Electronic Level Indicator using LM3914

MrCarlos

Joined Jan 2, 2010
400
Hello RodneyB

In the diagram Your post #18.
By the transistor Q1, no current will flow between its collector-emitter junction.
The base electrode is not polarized to achieve this.
 

Thread Starter

RodneyB

Joined Apr 28, 2012
697
Hello RodneyB

In the diagram Your post #18.
By the transistor Q1, no current will flow between its collector-emitter junction.
The base electrode is not polarized to achieve this.
Thank you, what must I do to achieve this, unless I get the Opto to switch a Mosfet?
 

MrCarlos

Joined Jan 2, 2010
400
Hello RodneyB

Connects the collector of transistor in the OPTO to the base of Q1 and the resistance, of the same collector, to VCC.

The resistance value is calculated according to the gain of transistor and the current required at the collector of Q1.
Analyzes data sheets I am enclosing you.

You should also take into consideration the electrical characteristics of OPTO you're using.
The transistor in the OPTO should be able to drain to ground the base current of Q1.
 

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Thread Starter

RodneyB

Joined Apr 28, 2012
697
Hello RodneyB

Connects the collector of transistor in the OPTO to the base of Q1 and the resistance, of the same collector, to VCC.

The resistance value is calculated according to the gain of transistor and the current required at the collector of Q1.
Analyzes data sheets I am enclosing you.

You should also take into consideration the electrical characteristics of OPTO you're using.
The transistor in the OPTO should be able to drain to ground the base current of Q1.
I am wanting to test the opto switching a 10 Watt 12 Volt LED. I don't understand the datasheet as I cant see anywhere the If or Vf for the led in the opto 4N35 data sheet. Looking on other projects it shows a 330R resistor between pin 2 and ground. I want to know how they work that out and where they get the information about the LED from
 

MrCarlos

Joined Jan 2, 2010
400
Hello RodneyB

Well, yes they are a bit confusing the parameters in the data sheets 4N35.
I attached the data sheets Motorola 4N35 OPTO.
I highlighted in yellow the text of interest to see.

On page 1 in the INPUT LED block are:
The IF parameter under the Symbol column whose value is 60 mAmp.
But note that this section (p. 1) are the MAXIMUM RATINGS (TA = 25 ° C unless otherwise NOTED)

Now, see page 2:
In block: INPUT LED see that when a 10mAmp current (IF) flows through the LED, the voltage drop (VF), at its terminals is 1.15 Volts.

In block: OUTPUT TRANSISTOR see that the transistor has a gain (hFE) of 400 when the IC = 2 mAmp. And his VCE = 2 Volts.

In block: COUPLED we find that the collector current @ 20 ° C is only 30 mAmp.
When the LED IF is 10 mAmp. And the collector-emitter voltage VCE in the transistor is 10 volts.

Note that the maximum collector current (IC) to the transistor is 150 mAmp.
Therefore:
The transistor in the OPTO can not handle a load of 10 Watt @ 12 Volt.

Moreover, it notes that the current to the LED in the OPTO is provided by the LM3914 as mentioned by bertus in his message #19.
 

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Thread Starter

RodneyB

Joined Apr 28, 2012
697
Hello RodneyB

Well, yes they are a bit confusing the parameters in the data sheets 4N35.
I attached the data sheets Motorola 4N35 OPTO.
I highlighted in yellow the text of interest to see.

On page 1 in the INPUT LED block are:
The IF parameter under the Symbol column whose value is 60 mAmp.
But note that this section (p. 1) are the MAXIMUM RATINGS (TA = 25 ° C unless otherwise NOTED)

Now, see page 2:
In block: INPUT LED see that when a 10mAmp current (IF) flows through the LED, the voltage drop (VF), at its terminals is 1.15 Volts.

In block: OUTPUT TRANSISTOR see that the transistor has a gain (hFE) of 400 when the IC = 2 mAmp. And his VCE = 2 Volts.

In block: COUPLED we find that the collector current @ 20 ° C is only 30 mAmp.
When the LED IF is 10 mAmp. And the collector-emitter voltage VCE in the transistor is 10 volts.

Note that the maximum collector current (IC) to the transistor is 150 mAmp.
Therefore:
The transistor in the OPTO can not handle a load of 10 Watt @ 12 Volt.

Moreover, it notes that the current to the LED in the OPTO is provided by the LM3914 as mentioned by bertus in his message #19.
Thank you so much for the comprehensive reply. I should have been a bit clearer in my question.

The information regarding the LED is crystal clear now. The transistor section is also clear. What I will be doing is switching the 10 Watt 12 Volt LED through a TIP41 transistor, as per your advice in post number #23.

I wanted to work out the resistor so I could trigger the opto independent of the LM3914

Once again Many thanks

Rodney
 

MrCarlos

Joined Jan 2, 2010
400
Hello RodneyB

Well now we have to play with the data sheets TIP41 transistor.
Open the file: TIP41.PDF in my message # 23.
Select page 5.
Analyzes Figure 9.
At IB = 100 mAmp. the collector current can be 1 Amp.
The load you intend to connect to the collector of TIP41 has the electrical characteristics: 10W @ 12V.

Calculate the current required for such Watts: I = W / V = 10/12 = 0.833 Amp. Close to 1 Amp.
I will assume that these LED’s you intend to connect to the collector of transistor already bring with internal current limiting resistor.
So when connected to 12 volts a current of 833 mAmp. Will flow and the power dissipation is 10W.

So if you apply to the base of the TIP41 a current of 100mAmp. By its collector will circulate a current of about 800 mAmp.

The value of the base resistance is calculated as:
(Vcc - VEB) / IB = (12 to 0.7) / 0.1 = RB.

The value for the current limiting resistor used for the LED in the OPTO was previously discussed.
 

Thread Starter

RodneyB

Joined Apr 28, 2012
697
Hello RodneyB

Well now we have to play with the data sheets TIP41 transistor.
Open the file: TIP41.PDF in my message # 23.
Select page 5.
Analyzes Figure 9.
At IB = 100 mAmp. the collector current can be 1 Amp.
The load you intend to connect to the collector of TIP41 has the electrical characteristics: 10W @ 12V.

Calculate the current required for such Watts: I = W / V = 10/12 = 0.833 Amp. Close to 1 Amp.
I will assume that these LED’s you intend to connect to the collector of transistor already bring with internal current limiting resistor.
So when connected to 12 volts a current of 833 mAmp. Will flow and the power dissipation is 10W.

So if you apply to the base of the TIP41 a current of 100mAmp. By its collector will circulate a current of about 800 mAmp.

The value of the base resistance is calculated as:
(Vcc - VEB) / IB = (12 to 0.7) / 0.1 = RB.

The value for the current limiting resistor used for the LED in the OPTO was previously discussed.
Sorry MrCarlos,

I just seem to be going round and round in circles. I have attached the circuit of the opto switch I am trying to get to work.

To test the switch I have used a 23 Watt 12 Volt Lamp.

I connected the one side of the lamp to the positive and the other side of the lamp to "A" on J2

I connected "c" and "d" to the ground of the 12 volt supply.

Between pins "b" and the positive I connected a 680R resistor. expecting the light to come on.

This did not happen.

When Testing between Pin "b" and ground I get 1. 229 Volts. on R15 and on the base of q1 there is zero voltage. I am not sure where to go from here
 

Thread Starter

RodneyB

Joined Apr 28, 2012
697
Hello,

Have a try with the following schematic:

View attachment 84696

Bertus
Hi Bertus

I have been trying to build this circuit to day but can not get it to work

My Input signals are 3V, 6V and 9V

When applying 9 Volts through the 15K and 1 K to ground voltage divider I get 0.576 Volts on the SIG pin 5.
When applying 6 Volts through the 15K and 1 K to ground voltage divider I get 0.377 Volts on the SIG pin 5.
When applying 3 Volts through the 15K and 1 K to ground voltage divider I get 0.190 Volts on the SIG pin 5.

The LED on Pin 11 remains permanently on, even when I remove the signal voltage.

I then found that I had not connected Pin 8 to ground. After I did this

at 3 volts on the signal before the divider into pin 5 LED's 1 and 11 illuminated
at 6 volts on the signal before the divider into pin 5 LED's 12,3 and 11 illuminated
at 9 volts on the signal before the divider into pin 5 LED's 1,2,3,4 and 11 illuminated

I then put a jumper between pins 3 and 9 and the circuit continued to work in bar graph mode. I thought by leaving this open it operated in dot mode.

Any advice and assistance will be greatly appreciated

Thanks

Rodney
 

Thread Starter

RodneyB

Joined Apr 28, 2012
697
I Am trying to get this opt transistor to switch on a 21 Watt 12 Volt lamp through a tip41 Transistor.

If I remove the Opto and fit the switch from positive to the 120R resistor push the button and the Light comes on. HOWEVER it will not switch through the Opto and not sure how to get it to or to test if my Opto is flat.
 

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ebeowulf17

Joined Aug 12, 2014
3,307
Shouldn't R1 be joining the opto's collector to +12V instead of the Q1 base, and then feed the opto's emitter to Q1's base? Right now there's no way to get positive voltage to Q1's base to activate it.
 

Thread Starter

RodneyB

Joined Apr 28, 2012
697
Shouldn't R1 be joining the opto's collector to +12V instead of the Q1 base, and then feed the opto's emitter to Q1's base? Right now there's no way to get positive voltage to Q1's base to activate it.
I have changed the diagram I cant see on the datasheet the collector emitter current, I don't want to build it until I can confirm that the current is not going to destroy the opto
 

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AnalogKid

Joined Aug 1, 2013
10,986
Change R1 to 470 ohms so you don't overcurrent the input LED. Also, there is no load in the TIP41 collector, so it will try to short out the battery and go boom.

To switch a 1.75 A load with a TIP41, you will need at least 20 mA of base current and would like around 100 mA. Make R2 330 ohms and see if that passes enough current. The rule of thumb for a hard-saturated transistor switch is that the base current should be 10% of the collector current. In your case that would be 175 mA, way too much for the opto. One solution is to change to a power darlington transistor or grow a darlington with an additional discrete transistor. But the TIP transistors have pretty high gain for old power devices, and you might be ok with what you have.

ak
 

ScottWang

Joined Aug 23, 2012
7,397
What are you trying to do for this circuit?
What's the load, are you trying to burn the Tip41?
The values of R1 and R2 are too small.
 
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