How are the V/I of 10W LED?
Why you need to used the opto transistor?

The voltage is 12volts the current is 1 amp

I am using the Opto transistor to switch the transistor because eventually I am wanting the switch to the Opto to come from a LM3914N.

I am thinking its best to get the Opto to switch in a relay rather than the transistor. at this stage I will take any advice

 

Earlier it was a 21 watt lamp? Either way the concepts are the same, but it does get easier if we're dealing with half the current!

 

Maybe you don't need the opto transistor.
What I thought is like this way, the arrows are showing as the direction of current flowing:
LM3914 output sink 10mA ← Resistor ← b,2SA684(P) 100mA, c to TIP42(P) c,e ← b,TIP42(P) 1A, e to 12V,c → Resistor → 12V/10W LED → Gnd.
Or if you want, maybe you can try the mosfet.

 

Earlier it was a 21 watt lamp? Either way the concepts are the same, but it does get easier if we're dealing with half the current!

Yes sorry that is correct it is a 21 Watt Lamp, I also have a 10 Watt LED that I wanted to Try

 

I have Built the attached circuit.

Firstly it only works in Bar graph mode

Secondly LED 9 is always on.

On the signal Input I use three voltages 3, 6, 9 volts.

Not sure why I am getting this result.

 

If you using the N mosfet will be as following, the arrows are showing as the direction of current flowing:
LM3914 output sink ← Resistor, 1 mA ← b,2N3906(P) e to 12V, c → G(10K to Gnd), N mosfet, S to Gnd, D ← Resistor ← 12V/10W LED ← +12V

 

The schematic looks ok, so perhaps you have a wiring error somewhere?

 

Hello RodneyB

I see you're having a very hard time designing your Electronic Level Indicator using LM3914
I say because the threads that you've started:
http://forum.allaboutcircuits.com/threads/switching-a-tip41-from-an-opto-transistor.111401/
http://forum.allaboutcircuits.com/threads/lm3914n-wont-go-into-dot-mode-and-led-9-always-on.111404/
and the original this:
http://forum.allaboutcircuits.com/threads/electronic-level-indicator-using-lm3914.110254/

Before continuing I would like to know what the simulator you are using.
It seems ISIS Proteus but have not figured out what version you are using.
I wonder which simulator are using to provide more effective assistance.


.

 

A little speculation: Substitute 1k resistors for LED's, connect gate of P CH FET's to IC pins to directly drive 10 W LED's. Or use PNP darlington's. Asuming 12 V supply to IC.

 

Yes sorry that is correct it is a 21 Watt Lamp, I also have a 10 Watt LED that I wanted to Try

If the LED is 10W/12V then the current I = W/V = 10W/12V = 0.833A

 

What is the voltage range (min-max) being measured at input Pin5?
What is the desired current load at each LED output pin?
Will all LED outputs be used?
The input signal has to be 1.5v less than the supply voltage, so if the supply voltage is 12v, the input can be as high as 10.5v

 

Sorry too busy to draw anything up tonight, but I think you should be able to get good results without using a mechanical relay. If this is for your see saw level indicator project you're not going to want piles of relays clicking and taking lots of space.

Have you tried my earlier suggestion, Scott's, or any of the recent ones from your other thread? The way you've got the opto wired to the TIP41 can't work, but once it's rewired the TIP41 should work, at least partially. It may be difficult to get enough lamp current through it because your opto can't provide enough current through the base, but wired properly it should work enough for you to at least see that it's wired properly!

Getting the full current through your lamp should be much easier with a Darlington transistor or a MOSFET instead of your TIP41. The Darlington (TIP120 is a good example) offers a current gain of around 1000 if I remember right, so you'd only need a couple milliamps through your opto (or directly from the bar graph driver) to control it. A MOSFET might be even better, but I'm too sleepy to get the explanation right at this point.

 

If a project separate to three threads then the helper may not easy to get the whole picture, so we decided to merged them to the beginning one.

 

If a project separate to three threads then the helper may not easy to get the whole picture, so we decided to merged them to the beginning one.

Thank You. I broke it up into sections as I was trying to learn about the sections separately. All I achieved was total confusion.

 

Thank You. I broke it up into sections as I was trying to learn about the sections separately. All I achieved was total confusion.

You still can learn the skills when they put together, you just label which part as the part of LM3914 or the part of drivers.

 

Hello RodneyB

I see you're having a very hard time designing your Electronic Level Indicator using LM3914
I say because the threads that you've started:
http://forum.allaboutcircuits.com/threads/switching-a-tip41-from-an-opto-transistor.111401/
http://forum.allaboutcircuits.com/threads/lm3914n-wont-go-into-dot-mode-and-led-9-always-on.111404/
and the original this:
http://forum.allaboutcircuits.com/threads/electronic-level-indicator-using-lm3914.110254/

Before continuing I would like to know what the simulator you are using.
It seems ISIS Proteus but have not figured out what version you are using.
I wonder which simulator are using to provide more effective assistance.


.

I have Proteus 8 but have no idea how touse the simulator

 

I drawn two LED drivers circuit for your reference, the one used bjt and another one used n type mosfet, if the 12V LED can be connected to 12V directly then you no need to use the current limiting resistor.

What I calculated the current from LM3914, the first is to simulates a 2V load to replace the 2V led, and calculated from the real load, how big the current is the LED needed and the PNP bjt or N mosfet and to the LM3914.

 

Thank you for this. For M2 I have IRF730 which will easily switch this current. R8 does this need to be a 10 Watt resistor?

I can not get a 2SA684 from our local electronics supplier but can get a BC557 will this do the Job. I have Drivers for the LED's if this will be better to eliminate such a large resistor for current limiting for the LED

Using this circuit then I wont have to use the opto is that correct

I drawn two LED drivers circuit for your reference, the one used bjt and another one used n type mosfet, if the 12V LED can be connected to 12V directly then you no need to use the current limiting resistor.

What I calculated the current from LM3914, the first is to simulates a 2V load to replace the 2V led, and calculated from the real load, how big the current is the LED needed and the PNP bjt or N mosfet and to the LM3914.

View attachment 86070

View attachment 86069[/QUOTE

 

I drawn two LED drivers circuit for your reference, the one used bjt and another one used n type mosfet, if the 12V LED can be connected to 12V directly then you no need to use the current limiting resistor.

What I calculated the current from LM3914, the first is to simulates a 2V load to replace the 2V led, and calculated from the real load, how big the current is the LED needed and the PNP bjt or N mosfet and to the LM3914.

View attachment 86070

View attachment 86069

I Had searched the Internet and found this circuit which I adapted. Sorry it is Hand drawn but we are in load sheading today so cant use the computer I usually use.

 

I have tested the attached circuit. It does not work despite there being 7 Volts on the gate.

However when I replace it with a relay it works. So I must be doing something wrong on the Mosfet side that I just can not see.

Really would appreciate being steered in the right direction