Is this schematic right?

hi i´m micheno.
I´m doing a led flasher on a soldarble breadboard. I use this schematic. It´s not working. I think i have a mistake in schematic. Please help.
Thanks for time and effort.
Michal aka micheno

ledFlasher.png

ledFlasher.png

More like this :wink:

thank you so much
i hope it will work. :ballot_box_with_check::ballot_box_with_check:

larryd:
More like this :wink:

The capacitors are backwards. The positive side of C1 goes to the base of Q2, and the positive of C2 goes to the base of Q1.

@Larry is right. It is a negative going sawtooth wave on Q1/Q2 base relative to ground, so the capacitor (-) should be connected to the bases.

For example, when Q1 is biased "off", C1 charges through R1, L1 and Q2 B-E junction. When approaches full charge, Q2 current approaches zero, Q2 turns off and Q1 turns on. When that happens, the (+) terminal of Q1 is near ground potential, so the (-) terminal has a negative voltage referenced to ground.

There will be 9V negative voltage spike on the base when the transistor and led go from conducting to not conducting. Your transistors will wear down in time and stop working. (some transistors are ok with that voltage but some are not). A diode from base to emitter will protect transistors. Cathode to base.
If you use higher supply voltages, your transistors stop working sooner.

Edit: Are you positive about which way the capacitors should be?

If you are building a flasher for practical not educational reasons, CMOS oscillators are much simpler and have more flexible design options.

LMI1:
Edit: Are you positive about which way the capacitors should be?

Not sure who this is addressed to, but I believe my explanation is adequate if not comprehensive.

Sure. There are many ways to blink a led. Usually the hobbyist will wear out faster than the transistors.

Capacitors. This is a difficult job for a electrolytic capacitor. The base will newer be more positive than about 0.7V. Collector will be about 9V-Uled, so I would put + terminal to collector.

LMI1:
Sure. There are many ways to blink a led.

One way, is to use an Arduino. :slight_smile:

This circuit was the basis of a contest when I was in school.

The company (I think it was either Philips or Mullard) provided the two transistors, being the expensive and hard-to-get components and one had to build it and submit it for judging. I think it may have had a third transistor to switch the incandescent bulb.

LMI1:
Sure. There are many ways to blink a led. Usually the hobbyist will wear out faster than the transistors.

Capacitors. This is a difficult job for a electrolytic capacitor. The base will newer be more positive than about 0.7V. Collector will be about 9V-Uled, so I would put + terminal to collector.

No.
When the LED is on, the transistor is on, making the collector effectively at ground. C1 charges through R2 until it reaches about .7v to turn on Q2. When Q2 is turned on, the collector is at ground and C2 offers low impedance to the R3 C2 junction, which pulls the base of Q1 down, turning it off.
When the LED is OFF, the collector is high and the capacitor discharges through R2. R1, L1 until the base of Q1 gets high enough to turn on.
Lather, rinse, repeat.

The + of the capacitor goes to the base of the transistors.

I became curious and simulated this with LTSpice and this is what I got.
https://www.analog.com/en/design-center/design-tools-and-calculators/ltspice-simulator.html
I tried to include simulation file too, but asc-files cannot be included.
Edit: Base resistors are not equal because this circuit has one stable state, when everyting is perfectly symmetrical. That is this may not start oscillate in simulator if everything is symmetrical.
A lot of post about a simple circuit.

LMI1:
Base resistors are not equal because this circuit has one stable state, when everything is perfectly symmetrical. That is this may not start oscillate in simulator if everything is symmetrical.

That may occur in LTSpice, but never in real life! :astonished: The circuit has positive gain.

SteveMann:
The + of the capacitor goes to the base of the transistors.

Sorry, but no, it doesn't. :roll_eyes:

It should be pretty obvious that when either transistor is turned off, the collector voltage rises, at the very least to the supply voltage less the LED drop, so for a 9 V supply, no less than 6 V. The transistor base at the other end of the capacitor is constrained to be at no more than 0.7 V as you have specified, so is very much negative.

SteveMann:
When the LED is on, the transistor is on, making the collector effectively at ground.

And for a bipolar transistor, this is not the case. As is so often explained here, to drop the collector voltage lower than the base voltage, you must saturate the transistor with extra base current. While this does happen in this circuit due to the current fed through the capacitor, the collector voltage drops only slightly below the base voltage, and the effect is brief.

A very minor voltage difference compared to the 5.3 V described above and described by the tiny "S-wave" notch in LMI1's tracing.