Need help in driving a 1 Watt Led using LDR

[pmadithya]

Hi,

I found a circuit on this blog to make a sensor to light up LED's using LDR( dark light sensor).
https://www.buildcircuit.com/darklight-sensor-using-transistor/
I replaced the BC 547 transistor with BD 679 and also with 2222A npn transistors.

The problem is that i want to light up a 1 Watt Led which runs on almost 300mA, however this circuit is providing only close to 10mA.

 

[phobos1]

sorry its actually 7ohm 1 watt in the previous one i kept on voltage or led as 6 this is right i.e (Vcc-Vled)/Amps so 6-3.5=2.5 2.5/.35= 7 ohms and .35^2*7= .8 so 1watt thus 8.2ohms/1watt is good

 

[BradtheRad]

You need to reduce the ohm values of R1 and R2.

R2 is the bias resistor. It turns on the transistor to pass greater or lesser current to the led.

R1 is a safety resistor. It is there in case the transistor were ever biased too intensely, to the point where it permits overcurrent to the led.

 

[Eshal]

I agree with BradtheRad.
Go with him.

 

[pmadithya]

Hi,

So what should be my ideal values of R1 and R2.

I used R1= 10ohms and R2= 100k Ohms. Still my current reads only 12 mA.

 

[Eshal]

I think, you can calculate your resistor value with your required load current which is 300mA.
You can use combination of several formulas.
Do you know how to do analysis of the circuit?

 

[betwixt]

As I've posted elsewhere, it will work if you use the darlington BD679 device without changing the LDR or it's series resistor, BUT the value of R1 must be dropped. The calculation for it's maximum value is:

R = (Battery Voltage - LED forward voltage - VCEsat of the BD679) / LED current.

So if we assume LED Vf is 2.5V and the battery is 6V and from the data sheet he worst case VCEsat of the BD679 is 2.5V, R1 = (6 - 2.5 - 2.5)/0.3 = 3.3 Ohms.
I suspect your LED has a higher Vf though and the only way to make more current flow will be to increase the battery voltage.

Brian.

 

[Eshal]

Darlington can increase current rating. Is it?
Darlington pair beta is given as beta of one transistor is multiplied by beta of second transistor i.e. beta=beta1*beta2
Is it possible to use one transistor which could have beta equal to the multiple of both?
Well, why I didn't think of darlington. Good point.

 

[betwixt]

Darlington transistors don't have higher current rating but they do have lower base current requirement. Using a single transistor would cause you problems because R2 would not allow enough base current to flow to turn most transistors on enough to conduct 300mA. The problem with dropping R2 to allow more base current is that the effect of the LDR becomes less until you reach high light levels but a Darlington doesn't need that extra current anyway. The drawback to a Darlington is it has a higher Vbe (two B-E junctions in series) and it has a poorer saturation voltage (VCEsat). You have to deduct VCEsat from the available voltage available for the LED.

In theory you could use a single transistor if it had enough gain but you would need a Hfe of more than 5,500 which would be difficult to find. You could make a Darlington out of two transistors, the first a high gain general purpose transistor and the second a smal power transistor as long as their beta's multiplied to 5,500 or more.

Brian.

 

[Eshal]

Nice explanation.
@pmadithya
Need further help? I think you want us to solve your circuit mathematically. Huh?????

 

[pmadithya]

Hi,

Let me just try with R1 = 3.3. Thanks for the excellent explanation Brian. Thanks Eshal.

 

[pmadithya]

Hi,

I've tried the same with 5 W resistors and it doesn't do sensing. Please suggest me a better circuit.

 

[betwixt]

The power rating of the resistors is irrelevant.

Try wiring it as I suggested then short out the collector and emitter pins of the BD679. If it doesn't pass 300mA (I suspect it wont) your ONLY option is to increase the supply voltage. My reasoning is that many 'high power' LEDs need more voltage across them than your circuit may be able to provide. If the voltage is less than the LED needs to make it pass 300mA there is no way you can change values to make it work.

If it does pass 300mA then you need to increase the bias current available to the transistor by dropping the value of R2 but this will have the side effect of making the LDR less sensitive. I would not suggest dropping it below say 10K, keeping it's value high is preferable.

Brian.

 

[kam1787]

My reasoning is that many 'high power' LEDs need more voltage across them than your circuit may be able to provide. If the voltage is less than the LED needs to make it pass 300mA there is no way you can change values to make it work.

Can you check the specs for the LED? Or wire it in series with a resistor [start with say 18 ohms, then decrease] directly to the 6V.

Does the LED have a good heat sink?