Help with a circuit for making LED lamp

depaul73

Joined Jan 7, 2017
2
I want to make a circuit for switching on LEDs as shown in the attached picture. The circuit shown is to just depict what i want to achieve(I know this is not right). A slider which on sliding up will switch on the LEDs. If it is on the first level only one LED lights on. On the second level it should switch on both the lower and the corresponding LED and so on. I have no idea as to how to achieve this. Maybe a circuit as shown below. I really don't know anything about electronics. So a little help will be really appreciated. Thank you in advance.

AlbertHall

Joined Jun 4, 2014
12,396
Each diode will drop at least 0.3V (for schottky diodes) so the bottom LED will see 1.2V loss from the battery voltage when the switch is at the top. That bottom LED will get dimmer and dimmer as more LEDs are switched on. You can minimise that effect by using a higher supply voltage (1.2V lost from 12V, for instance, is a relatively small proportion. Another way would be to use mode diodes. The third switch from the bottom would have a diode to each of the first two, the fourth switch would have three diodes, one to each lower Led and similarly for the top switch. This would mean that each LED would only lose one diode drop from the supply voltage but would require ten diodes.

dannyf

Joined Sep 13, 2015
2,197
I really don't know anything about electronics.
I think you will find it much easier to then start learning the basics of electronics and graduate to building what you wanted to build here.

Even if someone can provide you with a schmatic or two, without knowing anything about electronics, your chance of realizing it will be slim at best.

depaul73

Joined Jan 7, 2017
2
Each diode will drop at least 0.3V (for schottky diodes) so the bottom LED will see 1.2V loss from the battery voltage when the switch is at the top. That bottom LED will get dimmer and dimmer as more LEDs are switched on. You can minimise that effect by using a higher supply voltage (1.2V lost from 12V, for instance, is a relatively small proportion. Another way would be to use mode diodes. The third switch from the bottom would have a diode to each of the first two, the fourth switch would have three diodes, one to each lower Led and similarly for the top switch. This would mean that each LED would only lose one diode drop from the supply voltage but would require ten diodes.
Thanks a lot!

Alec_t

Joined Sep 17, 2013
14,409
The LM3914 integrated circuit in 'bar mode' responds to a rising voltage (which could be derived from a sliding or rotary potentiometer) to drive a column of LEDs as you describe, but additional components would be required to complete the circuit.