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How does this weird V-Booster work?

Pommie

Well-Known Member
Most Helpful Member
Google "monostable multivibrator" and find a page that describes how it works and compare to your circuit.

Mike.
 

Miracletech

Member
My confusion is how Q2 Turns off. How in the world does Q2 Turn off?
I don't even know if the circuit works.
How does it work?
 

rjenkinsgb

Well-Known Member
Most Helpful Member
It's an odd arrangement of astable multivibrator.

As Q2 turns on, Q1 turns off. That causes a "flyback" effect in the inductor and passes current through the LED (at a higher voltage than the battery).
Q1 turning off also add positive feedback to Q2 turning on, via C1.

Once the inductor demagnetises, the voltage at the junction of that and the LED drops again - and that reduction is passed via C1 to the base of Q2, turning it off and turning Q1 on, which continues to force the inductor output towards 0V, again providing positive feedback via C1.

R2 sets the delay before Q2 starts to turn on again.

And so on, repeating.
 

ronsimpson

Well-Known Member
Most Helpful Member
1603814910853.png
At power up R1 current (about 1mA) turns on Q1.
Q1=on causes current to flow (ramp up) in L1. The current gets larger and larger until Q1 can not hold it any more. Q1 tears open a little. Q1 collector starts up in voltage. This voltage drives current through C1 into Q2 and turns it on.
Q2=on, this turns off Q1.
Q1=off, L1 flys up into D1. (maybe 3V)
Now current goes from V1 through L1 into D1. The current ramps down to zero.
L1=0 current. The LED voltage drops down to 1.5V. (rings back) (L1 & C1 causes a LC resonant)
The LED voltage dropping will push down on Q2-base (through C1) causing it to turn off and allow Q1 to turn on.

R2 is sized to when Q1 is on there is no turn on current in Q2-Base. When Q1 is off and Q1-Collector is at 3V there is good current in Q2-Base.
--edited--
This circuit is very dependent on (voltage B-E of transistors) and (current gain of Q1).
 

Miracletech

Member
Thank you very much sir.
I appreciate for your kindness of explaining the circuit for me.
One question sir.
C1 Is a "Speed-Up" Capacitor, which is used to drive the Base of Q2.
Will removing it cause harm?
And can I also add another speed up capacitor through R1?
 

ronsimpson

Well-Known Member
Most Helpful Member
C1 Is a "Speed-Up" Capacitor, which is used to drive the Base of Q2.
Will removing it cause harm?
And can I also add another speed up capacitor through R1?
C1 makes the turn on/off faster. (switching losses) When a transistor is on hard there is very little voltage and much current. Power loss=voltage x current. When the transistor is off there is very little current and lots of voltage. Also low power. During switching you have both the high current and much of the voltage at the same time. Lots of power loss. We want to get through the switching time fast.

Capacitor across R1. This is the same as a capacitor Base to Emitter of Q1, or C-E of Q2. This will slow down the switching time.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
Without the cap it will not oscillate. That passes the rapidly changing voltage from Q1 collector & teh inductor to Q2 base, overriding the bias from R2.

Bypassing R1 with a cap would slow down the switching of Q1 and again probably stop it from oscillating, or at least reduce the output.
 

ronsimpson

Well-Known Member
Most Helpful Member
I made changes to Q2 base resistor(s) to get it to run.
Green is Q1 collector voltage. Red is inductor current. (LED current will be 100mA ramping down to 0 about 50% duty cycle) Blue is Q2 base voltage.
1603842035460.png
 

Miracletech

Member
Thanks for the simulations sir. I'm grateful.
C1 Helps drive the base of Q2 To oscillate.
Adding a Capacitor Across R1 Would reduce the switching speed.
But what of using the RC Time constant to calculate?
 

Diver300

Well-Known Member
Most Helpful Member
Since the switching frequency is Based on An RC Time constant, Is it possible to calculate the switching frequency?
The switching frequency is very dependent on the value of the inductor.

The inductance limits the rate at which the current rises. When the current gets large enough, the collector voltage of Q1 will start to rise, which triggers Q2 to turn on.

Also the inductor may start to saturate. That probably isn't shown in the simulation. As the inductor starts to saturate, it's inductance falls and the current will shoot up, triggering the switching.

The maximum current that Q1 can carry before the collector voltage goes up fast enough to trigger Q2 depends on the exact characteristics of the Q1.

The time that Q1 is turned on is mainly dependent on the inductor characteristics and the supply voltage. The time that it is turned off for depend on the characteristics of Q2 and the resistor values. The capacitor will have an effect on that time as well.

It's not a simple RC formula.
 

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