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!

 

[For The Popcorn]

At 1.5K Ω it was drawing 0.02A, resistor got hot, LED lights were a bit dim (same as with the battery).

As I said it would be, and the current the LED string was designed to use.

I would like to run the LED lights at around 0.10A, and because the circuit assembly will be in a confined space with very little air flow /Cooling, what can be done to reduce how hot the resistor gets?

If you need brighter LEDs, replace the fairy lights with a strip of LED tape lights.

 

[AnalogKid]

When a 1 W resistor dissipates 1 W, its temperature is over +70C, depending on lead length and ambient air temperature. Some are designed for continuous operation at those temperatures, but not generic parts.

When a 1 W resistor dissipates 2.2 W, it is hot enough to melt fingerprints.

For long term reliability, operate resistors at 50% *or less* of their rated power dissipation.

ak

 

[Dave001]

If you need brighter LEDs, replace the fairy lights with a strip of LED tape lights.

With how I need to use them, I cant use LED strips. Only the fairly lights fit into the application. And with how cheap they are, Im not worried about lifespan.

View attachment 143022

So i need ≈2W resistor correct?


My understanding of resistors is just not that good yet....

 

[Diver300]

So i need ≈2W resistor correct?

If the resistor is expected to produce 2.2 W of heat, use a resistor rated at 5 W or more.

Even at 2.2 W, a 5 W resistor that isn't on a heat sink will get too hot to touch after a few minutes.

 

[Dave001]

If the resistor is expected to produce 2.2 W of heat, use a resistor rated at 5 W or more.

Even at 2.2 W, a 5 W resistor that isn't on a heat sink will get too hot to touch after a few minutes.

Thanks!

 

[Dave001]

For The Popcorn Diver300 AnalogKid

So just to clarify a few things:

- the 24V is coming from an electronic system which gives the 24V AUX output power at specified times/triggers.
- The power supply to the electronics system is 24V @ 2.5A
→ Is the resistor burning up (heat) all the remaining amps in the aux output, minus what the system and LED uses??....

I can exchange the power supply for a 24V@ 1.5A, would that put less heat on the resistor?

 

[Diver300]

- the 24V is coming from an electronic system which gives the 24V AUX output power at specified times/triggers.
- The power supply to the electronics system is 24V @ 2.5A
→ Is the resistor burning up (heat) all the remaining amps in the aux output, minus what the system and LED uses??....

I can exchange the power supply for a 24V@ 1.5A, would that put less heat on the resistor?

The resistor is limiting the current, and burning ("dropping" might be a better term) the remaining voltage. The LED and resistor take the same current, because they are in series. The amount of heat in Watts is equal to the voltage across the resistor multiplied by the current.

The current taken is nowhere near the power supply rating 2.5 A. If you are going by the calculation in post #41, you are taking 100 mA from the supply. A 24 V, 2.5 A supply will put out 24 V, if any current from zero to 2.5 A is taken from it. When you take 100 mA from it, that's fine, the voltage will be 24 V, but there isn't a current of 2.5 A anywhere.

It is like when you have a 15 A or 20 A outlet in your house, that is the maximum rating, not the current that is actually taken. When you plug in a fan that takes 0.5 A, it works the same whether plugged into a 15 A outlet or a 20 A outlet. The voltage will still be 120 V, the fan take 0.5 A, and neither rating of outlet is running anywhere near its maximum.

 

[Dave001]

Got it. Thanks a ton Diver300
I also did some more research and understand it all better now!