Emergency lamp

[Bumblebee001]

Sounds like your ammeter is not working properly. Did you check it against a known current?

I would not know how to do that. I am a hobbyist with limited electronics knowledge. I can try a LED with known specs connected to a battery and the ammeter and see what it reads. Will that help? Do I attach a resistor?

 

[Harald Kapp]

A resistor is the better choice. A LED should never be operated without some form of current limiting.

 

[Bumblebee001]

A resistor is the better choice. A LED should never be operated without some form of current limiting.

Sadly my ammeter (if I'm using it correctly) is not registering any current in a circuit using a LED, batteries and a resistor that came with the LED. The LED does light up. :-(

 

[Bumblebee001]

I noted that there are 2 boards in the unit, one with 21 lamps connected to another with 42 lamps. I suspect the LEDs in the emergency lamp are wired up in 3 groups of 21 lamps each. I'll have to open it all up again and follow the tracks on the board they are soldered to.

 

[Bumblebee001]

Sounds like your ammeter is not working properly. Did you check it against a known current?

Ok... there was a burnt fuse inside the ammeter. The reading now is around 0.30A (300mA)

 

[Bumblebee001]

Ok... there was a burnt fuse inside the ammeter. The reading now is around 0.30A (300mA)

Since the charger delivers 360mA to the 3.6V battery pack, then I guess we have to work on the 360mA as the max current value.

 

[Harald Kapp]

The FDC658 is good for 30V and 20 A, plenty of headroom for your application while still being rather small.
Or use an NDP6020 which is through hole and may be easier to solder.

 

[Bumblebee001]

The FDC658 is good for 30V and 20 A, plenty of headroom for your application while still being rather small.
Or use an NDP60so which is through hole and may be easier to solder.

The first has 6 leads and the second has only 3. Besides is this the component?

 

[BobK]

when I turn the switch of the meter to the 200mA setting I get 0.00mA when I should be getting 120mA)!

The fuse in your multimeter is blown. There are two of them, one for the 10A and one for the mA range. The mA one is blown.

Bob

 

[Harald Kapp]

The first has 6 leads

Becaus 1+2 and 5+6 are connected in pairs to source and drain, respectively.

Besides is this the component?

It is.

The SMD component is less expensiove, though.

 

[Bumblebee001]

Hi Harald,

I finally received the components you recommended. May I kindly ask you now to guide me on how to translate the circuit diagram into an actual breadboard with components I can solder correctly?

I am grateful for your patience and assistance in anticipation.

 

[Harald Kapp]

Find the connections onthe transistors (base, emitter, collector or source , drain, gate, respectively) from their datasheets. Arrange the components on a prototyping board like this one:

The arrangement should follow the layout of the schematic (the way I drawed the schematic all components are connected by short stubs of wire only). You may need to rotate components depending on their pinout (datasheet). Then connect them as per the schematic by soldering between the pins using insulated wire (magnet wire will work, too, you only have to remove the insulating laquer at the end of the wire).

Watch out for the view whether the component is shown in top view or bottom view when assigning the pins from teh datasheet pictures. It is easy to get confounded. Double check to have the right leads in the right positions.

Here's a tutorial that may get you started: http://www.instructables.com/id/How-to-Prototype-Without-Using-Printed-Circuit-Boa/
You may like this one even better: http://makezine.com/2010/06/24/circuit-skills-perfboard-prototypin/

Does this help?

 

[Bumblebee001]

I've come up with another solution for replacing the relay:
View attachment 34485
When external power is present, Q2 is on, pulling the base of Q1 to 0 V. Q1 is therefore off. The gate of the P-MOSFET M1 ist pulled to source potential via R4. Vgs= 0 V and therefore M1 is off. The lamp is off.
When external power fails, R5 pulls the base of Q2 to 0 V. Q2 is offf, tehrefore Q1 is on via the base resistor R3. The gate of M1 is pulled to 0 V via R1, Q1 and M1 is on. The lamp is on.

The switch in series with M1 is used to disable the automatic.
The swizch in parallel to M1 and the other swithc is used as a manual override to turn on the lamp regardless of the state of the external power.

Can you please explain the 3 leads coming out of the "charger"?

 

[Bluejets]

It is symbolic.(block diagram) Most chargers require an input, an output, and a ground to operate any onboard regulator and sensing system.

 

[Harald Kapp]

Most chargers require an input, an output, and a ground

Your charger may have 3 leads as described by Bluejets or 4 leads, in which case you have 2* input and 2* output. Typically one of the 2 leads in input and output is cmon ground, so in effect there are 3 different potentials (having 4 leads is simply more convenient to wire up).

 

[Bumblebee001]

Your charger may have 3 leads as described by Bluejets or 4 leads, in which case you have 2* input and 2* output. Typically one of the 2 leads in input and output is cmon ground, so in effect there are 3 different potentials (having 4 leads is simply more convenient to wire up).

Is there a way to identify the ground? What should I look for?

 

[Harald Kapp]

Ground is '-', a blue or black wire or a label on your charger.