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Multivibrator of varying frequency

Akshatha Venkatesh

Jan 14, 2017
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Below I have the circuit of an astable multivibrator with variable frequency, this is just an example , I have not yet decided on the resistor and capacitor values. Can anyone explain how the capacitors are being charged and discharged and in turn how the frequency is changing. The voltage vmid is a capacitor charging and discharging waveform , constantly varying and that's how the output frequency is changing, but I have a doubt , aren't the capacitors being charged by R1 and R4 ? Which is the transistor that turns on first if R2 and R3 values are different, the way that I understand I feel that the transistor Q3 turns on first , is this right ?
 

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duke37

Jan 9, 2011
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Q2 and Q3 are running as a mutivibrator, you should look up how this works.
Each base is capacitor coupled to the opposing collector. When the base is pulled positive, base current flows to prevent the base rising above about 0.7V. When there is insufficient base current, the action switches and the base is switched off with a negative voltage, in this case up to about 5V.
R3 then raises the voltage towards the positive and switching will occur when the base is raised above about 0.7V.
R2 and R3 are provided with a variable voltage so the rate of voltage rise and hence frequency will depend on this voltage.

Do not run this circuit above 5V since the transistors will not like much more than 5V on the base/emitter junction.
 

ramussons

Jun 10, 2014
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When Q2 is in Saturation, C3 has to charge thru' R2 to finally switch on Q3.
When Q3 is in Saturation, C2 has to charge thru' R3 to finally switch on Q2.

R2 and R3 control the charge timing and hence the frequency, not R1 and R4
If the hFE of both Q2 and Q3 is same, yes, Q3 turns on first.
 

Akshatha Venkatesh

Jan 14, 2017
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To
Q2 and Q3 are running as a mutivibrator, you should look up how this works.
Each base is capacitor coupled to the opposing collector. When the base is pulled positive, base current flows to prevent the base rising above about 0.7V. When there is insufficient base current, the action switches and the base is switched off with a negative voltage, in this case up to about 5V.
R3 then raises the voltage towards the positive and switching will occur when the base is raised above about 0.7V.
R2 and R3 are provided with a variable voltage so the rate of voltage rise and hence frequency will depend on this voltage.

Do not run this circuit above 5V since the transistors will not like much more than 5V on the base/emitter junction.
To calculate R1 , I'm using the formula Vcc-Vce / Ic , vce min and Ic min I have taken from data sheet , is this right ? To calculate R2 and R3 , I calculated the Ibmin and then calculated a safe Ib and and then calculated the resistor values , however when I simulate the circuit in lrspice with the values I have got I'm not getting the output.
 

AnalogKid

Jun 10, 2015
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Because R1 and R4 are much smaller than R2 and R3 (this is correct), it is R2 and R3 that set the charging currents for the capacitors. That is why changing Vmid will change the frequency or oscillation.

ak
 

duke37

Jan 9, 2011
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If you are getting no output using a simulation, this may be because the two sides of the multivibrator are identical and the program cannot decide which to turn on first. I have seen where an oscillator circuit is given a kick to start. If you make R2 slightly different to R2, it may go.
In practice no circuit would have identical sides, there will be differences in the resistors, capacitors and transistors.
 

Akshatha Venkatesh

Jan 14, 2017
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If you are getting no output using a simulation, this may be because the two sides of the multivibrator are identical and the program cannot decide which to turn on first. I have seen where an oscillator circuit is given a kick to start. If you make R2 slightly different to R2, it may go.
In practice no circuit would have identical sides, there will be differences in the resistors, capacitors and transistors.
Thank you , could you please tell me the use of R5 in the above circuit, C1 and R6 form the RC circuit , what role is R5 playing in the circuit ?
 

AnalogKid

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R5 works with R6 to form a voltage divider. This sets the voltage at the Q1 emitter, which affects the oscillator frequency. Also, R5 and R6 work with C1 to form a low pass filter. The corner frequency is 7.5 kHz, which seems a bit high to me. It changes the edges of the V2 square wave into a trapezoid-ish waveform.

ak
 

Akshatha Venkatesh

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R5 works with R6 to form a voltage divider. This sets the voltage at the Q1 emitter, which affects the oscillator frequency. Also, R5 and R6 work with C1 to form a low pass filter. The corner frequency is 7.5 kHz, which seems a bit high to me. It changes the edges of the V2 square wave into a trapezoid-ish waveform.

ak
How is R5 and R6 forming a voltage divider?R5 is connected to a supply of 5V and R6 is also connected to a pulse supply. How did you Calculate the corner frequency ? Could you please explain.
 

AnalogKid

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R5 and R6 are two resistors in series between two voltage sources. If the two voltages are not equal, the voltage at the center node formed by the two resistors can be calculated using Ohm's Law.

The corner frequency is calculated using the equation for the corner frequency of a single-pole R-C circuit. For the R, use the Thevenin equivalent or R5 and R6, which is the calculated value of R5 and R6 in parallel.

ak
 

Akshatha Venkatesh

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I have been watching this thread and I think the important question to ask yourself is, WHAT IS THE PURPOSE OF c2 and c3.
The purpose of C2 and C3 is to charge and give the biasing voltage to the transistors to switch them on and off. Is my understanding right ?
 

duke37

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"The purpose of C2 and C3 is to charge and give the biasing voltage to the transistors to switch them on and off. Is my understanding right ?"
Yes, the circuit is an oscillator with each transistor driving the other. The gain of the loop is very, very high and the amplitude slams the collectors between apoximately zero and the power supply.
 

Cannonball

May 6, 2017
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The sole purpose of the caps is to turn off the transistor that the - lead is connected to the base and the + lead is connected to the collector of the transistor that is turned on. It will keep it turned off until this capacitor is discharged.

If there is any doubt build the circuit on a bread board and while it is running pull either cap out of the circuit and as soon as the other cap discharges both transistors will turn on and stay on.
 

Audioguru

Sep 24, 2016
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The other sole purpose of the caps is to turn on the transistor to which's base it is connected. Each capacitor is in series with a base-emitter junction, so 100% of the energy into and out of the base of that transistor comes from / passes through the capacitor.

ak
I disagree.
The capacitors quickly turn a transistor off and the base resistor turns it on then the capacitor has discharged and begins charging in reverse to about +0.7V.
 

AnalogKid

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For absolutely no reason, I didn't see the ***two*** base resistors in a circuit I first built in the 60's. Doh. Deleted the post.

ak
 

Cannonball

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AnalogKid

Would you agree that without the caps both transistors would both turn on?
 

Cannonball

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No feed back. It takes a - voltage to turn the transistor off. That is the purpose of the caps.
 
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