Why do they use a voltage divider in this instance?

[cramar]

Hi everyone, this is my first post so be kind

As I approach retiement I thought I would learn a new hobby. Electronics and Microcontrollers it is.
I am self taught (using resources from the internet) so I hope I am using the corect terminology.

Below is a circuit I have modified for my own needs - It is used to back light an instrument panel.
The backlighting will be turned on by a microcontroller but I know that the maximum current of a GPIO pin is in the range of 40mA. Therefore I am using a 12 volt supply to power the back lighting.

I have included my calculations for resistors in the image below - I have been doing this in my schematics for the time being just to force me to think about what i am doing.

Anyway to my question...

What is the purpose of the voltage divider in the reference circuit.
If I omit the voltage divider (R1 & R2) it simply raises the value of R3 - and saves two resistors.

I suspect there is another logical reason this was included in the reference circuit? Is it to reduce the load on the microcontroler pin? If so how do you go about working out the best values to use for R1 & R2 or are they somewhat arbitary?

Thanks for your time.
Craig

 

[bertus]

Hello,

See the comments in the drawing:

As the voltage drop accross the leds need to be 16 Volts.
There is also some voltage needed for Re and accross the transistor.
I estimate that the circuit will work from 18 Volts.

Bertus

 

[Alec_t]

What is the purpose of the voltage divider in the reference circuit.

It is to provide a reference voltage for the circuit, which is providing a nominally constant current through the LEDs.
Welcome to the forum!

 

[Harald Kapp]

If I omit the voltage divider (R1 & R2) it simply raises the value of R3 - and saves two resistors.

This will increase the voltage drop across R₃. Thus there will be less voltage available to power the LEDs - which is too low anyway, see @bertus ' comment in post #2.
It will also increase the power dissipation in R₃ unnecessarily.

 

[cramar]

Thankyou everyone for the time you spent on your replies. It all makes perfect sense and I have learnt something new today.

I am thinking I need to change my design approach. I have currently 2 power rails:

1) the boats nominal 12V power - (I guess this can have quite a wide range maybe 8-40V) and;
2) a 3.3V rail for board from a 8-32 V switching mode power supply IC (Silergy SY8401) which has a max output of 0.8 A.

So in my design how would people approach powering the backlighting?
Project constraints:
1) I need 5 blue LED's for backlighting as specified above.
2) I have a constraint on MCU pins available so need to control from one PWM pin (for brightness control).
3) The controller is a ESP32


So should I
1) Create a basic boost circuit to create something like an 18V rail from the 12 volt rail and power the LED's from that?
2) Find a constant current LED driver IC with a 12V input and 20mA output that meets the power requirements?
3) Some other approach?

I thought driving 5 LEDS for the backlighting would be the simplest part of the project so didn't give it a lot of thought initially, but as I have found there are a lot of gotcha's. I could use some guidance. I am not after a design, but would like your views on how you would approach the design requirement.

 

[roughshawd]

The new LED's are probably the most inconsistent and untrustworthy peice of electronics junk that has ever been built. Most guys I know use them as regular diodes so they know if one burns out or not...
The problem with them is they require a special power. It must be a certain amount of volts from the + and a certain amount of volts from the negative, and it cannot be of higher volts that both + and - combined, and its amps cannot be higher than the bulb is rated. I was told to use 1 amp for a standard LEDs, but the power requirements can vary. (those LEDS were big, almost 7/16" across) and they were marine. you said boat... you meant pleasure craft right??!!! Just buy a string of bulbs (make sure you get the spec requirements) and build a system to turn them on and off with... Or drop the dime and get a new control board!

 

[Harald Kapp]

@roughshawd : Please reconsider your post. LEDs are rather reliable when correctly used. The datasheet will tell you its requirements in terms of voltage and current. There ain't no such thing as "1 amp for a standard LED".

 

[Harald Kapp]

3) Some other approach?

That's what I'd recommend. Use 6 LEds instead of 5 in a 3s/2p configuration:

3 LEDs in series = 9.6 V, perfect for a 12 V supply (you might even use 4 LEDs in each leg).
Of course not the current from the current source (NPN transistor) is split between the 2 parallel legs. To achieve the same brightness you'll have to double the current from the current source.
R2 and R3 are meant to equalize the currents through the 2 legs to account for unequal parameters of the LEDs.

 

[cramar]

That's what I'd recommend. Use 6 LEds instead of 5 in a 3s/2p configuration:
View attachment 55989
3 LEDs in series = 9.6 V, perfect for a 12 V supply (you might even use 4 LEDs in each leg).
Of course not the current from the current source (NPN transistor) is split between the 2 parallel legs. To achieve the same brightness you'll have to double the current from the current source.
R2 and R3 are meant to equalize the currents through the 2 legs to account for unequal parameters of the LEDs.

Thankyou so much. That keeps me going.

 

[cramar]

Hello,

See the comments in the drawing:
View attachment 55955
As the voltage drop accross the leds need to be 16 Volts.
There is also some voltage needed for Re and accross the transistor.
I estimate that the circuit will work from 18 Volts.

Bertus

Bertus, I am getting a chance to get beck to this and just having a think about the two diodes in series (I have gone down another rabit hole but that is good for learning).
I want to make sure I am not missing something here. Wouldnt the voltage at Vb be 1.9V and not 1.4V? {3.3V - (0.7Vx2), Therefore Re in this case will be (1.9 - 0.7)/0.02 = 60ohms?

Thanks again

 

[Bluejets]

Wouldnt the voltage at Vb be 1.9V and not 1.4V? {3.3V - (0.7Vx2),

There is a resistor in the base which has to be accounted for as well..... the 1.4 is the 2 forward biased diode voltage

 

[cramar]

There is a resistor in the base which has to be accounted for as well..... the 1.4 is the 2 forward biased diode voltage

Of course! - thanks

 

[danadak]

A quick and easy way is (each leg) :



Its short circuit protected, thermal protected, line ripple rejection,
can get in SOT package or TO220 or TO92 package. Also very
close leg matching current due to decent Vref accuracy and T
stability. Note min load current to stay in regulation is 10 mA.



Regards, Dana.

 

[danadak]

Or a two terminal constant current diode regulator (each leg) :

https://www.diodes.com/assets/Datasheets/AL5809.pdf

It has lower droput V, 2.5V, vs 3.0 of the 317.


Regards, Dana.

 

[cramar]

Or a two terminal constant current diode regulator (each leg) :

https://www.diodes.com/assets/Datasheets/AL5809.pdf

View attachment 56067
It has lower droput V, 2.5V, vs 3.0 of the 317.


Regards, Dana.

Thankyou - simple and elegant

 

[roughshawd]

@roughshawd : Please reconsider your post. LEDs are rather reliable when correctly used. The datasheet will tell you its requirements in terms of voltage and current. There ain't no such thing as "1 amp for a standard LED".

Sorry HK. The transistor in the circuit hes doing is an amp to make the 12v look like 16 so the LEDS will work. If he removes the resistor, the transistor will fire too soon and the lights won't work because there is not enough power... sorry again!

 

[Harald Kapp]

The transistor in the circuit hes doing is an amp to make the 12v look like 16

Sorry, too. This is incorrect. Nowhere in the op's circuit are 16 V.
Study @bertus ' post #2.