Darlington Saturation Problem BCX38C - TIP120

[joshzstuff]

I acquired the BCX38C Darlington transistors used in the manual to power a bank of LEDs in series-parallel. (5V power supply)
I couldn't get a satisfactory current to flow to LEDs with the already low value current limiting resistors.

At first I though because of the duty cycle I was using in PWM from my PicAxe, however even with a constant current applied to the base I discovered I am only getting ~1/2 the current available (compared to when I bypass the transistor)
Even when I remove some of the load, It still only passes 1/2 the current, behaving as though the base is not saturated.

I used the suggested diagram on Pg. 6 (below) (minus the diode for inductive loads)


I've tried everything I can think of to get better current flow!
1) I used the recommended 1K base resistor between the Picaxe output.
2) I removed the resistor and ran the output directly to the base
3) I tried several BCX38Cs and when they all behaved the same I even tried a beefier 5A TIP120
4) I even used every setting on a 5K pot from the 5V supply to the base!
5) I have tried 3 different LED bank loads.

I don't want to have to use a relay to switch the LEDs, can anyone tell me If I'm doing something wrong, or will Darlingtons simply not work for this application?
Should I use a Mosfet instead?

Thanks!

 

[inglewoodpete]

What is your Vce when the BCX38C is turned on with a load connected? The datasheet says Vce(sat) = 1.25 volts maximum

Perhaps you can post a photo of your circuit.

 

[srnet]

And you need to tell us the actual current the load is drawing, this information is required to understand what is going on.

 

[Dippy]

Absolutely.

You need to post the exact circuit INCLUDING the wiring configuration and specification of the LEDs.

You need to measure current with your load and Vce, and repeat the measurement with a low and simple resistive load.
Also details of your power supply.

You need to double check pinouts and that you've got things the right way around.

Remember that your I will be (Vsupply-Vce-Vftotalled) / R.
It's possible that you are confusing yourself wrt expectations.
One of the reasons I like MOSFETs and dislike Darkingtons is the simplicity and lower losses of MOSFETs; Rds is so much nicer than Vce .

You could perhaps use a tiddly MOSFET with low gate charge and poss be able to PWM directly.
Depends on your circuit spec...

 

[gengis]

SERIES and parallel?

I acquired the BCX38C Darlington transistors used in the manual to power a bank of LEDs in series-parallel. (5V power supply)

With a 5 volt supply, that doesn't leave much for even two red leds in series. Red/Yellow run ~1.8 volts or so, two in series 3.6. Drop 1.2 in the darlington and you've accounted for nearly all the voltage you have. High power red leds might be up to ~2.3 volt (assuming nothing really hinky like series wired led modules). Blue or white leds? forget about it. You could always use the axe with PWM to boost the voltage a little - depending on how much current you can pull from your source - or the tried and true square wave voltage doubler circuit.

 

[g6ejd]

The members above have answered the question, your easiest solution is to replace the darlington with a single transistor device. Unless you have many LED's in series-parallel, then just about any device of the correct type (NPN) would do.

 

[joshzstuff]

the Devil is in the details . . .

Thank You all for your comments.
Sorry I could not get back to the forum sooner, but I have the information you requested. I will reply to each comment.

Power supply:​

2.1Amp (poorly) regulated.
(Has an IC chip for feedback regulation but see stats below to see voltage drop)
-starts off @ 5.35V w/ no load

With a 5 volt supply, that doesn't leave much for even two red leds in series.

Sry, I didn't mention that they are low voltage IR LEDs (please see description below)

LED Array​

The array is made of 5 series-parallel groups of 6 (30LEDs total)
-I am using 1/4 watt 1.8 ohm resisters with the array pictured above. (1 resistor shared for 2 series chains of 3) (6 leds/resistor)
-They form an IR illuminator 875nm TSHA4401 t-1 package 100mA Peak forward current (constant current)
-Vf = 1.5 Typ / 1.8 Max However after much experimentation with my bench power supply in the series configuration described above, they are only 'happy' with 1.29V when driven to ~80mA -100mA Per 3-LED series chain.
{Even though I plan on pwm them [2A pulse available] I do not exceed the constant current specs here}
(I also have tested the transistors with a slightly different array w/ 3.6 ohm resistors but both are electrically connected identical)

What is your Vce when the BCX38C is turned on with a load connected? The datasheet says Vce(sat) = 1.25 volts maximum
Perhaps you can post a photo of your circuit.

And you need to tell us the actual current the load is drawing, this information is required to understand what is going on.

Absolutely.

You need to post the exact circuit INCLUDING the wiring configuration and specification of the LEDs.

You need to measure current with your load and Vce, and repeat the measurement with a low and simple resistive load.
Also details of your power supply.

You need to double check pinouts and that you've got things the right way around.

Remember that your I will be (Vsupply-Vce-Vftotalled) / R.
It's possible that you are confusing yourself wrt expectations.

Brief circuit description:
I'm using a constant output (not pwm) from a PicAXE 08m2 w/ 1K resistor to the base of the darlington
- Emitter connected to ground (power supply)
- Collector connected to the negative feed of the array
- Positive array feed connected to amp meter - meter to positive power supply

Picaxe has the benefit of a low-dropout regulator @ 4.1V


Ok here are the specs on the load when running on constant current from the darlingtons:

Darlington_Test​

Note: readings are @ initial connection, (cold) after LEDs heat up the current rises until the temp/current stabilizes

BCX38C (800mA Capacity)
5.19V @270mA @Vce = 800mA

TIP120 (5A Capacity)
5.18V @300mA @Vce = 800mA

Direct connection Stats: (no transistor)
If I move the negative conductor of the array from the Collector directly to the negative of the PS (bypassing the transistor and only using the LED series resisters control the current)
5.02V @ 600mA

Note: reading @ initial connection, (cold) after LEDs heat up the current rises until the temp/current stabilizes
-This configuration does not stabilize until it is ~ 1Amp.
-This is due to the thermal operation, and the darlington configurations are Under-driving the LEDs so they only rise ~ 10-15% above the initial connection readings.
(If you wish to know the exact stabilized current and voltage let me know, however this is highly dependent on ambient temperature and ventilation)

One of the reasons I like MOSFETs and dislike Darkingtons is the simplicity and lower losses of MOSFETs; Rds is so much nicer than Vce .
You could perhaps use a tiddly MOSFET with low gate charge and poss be able to PWM directly.
Depends on your circuit spec...

Amen!

I replaced the darlingtons with a Mosfet.
I know the one I used was like 10X overkill but I had it on-hand and it also was featured in the manual.

Mosfet:​

IRL520NBF (10A Capacity)
5.02V @ 600mA

The Mosfet behaved like I hoped the Darlingtons would. Full current transfer!

The members above have answered the question, your easiest solution is to replace the darlington with a single transistor device. Unless you have many LED's in series-parallel, then just about any device of the correct type (NPN) would do.

Sorry for the poor LED load description, Is the theory below what you mean?

My theory: (sorry if you guys tried to tell me this above)​

Am I right to think that the compounded forward voltage drop of the darlington pair is the culprit and essentially is having the same effect as decreasing the power supply voltage (thereby making the series resisters too large)
The mosfet as I understand it being voltage controlled,(?) does not seem to have this problem.. . . Is that right?

Thank you, everyone who contributed!

 

[boriz]

"... Per 3-LED series chain ... "

Your description and photograph differ from this. Do you have 3 LEDs in series or do you have 6? The picture shows 6 (which will never work @ 5v). Crucial information.

"Am I right to think that the compounded forward voltage drop of the darlington pair is the culprit" - No.

As Flooby pointed out above, the the total Vf of the series LEDs plus the Vce(sat) of the transistor must be subtracted from the total available voltage (5v) in order to calculate how much voltage is left over to drop across the ballast resistor.

EG: 3 series LEDs with a Vf of 1.5v, plus a transistor Vce(sat) of 1.25v = 5.75v !!! That's 0.75v ABOVE your available supply. No wonder so little current is flowing. No current should flow at all really. Even only 2 series LEDs comes to 4.25v, leaving only 0.75v across your ballast resistor. This is really too small for any consistent brightness over several parallel strings. Of course 6 in each string is right out.

You definitely need use a higher supply voltage OR a logic level MOSFET with two LEDs in series max. The ballast resistor can then be calculated, once the Vds(on) is known.

 

[Dippy]

I confess I'm struggling with your descriptions, but to the question:
"Am I right to think that the compounded forward voltage drop of the darlington pair is the culprit"
I have to say, "partly" as it is contributory. Sorry Boris I can't give a 100% "no".

I'm afraid I can't follow from the photo, I haven't time , so I may have got that wrong.- I need schematics.

As I said ealirer in post#4.
"Remember that your I will be (Vsupply-Vce-Vftotalled) / R." (and reiterated by Flooby), you have to look at the total voltage drops in the system.
Personally, I can't see how you are getting significant current with 6 series IRLEDs.
Thus if your LEDd Vf , forward voltage drop, was 1.5V each and your Darlington Vce = 1V the equation would be.
I = (Vsupply - Vce - (6xLEDVf)) / R

On occasions when total LEDVf is 'on the edge' of the Vsupply then any BJT device can ruin it.
Hence my preference for MOSFETs.

But a number of things in your description are confusing.
Example:
"BCX38C (800mA Capacity)
5.19V @270mA @Vce = 800mA "
- this makes no sense to me, sorry.
Do this mean:-
a) It should have been 800mA
or
b) I measured it wrong.
???

Unlike a BJT/Darlington etc. a MOSFET doesn't pass current via a PN junction, it can be thought of as a voltage-controlled-resistor at this level. It's not really to do with being voltage or current 'controlled', that is merely a side-effect of it's construction and operation.

Anyway, I'm glad you have it working. I'll leave you to it.

 

[joshzstuff]

1 pic = a KilloWord

"... Per 3-LED series chain ... "

Your description and photograph differ from this. Do you have 3 LEDs in series or do you have 6? The picture shows 6 (which will never work @ 5v). Crucial information.

I confess I'm struggling with your descriptions, but to the question:
"Am I right to think that the compounded forward voltage drop of the darlington pair is the culprit"
I have to say, "partly" as it is contributory. Sorry Boris I can't give a 100% "no".

I'm afraid I can't follow from the photo, I haven't time , so I may have got that wrong.- I need schematics.

Sorry that the photo is confusing. I drew up a Schematic, let me know of any important information I have left out.

2 important details that I hope the schematic clears up:​

1) I am using only 3 LEDs in series not 6 (series parallel - please see schematic)
The idea here is to lower the component count of my circuit by "sharing" a resister between 2 "sets" of 3 series LEDs. (1/4 watt vs. 2 - 1/8watt resistors)
I think I read that possibly this method is a little more risky in the event of a resistor failure, but I think the risk is minimal. Please let me know if you have other reservations about this array design. (especially if it effects the circuit dynamics)
2) Although the Data Sheet claims 1.5V Forward Voltage (even up to 1.8V!) IN Practice I have not found this to be true (Please see Bench Power Supply LED tests below.)

Sorry for all of the newbie questions, but being a beginner I tend not to trust my understanding of the math (or Data sheets) so when I started this project I wanted to see the Forward Voltage and current for myself before I started soldering things together.
I know this is not the way things are commonly done, so let me know if my conclusions are flawed in some way

It appears my concern was well justified, because my discoveries with my power supply changed the development of my circuit.
I know this is not the way things are commonly done, so let me know if my conclusions are flawed in some way.

Bench Power Supply LED tests​

I am looking for getting the most from my LEDs without overloading/overheating them, so I monitored the readings of each scenario as I increased the current to the data sheet specs

In the tests below I used no series resistor, but instead controlled the current via my bench power supply.
Test 1: Single LED
@ 100 mA the Voltage = 1.356V (NOTE: this is far less than the 1.5V stated in the Data sheet)
@ 200 mA the Voltage = 1.4V (NOTE: this is over-driven for constant current and the heat will "run away" at this current but STILL <1.5v Data Sheet!)

Test 2: 2 LEDs in series
@ 100 mA the Voltage = 2.7V
@ 200 mA the Voltage = 2.8V

Test 3: 3 LEDs in series
@ 100 mA the Voltage = 4.06V (at my desired current there 'should' be plenty of headroom for 3 LEDs in series {as per my experiment NOT the DataSheet})
@ 200 mA the Voltage = 4.21V

Test 4: 2SETS of 3 LEDs in series
@ 100 mA the Voltage = 4.00V
@ 200 mA the Voltage = 4.10V
@ 300 mA the Voltage = 4.12V
@ 400 mA the Voltage = 4.21V

(NOTE: I think "Test 4" numbers may be a little skewed as I think the temperature was starting to rise which results in LOWER voltages)

Please Note: (with the exeption of "TEST 1") the measured Forward Voltage of each LED (voltage drop?) = ~1.29V to 1.3V (NOT 1.5V)
I don't know the reason for this phenomenon or why this differs from the DataSheet (but to be fair, the DataSheet says nothing about series LEDs)

As Flooby pointed out above, the the total Vf of the series LEDs plus the Vce(sat) of the transistor must be subtracted from the total available voltage (5v) in order to calculate how much voltage is left over to drop across the ballast resistor.

EG: 3 series LEDs with a Vf of 1.5v, plus a transistor Vce(sat) of 1.25v = 5.75v !!! That's 0.75v ABOVE your available supply. No wonder so little current is flowing. No current should flow at all really. Even only 2 series LEDs comes to 4.25v, leaving only 0.75v across your ballast resistor. This is really too small for any consistent brightness over several parallel strings. Of course 6 in each string is right out.

You definitely need use a higher supply voltage OR a logic level MOSFET with two LEDs in series max. The ballast resistor can then be calculated, once the Vds(on) is known.

Personally, I can't see how you are getting significant current with 6 series IRLEDs.
Thus if your LEDd Vf , forward voltage drop, was 1.5V each and your Darlington Vce = 1V the equation would be.
I = (Vsupply - Vce - (6xLEDVf)) / R

Please see actual test voltages above

But a number of things in your description are confusing.
Example:
"BCX38C (800mA Capacity)
5.19V @270mA @Vce = 800mA "
- this makes no sense to me, sorry.
Do this mean:-
a) It should have been 800mA
or
b) I measured it wrong.
???

Sorry for not making this clear, I do not know the convention for expressing this but let me explain:

"(800mA Capacity)" simply refers to the MAX capacity according to the DATASHEET
This merely is describing the characteristics of the darlington, not the circuit condition.
The circuit current should start out @ ~600mA and climb to ~ 1Amp as the LEDs heat up (at which point the BCX38C should fail! However it never even gets close to this level because of the problems listed)

The same applies for the TIP120, It is capable of 5A which is why it's "overkill"

On occasions when total LEDVf is 'on the edge' of the Vsupply then any BJT device can ruin it.
Hence my preference for MOSFETs.

That seems to be what is going on here.
It makes me wonder about the viability of the BCX38C (or any Darlington really) wouldn't this property make it undesirable for many applications?
I wonder why the manual is so keen on promoting it, what are it's advantages??

Unlike a BJT/Darlington etc. a MOSFET doesn't pass current via a PN junction, it can be thought of as a voltage-controlled-resistor at this level. It's not really to do with being voltage or current 'controlled', that is merely a side-effect of it's construction and operation.

Anyway, I'm glad you have it working. I'll leave you to it.

I'm glad to have it working too, but as this is partially an educational exercise (as most projects are) I am interested in knowing 'why' a particular configuration either works or doesn't work.

Thank you again for your insight!

 

[Dippy]

"It makes me wonder about the viability of the BCX38C (or any Darlington really) wouldn't this property make it undesirable for many applications?"
- of course, that's obvious to anyone in electronics and that's why God gave us Data Sheets, invented the calculator, and electronics teachers.

"Horses for courses" is the usual answer for all components and devices.
I say it a lot (and no-one listens).
Not everyone will be driving near the V limit like you are here.
People experienced in electronics will know all this.
You are at the start of your electronics education and you have learnt something important which, hopefully, you can take forward into your future designs and calculations.

Speaking for myself, I never use Darlingtons, but they are an easy way of driving straight from a wimpy output driver where and when appropriate.

On the subject of Vf; well, this can vary batch to batch and manuf to manuf.
At least you have measured the characteristic and , no doubt, your £300 calibrated Fluke voltmeter has given you accurate figures for voltage. I take it you ARE using a quality DMM when questioning Data Sheets?


I haven't seen the Data Sheet you are using, but you have also learnt that Vf changes with load/heat/temp.
A useful result to put into your Grimoire, and something most people don't know about, which is why I cringe when people use 'normal' LEDs and diodes as (supoosed) accurate V references.

 

[John West]

One problem with LED's is that the actual voltage they drop (as opposed to some nominal datasheet value) depends on both how much current is flowing through them and their temperature. Because of this, they are more easily controlled by regulating the current flowing through them rather than the voltage. Thus the use of a current limiting (or ballast) resistor in series with the string. The resistor's resistance changes very little with temperature fluctuations and acts as a "poor man's" constant current supply.

To do this, the resistor has to be dropping a more significant portion of the voltage than you are attempting to have it do with your circuit configuration. The more the resistor drops, the more stable the current being fed to the LED's. The small voltage drop fluctuations from the LED's need to be offset by a large "fixed" voltage drop across the resistor in order for the overall circuit current to remain relatively stable.

A more typical arrangement of LED's and resistors to do this would use no more than 2 such LED's in series with the resistor on a 5V power supply, so that a larger proportion of the voltage drop was performed by the resistor, creating more stability. You are trading off wasted energy (heat) for more current stability.

Also, a FET is a better choice for an on/off element in a low voltage circuit such as this because one can be selected that has a near zero voltage drop when it is turned on, whereas a Darlington transistor has a junction voltage drop of 2 semiconductor junctions. That could be over a Volt no matter how much or how little current is flowing. But worse, just like the LED's, that voltage drop too will vary somewhat with the temperature and amount of current flowing through it, adding further unknowns to your voltage drop calculations.

Darlingtons are very good for switching on a hefty 12 or 24 Volt motor from a small control signal, a situation where you have a bit more voltage you can afford to waste, and slight fluctuations in the actual voltage dropped don't much matter. They aren't so good in low voltage circuits such as this one where small voltage variations have a significant impact on the amount of current flowing through the circuit.

If keeping good control of the light output of the LED's is of any importance, and 5 Volt operation is required, I'd advise using no more than 2 LED's in series with an appropriately sized resistor, and switching them with a reasonably low "on resistance" FET. Be sure to use one that will turn on fully from the voltage being supplied by the PICAXE. Many common FET's require 7 to 10 Volts for effective turn-on. The type you want often has an "L" in the designator, indicating it is designed for "logic level" turn-on. To be certain, check the gate-source turn-on voltage curve in the datasheet.

If you use a FET, I'd suggest you use an N channel device hooked up with the Drain connected to the LED string and the Source to ground. Depending on the device selected and how you drive it you may also want to mount it on a small heatsink, as you're reaching a point with all of those LED's where component heating needs to be considered.

BTW, the transistor symbol you use in your schematic is not of a Darlington transistor, (a double transistor package) nor is it shown hooked up correctly for a single NPN transistor. The
emitter and collector leads are reversed.

 

[srnet]

At least you have measured the characteristic and , no doubt, your £300 calibrated Fluke voltmeter has given you accurate figures for voltage. I take it you ARE using a quality DMM when questioning Data Sheets?

So true.

I guess most people buy a multimeter and it never get calibrated. The cost of calibration may be more than a lot of multimeters cost anyway.

But what you really want is your meter adjusted & trimmed, which is different to calibration.

 

[John West]

OT, but worth mentioning:
The price of equipment calibration is a very good reason to have both a voltage standard and a frequency standard in your shop to do your own calibration with. Good enough standards are cheaper than the cost of a single cal on even one piece of gear.

I used to get by with a surplus Vref bd, a couple of .01% resistors and a home-brew WWVB receiver - until I could get better gear. They were inconvenient for cal work, but they did work, and cost next to nothing.

 

[srnet]

For a frequency 'calibration' standard the UK MSF signal transimision is hard to beat, 60 kHz carrier to within 2 parts in 10 to 12. A reciever is easy to build, ferrite rod antenna and a few audio op amps.

 

[eclectic]

For a true
Forum-style ramble,
see:
http://www.picaxeforum.co.uk/showthread.php?11269-Calibration.&highlight=sunday+wet

e