Hi,

I made circuit following directions from this link http://www.circuitstoday.com/battery-charging-indicator-circuit

I needed indicator for battery charger on my Bosch cordless drill/driver, because I want to know when the battery is charged.

It says that charger operate on 12V and the current is 400mA, but when measured voltage is near 18V when no battery is inserted, and don't know for the current.

The problem is, when circuit was finished and all plugged in, the IC burn out.

Does anybody have any idea?

Thanx.

 

I believe you made a mistake assembling it because the circuit is valid.

The voltage with no battery plugged in is not very important. It is "enough". Do not be concerned about that. However, you must determine what the current is when the battery is full by measuring it, then change the 1 ohm resistor to the right amount of ohms.

 

Are you sure the positive source voltage was connected to pin 8, that the negative of the charger to pin 4, and that the diode's black band was closest to ground? Are you sure that the positive of the battery was connected to the positive of the battery?

Can you post a pcture of the circuit you built.

 

Can you post a pcture of the circuit you built.

Just imagine a small pile of charred rubble. With smoke coming out.

 

Just imagine a small pile of charred rubble. With smoke coming out.

He did say he toasted the IC, I suppose I was asking for too much...
But plastic can take so many interesting forms, it still may be worth while!


@ gingko master,

Perhaps build it again and take a couple of pictures before you hook it up.
Maybe you can video the initial hookup for Sgt. and I...

P.S. Not picking fun of you, gingko master, actually the fun is at me and my assumption that there would be anything to decipher after the mishap to take a picture of.

 

Thank you for yours very fast replies, folks.
Will post pictures, but don't make laugh with it, ok?

@#12
Is it possible that high current burned out D1? And how to calculate the right value for the R3?

Thank you.

 

Anything is possible to burn out from high current when you make a mistake in assembling.

The right value for R3 is found with Ohm's Law. The voltage needed across R3 to signal finish is about .026 divided by the charging current that is still flowing when the battery is full.

The .026 comes from: finish voltage of battery times (2200/1002200)
again, Ohm's Law.