TIP120 reducing LED brightness by a lot. Help. Moderators: adafruit_support_bill, adafruit

[Ninjared]

I have a very bright LED that I am controlling with an Arduino through a TIP120 transistor. The LED is 3.4Vf @ 1A. For some reason, when I trigger the LED through the transistor it ends up being significantly less bright than when I power the LED alone. Here is a basic description:

Test 1: 8.4V battery > 3.3V 1A regulator > 10ohm 3W resistor > LED anode > cathode to GND. The LED is VERY bright, hard to look at. Perfect!
Test 2: 8.4V battery > 3.3V 1A regulator > 10ohm 3W resistor > LED anode > cathode to TIP120 collector > TIP120 emitter to GND. The TIP120 base is connected to an Arduino digital pin through a 1Kohm 1/2W resistor (the Adruino is also powered by the battery.) The LED is nowhere near as bright, probably close to half the intensity.

Did I do something wrong? I thought for sure there would be no current reduction through the TIP120 since it's rated for 5A constant. Any ideas?

[westfw]

A TIP120 is a darlington transistor, and it has a Vce voltage drop (even when fully "on", the Vce(sat) spec from the datasheet) of about 2V. If the power supply is 3.3V, and the voltage drop of the transistor is 2V, that leaves 1.3V for your 3.4V LED, so I'm surprised that it lights at all.

Get rid of the voltage regulator and maybe increase the value of your resistor.
(8.4V-3.4V-2V)/1A should mean ... your 10ohm resistor should still be OK (I get 0.3A, if I did my math right.)
(0.6W dissipated in the transistor and a bit over 1W in the LED. Both might want heat-sinks.)

[Ninjared]

A TIP120 is a darlington transistor, and it has a Vce voltage drop (even when fully "on", the Vce(sat) spec from the datasheet) of about 2V. If the power supply is 3.3V, and the voltage drop of the transistor is 2V, that leaves 1.3V for your 3.4V LED, so I'm surprised that it lights at all.

Get rid of the voltage regulator and maybe increase the value of your resistor.
(8.4V-3.4V-2V)/1A should mean ... your 10ohm resistor should still be OK (I get 0.3A, if I did my math right.)
(0.6W dissipated in the transistor and a bit over 1W in the LED. Both might want heat-sinks.)

Thank you for the reply!
Huh, that's super crucial information; funny that I was never aware of this. So it looks to me like I should be able to simply swap out the 3.3V regulator for a 5V one, and it should work just fine. I'll test it with a multimeter to be sure. Since the NiMH battery pack will range from 9.8V to 8.4V during discharge, I'd much rather rely on a regulator than power it directly.
I'm looking at the datasheet for the PN2222, and it looks like the voltage drop is pretty much 1.0V. That sound right? I've been using those forever and never even noticed.

[adafruit_support_bill]

That sounds about right. You might want to look at MOSFETs also. These have a very low Rds(On), so the voltage drop is typically quite low as well.

[westfw]

I'm looking at the datasheet for the PN2222, and it looks like the voltage drop is pretty much 1.0V. That sound right?

For high currents, yes. Normally you can expect <0.5V for a "typical" operation of a single transistor. Transistors tends to be much less "constant" in their parameters than most people expect. :-(
here's a piece of a pn2222a datasheet (Note that 1V is the MAX specified Vce drop):

Screen Shot 2022-05-16 at 3.14.39 PM.png (36.18 KiB) Viewed 61 times

Darlington transistor pairs are significantly worse, because the main power-passing transistor of the pair never quite makes it into "saturation." (I remember being a bit stunned when analysis showed that in some long-ago EE class.)

Note that this also makes darlington-based motor drivers like the L293d problematic. If you don't account for the ~1.5V drop in BOTH the high and low-side outputs, your motor can under-perform - thus the popularity of MOSFET drivers (and their lesser heat-sinking requirements.)

[Ninjared]

Thank you for taking the time to help me out and for using simple terms that I can follow. I thought I would share the results of some of the testing I did:

Voltage readings between the positive and negative connections of the load device (or the point where the load would be):
3.3V 1A regulator alone: 3.29
No load with PN2222: 3.27
No load with TIP120: 2.72 (ouch!)
3W LED, 2ohms resistor, no transistor: 2.87 (not bad considering the current load is right at the max of the regulator)
3W LED, 2ohms, PN2222: 2.83 (almost negligible)
3W LED, 2ohms, TIP120: 2.60 (Less drop than expected, but still significant)

5V 1A regulator alone: 4.98
With 5V vibration motor: 4.10
Motor with PN2222: 3.98 (not too bad)
Motor with TIP120: 3.51 (motor noticeably slower, oof!)

So any benefit I am getting with the higher current from the TIP120 transistor is being negated by the large voltage drop. I think a better use of darlington transistors would be a device that could operate on a wider range of volts and could therefore be powered directly by a battery. Devices like LED's, which will go "poof" very quickly if you exceed the max voltage, are not such a good fit. If I ever need to really max out that LED's brightness, I could probably make it work by taking a 5V 2A regulator and calculating the proper resistor to bring it down to 3.3V, including the voltage drops. For now, I'll stick with the PN2222.

ALSO:
I do have one more question. In my testing I tried using different resistors at the transistor's base. The higher the resistance I put there, the more voltage drop I saw. Why is this? Is it because the resistor is preventing the transistor from allowing the 5V power of the Arduino's digital pin pass through? Is there a proper way to calculate how much resistance to put on the transistor's base? I've only been using 1Kohm there because that's what I read in an Adafruit tutorial years ago and I've just stuck with it. Thanks!