I've been reading about transistor saturation to drive 3pcs of 1W LEDs in series from a 12V supply via a TIP31C. I was assuming I would need a Power resistor to balance the voltage drops as the LED will have a FB voltage of 3.3V.

But I ended up using the transistor in the active region, thus eliminating the need for the resistor, but the transistor runs hot.

Problem for me is I had to use a multimeter and measure to find out the Vce, Vbe values for that given current (325mA), I felt the datasheet was useless. The only thing I took out of the data sheet was the Hfe value. my test circuit

All of the values in red are measured.

TIP31C data graphs

These are data graphs from the datasheet.

My question is, since none of the values I have measured are in the graphs, when I'm doing design work in the future will I have to measure the parameters like I did here? Or have I missed how to read the data sheet properly.

  • 2
    \$\begingroup\$ Please provide a schematic of your circuit. Edit your post and click on the 7th icon from the left, the one that has a diode in it, or else just hit ctrl-m. \$\endgroup\$ – WhatRoughBeast Mar 6 '16 at 12:48
  • \$\begingroup\$ Without a schematic, this is just a bunch of hand waving and incomprehensible babbling. Closing. \$\endgroup\$ – Olin Lathrop Mar 6 '16 at 12:53
  • \$\begingroup\$ @OlinLathrop: Instead of damning the OP because he's a newbie and English isn't his first language, wouldn't it be more in the spirit of this site to cordially elicit more detail from him as WhatRoughBeast has done? BTW, since English appears to be your first language, why do you consciously make people stumble over your a/an gaffe? \$\endgroup\$ – EM Fields Mar 6 '16 at 14:56
  • 3
    \$\begingroup\$ @EMFi: First, anyone's language abilities are irrelevant here. It only matters that they post readable text that is clear, not irritating to read, and respects those they ask to read it. It may not be someone's fault that they can't write English well, but it's not our fault either, so we don't have any obligation to put up with crap. However, the problem here is not English usage, but lack of clear explanation by not keeping the context of the reader in mind. Much context is assumed that we can't know. And yes, people need to be kicked for posting crap like this. Not my job to fix it. \$\endgroup\$ – Olin Lathrop Mar 6 '16 at 15:07
  • \$\begingroup\$ @OlinLathrop: 1. Since "it only matters" that they post text that is clear, readable, non-irritating, and respectful of the reader, and all of that depends on language skills, the poster's language skills are clearly not irrelevant. Nor are the reader's, since non-existent nuance read into a sender's post can often be mistaken for "crap" and lead to misunderstanding and conflict, especially if the reader lives with a chip on his shoulder. \$\endgroup\$ – EM Fields Mar 6 '16 at 17:03

If you have a supply voltage of 12VDC and your LEDs are rated for a forward voltage \$ ( V_F, not FB)\$ of 3.3V with 325mA through them, then each one will need a series ballast resistor with a value of:

$$ R_S = \frac {V_S - (V_F+ V_{CE}(_{SAT}))}{I_F} = \frac{12V - 3.3V+1V}{0.325A} \approx 24\text { ohms.}$$

the ballast should be connected in series between the supply, the LED, and the TIP31 collector, like this:

enter image description here

In a circuit like this, you'll want to run the driver transistor in saturation in order to have it dissipate only a small amount of power, while the ballast does the bulk of the work.

In order to do that, it's common practice to run the transistor with a "forced beta" of 10 by making its base current 10% of the collector current, which will force the transistor into saturation if its natural beta (Ic/Ib) is greater than 10 with the load current through the collector. In this case we have 325 mA into the collector, so to get a forced beta of around 10 we'll have to push about 30 mA into the base.

In order to do that, go to the data sheet and note the base-to-emitter saturation voltage with 325 mA into the collector, and then select the base resistor like this:

$$ R_B = \frac {V2 - V_{BE}(_{SAT})}{0.1\times I_C} = \frac{12V - 1V}{0.033A} \approx 330 \text{ ohms}$$

The plot above shows the collector current VS base current for the circuit shown, and I've offset the green trace a little for clarity.

In reality, it starts at zero volts just like the yellow trace.

Finally, here's the LTspice circuit list so you can simulate the circuit and get a feel for what's what, if you want to.

If you don't have LTspice it's available, free, here.

Version 4
SHEET 1 880 744
WIRE 352 112 -16 112
WIRE 352 176 352 112
WIRE 352 304 352 256
WIRE 352 400 352 368
WIRE 176 448 112 448
WIRE 288 448 256 448
WIRE -16 496 -16 112
WIRE 112 496 112 448
WIRE -16 624 -16 576
WIRE 112 624 112 576
WIRE 112 624 -16 624
WIRE 352 624 352 496
WIRE 352 624 112 624
WIRE -16 688 -16 624
FLAG -16 688 0
SYMBOL npn 288 400 R0
SYMBOL LED 336 304 R0
SYMBOL res 336 160 R0
SYMATTR Value 24
SYMBOL voltage -16 480 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR Value 12
SYMBOL voltage 112 480 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR Value PULSE(0 12 1 1u 1u .5 1)
SYMBOL res 160 464 R270
WINDOW 0 64 56 VTop 2
WINDOW 3 64 56 VBottom 2
SYMATTR Value 330
TEXT -2 656 Left 2 !.tran 10
  • \$\begingroup\$ Thank you for your efforts, given my unclear question. I didn't think that we should try to run the transistor in the forced Beta . I was thinking to drive the TIP31 directly from the microcontroller. Since PIC chips are recommend to draw maximum of 25mA per pin and 125mA from the total outputs, I thought the lesser I draw the better it will be. But however with your suggestion we are increasing the part count, addition of the resistor and an additional stage of transistor. Given the choice to reduce the cost of this circuit, would you recommend to run the transistor in the active region? \$\endgroup\$ – Thilina T. Mar 7 '16 at 16:10
  • \$\begingroup\$ @EMFields With the updated OP post indicating series LED wiring, this answer may need updated due to appearing to be based on parallel LED wiring. \$\endgroup\$ – Robherc KV5ROB Mar 7 '16 at 17:42
  • \$\begingroup\$ @ThilinaT. Using my method will increase the part count by one resistor, and my answer merely showed the way to calculate the ballast resistor using a saturated transistor switch and a single LED. For series-connected LEDs the procedure would stay the same, but the parameters would change. \$\endgroup\$ – EM Fields Mar 7 '16 at 19:33
  • \$\begingroup\$ @ThilinaT. Additionally, if you're planning to build a lot of these widgets it would be better if you didn't have to limit the collector current by adjusting (beta-biasing) each widget's base current manually before it went out the door. A far better way is to force the transistor into saturation and to limit the current through the load with a resistor large enough to preclude the possibility of over-currenting the LEDs and be done with it. \$\endgroup\$ – EM Fields Mar 7 '16 at 19:39
  • \$\begingroup\$ @RobhercKV5ROB How did you manage to come up with "parallel LED wiring" given the single ballast, the single-LED \$ I_F \$ of 325mA, abd the single transistor, as shown on the schematic? \$\endgroup\$ – EM Fields Mar 7 '16 at 19:51

Are the LED´s parallel or serial connected? Assuming that the three white LED are connected in series (you said that the feedback voltage is 3.3 V), the transistor dissipates 3.3v x 325mA = 1.07W; this one watt is heating the transistor, but a TIP31C can handle this one watt without struggle.

Using the transistor in the active region is the reason because the transistor runs hot; the transistor dissipates the heating that, otherwise, would have to be dissipated in the power resistor.

Vce and Vbe aren't directly related to hFE; TIP31 is a bipolar transistor, which hFE is the result of colector current divided by base current (hFE = iC / iB). Vbe, by the way, should be invariable at about 0.6v.

If you want to confirm the data from the datasheet, you should measure iC and iB. A easier way to measure hFE is using a multimeter with hFE measure.

You made two questions, which can be answered independently:

Do I have to do the same again when I'm doing another project

It's hard to say what you'll have to do in another project...

or is there a way to intepret the data I measured from the transistor data sheet.

Taken yours as a generic question, the answer is YES: all the data that someone can measure in a circuit always can be interpreted and confirmed in the component's datasheet.

In the case you said, if you want to confirm the data from the datasheet, you should measure iC and iB, in order to calculate hFE.

  • \$\begingroup\$ I have added a schematic and the appropriate data graphs to my question. I really would like to reduce the component count so that is why I was okay with the transistor being in the active-region. But does this lead to short term / long term problems when it comes to designing simple LED drivers that would run from a direct microcontroller output? \$\endgroup\$ – Thilina T. Mar 7 '16 at 16:17

My question is, since none of the values I have measured are in the graphs, when I'm doing design work in the future will I have to measure the parameters like I did here? Or have I missed how to read the data sheet properly.

  • Firstly, depending on the level of accuracy needed in your "final product," it's never a bad idea to "double check" the circuit & include 1+ adjustable element(s) (pots are fairly inexpensive, especially when compared to the hassle of rework). The datasheet should provide fairly reliable values, but every component has some degree of "tolerance," so final performance rarely perfectly lines up with "ideal theoretical" performance without any tweaking (at least not for me).

  • Secondly, if you double-check the Hfe graph you posted in your image, it states in the upper-right corner that the values charted are for a Vce of 4V. Since your Vce measured in your circuit is only 2.19V, I wouldn't consider your observed Hfe of 81.25 to be "out of spec" when compared with the ~160 Hfe value predicted @ 325mA Ic on that chart.

    • Calculating based on an "ideal" trans-resistance (i.e. assuming the transistor to have a fixed Rce, based only on Ib), I multiplied your observed Hfe=81.25 @ 2.19Vce, by 4Vce / 2.19Vce, gave an 'adjusted Hfe' of ~148.4. Considering that my visual estimation of the Hfe=160 from the graph could easily be off by +/-10, I'd consider that to be really close.

My personal opinion:
If you want to build a "constant current" LED driver where the current is controlled by a transistor, rather than using series resistor(s), a FET is designed to control Id based on Vgs. Thus, while I prefer BJTs for many things, a FET paired with a voltage-divider across the gate, would generally be easier to design/adjust for controlling your current through the load.

  • \$\begingroup\$ I have no experience with FETs, because its hard to source parts from local shops. But from what I gathered to drive a MOSFET you need about 10V?? Furthermore, I want to run this from a micro-controller output directly, IMHO I would still prefer to use the BJT. To limit the current I have put one 2SC828 BJT at the base of of the TIP31 and made it turn on using ballast resistor. \$\endgroup\$ – Thilina T. Mar 8 '16 at 7:24
  • \$\begingroup\$ @ThilinaT. If you're interested, th BUK98150-55A looks like a very suitable FET for this use. I found it on digikey, for $0.45USD at: digikey.com/product-detail/en/nxp-semiconductors/… but I'm sure you could find it on mauser, ebay, or other sites if you prefer. - At Vds=2.19V it's well within its stable-current area for 325mA @ around 2.2Vgs, according to the graph on figure 6 (bottom of page 6) of its datasheet: nxp.com/documents/data_sheet/BUK98150-55A.pdf \$\endgroup\$ – Robherc KV5ROB Mar 8 '16 at 14:34
  • \$\begingroup\$ ...so you could use a 10K pot from your mcu to the FET's gate, then a 2.2K fixed resistor from gate to GND for the voltage divider. Adjust the 10K pot until you get your wanted 325mA of current (somewhere near 2.8K on the pot), then the FET should "hold it steady" very near that same current, regardless of whether you use 1, 2, or 3 LEDs in series for your load. :-) \$\endgroup\$ – Robherc KV5ROB Mar 8 '16 at 14:38
  • \$\begingroup\$ I'll get some FETs and play around. As you said rightly, using a FET is a much better idea. \$\endgroup\$ – Thilina T. Mar 9 '16 at 15:50
  • \$\begingroup\$ ... I checked your suggestion, the packaging was a problem for me since I'm using a BreadBoard for development work. After searching in eBay found this, BUK9508-55A. It has a through hole packaging. Cheers \$\endgroup\$ – Thilina T. Mar 9 '16 at 16:51

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