Hello friends, I made this design for the use of a LM2596.
The idea is to use a VIN of 12V that then with the help of the LM2596 we will get a VOUT of 6.5 V .

Any one have an opinion of what should be changed?

 

The -12 Version will give you 12V output. If you want 6.5V output you need the -ADJ version, and a potential divider, but your potential divider has to give 1.2V when Vout is 6.5V, so it divides by 5.41. Your divides by 11.
You can use this website
http://sim.okawa-denshi.jp/en/teikokeisan.htm
to calculate the values for you.
Type in 6.5V to V1 and 1.2V to V2.
By the way, how are you limiting the current to the LED? If you drive it with a regulated voltage, you'll blow it up.

 

Hello Friend, thanks for the help.
Ok, for the divider im using a pot of 100K (r6) with a resistor of 10k(r7).
And about the current of the led, im using an resettable fuse (F4) SMD1206 Series . This guy holds up to 2A.

 

Hello Friend, thanks for the help.
Ok, for the divider im using a pot of 100K (r6) with a resistor of 10k(r7).
And about the current of the led, im using an resettable fuse (F4) SMD1206 Series . This guy holds up to 2A.

If you're using the pot, you need to connect the wiper to one end.
An overcurrent device won't do you any good. You must regulate the current to the LED.

 

If you're using the pot, you need to connect the wiper to one end.
An overcurrent device won't do you any good. You must regulate the current to the LED.

About the connection of the pot could you explain a little bit please ?
you are telling me to connect 1st pin to gnd, second to the divider and third to 12v.
im right?

Is adding this little wire to the system



About the current of the LED, what would you recommend?

 

A the moment, you have just connected the ends, so you have a fixed 100k resistor. Moving the wiper will have no effect at all on your circuit, it will still be a fixed 100k resistor.
So you should connect the wiper terminal and one of the end terminals.
It is then good practice to connect the other (unconnected) end terminal to the wiper, just in case the wiper loses contact with the track.
It is even better practice to avoid potentiometers and presets whenever possible. Is there really a reason that it can't be done with fixed resistors?

 

A the moment, you have just connected the ends, so you have a fixed 100k resistor. Moving the wiper will have no effect at all on your circuit, it will still be a fixed 100k resistor.
So you should connect the wiper terminal and one of the end terminals.
It is then good practice to connect the other (unconnected) end terminal to the wiper, just in case the wiper loses contact with the track.
It is even better practice to avoid potentiometers and presets whenever possible. Is there really a reason that it can't be done with fixed resistors?

the reason im making this with a POt is beacause my VIN can be 12, 13.5, 15.5, 16 or 18 v . But it is important that my VOUT stays in 6.5 V.
This because there is another component that could be in trouble is the VOUT goes beyond 6.5V but no less than 5V

 

the reason im making this with a POt is beacause my VIN can be 12, 13.5, 15.5, 16 or 18 v . But it is important that my VOUT stays in 6.5 V.
This because there is another component that could be in trouble is the VOUT goes beyond 6.5V but no less than 5V

If I make this connection I could fix the issue with the POt?

 

R6/R7 = 6.5V/1.2V-1 regardless of the input voltage.
And the -12 version will give you 12V output, that's why it's called the -12 version.
You pot is now correctly connected.

 

R6/R7 = 6.5V/1.2V-1 regardless of the input voltage.
And the -12 version will give you 12V output, that's why it's called the -12 version.
You pot is now correctly connected.

ok let me undertand,
even my input is bigger than 12V i still be getting a VOUT of 6.5?

 

ok let me undertand,
even my input is bigger than 12V i still be getting a VOUT of 6.5?

I dont want to abuse, but what you recommend to do with the LED ?

 

he reason im making this with a POt is beacause my VIN can be 12, 13.5, 15.5, 16 or 18 v .

This is a "regulator". which means it keeps the output voltage basically constant for a change in both the input voltage and the output current.
As Ian0 noted, the output voltage is determine by the voltage divider voltage into the 1.2V reference.

And add a resistor in series with the LED to avoid zapping it.
LEDs must have their current limited by some means otherwise they will draw a very large current.
The resistor value should be the output voltage minus the LED voltage divided by the desired LED current (not necessarily the maximum rating).

 

ok let me undertand,
even my input is bigger than 12V i still be getting a VOUT of 6.5?

Yes.

I'm assuming that your LED has two chips of ~3.25V in series, giving ~6.5V forward voltage drop. Have a careful look at the datasheet, and see how much spread there is on this specified voltage. You'll probably find that it varies quite a lot, possibly from 5.8V to 7V.
So what happens if you have a 7V LED and you drive it with 6.5V? It doesn't light up.
Or if you have a 5.8V LED and you drive it with 6.5V? It lights up very brightly, for a short time and then fails.
And what happens as it gets warm? The forward voltage reduces.
All these things will be in the datasheet.

My suggestion would be that you drive the LED separately from whatever is also connected to the 6.5V supply, using a device designed for driving high powered LEDs. Or, if you don't mind a very warm resistor, you can put the LED where the Pot is in the circuit, and change R7 to equal 1.2V/[LED_current]. That will make a constant current driver, which won't blow up your LED.

 

Yes.

I'm assuming that your LED has two chips of ~3.25V in series, giving ~6.5V forward voltage drop. Have a careful look at the datasheet, and see how much spread there is on this specified voltage. You'll probably find that it varies quite a lot, possibly from 5.8V to 7V.
So what happens if you have a 7V LED and you drive it with 6.5V? It doesn't light up.
Or if you have a 5.8V LED and you drive it with 6.5V? It lights up very brightly, for a short time and then fails.
And what happens as it gets warm? The forward voltage reduces.
All these things will be in the datasheet.

My suggestion would be that you drive the LED separately from whatever is also connected to the 6.5V supply, using a device designed for driving high powered LEDs. Or, if you don't mind a very warm resistor, you can put the LED where the Pot is in the circuit, and change R7 to equal 1.2V/[LED_current]. That will make a constant current driver, which won't blow up your LED.

I really appreciate all the help friends, believe me this is strongly good from your side.
Considering all the recommendations you have been making, what would you think of this circuit.
It will be getting a VIN of 12V or more, but the VOUT will be of 6V (If im correct).
The LED min voltage is of 5V 220mA according to the datasheet.

 

the VOUT will be of 6V (If im correct).

How did you figure that?
With a 1.2V reference, it will take 2.4v at Vout to give that voltage at the junction of the two 10k ohm resistors, R6 and R7, which is then the regulation value.

A mild suggestion:
Next time you do a schematic, try to make it a little more readable.
Avoid kinks in your lines as much as possible (many in your schematic have 2 or 3 kinks, several have 4 kinks and a few have 5 kinks [C19 left and R7 top for example]).
Don't have component designations covering connection lines.
Draw the ground as one straight line.
If you have two resistors connected in series to ground (like R6 and R7) draw them in line vertically.

All that may spread out the schematic a bit, but it will be a lot easier to follow and understand.
It's rather like good grammar versus bad (missing punctuation, no capital letters, rambling sentences, etc.)
You can likely understand both but good grammar is a lot easier.

Look at commercial electronic schematics and you'll see what I mean.

 

How did you figure that?
With a 1.2V reference, it will take 2.4v at Vout to give that voltage at the junction of the two 10k ohm resistors, R6 and R7, which is then the regulation value.

A mild suggestion:
Next time you do a schematic, try to make it a little more readable.
Avoid kinks in your lines as much as possible (many in your schematic have 2 or 3 kinks, several have 4 kinks and a few have 5 kinks [C19 left and R7 top for example]).
Don't have component designations covering connection lines.
Draw the ground as one straight line.
If you have two resistors connected in series to ground (like R6 and R7) draw them in line vertically.

All that may spread out the schematic a bit, but it will be a lot easier to follow and understand.
It's rather like good grammar versus bad (missing punctuation, no capital letters, rambling sentences, etc.)
You can likely understand both but good grammar is a lot easier.

Look at commercial electronic schematics and you'll see what I mean.

For the values of R6 and R7 I used this calculator:
Voltage Divider Calculator (ohmslawcalculator.com)
I wanted to have a 6V in Vout with 12V in VIN.
The calculator gave me those values, so I used them.
Also at the start of the question "Ian0" told me to use an online calculator to get the values I wanted.

About your suggestions, I strongly appreciated them, believe I will get better.
Like this is a little bit more clear?

 

I wanted to have a 6V in Vout with 12V in VIN.

As has been stated, the input voltage has no direct relation to the output voltage.
I thought you understood that.

The output voltage is determined by the voltage at the FB input.
The values of R6 and R7 are selected to give a voltage of 1.2V at the FB input with the desired output voltage.

Here's a simple way to calculate the required values:
We know that the voltage from the FB input to ground is 1.2V for the desired output voltage, thus the current through R7 is 1.2V / 10kΩ = 0.12mA.
For the desired 6V output, the voltage across R6 is 6V-1.2V = 4.8V.
Since there is no appreciable current into the FB input, R6 must have the same current through it as R7.
Thus R6 = 4.8V / 0.12mA = 40kΩ.

Make sense?

That schematic is much better.
But I would make the ground connection one straight line, with no kinks, and make the LED vertical with no kinks.
It would also help if you move the output above and to the right of U8, rather that below, which is less conventional.
That way D3 and C10 can be vertical with the ground connections at their bottom which is how it is usually shown.
Try to make most ground connections at or near the bottom of the schematic.

 

As has been stated, the input voltage has no direct relation to the output voltage.
I thought you understood that.

The output voltage is determined by the voltage at the FB input.
The values of R6 and R7 are selected to give a voltage of 1.2V at the FB input with the desired output voltage.

Here's a simple way to calculate the required values:
We know that the voltage from the FB input to ground is 1.2V for the desired output voltage, thus the current through R7 is 1.2V / 10kΩ = 0.12mA.
For the desired 6V output, the voltage across R6 is 6V-1.2V = 4.8V.
Since there is no appreciable current into the FB input, R6 must have the same current through it as R7.
Thus R6 = 4.8V / 0.12mA = 40kΩ.

Make sense?

That schematic is much better.
But I would make the ground connection one straight line, with no kinks, and make the LED vertical with no kinks.
It would also help if you move the output above and to the right of U8, rather that below, which is less conventional.
That way D3 and C10 can be vertical with the ground connections at their bottom which is how it is usually shown.
Try to make most ground connections at or near the bottom of the schematic.

Ohh okok I got you now
You get the 1.2 from this formula of the datasheet


in fact r6 should be of 40K; But i will change r7 to 1k and then r6 to 4k .
what does you think?

 

in fact r6 should be of 40K; But i will change r7 to 1k and then r6 to 4k .

That would be fine.
As stated, R6 can be anywhere from 1k to 5kΩ.
With 5kΩ, R7 would be 20kΩ.

 

According to your teachings, this array should give me a VOUT of 6.69 V im right?
R11 is of 2.7K and R9 of 12k
VOUT= 1.23(1+(12K/2.7K)) -> 6.69 V