Astable Multivibrator help!

Sometimes an AC coupled multivibrator will not startup if the circuit is balanced too perfectly. In this case the circuit could very well end up with both transistors on and both LEDs lit.

One way to check if it would start to oscillate would be to momentarily ground one of the transistor base connections. If this causes the flasher to start then you know you are too well balanced. One way to imbalance the circuit at startup would be to add a small

mkaras

Thanks for the quick reply. I tried that, and if I ground the base on the left transistor, the left LED gets brighter (only a .15v increase across the LED). If I ground the base on the right transistor, nothing happens at all. This is so weird.

You know, I guess I should mention my purpose here, in case there is a better way. I want to make two 12v light bulbs blink alternately. They don't have to be 12v bulbs, as I'm sure I can rig up what I need with LED's, but it would be nice.

Maybe something with a 555? I sure wish I could find my notebooks from school.

Railroad crossing light effect.

I suspect any digital oscillator circuit that can sink and source current. Current levels? LED1 on when sinking. LED2 on when sourcing. _LED1___ ____| |____LED2___

A 555 set up as an astable might be ok if you stay within it's sink/source specs.

Keep in mind that there are guys on here that work with stuff 4000x more complicated. So it's kinda like asking Einstein to add 2+2...

D from BC British Columbia Canada.

Are you breadboarding this? I am wondering if you've managed to punch through the bottom of the breadboard and one or more of your components are shorting out on the metal backing plate.... Trust me on this. Breadboards do wear out!

It's a simple enough circuit to move to a different area of the board.

-mpm

Are you using NPN transistors? If you ground the base of one of those transistors, the corresponding LED should go _off_. Take out the capacitors, and see if (a) the LED's light with the 39k resistors to +9 and (b) if they go out with the 39k resistors removed or the transistor bases shorted to ground. If the answer to either of those is 'no', then there's something more basic than the multivibrator going wrong.

Hello,

I think your multivibrator is too fast.

Replace 39k with 220k.

@+

Vincent

"Gabe Knuth" a écrit dans le message de news: snipped-for-privacy@v4g2000hsf.googlegroups.com...

"vincent.thiernesse" a écrit dans le message de news:479866e5$0$888$ snipped-for-privacy@news.orange.fr...

No, replace 10microF with 47 microF

If you ground the base of the left transistor, the left LED should turn OFF. What is the code on the transistors? They may be PNP, or the sequence of the leads may not be what you think; the sequence may not be the usual EBC.

Tam

If you ground the base of one transistor and the led in that transistor's collector circuit gets brighter that transistor is either bad or connected wrong or the wrong gender.

Grounding the base should be switching the transistor off - no current in the collector.

Hi Gabe. My name is Eddie. I have basically the same schematic and the difference is that the 1N914 small signal diodes are not in the circuit. The emitters are directly tied to ground on my schematic. Not to be alarmed. Your multivibrator is working but much too fast for the human eye to detect. Since the 2N3904 transistors are basically switches and the 10uf capacitors charge and discharge alternately to light the leds I would remove the 1N914's and let the Transistors do its intended job. Just my opinion. Good luck.... Eddie

Gabe-- With a +9 volt supply, the diodes are required to protect Q1, Q2 from emitter-base breakdown.

The circuit is marginal for starting because the base drive is a bit high--the circuit can lock up unless R(b) >= hfe * R(c). If you happened to get a pair of high-gain 2n3904s, the circuit might "hang" with both transistors on.

In practice, though, I find saturated 2n3904 transistors generally have enough gain left for the basic astable circuit to start as long as i(c) >= 20 * i(b). I much prefer i(c) >= 50 * i(b). Other topologies avoid this startup problem, e.g., connecting the timing resistors from Q_(b) to Q_(c).

Most likely, though, is a wiring or component error. When you ground the base of either transistor, the grounded-base transistor's LED should go OFF, and the other LED should turn fully ON. If not, check the voltages at the bases of the transistors, etc.

The flash rate, as drawn, should be visible. The time constant is 39k x 10uF = 390 milliseconds. Knowing the threshold voltage to be Vbe+ a diode drop = +1.2v, the initial voltage on the timing cap of -(9v -

HTH, James Arthur

Why not use a microcontroller? They're cheap enough now, particularly the Arduino bases stuff.

They have 16 digital I/O ports of which 6 can act as PWM ports, another

They use an IDE called Processing which is very much like C and very easy to understand.

All you'd have to do is drive the transistor from one of the digital outputs and use some current limiting, but you could do it easily.

I use both the Arduino Diecimila (It's about $34), and the Bare Bones Board ($17)

The Arduino board site is:

And the Bare Bones Board site is:

On Wed, 23 Jan 2008 22:19:03 -0800, Gabe Knuth wrote: [about astable]

When you troubleshoot this, and find the problem, you'll probably go "D'oh!", but you'll never make that same mistake again. ;-)

Good Luck! Rich

On Jan 23, 10:19 pm, Gabe Knuth wrote: [....]

LED LM555 ---/\\/\\--!!---GND ! ---------- ! +------/\\/\\--------- ! === ! GND

The circuit may be locked up with both transistors stuck in saturation.

In order to guarantee oscillation, you need to add a diode logic OR function which provides base drive only if one OR the other of the transistors is OFF, like shown below, a lock-up cannot occur because there will be insufficient base drive. You get better switching by placing the LEDs in the emitters, and you get around the breakdown problem by reducing the voltage through a common resistor in series with the battery lead, in this case it is a R1xIled=150x0.015=2.25V drop. The LED currents are the same, about 15mA, because the total resistance in series with Vbatt is the same. The frequency is about 2Hz: View in a fixed-width font such as Courier.

. . Vbatt . | . [150] . R1| . .----------+----------. . | | . [330] [330] . R2| D1 D2 R3| . +---|>|-+-----+-|

Yes that sure looks like it;

There is a simpler circuit that does it.

Modified:

If you remove the positive feedback both sides will settle with just about 2V on the collector of the transistors, depending on the HFE. As a result both sides will have a gain well over unity. With the positive feedback, it just takes a little noise in the transistors to get things started.