[SOLVED] Signal distribution using transistor .................

[KerimF]

What is the resistance of the series resistor with each LED?

Isn't it 47 Ohm?

 

[kdg007]

Rc = (5-3.3-1.1)/20mA = 30 ohms(for each LED)

 

[godfreyl]

i want to switch the LEDs on and off using some other circuit .but for now i am concentrating on this circuit.

Then the base current has to be more than enough, so don't design based on "how much will it typically need", but rather "What's the most it could possibly need".

Part of your problem is that the total voltage drop = Q1's Vbe + Q2's Vce. 290mA (10* 29mA) is a lot if current for a 2n4401. You'll probably find that Q1's Vbe is higher than you'd expect, more than 1V.

You can get less voltage drop with a circuit like the one below. With this you only lose Q1's Vce(sat). R1 limits the current through Q2 into Q1's base to about 30mA, which should be enough.

 

[kdg007]

the current through each LED is 40 mA according to your circuit..think its too current current through an LED ?

 

[godfreyl]

You can use use higher resistors for less current. At least now the voltage loss in the switch is minimized.

 

[kdg007]

what shall i do with the Icbo current ? usually i kept 5.6k between base and ground (for Q1 and Q2)..... i can keep for Q1 but what about Q2 ?

 

[godfreyl]

Icbo will be very small. For Q2, you could ignore it, or put 100K between base and ground to be safe. For Q1 it's probably a good idea to put 5.6K (or 10K or...) between base and ground, as it also has Q2's leakage current flowing into it's base.

 

[KerimF]

Rc = (5-3.3-1.1)/20mA = 30 ohms(for each LED)

This is right ... but you found out
I_led= 11 mA.

I think you know how to measure I_led. If you know the value of Rs (the resistor in series with the LED), the voltage over it is equal to:
V_rs = I_led * Rs

I know theorically Rs=30 but what is its value on the real circuit?

Sorry... I see it now... it is 33 R

The simplest way is to increase its value in order to decrease the LED current... V_led will decreases slightly when the LED current decreases, and it may be differrent for different LEDs unless they are from a well matched lot.

 

[kdg007]

**broken link removed**
so i kept 100k for q2 and 10k for q1..that should be ok for Icbo.
-
theoretically current for LED should be 20mA but in simulation it is showing more than that ...hmm

 

[KerimF]

theoretically current for LED should be 20mA but in simulation it is showing more than that ...hmm

This is natural mainly in your situation since the model of the LED is very approximated and it is supplied with a voltage very close to its typical voltage (3.3 versus 5-1.1=3.9V)
If you cannot get an LED model close to what you use in your real circuit you can create a simple one by a test as follows :

The idea is to add the dynamic internal resistance of the actual LED and adjust the value of the its forward voltage.

In a way convenient to you, you measure the LED voltage (Vnominal) @20mA by using a DC voltage supply and a suitable current limiting resistor. We can assume the measured voltage is reverse voltage source in its model.

Now you measure again its voltage (Vmin) but at a lower current, say 15mA (Imin). The current value could be any value around 15mA.

You do the same for a higher current, say 25mA (Imax) to get Vmax.

The dynamic internal resistance of the LED could be calculated:
Rled = (Vmax - Vmin) / (Imax - Imin)

So the LED model will consist of Vnominal in series with Rled. Naturally there will be the external limiting resistor as well.

If you manage getting Vnominal and Rled, I think you will be pleased with the result you get from your simulator.

Kerim