Trying to figure out how this oscillator circuit works

- Dirk -

Joined Aug 8, 2019
7
Hello everyone,

When I was still a kid I got a "ScienceFair 200 in 1" board .... I didn't understand electronics but just wiring according to the numbers and playing with it was big fun! I rediscovered it 27 years later ;-) I recently started studying electronics in my spare time. It's a lot of fun but from time to time also hard as I don't know each component yet and the interactions occuring.

My question is about a basic oscillator circuit (in attachment). Based on the variable resistance it produces a short <beep> at a certain interval. What I don't understand is what triggers the discharge of the 3.3uF capacitor. I know when it discharges at the + side the voltage at the - side goes down as well (as is the case with e.g. a multivibrator circuit) and causes the transistor to switch off. When the capacitor is discharged the transistor base is again at 0.6V, it starts to conduct current again from collector to emitter.... and then I suppose the 3.3uF cap starts to charge... and when it's fully charged, what triggers the discharge? Is it due to the transformer which induces some reverse current when charging stops ???

Many thanks for taking the time to help me!

Best regards,

Dirk

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Jony130

Joined Feb 17, 2009
5,155
When BJT is on we will have more or less this situation:

Notice that I mark the voltage polarity across the transformer windings and across the capacitor and the current direction ( conventional current direction).

Also, notice the capacitor will be charged to 8.3V = 9V - 0.7V very quickly - no current limiting resistor for the base current, only battery resistance, and winding resistance.
Hence at this phase BJT' will be in hard saturation (Vce ≈ 0.1V).

At the end of a charging phase, the capacitor no longer provides any current to the transistor base.
The base current is now only coming from a base resistor (22k + 51k).
And this current (green on the diagram) is too low (4.5V/(22k + 51k) ) to maintain saturation is a transistor.
And this is why the transistor immediately after capacitor was fully charged comes out from the saturation region into the linear region.
And the current and the voltage in the transformer windings will die-out to zero. But we have a charged capacitor (8.3V) and as a voltage across the upper transformer winding changes from 4.5V to 0V. the same think will be feedback via capacitor into transistor base - the voltage at the base will drop by the same amount, from 0.7V to -3.8V (0.7V - 4.5V). And negative voltage at the base will cut-off the transistor.
And now capacitor will start the discharge phase via base resistance (22k + 51k).
The situation will look like this:

As the capacitor discharge process via base resistance (22k + 51k) goes on the voltage at the base terminal will rise exponentially from -3.8 to 0.6V
And as this voltage reaches transistor forward voltage the transistor will start to conduct the base current opens again (capacitor voltage is around 4V at this ). And the cycle repeats itself.

MrAl

Joined Jun 17, 2014
7,466
Hello everyone,

When I was still a kid I got a "ScienceFair 200 in 1" board .... I didn't understand electronics but just wiring according to the numbers and playing with it was big fun! I rediscovered it 27 years later ;-) I recently started studying electronics in my spare time. It's a lot of fun but from time to time also hard as I don't know each component yet and the interactions occuring.

My question is about a basic oscillator circuit (in attachment). Based on the variable resistance it produces a short <beep> at a certain interval. What I don't understand is what triggers the discharge of the 3.3uF capacitor. I know when it discharges at the + side the voltage at the - side goes down as well (as is the case with e.g. a multivibrator circuit) and causes the transistor to switch off. When the capacitor is discharged the transistor base is again at 0.6V, it starts to conduct current again from collector to emitter.... and then I suppose the 3.3uF cap starts to charge... and when it's fully charged, what triggers the discharge? Is it due to the transformer which induces some reverse current when charging stops ???

Many thanks for taking the time to help me!

Best regards,

Dirk

Hi,

There are actually two modes of operation either or both could play a part in the oscillation.
There is a competition between the 3.3uf cap, and the transistor itself combined with the primary inductance.

First, if the 3.3uf cap controls the oscillation alone then the cap charges and while it is charging it
provides plenty of current for the transistor base which turns it on and keeps it on.
Once on, the cap charges up all the way and eventually starts to discharge. Once it starts to discharge
the current reverses and that turns the transistor off. The falling secondary voltage through the cap
keeps the transistor off for some time. Eventually the cap discharges enough where the resistor starts
to turn the transistor on again and soon the secondary voltage starts to rise again which causes the
cap to start to charge again and thus turn the transistor on again.

Now if the transistor itself combined with the primary inductance controls the oscillation then
the the gain of the transistor and primary inductance work together to turn the transistor off.
With the transistor on the rising current in the primary winding causes the transistor to start
to pull out of saturation and then the collector voltage starts to rise. Once that happens
the falling secondary voltage through the cap turns the transistor off completely.
Of course if the primary begins to saturate that will start the turn off action and that is the primary
mode of operation for some circuits similar to this one.

So how do we know what is in control, the cap or the primary winding inductance and transistor gain?
We look for a current spike in the collector. If we see a spike when the transistor turns on that is
normal, but if we also see a spike when the transistor turns off that means the gain and inductance
started the turn off action.

The best operation is when the cap controls the oscillation because then the transistor heating is
reduced. It depends how long it takes to get it out of saturation as to how much more heating
takes place when the gain and inductance are in control.

Zeeus

Joined Apr 17, 2019
521
When BJT is on we will have more or less this situation:

View attachment 183478

And the current and the voltage in the transformer windings will die-out to zero. But we have a charged capacitor (8.3V) and as a voltage across the upper transformer winding changes from 4.5V to 0V. the same think will be feedback via capacitor into transistor base - the voltage at the base will drop by the same amount, from 0.7V to -3.8V (0.7V - 4.5V). And negative voltage at the base will cut-off the transistor.
And now capacitor will start the discharge phase via base resistance (22k + 51k).
The situation will look like this:
View attachment 183486.
Hi. Not really clear

The upper transformer winding from your diagram is 4.5V or 9V?
Voltage at base why drop to (0.7 - 4.5) and not (0.7 - 9)?

Thanks
Also, not really relevant but why input impedance 51k?

Edit : lol sorry : can't be 9V

Last edited:

Jony130

Joined Feb 17, 2009
5,155
vo
The upper transformer winding from your diagram is 4.5V or 9V?
The upper winding voltage will be around 4.5V. Hence the capacitor will see Vc = Vbat + Vwind2 = 4.5V + 4.5V = 9V.
But during normal operation notice that the cap will be discharged to around 4V (when BJT is off) and charged up again to 9V.

Voltage at base why drop to (0.7 - 4.5) and not (0.7 - 9)?
Because I assumed that the upper transformer winding is 0V when BJT is off. It could be negative as well, I simply do not now the transformer parameters and reflected (from secondary winding) voltage. So, the voltage at the base is Vb = Vbat -Vcap = 4.5V - 8.3V = -3.8V

Also, not really relevant but why input impedance 51k?
We could ask why 3.3μF? It's a designer choice based on some technical requirement.

Look here for a similar circuit in working principle.
https://electronics.stackexchange.com/questions/439255/understanding-a-pnp-npn-led-schematic/439375#439375

Last edited:

- Dirk -

Joined Aug 8, 2019
7
Hi,

Thanks all for taking the time to explain in so much detail. Some things are more clear now! I'll probably do some experiments and read more about transformers... I didn't know the voltages also add up at the left side of the transformer (ratio is 900:8 here) but it is indeed reflected by a voltage across the capacitor which is higher than 4.5V (multimeter).

"With the transistor on the rising current in the primary winding causes the transistor to start
to pull out of saturation and then the collector voltage starts to rise."
? Not sure I understand this... how does the current in the primary winding affect the saturation state?

Even though the experiments seem to build up from simple to more advanced, from time to time it's quite challenging for me ... but very interesting to learn from and sort out misunderstandings!!

Have a nice evening!

Dirk

Jony130

Joined Feb 17, 2009
5,155
Not sure I understand this... how does the current in the primary winding affect the saturation state?
First, do you know that when you connect an inductor directly across a DC voltage source the current in the inductor will start to increases linearly with the rate of change equal to di/dt = V/L?
For 4.5V and L = 1H the current will rise with the rate of 4.5A per every second.

And now how this rise in collector current (4.5A/s) will affect the transistor in saturation if you assume the base current is at a fixed value.

MrAl

Joined Jun 17, 2014
7,466
Hi,
"With the transistor on the rising current in the primary winding causes the transistor to start
to pull out of saturation and then the collector voltage starts to rise."
? Not sure I understand this... how does the current in the primary winding affect the saturation state?
Dirk
The saturation state of a transistor depends on at least three main things:
1. The base current
2. The current gain
3. The collector current
There are other things but these are most significant.

Saturation is reached only when the base current reaches a certain level with some given level of collector current. A rough rule of thumb that a lot of people accept is the base current should be at least 1/10 of the collector current so we assume a minimum current gain of 10.
But if either of those things changes, the transistor can come out of sat and that means the collector voltage rises.
For example, say we have 10ma in the base and 100ma in the collector and we have saturation. That obeys the 1/10 rule.
But when something external raises the collector current (a winding, a power supply, etc.) the base current is no longer 1/10 of the collector current so we loose saturation. You could test this using a power supply at the collector and a series resistor, and raise the collector current little by little. Eventually the transistor comes out of saturation and the collector voltage starts to rise.
Note however this mode only occurs if the inductance is low enough to make that happen (or the cap is very large) because if it is high enough then the cap determines the cycle time. The main question is does the winding saturate before the cap starts to cut off the transistor.
Some circuits like this use that lower inductance feature to oscillate rather than an RC time constant.
It is interesting that when the transistor STARTS to come out of sat the upper winding voltage immediately starts to decrease, which couples to the base via the cap. That means we get a snap action that turns off the transistor more rapidly then the slight loss of transistor sat could cause by itself.

- Dirk -

Joined Aug 8, 2019
7
Was very useful information! I was not aware the collector current could pull it out of saturation. There is still a lot to be learned

I have another oscillator circuit for which I would like to ask some help. I’ll create a new post and link here for those interested.

Thanks again
Best regards,
Dirk