WOW! Two transistors and a couple of passives.
A perfectly simple circuit.

Nowadays this circuit would require an Arduino, as a bare minimum.

 

WOW! Two transistors and a couple of passives.
A perfectly simple circuit.

Nowadays this circuit would require an Arduino, as a bare minimum.

Actually, there is a 1-transistor version.

 

Actually, there is a 1-transistor version.

Yeah, but good luck finding a young'n who knows what a UJT is.

OTOH, this has been covered in other threads, but I'll throw it in here as a "true" 1-transistor design:

http://www.cappels.org/dproj/simplest_LED_flasher/Simplest_LED_Flasher_Circuit.html

ak

 

And of course, the zero transistor version;

 

Hello,

An other version will be the PUT oscillator:

See this page for more info (scroll down about halfway the page):
https://wiki.analog.com/university/courses/alm1k/alm-lab-scr

Bertus

 

Here is another zero transistor oscillator circuit.

 

When capacitor is charged, current through capacitor and base Q1 dramatically drops,

how dramatically?
thx first.

 

how dramatically?

We are interested in violet marked region, where collector voltage V(3) starts drop:



Let us see on base Ib(Q2) and capacitor I(C1) currents under microscope:

 

We are interested in violet marked region, where collector voltage V(3) starts drop:

View attachment 296121

Let us see on base Ib(Q2) and capacitor I(C1) currents under microscope:

View attachment 296122

here is my multisim simulation. no oscillation. q2 collector is always about 3v. and q1 base is always about 0.7v. I guess it's from speaker. because when I use resistor. it oscillate well. and when I use inductance. it oscillate differently.

 

here is my multisim simulation. no oscillation. q2 collector is always about 3v. and q1 base is always about 0.7v. I guess it's from speaker. because when I use resistor. it oscillate well. and when I use inductance. it oscillate differently.

Slightly changed circuit works with all your loads.
Use diode parallel to inductive load else transistors will breakdown by hundreds volt spikes.

 

Slightly changed circuit works with all your loads.
Use diode parallel to inductive load else transistors will breakdown by hundreds volt spikes.

View attachment 296401

View attachment 296403

View attachment 296404

View attachment 296406

what simulation software do you use?I think multisim's speaker is not so well for simulation.
thx.

 

what simulation software do you use?I think multisim's speaker is not so well for simulation.
thx.

It is free simulation software LTspice XVII.
Download from: https://www.analog.com/en/design-center/design-tools-and-calculators/ltspice-simulator.html
Additional library by @Bordodynov (where speaker model is): http://bordodynov.ltwiki.org/.

ADDED:
XLV1 is not speaker model, but it is LabVIEW instrument. Read manual here:
https://knowledge.ni.com/KnowledgeArticleDetails?id=kA03q000000YI4sCAG&l=ru-KZ

 

In LTspice it is possible to record the signal obtained in the simulation in WAV format. The resulting WAV can be played back on the computer. But you have to make this file from a signal that is less than 1 V in amplitude. The signal must be made without a constant component.

 

We are interested in violet marked region, where collector voltage V(3) starts drop:

View attachment 296121

Let us see on base Ib(Q2) and capacitor I(C1) currents under microscope:

View attachment 296122

here is what I leaned,have a look:
first. let's put off c1, to see what is the stable state:let voltage of q2 to be Vo. then vo=((vcc-vbe1)/r1)*beta1*beta2*rload. let r1>>r2,we could let Vo less than vcc.it's better to let q2 to be in amplify area.
now put c1 on. as all know s. vcc charge c1 through r1 and load. as q1 base voltage goes to about 0.6v. q1 start on. and this makes q2 on. q2 on with load makes q2 collector voltage goes up. this feedback to q1 base through c1. which makes q1 conduct more. by this posative feedback. q1 base will be 0.7v. q2 collector will go to nearly vcc.

points come now:will q2collector voltage go high? stay there? or go down? as I have shown above. amplify region is the stable state. so q2 collector will go down. and there's road: right plate of c1--load--source--r1--left plate of c1.and as q2 collector voltage go down. c1 couple it to q1 base. which makes q2 base go up. which makes q2 current go down. so Vo go down more. this is still a positive feedback. Vo will go to zero. q1base -(vcc-0.7).

now vcc will charge C1 again. cycle.

 

The signal must be made without a constant component.

is constant component like function generator?

 

here is what I leaned,have a look:
first. let's put off c1, to see what is the stable state:let voltage of q2 to be Vo. then vo=((vcc-vbe1)/r1)*beta1*beta2*rload. let r1>>r2,we could let Vo less than vcc.it's better to let q2 to be in amplify area.
now put c1 on. as all know s. vcc charge c1 through r1 and load. as q1 base voltage goes to about 0.6v. q1 start on. and this makes q2 on. q2 on with load makes q2 collector voltage goes up. this feedback to q1 base through c1. which makes q1 conduct more. by this posative feedback. q1 base will be 0.7v. q2 collector will go to nearly vcc.

points come now:will q2collector voltage go high? stay there? or go down? as I have shown above. amplify region is the stable state. so q2 collector will go down. and there's road: right plate of c1--load--source--r1--left plate of c1.and as q2 collector voltage go down. c1 couple it to q1 base. which makes q2 base go up. which makes q2 current go down. so Vo go down more. this is still a positive feedback. Vo will go to zero. q1base -(vcc-0.7).

now vcc will charge C1 again. cycle.

Below is demo:

 

Below is demo:
View attachment 296530

ok. thx.

Below is demo:
View attachment 296530

see VB. it goes rapidly high. means positive feedback.then goes down exponentially(?). then goes down rapidly. then goes up linearly.

 

see VB. it goes rapidly high. means positive feedback.then goes down exponentially(?). then goes down rapidly. then goes up linearly.

1. Capacitor C1 is charging through diode B-E of Q1,
so curve of voltage on base Q1 is defined by AV characteristic
of forward biased diode B-E of Q1.
2. Discharging curve is almost linear, because
discharging current through resistor R1 is almost constant .

 

can't get answer there

The answer is certainly provided if you read the whole thing. AND NOTE that it usually requires a weak battery. Every bit of the critical comment does apply. It IS A JUNK CIRCUIT.
It is not really a "relaxation oscillator"

 

It IS A JUNK CIRCUIT.

why?