voltage stepdown chip and diode location

[Dave001]

I have a simple LED strip light setup as shown in the pic. the source is a trigger / switch from a accessory output port, so no issues on that end.

I need help with this:
1- Is there is suitable IC chip for this job?
2- I need to place a diode in there somewhere to keep juice from flowing backwards to the other source. Where do I place it? (yes Im fairly new to this stuff.)
3- What diode would be suitable for this?

Thanks in advance!

 

[danadak]

Place Schottky diode, 1 in each positive leg of stepdown output. Pick a diode with worst case
total load current and one with reverse breakdown V at load when one supply is off, grounded.
Use ratings that are 30 - 50% greater than needed for margin. Also compute worst case power
in diode, its Vdropmax x I loadmax to see if it needs heatsinking.

http://personal.strath.ac.uk/barry.williams/Book/Chapter%205X.pdf

Heat Sink Calculator - Engineering Calculators & Tools

This calculator helps determine the maximum temperature a device can withstand

www.allaboutcircuits.com

The led strip is only 20 mA of current draw ? That does not seem right.


More sophisticated approaches :

https://www.analog.com/en/technical-articles/novel-current-sharing-ic-balances-two-supplies-with-ease.html

https://www.analog.com/en/products/ltc4358.html

Regards, Dana

 

[ronsimpson]

There would be less power loss if the diodes were on the 24V,

 

[danadak]

Excellent point Ron and the accuracy of Vload improved due to lack of additional
V drop outside control loop and its Vdrop T sensitivity, eg. the diodes.

Regards, Dana.

 

[Dave001]

Place Schottky diode, 1 in each positive leg of stepdown output. Pick a diode with worst case
total load current and one with reverse breakdown V at load when one supply is off, grounded.
Use ratings that are 30 - 50% greater than needed for margin. Also compute worst case power
in diode, its Vdropmax x I loadmax to see if it needs heatsinking.

View attachment 142473

http://personal.strath.ac.uk/barry.williams/Book/Chapter%205X.pdf

Heat Sink Calculator - Engineering Calculators & Tools

This calculator helps determine the maximum temperature a device can withstand

www.allaboutcircuits.com

The led strip is only 20 mA of current draw ? That does not seem right.


More sophisticated approaches :

https://www.analog.com/en/technical-articles/novel-current-sharing-ic-balances-two-supplies-with-ease.html

https://www.analog.com/en/products/ltc4358.html

Regards, Dana

Dang your smart. That is above my level of understanding tho and I would need help with picking the right diode.

Its a 3.3 ft of fairy light led strip inside of a 3D printed custom logo/ sign. The light comes powered by 2x CR2030 batteries and Im modifying the power source to use for my project. So its probably even less than 20ma, I just rounded up cuz batteries it came with are cheap and dont put out much.

 

[Dave001]

There would be less power loss if the diodes were on the 24V,
View attachment 142477

As long as there will be no damage to the voltage stepdown chip, I can put it there.
I just assumed it was best to place it between the load and stepdown chip.

For either setup, need helping picking the correct diode.

 

[Diver300]

You need a diode that can take the current of 0.02 A and the voltage of 24 V.

For standard diodes almost any diode will be OK. For Schottky diodes, some won't be rated at 24 V.

A safe bet would be a 1N4007, which is rated at 1A and 1000 V. That may seem a bit excessive but they are one of the cheapest diodes available.

 

[ronsimpson]

1N4007

There is a series of diodes 1N400x where 7 is the 1000V and 1 is the 50V diode. The higher voltage diode has more power loss and voltage drop. So in this case the 50 and 100V parts will work best.

1N5819 and SB140 are common Schottky diodes. 40V 1A They will have less power loss than silicon diodes.

 

[Nigel Goodwin]

There is a series of diodes 1N400x where 7 is the 1000V and 1 is the 50V diode. The higher voltage diode has more power loss and voltage drop. So in this case the 50 and 100V parts will work best.

1N5819 and SB140 are common Schottky diodes. 40V 1A They will have less power loss than silicon diodes.

To be fair it will make no difference regardless, as long as the diodes are in the 24V feed - feeding to a 6V converter, you've got 18V headroom to play with, so no need for Schottky diodes at all.

 

[Dave001]

Ok I will use the 1N4001 between the 24V and the step-down chip.

I'm still having trouble finding a 24V max to 6v step down chip that is DIP style. Please help.

 

[For The Popcorn]

I'm still having trouble finding a 24V max to 6v step down chip that is DIP style. Please help.

I'd suggest a switch-mode buck converter. You won't have to worry about heating of a linear regulator - dropping 24v to 6v generates a lot of heat. A module like this one is convenient because it has terminal blocks for Vin and Vout, and a voltage display to make setting the voltage easy.

 

[ronsimpson]

I'm still having trouble finding a 24V max to 6v step down chip that is DIP style. Please help.

small simple not dip
8 pin dip
Good choice
link Getting a working board from China is a good choice.

 

[rjenkinsgb]

Self contained three pin switched mode regulator:

https://www.digikey.co.uk/en/products/detail/cui-inc/P7805-Q24-S6-S/4009633

6.5V just add a diode in series with the output if that's too much.

 

[Dave001]

Self contained three pin switched mode regulator:

https://www.digikey.co.uk/en/products/detail/cui-inc/P7805-Q24-S6-S/4009633

6.5V just add a diode in series with the output if that's too much.

Awesome!! Huge thanks!!!

 

[Dave001]

would this be a correct diagram then?

 

[rjenkinsgb]

The ground (negative) is common between the input and output on those, so the 24V ground / negative must also be connected between the two power sources.

Which means you can just use one regulator & feed its (positive) input via a diode from each 24V source.

If the two 24V sources must remain totally isolated, you would need to use a fully isolated 24V to 6V module for one of them, then have the diodes at the 6V side as in your diagram.

 

[Dave001]

The ground (negative) is common between the input and output on those, so the 24V ground / negative must also be connected between the two power sources.

Which means you can just use one regulator & feed its (positive) input via a diode from each 24V source.

If the two 24V sources must remain totally isolated, you would need to use a fully isolated 24V to 6V module for one of them, then have the diodes at the 6V side as in your diagram.

Thank you!!

 

[Dave001]

So turns out I also need to limit the current for this setup... with how the lights came it has no chips or resistors, I guess current was limited to what the batteries could put out.

With the stepped down voltage, the current is still very high from my power source and it burned up the LED in seconds. How can I step down / limit the current draw? Can I do that with a few resistors? If so which ones and how many?

 

[Diver300]

What had been missed in this thread is that the CR2032 batteries will have a resistance of maybe 40 Ohms each. Two on series will have a resistance of somewhere around 80 Ohms.

If you have constant current supply, set it to 10 mA. If the LEDs are too dim, increase it a bit.

If you have a constant voltage supply, set it to 6 V and put 200 Ohms in series. What is probably better to do is to get a bunch of 1 kOhm resistors. Put about 5 in parallel and use them in series with the LED strip. The more resistors you put in parallel, the brighter the LEDs will be.

 

[rjenkinsgb]

I was going to suggest a 100 Ohm resistor to start with.

If the LEDs are white they will probably drop around 3V, so 100 Ohms from a 5V supply (2V across the resistor) will give a current of around 20mA as given in the OPs first post.

If the supply is actually 6V, then 150 Ohms for ~20mA