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led control transistor

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yalect:
Hi,
I would like to ask you that I found this two circuit, I want to determine which are better and safty for current and load and transistor working
thank you

Zero999:
I Goolged the part number for the LED and it requires something to limit the current.
https://www.mouser.co.uk/ProductDetail/Lite-On/LTL-307EE?qs=Yz4wJs0d%252BpgyXm%2FpkMp2pg==

Neither of those circuits will limit the current to the LED. The first one is a common emitter amplifier. The hFE of a transistor isn't a tightly controlled parameter. It can vary from device to device of the same type and has a positive temperature coefficient. The second one is an emitter follower, which outputs a constant voltage, less the base-emitter voltage of the transistor.

If the power supply voltage doesn't vary much, just use a resistor. Some power can be saved by connecting the LEDs in strings of two in series.

R = (V-VF)/(I)

Were V is the supply voltage.
VF is the forward voltage.
IF is the forward current, in amps.

Example
V = 6V
VF = 2V
IF = 10mA = 0.01A.

R = (6-2)/0.01 = 4/0.01 = 400R, so use 390R or 430R, the two nearest standard E24 values.

mariush:
Both circuits are bad in different ways.

In the first example, you may damage your leds because the current on the leds may not be limited in any way (depends on the hFe of the transistor, if it's fully open or not)

In the second example, you're gonna damage the transistor because you're not limiting the current going into the base pin.

Here's a very good tutorial about npn transistors, that explain how you'd use them as on/off switches. It's best not to use them in the in-between mode, and rely on that to limit the current :



If you want to use the transistor to turn leds on and off, the 2nd circuit will work, provided the base current is high enough (calculate the resistor value by following the video above, but pick a current that is low enough but guarantees the transistor will be fully on)

Then, add a resistor in series with the leds to limit the current. You have the formula V = I x R so with transistor in circuit, this becomes:

Vin - number of leds in series x Forward Voltage - voltage drop on transistor =  I x R

If you have a bunch of leds in parallel, you put 1 in the formula, but your current is  number of leds x individual current
If you have a bunch of leds in series, you put number of leds in the formula
The voltage drop on the transistor is something like 0.4v .. 0.6v, depends on transistor and it's in datasheet (collector-emitter drop) ...for SS8050 it's 0.5v

So let's say 5v input, 4 leds in parallel each with 10mA (0.01A) with a 3.2v forward voltage (white led) you'd have 

5v - 1x3.2v - 0.5v = (4x0.01) x R  => R = 1.3 / 0.04 = 32.5 ohm , so you'd use standard R24 value of 30 ohm, or 33 ohm

If you want 4 leds in series, you'd need at least 4x3.2v + around 1v for safety, or  let's say 14v.
Then formula becomes

14v - 4x3.2 - 0.5 = 0.01 x R => R = 0.7/0.01 = 70 ohms, so you'd probably go with 68 ohm or 75 ohm resistors that are easier to source.
 
   

For LTL-307EE you have datasheet here: https://www.digikey.com/product-detail/en/lite-on-inc/LTL-307EE/160-1701-ND/140833
It says typical forward voltage 2.0v, maximum 2.5v  and maximum continuous current 30mA
If you go with the maximum forward voltage, then I'd suggest staying to let's say maximum 25mA of current.

How much current you decide on it's up to you, brightness doesn't scale linearly with current. For some applications even 5mA will be plenty.

yalect:
thank you for reply
  the example above with the resistor connected to the base you said, may I will damage my leds but when I measure the combined led resistance I found that is high more then 2M ohm which means very small current of collector.
  for the example of no resistor in the base of transistor, the collector will adapt the current of the base?
thank you   

mariush:
With no resistor on the base of the transistor, you'll damage the transistor (unless you power the circuit from a battery with high internal resistance like a CR2032 battery for example)

With a 100kohm resistor, the current and voltage on the base are so low that the transistor doesn't open up.

You want a resistor that will cause the transistor to open up fully and work as an on/off switch. You don't want to open it partially and rely on that to limit the current, because hFe (amplification) varies from transistor to transistor (for example could be typical 100x , maximum 300x ... you get a bag of 100 transistors and this hFe will be all over the range on each transistor) and will also vary with temperature of the transistor. 
So for example, if the transistor is rated for maximum 800mA of collector-emitter current and the minimum hFe is 100, you'd need 800/100 = 8mA on the base to have the transistor open, but you can put some tolerance there and set your base current to something like 10-12 mA... basically making sure the transistor will be fully open.

So as an example, with 10mA on base and 5v power supply, you have formula V = IxR so you have 5v - Vbe (~0.7v...1.2v, i'll use 1v)  = 0.01A x R => R = 4v/0.01 =   400 ohm ... so I'd choose either 390 ohm or 470 ohm (4v/80mA = 500 so even at 8mA current, 470 ohm is good enough)

Watch the video. It explains these things.

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