Hi all,

I am going to use this below ckt with the atmega328 P, There are 8 IR LED (940nm) and they are in parallel. I just If the condition is matched a pulse will be sent to this IR's and it will turn on things. So I want to know that will this ckt work or I have to do some changes.

If I use an NPN transistor instead of MOSFET in his ckt will it work? and Which transistor should I prefer?

I have attached the Image with this thread. Please give your valuable opinions and solutions.

The LED's in below Image are IR LED's 940 nm. It's due to my fault it is different in the diagram. Sorry about that.

 

Yes your circuit will work with a N fet, i would use a 12V supply for the leds in series and put a 1K across the G and S terminals to make the fet turn off better...

 

Yes your circuit will work with a N fet, i would use a 12V supply for the leds, and put a 1K across the G and S terminals to make the fet turn off better...

I can't provide 12V. I need to work with 5V only. So please tell me will this work?

 

Yes it will work, but put your leds in Parallel with a Series resistor in each led.

Each led takes 2V, so you have 3V to drop across the series resistor...

 

Yes it will work, but put your leds in Parallel with a Series resistor in each led.

Each led takes 2V, so you have 3V to drop across the series resistor...

So what is the solution to work it at 5V with NPN transistor? I am not Understanding it.

 

hi Raj,
What is the specified forward voltage drop of your IR LED's.?
E

 

hi,
The TS stated 8 IR LED (940nm), usually IR's are 1.2Vfwd
He needs to clarify the type.
E

EDIT:
A web search for the d/s shows RED.

 

hi Raj,
What is the specified forward voltage drop of your IR LED's.?
E

1.4 V 100mA for continuous 100% duty cycle
1.2 V 20mA was written without context.

hi,
The TS stated 8 IR LED (940nm), usually IR's are 1.2Vfwd
He needs to clarify the type.
E

EDIT:
A web search for the d/s shows RED.
View attachment 178310

I don't know how to find type. I just have white colored 940nm IR led.

 

hi Raj,
A wavelength of 940nm is Infra Red, so as you say approx 1.2Vfwd drop, so you can connect 3 LED's in series to 5V via a suitable resistor.
What is the application for this project.?
E

 

hi Raj,
A wavelength of 940nm is Infra Red, so as you say approx 1.2Vfwd drop, so you can connect 3 LED's in series to 5V via a suitable resistor.
What is the application for this project.?
E

I am going to use this circuit to control the TV and AC. So I need 8 LED's which will be in square positions as in circuit. so it will control the TV and the AC no matter where they are. So I can use this ckt with stated MOSFET and there will no issue.

And this circuit of IR array was drawn from the online LED array generator with the respective resistors.

 

hi,
What is a little confusing, is that the drawing you posted shows LTL307EE, which is a RED LED at 635nm, not a IR type.??
E

 

hi,
What is a little confusing, is that the drawing you posted shows LTL307EE, which is a RED LED at 635nm, not a IR type.??
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I had only this symbol in the online simulator so... that's actually my fault. Sorry for the inconvenience. So should I start making this ckt or there are any changes?

 

hi,
Recalculate the resistors , assuming 1.2V drop per LED at say 20mA.
E

 

3 IR leds in series @1.2V 20mA each is 70 ohms 125mW on a 5V supply...

 

hi,
Recalculate the resistors , assuming 1.2V drop per LED at say 20mA.
E

15 ohms it is.

 

hi Raj,
I would agree with post #14 ie: 70R, how did you calculate 15R.??

E

 

hi Raj,
I would agree with post #14 ie: 70R, how did you calculate 15R.??

E

Just put in LED array wizard and got this for 4 LEDs in series. for 3 LED's in series it is 82 ohms.

3 x 2 array, 2 extra LEDs
+----|>|----|>|----|>|---/\/\/----+ R = 82 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 82 ohms
+----|>|----|>|----------/\/\/----+ R = 150 ohms
The wiz sez: In solution 1
the resistors dissipate 125.6 mW
the diodes dissipate 192 mW
the total power dissipated by the array is 317.6 mW
the array draws a current of 60 mA from the source.

 

Anyone Please reply.

 

Looks Ok what supply voltage are you working with 5V?

 

Hi Raj,
Post #17 values look OK to me.
ie:
5V- [ 3*1.2Vfwd]= 1.4V/82R =17mA
5V- [2* 1.2Vfwd] =2.6V/150R = 17mA

Assuming that you are using a 5V source for the LED's.
E