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Topic: As fast as an NFL player? (Read 4525 times) previous topic - next topic

Techone

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The transistor is also a 2N3904


Than, on your picture, you did not have a base resistor - the pin 10 is directly to the base. That is very bad. It may damage the chip output pin. And you did not properly connect the current limiting resistor. I recommend you look at my picture and that will tell you how to connect properly a transistor controling a led. See the two resistor. One is for the base and the other is the current limiting of the Led.

In your case, the base resistor will be the same then mine, but the value of the current limiting resistor will be different value.

Techone

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I need to drive two separate lasers, though. I'm assuming it's better to use two separate transistors utilizing two separate pins to switch them?


Yes.

astrodad


Quote
The transistor is also a 2N3904


Than, on your picture, you did not have a base resistor - the pin 10 is directly to the base. That is very bad. It may damage the chip output pin. And you did not properly connect the current limiting resistor. I recommend you look at my picture and that will tell you how to connect properly a transistor controling a led. See the two resistor. One is for the base and the other is the current limiting of the Led.

In your case, the base resistor will be the same then mine, but the value of the current limiting resistor will be different value.


Actually, I did put a resistor on before the base. I didn't connect one from the transistor the the LED though. I didn't think I needed one since the batteries are connected directly to the laser in their housing.

Techone

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I didn't connect one from the transistor the the LED though. I didn't think I needed one since the batteries are connected directly to the laser in their housing.


In that case, you are correct. But the resistor connect to the base is incorrect. <-- For a switch operation ( on / off ). But you do connect a base resistor between the base and collector, only if you use the transistor to amplify a small signal. That configuration is call : a feedback resistor. It usaly a high value and set the transistor between Saturation and cutoff. For a switching, the transistor need to be in saturation or cut-off. Just see my post picture. The 4.7 K resistor connect between base and the output of the FDTI board.

astrodad

Ok, I'm not sure I'm comprehending what you're saying but I'll take a look at your picture some more. Can you send a circuit diagram? Or maybe a link to a site with a good explanation of what you're trying to tell me?

Thanks for the help!

Techone

Ok. Here a site about what I am trying to say.

http://www.kpsec.freeuk.com/trancirc.htm

And here a simulation program, this program is design for easy to understand.

http://www.new-wave-concepts.com/pr/spark.html

To buy it, you have to check "Sales" of the software company.

Here a store in Canada, where I got Circuit Wizard Pro from. The program is call Bright Spark. It will give you an idea how much it is. And it is not cheap. Easy to understand for someone with no electronics background

http://www.abra-electronics.com/products/Bright-Spark.html?setCurrencyId=1

Here a free one. http://www.falstad.com/circuit/. But not easy to understand if someone have no electronics background.

And here a picture of what I was talking about.

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