ni-Cd charger for 4 series cells. Charging voltage I applied is 6.30V. But in the battery pack, there is one 4007 diode is connected for safety. So there might be some voltage drop. The problem is as the charging is going on the output voltage also increased. I wondered y the current flow indicator doesn't stop. After some hours, I checked the output voltage it was 6.55 V! y this happens! Also what is the maximum charged voltage of a fully charged Ni-Cd AA cell.what is the voltage drop for 4007 I have to add for setting LM317?

Click to expand...

hi,
Please post the circuit diagram. so that we can see what you have made.

Hi,
https://www.electro-tech-online.com/custompdfs/2008/08/Circ01-Feb05.pdf
I have removed the auto cut-off section consisting of T1 together with T2. On the emitter of T3 I connected an LED for indicating current-flow. It was working fine for short duration.ie, for charging an almost charged pack. But when there is a longer duration the charge voltage is also going up and LED never turn off. I didnt use a heat sink for IC1. Is that causes instability?
Hope you understood my problem.

transistor495 said:

Hi,
https://www.electro-tech-online.com/custompdfs/2008/08/Circ01-Feb05-1.pdf
I have removed the auto cut-off section consisting of T1 together with T2. On the emitter of T3 I connected an LED for indicating current-flow. It was working fine for short duration.ie, for charging an almost charged pack. But when there is a longer duration the charge voltage is also going up and LED never turn off. I didnt use a heat sink for IC1. Is that causes instability?
Hope you understood my problem.

Click to expand...

hi,
If you have removed the auto cut off section that stops the charge when the battery is charged ,
WHY do you think it continues to overcharge....

If you have connected an LED is series with the T3 emitter, what effect do you think that will have on the current limit.?

Hi,
Removed the auto-cutoff section for minimizing it not bulky. I have replaced it with current flow indicator.When battery fully charged, it will go off. Thats all. I think there is no overcharging because when the batterypack voltage equalizes the charging voltage there will not be any current flow after that.
ie, Charge voltage-diode drop=Actual charge voltage
ie, 6.30-.8(not sure)=5.5
so, when battery pack attains 5.5v, LED turn off for indication so that it can be removed.

ericgibbs said:

hi,
If you have connected an LED is series with the T3 emitter, what effect do you think that will have on the current limit.?

Click to expand...

No effect. Just for indicating that battery is fully charged. Because current limit is determined by series resistor only. T3 based LED circuit is just for charging indication only.

transistor495 said:

Hi,
Removed the auto-cutoff section for minimizing it not bulky. I have replaced it with current flow indicator.When battery fully charged, it will go off. Thats all. I think there is no overcharging because when the batterypack voltage equalizes the charging voltage there will not be any current flow after that.
ie, Charge voltage-diode drop=Actual charge voltage
ie, 6.30-.8(not sure)=5.5
so, when battery pack attains 5.5v, LED turn off for indication so that it can be removed.

No effect. Just for indicating that battery is fully charged. Because current limit is determined by series resistor only. T3 based LED circuit is just for charging indication only.

Click to expand...

hi,
This increase in voltage happens partly because of the series diode on the output of the LM317. The voltage drop across the diode is a function of the current flowing thru it.
As the current into the battery falls so does the voltage across the diode, so the battery terminal voltage goes up.

With the original shut off T1/2 in circuit this couldnt happen.

In original circuit also this may happen I think. Because as the time goes up,the charging voltage on the output of the diode increases. The battery also tries to equalizes with that voltage. So there will be a current flow.LED will be on. When the battery attains its maximum charged voltage, it will go steady. But as the charging voltage is increasing, there will be still a current flow(trying to overcharge). So LED never turn off. Thus T2 will never turn on, or T1 will not go off, in original circuit. Actually what I want to know is y the voltage instability occurs for the IC?

ericgibbs said:

hi,
.......The voltage drop across the diode is a function of the current flowing thru it.
As the current into the battery falls so does the voltage across the diode, so the battery terminal voltage goes up.

With the original shut off T1/2 in circuit this couldnt happen.

Click to expand...

I fear not, except for the knee portion of diode characteristic curve, the diode drop is fixed irrespective of current within its applicable current limit. of course temperature pays a role.

mvs sarma said:

I fear not, except for the knee portion of diode characteristic curve, the diode drop is fixed irrespective of current within its applicable current limit. of course temperature pays a role.

Click to expand...

hi Sarma.
But isnt the circuit working on the knee portion of its Vdp versus Id conduction section.???

Also this statement is not correct the diode drop is fixed irrespective of current as you say,
please look at the characteristic curves for a 1N400x.

EDIT: refer this gif for a 1N4001

transistor495 said:

In original circuit also this may happen I think. Because as the time goes up,the charging voltage on the output of the diode increases. The battery also tries to equalizes with that voltage. So there will be a current flow.LED will be on. When the battery attains its maximum charged voltage, it will go steady. But as the charging voltage is increasing, there will be still a current flow(trying to overcharge). So LED never turn off. Thus T2 will never turn on, or T1 will not go off, in original circuit. Actually what I want to know is y the voltage instability occurs for the IC?

Click to expand...

hi transistor,
Look at the graph in the post to Sarma, ref the diode drop.

As you have modified the circuit, what exactly do you want the charger to do.?

Give us a brief specification, so that we can suggest changes to the circuit.

Hi,
I want to charge 4*700mah Ni-Cd cells using the above circuit.Key points are:
1. It should charge the cell pack into maximum.
2. An LED indicator should be there indicating that the unit is charging..
3. LED should off when the battery is fully charged.
4. Diode protection should be there for polarity changes.
4. I dont want to use any relay or triacs in the circuit for cut-off. When I see that the LED went off, I will manually remove the battery.
Also one question.
what is the fully charged voltage of a 700mah Ni-Cd cell? I'm using Eveready.

hi transistor,
Just seen you last post, I had prepared some notes for you which I will post.

I'll look at the spec you have posted.OK

This is how the designer intended the charger to work.

The voltage drop across R4 due to the charge current controls the conduction state of T3.
When the charge current is high, the voltage drop across R4 is sufficient to make T3 conduct.
The collector of voltage of T3 is close to the +V supply rail, so T2 is not conducting, so T1 base current
flows from the 0V rail via R2.
This makes T1 conduct and the relay is energised.
The closed contact of the relay keeps the mains supply connected to the transformer input.

When the battery charge current falls to a value such that the voltage drop across R4 is too low to keep T3 conducting, it turns off.
This causes T2 to conduct as current now flows thru R3 causing T2 collector voltage to be close to the +V rail.
So T1 stops conducting and the relay de-energises and the mains supply is switched off to the transformer input.

Your modification.
You have removed the relay, T1 and T2 so that there is no auto shut off when the battery charge current falls.
The effect of that is that the voltage drop across D6 falls due to the reduction in charge current, the voltage on the battery slowly rises.

You will have initially set the Vcharge voltage across the battery at 6.3V when the initial charge current was high.
Measure the Vout from the LM317 when the charge current is high with the battery voltage set at 6.3V.

NOTE:
If the voltage drop across D6 is 0.9V at 1 amp it will be close to 0.7V at very low charge currents.
Subtract 0.7V from Vout and that will be the voltage the battery will try to charge to,
it will be 6.3+ [0.9-0.7]= approx 6.5V..

Which is close what you have actually measured.

Hi,
There is one mistake. Vout from LM317 is set to 6.3v. D6 is actually connected inside the battery pack.
So 6.3v-.9v(diode drop)=5.4V is the actual charging voltage coming to battery!!

<quote>
You have removed the relay, T1 and T2 so that there is no auto shut off when the battery charge current falls.
The effect of that is that the voltage drop across D6 falls due to the reduction in charge current, the voltage on the battery slowly rises.
</quote>
I didnt understand that one. If that is correct, then if T1 and T2 is there also the same thing will happen because of D6.What you are telling is when the charging current falls as time goes up, charging voltage will increase because of the characteristics of D6.correct? So u mean I have to remove D6?

transistor495 said:

Hi,
There is one mistake. Vout from LM317 is set to 6.3v. D6 is actually connected inside the battery pack.
So 6.3v-.9v(diode drop)=5.4V is the actual charging voltage coming to battery!!

<quote>
You have removed the relay, T1 and T2 so that there is no auto shut off when the battery charge current falls.
The effect of that is that the voltage drop across D6 falls due to the reduction in charge current, the voltage on the battery slowly rises.
</quote>
I didnt understand that one. If that is correct, then if T1 and T2 is there also the same thing will happen because of D6.What you are telling is when the charging current falls as time goes up, charging voltage will increase because of the characteristics of D6.correct? So u mean I have to remove D6?

Click to expand...

Of D6 is built in , i wonder whether it is not a schottky device with lower drop than expected.

Hi,
Me only constructed that battery pack also . It's actually inside a little audio amplifier box (TA7368P,1W) I assembled whole as a re-chargeable ipod amplifier!

So 6.3v-.9v(diode drop)=5.4V is the actual charging voltage coming to battery!!

I didnt understand that one. If that is correct, then if T1 and T2 is there also the same thing will happen because of D6.What you are telling is when the charging current falls as time goes up, charging voltage will increase because of the characteristics of D6.correct? So u mean I have to remove D6?

Click to expand...

Hi,
The circuit you posted dosnt show where D6 is.
Look at the curve I posted for the 1N4001 diode, that explains whats happening.
I would not remove the diode if its already part of the battery pack.

Have a look at this modified dwg, check that it will do the job.

BTW: the nominal voltage for a NiCd is 1.2V
Look at this link:
**broken link removed**

Hi,
May be this is what I wanted.Thanks . What I have done is ,connected LED in series with a 1K resistor on the collector of T3(Big sorry! Previously I told that I connected LED to the emitter!). I didnt think that it will affect the operation. I think you hav attached the required LED driver. What's the use of 10K pot? For adjusting brightness?

transistor495 said:

Hi,
May be this is what I wanted.Thanks . What I have done is ,connected LED in series with a 1K resistor on the collector of T3. I didnt think that it will affect the operation. I think you hav attached the required LED driver. What's the use of 10K pot? For adjusting brightness?

Click to expand...

hi,
The pot is so that you can set the point where the charge current is low enough to switch off the LED.
If the +Vrail is at about 12V then the 820R will allow about 12mA thru the LED.

Connecting the top of the 820R to the end of the current sense resistor will give a little positive feedback to the LED.

Do you follow that part,?

Hi transistor,
You can use some other charging mechanisms using IC 7806.

ericgibbs said:

hi,
The pot is so that you can set the point where the charge current is low enough to switch off the LED.
If the +Vrail is at about 12V then the 820R will allow about 12mA thru the LED.

Connecting the top of the 820R to the end of the current sense resistor will give a little positive feedback to the LED.

Do you follow that part,?

Click to expand...

I wonder y u didn't connect that 820R before current sendor R4.