# Two bjts oscillator circuit scheme.

Discussion in 'Homework Help' started by lordeos, Nov 26, 2015.

1. ### lordeos Thread Starter Member

Jun 23, 2015
33
1
Hi all,

currently we are learning how to analyze and understand the working of electronic schema's. I got scheme " simple transistor tester circuit" (as an attachment in this post) which i have hard to figure it out how it works. I can identify and explain some parts of the scheme but the greater picture battles my mind.

If both transistor works (first one "npn" and second one pnp") the LED should flash (alternate between on and off).

- I believe that both transistors should be conducting before the LED goes on; Since the PNP transistor needs to get current out of the base the NPN transistor must be turned on, otherwise the base current from the PNP can't find his way to ground, correct ?
- I don't understand what the capacitator is doing at the bottom; The only thing i can think of is that the - plate of the ELCO charges up to 0.7 V trough the resistor of 330K and then the npn transistor switches on.
- the values of the resistors, especially the 330k is a mystery to me.

Thanks for the help

Jun 22, 2012
5,166
776
3. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
This circuit is not so easy to understand.
This circuit is nothing more then cascade amplifier (two stage amplifiers) with positive feedback add by C1 capacitor. So this multivibrator circuit will oscillate.

When we first connect this circuit into supply. The C1 capacitor is initial discharge.
So the voltage across capacitor is 0V (capacitor act just like a short circuit).
Immediately after we power up the circuit. The current start to flow via R1. But because C1 capacitor is discharge (0V across capacitor) all this current will flow through C1 and R3 to ground (Vcc--->R1 ---> C1--->R3--->GND).
No current will flow into T1 base and this is why T1 and T2 or OFF.
But as you know the voltage at charging capacitor start to rising. And when this voltage (at point X) reach T1 threshold voltage (about 0.6V) T1 start to conduct a current. If T1 start to conduct the T2 also will start conducting a current. This means the voltage on Q2 collector starts to rise (at point Y). And this rise in Y voltage is feedback to point X via capacitor C1.
But as you know the voltage at T1 base (point X) can not be higher then 0.7V because of the diode in the base-emitter junction T1. This means that there will be a huge T1 base current. Which will ensure that
T2 is in saturation region, and the voltage at point Y is close to 7.4V (9V - Vled = 9V - 1.6V = 7.4V).

Also at the same time C1 will quickly discharge then C1 will start recharged in the opposite polarity.
Vcc ---> D1--->emitter-collector T2--->C1--->Base-emitter T1--->Gnd.
And all this will happens very quickly because of a large T1 base current (charging current).
So the C1 was rapidly charged to V_C1 = Vcc - Vled - VbeT1 - Vce2(sat) ≈ 6.8V
So when C1 capacitor is full charged (cap no longer conducts any current). The T1 base current is provided only by R1 resistor. And R1 don't provide enough base current into T1 to keep T2 in saturation region.
And this is why T2 immediately after C1 was fully charged comes out from saturation region into linear region. The voltage at T2 collector start to fall. And this drop in collector voltage again is feedback via C1.
For example if voltage at point Y drops to from 7.4V to 7.3V, the voltage at Q1 base will also drop to 7.3V - 6.8V = 0.5V. Thanks to the positive feedback provided by C1 capacitor.
So eventually C1 that was previously charged to 6.8V "pulls-down" the voltage on the Q1 below ground(-6.8V). So T1 and T2 will immediately goes into cut-off.
And now C1 cap will act very similarity to a voltage source and start the discharge phase in the circuit:

C1 right plate ---> R4---> power supply ----> R1 ---> C1 "left" plate.

Eventually cap will be full discharge (0V across the cap) but the current will still flow in the same direction and capacitor will start the charging phase, as I described this at the beginning.
The cap will charge to 0.6V ( left plate more positive than the right plate) and Turn ON T1 and T2 .
That is one complete cycle.
So as you can see Q1 and Q2 are ON for the "very short" time ( C1 charging phase).
And they are OFF for "long time" (C1 discharging phase).

http://www.talkingelectronics.com/p...er/TheTransistorAmplifier-P1.html#OSCILLATORS

absf likes this.
4. ### lordeos Thread Starter Member

Jun 23, 2015
33
1
@Jony130

Many Tanks for your very detailed explanation.

However I'm struggling with the term with the term "feedback"

Does this mean that as the + plate of the cap increases in voltage, the - plate of the cap (already has 0.6V) also increases... for example + plate is charged up to 1V so - plate has 1.6V ?

i assume this means the + plate of C1 is charged up to 7.4V - 0,6V present on the - plate = 6.8 V over C1, correct ?

Still have some issues with the understanding of the Voltages of the + plate and - when charging and discharging

thx

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
Well, replace the cap with a resistor, and try analysis what will happens when T2 collector voltage increase.

Start from "Now let as change the type of the feedback from negative feedback to positive feedback and see what will happen. "
Yes, but there is nothing mysterious here, treat the cap as a voltage source. If the voltage across cap is 1V and the voltage at T2 collector is 5V.
The voltage at T1 base wants to be 4V. But T1 base-emitter junction do not like it. We almost have a short circuit and this is why very large current will start to flow. And this current will instantly charge the cap to the voltage difference between point Y and X .
So the voltage across the cap is 7.6V - 0.6V = 6.8V

6. ### ian field Distinguished Member

Oct 27, 2012
4,459
792
That circuit has been around since they figured out how to make NPN germanium transistors. Its quite difficult to get it wrong.

A lot of people don't bother with the 1k from first collector to second base, but it does mean the first transistor tries to dump the full Vcc across the second transistor B/E - adding a supply decoupling capacitor would then cause a problem.

Its usual to put all the load in the second transistor collector circuit - so move the LED there as well as the resistor.

That oscillator usually produces needle pulses - with 10uF feedback the pulses will be too few and far between for persistence of vision to catch them - you can widen the pulses by putting a resistor in series with the capacitor, try the 1k taken from the C/B but it may need to be just a few hundred.

7. ### lordeos Thread Starter Member

Jun 23, 2015
33
1
@Jony130

Hello Jony,

The last two final questions :

Is the discharge the 0.6 Volt on the - plate ? So the moment the transistors (T1 & T2) starts conducting the - plate with 0.6 V will be discharged to 0, correct ?

VbeT2 also has 0,7 V ... Why is this not included in the equation and only VbeT1 ?

Thx for all the help and your patience with me

Cheers Mike

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
Yes, but we have 0.6V across the cap and the left plate is at +0.6V and the right plate is almost at 0V (figure 1).
And as T1 & T2 starts conducting the voltage at point Y (on the C1 right plate) start to rise and reaches 7.4V. And as this voltage rise at point Y the C1 capacitor start charging process in the opposite polarity (the voltage at the right plate is now more positive (figure 2)) and this process ends when voltage across the cap reach 6.8V.
But why is should be included??

9. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
This is the circuit used in a lot of "Flasher Barricades", where the low duty cycle keeps them flashing for a very long time without battery replacement.

Here is a sim of the originally posted circuit, showing the default timing. Note that as stated by Ian, the current pulse in the LED is very narrow.

Note that the capacitor is incorrectly polarized in the published circuit, because it has reversed voltage across it.

10. ### ian field Distinguished Member

Oct 27, 2012
4,459
792

With no resistor in series with the feedback capacitor, the pulses were too narrow to have any visible effect on the filament.

AFAICR: it was 1k2, an LED doesn't have thermal inertia so the resistor should be somewhat lower.

11. ### lordeos Thread Starter Member

Jun 23, 2015
33
1
@Jony130

The max voltage at point X of 0.7 V is something i can't seem to understand

- Is this because the diode (base emitter) is directly connected to 0V ? Because if the left plate of the CAP by the positive feedback of the right plate rises to 6.8 V (7,4 - 0,6) and only 0.7 V drops over the diode ... where is the rest of the voltage being dissipated then ?

- In my simulator program if i put a resistor after the diode point X can be higher then 0.7V ?

thx Mike

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
Yes, but the BJT base-emitter junction is a diode, and when we analyze the circuit with diodes we usually assume constant voltage drop across the diode when diode is in forward biased. And no matter how large the diode current is the voltage drop will not change.