LM3429: Power Output Changes as Input Voltage Range Changes

Hello Lonnie,

May I ask a few questions? I see your chart but what exactly are they? Is "V" the power supply voltage reading, "A" the power supply current reading, and "W" the power supply power reading? Or are you measuring the V and A of the LED string itself? Pin will not equal Pout and measurement is important as well since there will be significant voltage drops across cables at higher currents.

 But this is also a very high power driver so there could very well be something else going on. Is it at all possible to get a schematic, BOM, and layout for your circuit? Any or all of them could be important so whatever you could share would help. Webench isn't always completely accurate. We can also do private messages if you do not feel comfortable with sharing in a public forum.

Regards,

Clint

Clint
V and A are the power supply values. W is the calculated from the power supply values. Bottom line this is the input power to the light fixture. The power cable is 16awg/2 stranded and approx. 3meters (9feet). The light fixture has two(2) strings of 5 LED's and one(1) string of 3 LED's. Each fixture has three(3) LM3429 devices. Target voltage and current for the two 5LED strings is 15.25V @ 1.6amps, the single 3LED string is 9.15V@1.6amps. As for sharing the schematic, BOM and layout we will have to go off line and Private message. Let me know our next step.

Thanks

Hello Lonnie,

The first things to do would be the following:

1. Check the actual LED current and not use the input power for output power (this will become more evident).

2. Check the FET drain waveform and make sure it is a solid square wave (to make sure operation is correct).

3. Measure the input voltage directly at the DUT so that you eliminate the voltage drop across the wires.

I say this simply because if it is operating correctly and regulating what you are seeing could be quite normal. There are two main factors that can affect the input power reading:

1. The efficiency of each regulator will change over the input voltage range. Between 15V and 30V it is not unrealistic to have an efficiency difference of 10%, maybe more. At that total power level that is a significant input power difference between the two voltages.

2. The wires themselves. For example, you have 9ft and two strands, so 18 total feet. Copper wire resistance is 4.016ohms per 1000ft (16awg). So when you are at 30V input at 4A input current you will have a voltage drop of about 0.3ohm across them and you will be burning about 1.16W in the wires (wasted power). But when you are at 15V input at 9.1A input current you will have a drop of about 0.66V and you will be burning around 6W in the wires. That is a 5W difference between the two right there.

Let me know what happens when you check the first 3 and we can determine if we need to get into the schematics, BOM, and layout. If we do you just need to send me a friend request and then we can private message or exchange direct contact information.

Regards,

Clint

ok.. I'll disassembly the light, make some measurements and get back to  you.

Thanks

Clint;

How do I send you a friend request?  I'm need the two of us to review the schematic, BOM and latest set of data.  Looking forward to your reply.

Thanks.

Lon

Hello Lon,

Click on my profile and click on "+Connect" to send a friend request. After it is accepted you can do private messages and attach things to that or we could exchange direct e-mail addresses, whichever you prefer.

Regards,

Clint