Electronics > Projects, Designs, and Technical Stuff

[LTSpice] Simulate Led Strip

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Zero999:

--- Quote from: Ed.Kloonk on August 09, 2020, 03:20:24 pm ---Think about factoring in any volt drop on the supply wires to the strip if it is far away from your power source.

--- End quote ---
Good point, but if it's a constant current supply, the voltage drop is less of an issue, as long as it can increase sufficiently increase the voltage to compensate.

Jay_Diddy_B:

--- Quote from: lfaborges on August 09, 2020, 02:19:56 pm ---What I've been doing is using a rectifier bridge and a capacitor, to try to light the LEDs. The problem is that the voltage has been below 12V.

The only thing I can't change about the project is the 2.8A current power supply

In the image, the green signal is the current source 2.8A and the blue signal is the measurement on the LED.

--- End quote ---

Are you measuring the voltage across D1, just the LEDs, or the voltage across C5?

The resistor R1, is part of the LED strip. Each three LED section of the strip has a 100 \$\Omega\$ resistor in series with three LEDs.

There will be voltage drop along the strip. This means if you feed the strip from one end the LEDs will be brighter at the end you feed than at the far end.

Please attach the .asc file to your posts.

You only need to run the simulation for 200ms, not 5 seconds.

Regards,
Jay_Diddy_B

lfaborges:

--- Quote from: Jay_Diddy_B on August 09, 2020, 04:51:56 pm ---
--- Quote from: lfaborges on August 09, 2020, 02:19:56 pm ---What I've been doing is using a rectifier bridge and a capacitor, to try to light the LEDs. The problem is that the voltage has been below 12V.

The only thing I can't change about the project is the 2.8A current power supply

In the image, the green signal is the current source 2.8A and the blue signal is the measurement on the LED.

--- End quote ---

Are you measuring the voltage across D1, just the LEDs, or the voltage across C5?

The resistor R1, is part of the LED strip. Each three LED section of the strip has a 100 \$\Omega\$ resistor in series with three LEDs.

There will be voltage drop along the strip. This means if you feed the strip from one end the LEDs will be brighter at the end you feed than at the far end.

Please attach the .asc file to your posts.

You only need to run the simulation for 200ms, not 5 seconds.

Regards,
Jay_Diddy_B

--- End quote ---

I am measuring after the capacitor.

Zero999:

--- Quote from: lfaborges on August 09, 2020, 05:33:07 pm ---
--- Quote from: Jay_Diddy_B on August 09, 2020, 04:51:56 pm ---
--- Quote from: lfaborges on August 09, 2020, 02:19:56 pm ---What I've been doing is using a rectifier bridge and a capacitor, to try to light the LEDs. The problem is that the voltage has been below 12V.

The only thing I can't change about the project is the 2.8A current power supply

In the image, the green signal is the current source 2.8A and the blue signal is the measurement on the LED.

--- End quote ---

Are you measuring the voltage across D1, just the LEDs, or the voltage across C5?

The resistor R1, is part of the LED strip. Each three LED section of the strip has a 100 \$\Omega\$ resistor in series with three LEDs.

There will be voltage drop along the strip. This means if you feed the strip from one end the LEDs will be brighter at the end you feed than at the far end.

Please attach the .asc file to your posts.

You only need to run the simulation for 200ms, not 5 seconds.

Regards,
Jay_Diddy_B

--- End quote ---

I am measuring after the capacitor.

--- End quote ---
What voltage did you think you'll measure?

It's what you should expect. M = length = 176, which means R1's value is divided by 176, leaving only 0.57 Ohms, so most of the voltage drop will be due to the LEDs' forward voltage. If you set length to 1, the voltage will increase to 192V.

lfaborges:

--- Quote from: Zero999 on August 09, 2020, 05:56:02 pm ---
--- Quote from: lfaborges on August 09, 2020, 05:33:07 pm ---
--- Quote from: Jay_Diddy_B on August 09, 2020, 04:51:56 pm ---
--- Quote from: lfaborges on August 09, 2020, 02:19:56 pm ---What I've been doing is using a rectifier bridge and a capacitor, to try to light the LEDs. The problem is that the voltage has been below 12V.

The only thing I can't change about the project is the 2.8A current power supply

In the image, the green signal is the current source 2.8A and the blue signal is the measurement on the LED.

--- End quote ---

Are you measuring the voltage across D1, just the LEDs, or the voltage across C5?

The resistor R1, is part of the LED strip. Each three LED section of the strip has a 100 \$\Omega\$ resistor in series with three LEDs.

There will be voltage drop along the strip. This means if you feed the strip from one end the LEDs will be brighter at the end you feed than at the far end.

Please attach the .asc file to your posts.

You only need to run the simulation for 200ms, not 5 seconds.

Regards,
Jay_Diddy_B

--- End quote ---

I am measuring after the capacitor.

--- End quote ---
What voltage did you think you'll measure?

It's what you should expect. M = length = 176, which means R1's value is divided by 176, leaving only 0.57 Ohms, so most of the voltage drop will be due to the LEDs' forward voltage. If you set length to 1, the voltage will increase to 192V.

--- End quote ---

I expect 12V. I tested an LM7812, but the input voltage was low and the output resulted in approximately 8V. I thought about using a TRIAC or a Zener Diode if the voltage went up too much. But I can never find the perfect model to find 12V on the LED.

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