transistor load calculation help

[Nigel Goodwin]

Those LEDs are connected in parallel not in series, a very big difference!
I am afraid this makes things worse.
Connecting two LEDs in parallel is much like connecting two diodes or Zener diodes in parallel. The slight difference in their V-I characteristic lets their currents be different noticeably, as Dana pointed it out (post #19).

It's pretty much a myth (other than in theory) - as long as you use the same type LED's, and parallel at least three (or more - the more the better) then the current shares out quite nicely. But don't order different LED's from different sources, and expect everything to work out well.

But regardless, Ron's suggestion in post #18 is what the OP should be looking at - if you've got enough supply voltage, then use it, and make everything more efficient.

 

[Kerim]

It's pretty much a myth (other than in theory) - as long as you use the same type LED's, and parallel at least three (or more - the more the better) then the current shares out quite nicely. But don't order different LED's from different sources, and expect everything to work out well.

But regardless, Ron's suggestion in post #18 is what the OP should be looking at - if you've got enough supply voltage, then use it, and make everything more efficient.

Before year 2011 (when free trading was allowed), I used to design and produce economical outdoor 1-line 1-color message signs (to display user's texts in English and/or Arabic which could be uploaded by an RF remote unit using an in-house transmission protocol). Each pixel was formed by 4 red LEDs as 2x2 (with 5V supply if red or 7V if green). Instead of adding a limiting resistor for each pixel, there are 16-bit shift register ICs whose output current could be set by one resistor only (these ICs are made for MCU SPI serial interface). The LED modules, 8 rows x 32 columns, were produced for me in China in which good LED grades were used (with an extra cost ). The signs look as new even after more than 10 years unless they were hit by something... you know

I mean, yes, as you said, using the same good type LED's, connecting them in parallel also works. And the effect of their sensitivity to temperature is rather very small because the set of LEDs in parallel usually have the same temperature.

 

[danadak]

LED Variations from paralleling them (not a myth) :

https://lumileds.com/wp-content/uploads/files/AB20-3.pdf

As a FAE we dealt with large signage and facility manufacturers and delivered LEDs either
screened for brightness matching, and/or made sure they were from same manufacturing
run, eg. better matching of dopants, phosphors, etc.. Same techniques in analog matching
back in the day.

Some folks can tolerate the non uniform brightness variation, the majors not so much.
The majors have specific specs, typically, for measured variation. Both luminosity and
spectral distribution.

Also critical in LCD backlight display allowed variation.

Human eye capable of seeing these variations : https://www.telescope-optics.net/eye_intensity_response.htm#:~:text=Human eye is capable of,10-billion-fold).


Regards, Dana.

 

[Kerim]

LED Variations from paralleling them (not a myth) :

In general, this is very true.
This reminds me when once, always in the golden years, I received 100 red LEDs as samples, also from China. I wonder why they took the trouble to send them! While I was testing them, I wished finding even a few LEDs which have the same V-I characteristics or similar brightness (as seen by eyes in the least) for the same current. So, it was out of question to take two of them at random and use them in parallel or in series, unless their visible different brightness was not important.

 

[EdStraker]

LED matching not withstanding, I see how Ron's diagram is more efficient. Thanks for that tip.
(I did notice that a lot of LED strips are wired exactly that way also)

I was just trying to confirm that my math was right in trying to select the Base R.

Assuming all the same resistors, R value / # of R's = actual resistance the Transistor is seeing. Then calculate base from there. Or am I still in error here?

 

[Nigel Goodwin]

LED matching not withstanding, I see how Ron's diagram is more efficient. Thanks for that tip.
(I did notice that a lot of LED strips are wired exactly that way also)

I was just trying to confirm that my math was right in trying to select the Base R.

Assuming all the same resistors, R value / # of R's = actual resistance the Transistor is seeing. Then calculate base from there. Or am I still in error here?

There's no need to attempt to calculate base at all - you want the transistor to operate as a switch - so you want it turned on as hard as possible. Assuming the IO pin can provide around 30mA? (check the data sheet), then calculate the base resistor for 30mA or a bit less, by choosing the next highest resistor.

What ever is in the collector of the transistor, has no bearing on it.

If the processor supply is 5V, then 180 ohm or 220 ohm would be good choices.

 

[crutschow]

Curve two shows 100mA, VCE=0.35 Base needs 7 to 10mA.

Those are, of course, nominal/average values, not worst-case (those with minimum gain).
Good design practice is to use a base current of 1/10 the collector current for good saturation (as the data sheet saturation-voltage measurement shows).

 

[danadak]

So in short Rbase = (Vout - Vbe) / (Icollector / 10 ) = (Vout - Vbe) / Ibase


Regards, Dana.

 

[Tony Stewart]

Ed:
If you do not understand anything I said, please ask

This is 2S5P only uses 4V of the 9V and wastes more than half the voltage on R's

This is a practical solution with 4S3P array

The 2N2904 vs 2N3904 was my oversight but this is irrelevant.

Using only 2 in series for the same total LEDs say 12 here uses twice the total power and current

 

[EdStraker]

I completely understand what you're saying about efficient power balance.
In my quicky calculation I am getting different base resistor values for your examples:

Diagram1: I get for Rb = 889Ω (1kΩ)

Diagram2: I get for Rb = 2k5Ω


What am I missing here??

 

[Pommie]

As others have said, for hard switch on (saturation) you need the base current to be ~1/10th the collector current. The collector current, in the case of three parallel strings, is ~20*3=60mA so your base current needs to be ~6mA. Assuming ~4V across the base resistor means it needs to be ~660R.

All done in my head so apologies for miscalculations.

Mike.

 

[Tony Stewart]

Don't worry about using Ic/Ib=10 if your load R is relatively high compared to the Rce value.
When using the switch with a higher resistance than optimal, and Vce > 0.7V Vbc please understand that Vbc ~ 0V is no longer conducting as much and reducing the current gain.

This is what I want you to understand. In this mode, you may consider Ic/Ib somewhere between 10 and hFE min. However when you cannot tolerate much switch voltage drop, then you stick closer to the Ic/Ib=10 guidelines

That means unlike what Pommie says above with Ic=60 mA and the R drop resistance is like 47 ohms as in my 1st sim. Ib=2 mA = Ic/Ib=30 works fine

 

[Pommie]

I don't see anything wrong with 6mA of base current.

Mike.

 

[Tony Stewart]

I didn't say it was wrong, it is just analog overkill.

 

[Pommie]

It's not like we're trying to conserve batteries.

Mike.