MAX6958, LED forward voltage > 2.4v

Hi,

I’ve got some MAX6958 chips I’ve been setting up a project with. So far, on my solderless board I used some basic normal intensity LEDs for debugging, but my deployed project will use higher intensity LEDs and I want to make sure it is safe, and won’t blow the MAX controller.

The datasheet http://datasheets.maximintegrated.com/en/ds/MAX6958-MAX6959.pdf only ever uses Vled = 2.4v as an example, but my high intensity LEDs are rated between 3-3.4v, at 20ma.

Will using these higher voltage/intensity LEDs be a problem, and should i be using a resistor between the LED cathode and ground?

Thanks,
Brian

Will using these higher voltage/intensity LEDs be a problem,

No. What will happen is that the power dissipation in the chip is lower than it otherwise would be, so that is good news.
What voltage are you going to use to power the chip?

and should i be using a resistor between the LED cathode and ground?

No.

Thanks, Grumpy Mike.

Grumpy_Mike:
What voltage are you going to use to power the chip?

My power supply is an atx computer PSU, and the 5V line comes in at about 5.15.

Grumpy_Mike:
What will happen is that the power dissipation in the chip is lower than it otherwise would be, so that is good news.

I certainly will take your word on this, but i don't understand why this is from an explanatory standpoint. I see in the MAX datasheet that

PD = (V+ ✕ I+) + (V+ - VLED) (DUTY ✕ ISEG ✕ N)

the power dissipation formula takes the difference between V+ and Vled, and this is returns as a smaller difference as the Vled increases. In turn this effectively downscales the next duty,segment,n term in the equation, and leads to a decrease in the summed total power dissipation.

But why?

If I might ponder, is it because one cost associated with controlling the constant-current demands are balanced by the chip somehow 'absorbing' and dissipating the difference in the voltage?

Regardless of the mechanism, if I calculate my expected power dissipation by
PD = (5.15 * 5.9) + (5.15-3.0)(63/64 * 23 * 9)

I+ is 5.9ma device operating current

Vled is 3.0 as lowest range which maximizes PD

ISEG is 23ma operating current

DUTY is largest 6bit value 63/64

n is 9 total segments

PD = 468.4 mW

The continuous power dissipation for 16-pin dip chip, the continuous power dissipation is 842mW, so I'm well in the clear.

all that sound right?

If I might ponder, is it because one cost associated with controlling the constant-current demands are balanced by the chip somehow 'absorbing' and dissipating the difference in the voltage?

Yes the excess power is being burned off by the chip. Power is the product of voltage and current, so if the voltage across the chip decreases because the voltage across the LED increases then more power is being dissipated in the LED and less in the chip. With a higher Vf for the LED there is less power that is needed to be burned off.

:grin:
much appreciate your help!

I have a similar project to Brian and I did encounter problems with higher voltage LED's rated at 3.3V typical forward voltage. Driving more than 4 segments on a 7 segment display digit caused the chip to crash. I added resistors to the cathodes and now everything works. The good news is that the chip shuts itself down if there is a problem which you can check by polling register 0x4. I still have to play with the resistors because there is some inconsistency in brightness due to the charlieplexing.

Hey rdagger - thanks for posting this. We're having similar troubles driving 4 or more segments and we're hoping to get it working by adding resistors to the cathodes. What value of resistor did you use?

You are asking someone who posted here 3½ years ago, did not have a login and and hasn't been seen since?

You're kidding! :astonished: