How does a 2 transistor astable multivibrator work?
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- Опубліковано 6 вер 2024
- In this video, I explain in detail how an astable multivibrator (two npn transistors) works. It's all about transistor switching and capacitor charging.
I am focusing on the voltage on either side of the capacitors, how the transistors get turned on/off.
I am assuming you already know about the npn transistor and how it can be used as an electronic switch. A forward bias voltage of 0.6v between base and emitter is needed to turn on the transistor. That's basically all you need to know.
Thanks to "Practical Electronics for Inventors" by Paul Scherz and Simon Monk for the picture and some of the explanations. I go much deeper than what's in the book (I even found a mistake).
Bravo
I have 2 questions.
FIRST..
I saw that usually C1 and C2 capacitors are electrolitics.
Lets take capacitor C1.
Plate A is positive and plate B is negative.
It is possible to charge C1 in reverse connection (plate A on ground and plate B to positive plate of supply)?
Because at one moment Plate A has zero potential and plate B has -5.4 potential.
When Q1 is conducting Plate A is on the ground and Plate B is on positive side of supply.
I understand that this is the phase of discharge of C1 (i saw that a charged capacitor in reverse connection discharge) but why potential on plate B (negative) begin to rise above zero?
Shouldn stay at zero?
So intead of that we have 0V on plate positive A and +0.7V on plane negative B.
So electrolitic capacitors can be charged in reverse connection?
SECOND
I dont understand at the beginning how C1 is charged.
Both plates (A and B, positive and negative) are connected to the positive plate of supply.
Q1 is blocked so plate A receive +Vcc (5 v) potential.
But why and how potential on negative plate B begin to rise?
And if begin to rise, plate A shouldn get some electrons from the plate B and loose some potential?
Because Plate B is negative.
Thankyou.
7:02. Lets not assume Q1 turns on at some point. It doesn't say much. :-)
Q1 can only turn on if it gets current from the cap at Q2 which means a cap that is discharging. Right?
But the cap at Q2 could not have charged since the collector at Q2 was at 0 volts due to current flowing through it. The cap can only charge if the transistor is off with its collector at 6V.
In the electronics for inventors book, the author makes the same assumption, which i think is very reasonable.
Sharkit kapoj ni velu nathi tejanavo
Thanks for your comment. No idea what it meams though