Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

How to reduce DC a bit without resistor or capacitor? Is there a simple solution?

Status
Not open for further replies.

stevenmahoney

New Member
I have a simple schematics with a number of LED drivers. LED strings have a certain number of LEDs that I cannot change. In order for LEDs to operate properly, I have to feed them with 108V and 1.3A current (combined).
If I rectify 120V after diode bridge voltage increases to approximately 168V. I have excess of 32 volts (28V drops on LED drivers).

I can use a resistor or a capacitor (on AC side) to get rid off excess of 32 volts, but both are bulky (and in case of the resistor hot) solutions.

Is there any other smarter way to solve this problem? Any ideas will be appreciated.

Steven

problem.png
 
What is the current and voltage in each string? There must be more than are shown in the schematic?
 
I agree with Ron....how many LEDs are per string?
You show 8...at about 3.5 to 3.8 volts per LED (assuming white ones) you are talking about 28 to 30.4 volt per string. There are more?


Additional thoughts; if you are planning to feed them from household current -nominal 120VAC in North America- you must consider at the very least +/- 10% voltage fluctuations for your calculations.

The constant current drivers you chose are nominally designed for vehicle applications, where the supply voltage is much lower. The power dissipation in that instance is acceptable, but in higher voltages the power lost by the drivers could negate the LED efficiency gains.

AC powered high wattage LED strings always, or almost always, use a SMPS solution. Many IC manufacturers, TI, Fairchild, OnSemi, STMicro, Sipex and others offer many LED-specific solutions...including datasheet, design software, app notes, eval boards, complete turnkey designs.....
 
I have a simple schematics with a number of LED drivers. LED strings have a certain number of LEDs that I cannot change. In order for LEDs to operate properly, I have to feed them with 108V and 1.3A current (combined).
If I rectify 120V after diode bridge voltage increases to approximately 168V. I have excess of 32 volts (28V drops on LED drivers).

I can use a resistor or a capacitor (on AC side) to get rid off excess of 32 volts, but both are bulky (and in case of the resistor hot) solutions.

Is there any other smarter way to solve this problem? Any ideas will be appreciated.

Steven

View attachment 73033


Hello,

Get rid of the capacitor. An 120vac line has an average voltage of 108v when rectified WITH NO FILTER CAPACITOR. With a filter capacitor the average goes up quite a bit. So without the capacitor it may work just fine. It depends if the current regulators can work correctly with a full wave rectified sine and no filtering, which they probably can. There will be minor blinking but it will be very fast around 120Hz so the eye should not be able to catch that.
 
dougy83, thaks for the idea. That is what I am contemplating right now... I wish I knew how to calculate circuit currents and voltages...
 
I know. I simplified the schematics so I do not have to list hundreds of LEDs. :) All I wanted to find out out how to get rid off the resistor. :)
 
MrAl, I tried that. LED brightness goes down... :-(

Hello,


Yes it should go down. That's because now it is working on the correct voltage when before it was operating at a voltage that was too high. So your sad face should be a happy one because the LEDs will last longer :)
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top