BJT as switch...

[alec_t]

.....and you could have found that out for yourself Ritesh, using the LTspice simulator which you obviously have

 

[audioguru]

You show one LED being driven with a square wave.
The peak current is 190mA but the average current is about 100mA.
Then the LED is turned off for half of the time and it looks dimmed.

If your supply is actually as high as 24V then you should use it to light LEDs in series instead of heating the resistor.

 

[koolguy]

Hi again,

Today i have taken Vf for Led for white it is 3V and for Red is ~2V. and it is working well in 4mA to more than 80mA.
so, my question is if i connect 2 Led in series and with 110Ohm V cc of 5V.
so, 5-4=1v across 110ohm collector resistance of 2watt. p=V x i 2=1xi
i=~2Amp so, if i make series of 2 Led so, how many parallel combination should be there??

 

[koolguy]

Hi i have taken this pic from Google and yes i have used Lt spice but not again.

 

[alec_t]

1v across 110ohm collector resistance of 2watt. p=V x i 2=1xi i=~2Amp

If a 2W 110 Ohm resistor has 1V across it that does not mean the current is 2A! Use Ohm's Law and tell us what the current really is, Ritesh.

 

[koolguy]

If a 2W 110 Ohm resistor has 1V across it that does not mean the current is 2A! Use Ohm's Law and tell us what the current

p=i2r
so, 2=i2 x 110
i=134mA

 

[alec_t]

No.
I=V/R = 1/110 = 9mA.
P = I^2*R = V^2/R = I*V = 0.009*1 = 0.009W = 9mW.

 

[large_ghostman]

i love these postings i get to learn loads! and i dont have to be the one asking the dopey questions!! brilliant

 

[Electroenthusiast]

Hi again,

Today i have taken Vf for Led for white it is 3V and for Red is ~2V. and it is working well in 4mA to more than 80mA.
so, my question is if i connect 2 Led in series and with 110Ohm V cc of 5V.
so, 5-4=1v across 110ohm collector resistance of 2watt. p=V x i 2=1xi
i=~2Amp so, if i make series of 2 Led so, how many parallel combination should be there??

Why did you take it as 5-4? What is 4V. Forward Voltage across diodes wont be 3v ot 2v always. You need to take the resistance of the diode in this case, and not the Vf.

 

[koolguy]

No.
I=V/R = 1/110 = 9mA.
P = I^2*R = V^2/R = I*V = 0.009*1 = 0.009W = 9mW.

so, for increasing current i have to increase voltage drop across it..?

 

[Ian Rogers]

so, for increasing current i have to increase voltage drop across it..?

OR!!!! Drop the resistance... See my early pots....

 

[koolguy]

OR!!!! Drop the resistance... See my early pots....

15 * .02A = 300mA ...

You will need about 2 watt (with heat sink) 15Ω resistor..... You'll need to generate about 371mA to run 15 LED's at about 20mA each..... You'll need a heat sink on the BC337 aswell..

this....?

 

[ericgibbs]

this....?

hi Ritesh,
Download and study this image.
Eric

 

[koolguy]

OR!!!! Drop the resistance...

OK, i was just checking and testing the whole experiment after taht i have some result it was good to study..
i have used series combination of two red led which has Vf of 2V and combined several parallel led with them near about 15+15=30 it was working fine as i increases the led the currents increasing very slowly in very few mA near 2mA. and the voltage across 110ohm was 1V, after that i have added total 12 parallel 110ohm one by one the voltage become across R was 0.78V and current also increases and there was not so much power diss across R which was achievement then earlier....
so, my red led project is over and i am thinking about white LED which has Vf of ~3V so, how to combine with 5V 2V across resistance so, should i use 9V for it making 3 or 2 in series??/

 

[ericgibbs]

OK, i was just checking and testing the whole experiment after taht i have some result it was good to study..
i have used series combination of two red led which has Vf of 2V and combined several parallel led with them near about 15+15=30 it was working fine as i increases the led the currents increasing very slowly in very few mA near 2mA. and the voltage across 110ohm was 1V, after that i have added total 12 parallel 110ohm one by one the voltage become across R was 0.78V and current also increases and there was not so much power diss across R which was achievement then earlier....
so, my red led project is over and i am thinking about white LED which has Vf of ~3V so, how to combine with 5V 2V across resistance so, should i use 9V for it making 3 or 2 in series??/

hi,
Have you considered having two White LED's each with its own resistor, using a 5V supply, connected in parallel from Vdd to the Collector.?

E

 

[koolguy]

Have you considered having two White LED's each with its own resistor, using a 5V supply, connected in parallel from Vdd to the Collector.?

sorry not getting you i have tested red not white.

 

[Ian Rogers]

Look!!! Its as simple as this..... 5v Rail.... Red LED...
Red LED consumes MAX 30mA ..
Red LED has about 1.8v forward voltage..

5v - 1.8v = 3.2v... Right!!!

If we want 30mA... 3.2 / .03A = 106 ohms( Nearest value = 110 ohm )
If we want to change the brightness to say half ( remembering its not linear ) we'll use 10mA..

3.2v / .01A = 320 ohms.... So!!! decrease the resistance, increases the current...

You want 9v to supply 15 LED's... assume you got a batch and want them to be half bright!!! you need 15LED's * 0.01A or 150mA

9v - 1.8v = 7.2v..

7.2v / 015A = 48 ohm..... or ( 47 ohm nearest value ).. 150mA * 7.2v = 1.08Watt..

BUT!!!! As audio has already informed you that running LED's at this level on a 9v cell will not last very well...

Everyone has said to use @ 4mA per LED.. ( Even thats too much for a battery project...)

 

[koolguy]

Everyone has said to use @ 4mA per LED.. ( Even thats too much for a battery project...)

It is not battery project it is professional market board.. with SMPS.

5v - 1.8v = 3.2v... Right!!!

that 3.2V across collector resistance will diss power and become to hot!

 

[koolguy]

professional market board.. with SMPS.

what is problem wit 9V SMPS??

 

[tvtech]

^^No problem with 9V SMPS. I actually like them and collect them. Want some?

Regards,
tvtech