# led flasher

Discussion in 'Homework Help' started by bug13, Mar 4, 2012.

Feb 13, 2012
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2. ### Wendy Moderator

Mar 24, 2008
21,838
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Several things actually. Where did you get that schematic?

The big item is you can never run an LED without some kind of current limiter. LEDs do not limit current the slightest. Neither do transistors. If you actually ran this circuit the current surge (due to lack of current limiting) has blown both Q2 and LED1. You need a resistor next to LED1, something like 330Ω.

The base oscillator may work, but I am not familiar with that configuration off the top of my head, which is why I asked where you got the schematic.

AAC has local albums hosting and also allows attachments, gifs and png file formats preferred. You can used them to do something like this...

I have a tutorial article about LEDs. It also has quite a few flasher circuits, though most of them are based on 555s.

LEDs, 555s, Flashers, and Light Chasers

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3. ### bug13 Thread Starter Senior Member

Feb 13, 2012
1,682
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Hi bill, this actually the first circuit I ever design myself, the main thing I am looking for generating a oscillating current/voltage, the led is just something visible I want to attach to the circuit.

4. ### Wendy Moderator

Mar 24, 2008
21,838
3,047
I've been analyzing this circuit. An LED is not a resistor, so the original comment is correct. For the moment, elliminate the LED and replace it with a resistor. You can detect the logic state by adding a third transistor and monitor the logic state with the following circuit.

Looking at it though, you have other problems.

The capacitor starts off with no charge. This effectively couples the base of Q1 to the collector of Q2, and puts a ground state on Q1 base, which turns off Q2. This is a stable condition. I see what you are thinking, R1 will charge C1, but not so, as the base of Q1 is going to be ground or 0.7V. You also have a problem with the size of R1, try dropping its value, a lot.

Basically this is a form of monostable I think. If you remove the capacitor, replace the LED with a resistor, and look objectively this is a non inverting amplifier. The basic digital instability is not there.

5. ### bug13 Thread Starter Senior Member

Feb 13, 2012
1,682
62
thanks for you analysis, now I can understand why it doesn't work.