4
\$\begingroup\$

I am trying to regulate the current across an LED so that I can tightly control its intensity since I will be using this light as a source of illumination for a sensor.

I am using a CAT4002A-D LED driver, in the configuration shown in the schematic below.

LED driver circuit

I am powering it using 5V from a lab bench power supply, so I used 3*100 ohm resistors to drop the excess voltage.

Now when I set the Rset to 3.74kohm, as per the datasheet I should be getting 20mA whereas I get only around 4mA which is strange since the current drop in series and the resistors shouldn't affect that!!

But when I remove that resistance and use 3.3V as VDD (to prevent the LED from blowing up) then I get current as expected.

I don't understand why should a series resistor cause the current to drop?

\$\endgroup\$
6
  • \$\begingroup\$ What precisely is your intent in using an LED as a "source of illumination?" You write "tightly control its intensity." Is it your intent that the LED emits a constant and predictable irradiance given a specific optical alignment and acceptance area? (And as a side-question, is this for a single LED or are you planning to create more than one of these?) \$\endgroup\$
    – jonk
    Commented Jul 27, 2017 at 17:18
  • \$\begingroup\$ @jonk Yes, I need the LED to emit constant and predictable irradiance. And this is for single LED. \$\endgroup\$
    – Mayank
    Commented Jul 28, 2017 at 4:58
  • 2
    \$\begingroup\$ I worked on a project trying to use LEDs as standard candles. To start, we used an expensive rack-mounted (paid for NIST-traceable calibration, too) 0.01% constant current power supply and put the LEDs into a thermal bath to maintain temperature precise to less than 50 mK over a 48 hour period. We found very few LEDs that would hold anything close to a constant flux. Instead, about 99.4% of the LEDs (and these were the best quality Siemens-Optonico devices we could buy) were tossed. Most just flittered around -- worthless. We did okay. But it was tough. I wish you the best of luck. \$\endgroup\$
    – jonk
    Commented Jul 28, 2017 at 6:06
  • \$\begingroup\$ thank you @jonk :-). Just a thought, what if I flash the LED for like half a second and take a reading and switch it off. Can that help in achieving constant irradiance? My error margin is not so tight, I mean I can't let it float around, but I also don't need 0.01% error margin. Fluctuation of 10uA or so is not an issue. \$\endgroup\$
    – Mayank
    Commented Jul 28, 2017 at 10:55
  • 2
    \$\begingroup\$ We didn't need 0.01% either. We just started out using one of the better equipment we could use in order to do an investigation. We did not want to have to track down equipment problems while studying LED behavior. Kept temps stable for that reason, too. The LEDs vary their flux a whole lot by themselves (easily more than 1%.) I just wanted to make sure you knew what you faced. \$\endgroup\$
    – jonk
    Commented Jul 28, 2017 at 19:10

4 Answers 4

Reset to default
5
\$\begingroup\$

Consider what you are doing from a simple perspective:

  1. Your power supply is 5 volts.
  2. Your series resistance is 300 ohms.

That means the most current you could possibly have through the LED is:

V = I * R
I = V / R
I = 5 / 300
I = 16.7mA

This is below your set point of 20mA. So is likely the reason the constant current supply is operating in an unexpected manner.

Also, the constant current chip specifications state it should operate up to 5 volts. And there are no LED limiting resistors in the specification's example application circuits. It follows you do not need them in your circuit when power it with 5 volts.

\$\endgroup\$
20
\$\begingroup\$

Lose the series resistors. It's not clear what you think they do for you, but they won't do anything useful. You have the LEDs connected to constant current drivers. Series resistors only get in the way and make the constant current drivers require a larger compliance range.

Also, do the math. (300 Ω)(20 mA) = 6 V. You say your are powering the complete string of resistors, LED, and current controller from 5 V. Obviously, that can't possibly result in 20 mA since the resistors alone drop more than the supply voltage at that current.

Again, lose the resistors. They make no sense at all.

\$\endgroup\$
0
3
\$\begingroup\$

The constant current driver in the IC takes care of it for you. The whole point is that it sinks the exact current you want without needing anything external to set the correct current.

When you put those resistors in series they are dropping so much voltage it doesn't leave enough headroom to drive the LED properly. You're getting confused between just using the resistor as what is used to set the current and using a constant current sink like this.

\$\endgroup\$
3
\$\begingroup\$

Since you are using a current driver there is no need for the series resistance. The driver controls the current up to the max open voltage of the driver. So remove the series resistors and you can use the driver up to the max of 5.5 V Therefore remove the resistors R4-7

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.