uln2803 + load (led strip), 100mA "disappears"?

Thread Starter

prometei

Joined Apr 13, 2008
74
Greets,

I've got a load (led strip segment) that consumes about 350mA when connected directly to the 12V PSU, but when I use a uln2803apg driver IC in between the DMM reads about 250mA, how is that possible? Does it have something to do with the -VCE(sat) Collector-emitter saturation voltage-? The datasheet reads that at 350mA collector current the Vce(sat) is typically 1.3V. Any ideas?

p.s. I've tried sinking the current through 1 and multiple (darlington collector) pins of the uln2803 connected together, but the current stays at ca. 250mA.
 

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AlbertHall

Joined Jun 4, 2014
11,239
It probably is because of the reduced voltage to the LED strip. The current through the strip will not vary linearly with the voltage.
You could avoid the voltage drop by using a MOSFET to do the switching. This will have much lower voltage drop.
 

TeeKay6

Joined Apr 20, 2019
572
Hello,

Did you have a look at the mosfet version?
The TBD6208X has 8 drivers for 500 mA per channel.

Bertus
@bertus
Caution: The TBD6208X has a high enough current rating but the output MOSFETs have Ron=3.25ohms. Thus, the worst case voltage drop is only slightly improved over the bipolar device. The power dissipation per MOSFET (Ron=3.25ohms) is rather high and the device would certainly require heatsinking for multiple "on" MOSFETs.
Of course, discrete MOSFETs of sufficiently low Ron are available.
 

AnalogKid

Joined Aug 1, 2013
9,191
Without knowing the construction of the strip (or even the number of LEDs you are driving), everything is a guess. My guess is that the strip has multiple parallel strings of LEDs and resistors. When connected directly to 12 V, there is a fixed voltage drop across the LED(s) and the remainder of the source voltage is dropped across the resistor. When you insert a second fixed voltage drop (Vcesat) into the circuit, the available voltage across the resistor is less, so the current is less. If you know the Vf of the LEDs, the resulting Ohm's Law calc should be close to predicting the current decrease.

ak
 

Thread Starter

prometei

Joined Apr 13, 2008
74
Just noticed this. What is the burden voltage of your DMM?
I don't know, but why would it matter?

@AnalogKid
It's a standard strip, each segment has three 2538 LEDs in series with a 150 ohm resistor.

I don't mind the slight drop in brightness with these 100mA "missing" so I'll just use the uln2803 that I have. I've tested them for a few hours with 250mA currnet going through two of the 8 transistors and the chip did not get hot.
 
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