Thank you, what must I do to achieve this, unless I get the Opto to switch a Mosfet?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.
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 fromHello 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.
Thank you so much for the comprehensive reply. I should have been a bit clearer in my question.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.
Sorry MrCarlos,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.
Hi Bertus
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 optoShouldn'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.
Hi MikeAre you really using a 9V battery like this?
Thank You the Load is 10 Watt LEDWhat 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.
How are the V/I of 10W LED?Thank You the Load is 10 Watt LED