Need help in change 6 volt to 12 volt circuit

[Les Jones]

I don't know how how you get 10 ohms. 0.6/1.68 = 0.357 ohms. If the sample of the LED you obtain is near the top of the voltage spread then you will not get the full output with a 12 volt supply. As you have no data on the LDR you will have to measure it's resistance in the dark and in bright light.

Les.

 

[audioguru]

The high power LED needs a current regulated supply, not just a resistor.
The current regulation circuit in series with the LED needs about 1V.
Some of those LEDs need 12.9V, you get whatever they have.
Then you need a supply that is at least 13.9V or some of the LEDs will not make any light.

The LED uses 1.68A and you guessed a 10 ohm resistor. But the formula for the resistor is 0.6V/1.68A= 0.357 ohms. You were very wrong.

 

[mrel]

audioguru
So is 10 ohms right or wrong?
The circuits Les and alec_t post would this led work with there circuit or do i need to build current regulated supply for the led than connect to the circuit?

audioguru
Do you have current regulated supply circuit for the led that you can post .
mrel

 

[audioguru]

Some or all available LEDs will not work from a 12V supply.
The circuits posted by Les and Alec have regulated current for the LED.

The resistor for LES's circuit is what I said before and you were VERY WRONG!
You calculated 10 ohms which would produce a current of only 0.6V/10 ohms= 60mA but the LED has a maximum allowed current of 1680mA (1.68A).

 

[mrel]

I don't know how how you get 10 ohms. 0.6/1.68 = 0.357 ohms. If the sample of the LED you obtain is near the top of the voltage spread then you will not get the full output with a 12 volt supply. As you have no data on the LDR you will have to measure it's resistance in the dark and in bright light.

Les.

Les
Can you explain how you arrive at 0.6 volt for led since led take 12volt.
I understand that 1.68 is the current goes to the led.
In dividing 1.68 into 0.6 volts you get 0.357 ohm that i understand.
The i don't understand where you get 0.6 volts come from.
Is there such a resistor valve that low 0.357 ohm on the market?
mrel

 

[Les Jones]

The 0.6 volts is about the base emitter voltage at which the transistor starts to conduct. When the transistor conducts it pulls down the gate voltage of the MOSFET which increases the source to drain resistance of the mosfet. This results in the circuit trying to maintain 0.6 volts across the resistor which means it is trying to maintain the led current at 1.68 amps. You will have to make the 0.357 ohm resistor from a combination of standard values. You could get 0.4 ohms using three 1.2 resistors in parallel.

Les.

 

[audioguru]

The LED does not take 12V. Some of them are 10.5V and others are 12.9V. Also the LED voltage changes as it warms up so you feed it with a current, not a voltage. The current must be regulated.
You need to make this Current Regulator circuit for the LED since you probably want ANY ONE of that LED to work.
The current for the LED flows through the 0.357 ohms resistor. When it reaches 1.68A then it has 0.6V across it and also from the base to the emitter of the transistor which causes the transistor to conduct which causes the Mosfet to stop increasing the current.

 

[mrel]

Here's a variation on Les' circuit, using the other half of the 393 (dual comparator IC) instead of the BC337 to provide current regulation, and with hysteresis added to the first comparator so that passing clouds etc don't cause false switching of the light. Trimmers allow adjustment of hysteresis, dark/light threshold and LED current.
View attachment 94418
The blue graph simulates LDR resistance. For 'kV' read 'kΩ'.
Note that, as with Les's circuit, a small heatsink will probably be needed for the FET (as well as a large heatsink for the LED). A 100uF cap across the supply, as Les shows, would be advisable.

Here's a variation on Les' circuit, using the other half of the 393 (dual comparator IC) instead of the BC337 to provide current regulation, and with hysteresis added to the first comparator so that passing clouds etc don't cause false switching of the light. Trimmers allow adjustment of hysteresis, dark/light threshold and LED current.
View attachment 94418
The blue graph simulates LDR resistance. For 'kV' read 'kΩ'.
Note that, as with Les's circuit, a small heatsink will probably be needed for the FET (as well as a large heatsink for the LED). A 100uF cap across the supply, as Les shows, would be advisable.

Alec-t
Might try to building your circuit.
Got few question (1) the circuit show lm339 chip you did not put pins number in the lm339 drawing you just got plus & minus sign in the drawing for the lm339.
(2) your post mention 393 chip which is correct chip lm339 or 393.
(3) Is there replacement for the IRFH5306 chip,the IRFH5306 chip is surface mount chip ,is there replacement to Dip mount chip.
mrel

 

[audioguru]

Look on the datasheet of the IC to see what it is and to see the pin numbers. An LM339 has 4 comparators in it (14 pins case) and an LM393 has only 2 of the same comparators (8 pins case).
Since the supply for the circuit is 12V then just about any N-channel Mosfet will work, even a common old IRF540 or IRL540. I have never seen the tiny package used on the IRFH5306 before.

 

[alec_t]

Audioguru has answered your questions. I used LM339 and IRFH5306 models in the simulation simply because they were to hand in the libraries I have on my PC.

 

[mrel]

Look on the datasheet of the IC to see what it is and to see the pin numbers. An LM339 has 4 comparators in it (14 pins case) and an LM393 has only 2 of the same comparators (8 pins case).
Since the supply for the circuit is 12V then just about any N-channel Mosfet will work, even a common old IRF540 or IRL540. I have never seen the tiny package used on the IRFH5306 before.

audioguru
Got the data sheet for the LM 339 chip picture show triangle in rectangular form see attachment file picture (2).
Picture (1) attachment file I circle picture asking where plus & minus sign are what pin are these ,the datasheet show triangle only have three pins not four pins.
mrel

 

[audioguru]

The VCC pin #3 connects to the positive supply voltage. The GND pin #12 connects to GUESS WHAT! These two pins power the IC and all the comparators in it.
When a comparator is already powered then it has two input pins and one output pin for a total of 3 pins, not 4.
The comparators in the schematic show 5 pins because the person who drew the schematic is stupid. Only one comparator needs to have its power pins shown.

Since the LM339 has 4 comparators in it then you simply pick the ones you want then learn how to disable the extra ones that you are not using.

 

[alec_t]

The comparators in the schematic show 5 pins because the person who drew the schematic is stupid. Only one comparator needs to have its power pins shown.

Thanks for the uncompliment, AG . Actually, LTspice needs 5 pins for each comparator. The schematic is that of a simulation; not a working drawing. Anyone building a practical circuit has a choice of which comparators to use for which function, so for that reason pin numbers are not shown either.

 

[audioguru]

Sorry, Alec. I did not know the schematic was yours from LTspice. Actually, many SIM programs do not show ANY power supplies for comparators and opamps.

 

[mrel]

Hello
Using this circuit see attach picture of circuit I use to light a 3 volts led the led turn on when no light hit cadmium sulfide photocell.
What need to do to change this 6 volts circuit to 12 circuit using 12 volt led as mention in attach file
mrel

I found new circuit see attach file .
Will this circuit work with 12 volt 1.68 amp led.
Don't look this circuit have a current regualor for led mention in this post.
mrel

 

[audioguru]

You do not have an LED that works when it gets 12V. Your LED might need 10.5V or it might need 12.9V or any voltage in between. If the supply voltage for the new circuit is about 15V then a current-limiting resistor can be added in series with the LED and the circuit will turn it on when the photo-transistor is in the dark and turn it off when there is light.

 

[mrel]

You do not have an LED that works when it gets 12V. Your LED might need 10.5V or it might need 12.9V or any voltage in between. If the supply voltage for the new circuit is about 15V then a current-limiting resistor can be added in series with the LED and the circuit will turn it on when the photo-transistor is in the dark and turn it off when there is light.

Audioguru
Are you commenting on my new post or my old post i post two different circuits?
mrel

 

[audioguru]

Audioguru
Are you commenting on my new post or my old post i post two different circuits?
mrel

New Circuit post 10-13-2015 shows a 12V LED that you do not have (yours is 10.5V to 12.9V) and a 12V power supply that will not light the LED if it needs 12.9V. The circuit will work if the supply voltage is about 15V and the LED has a properly calculated series current limiting resistor.

"Mr. electronics parts store clerk, can you test a few hundred of your "12V" LEDs and find one that lights at exactly 12.0V, plzzz?
Oh, you couldn't find one today but maybe next month?"