# Two Transistor Question

Discussion in 'General Electronics Chat' started by raghavanunm, May 24, 2013.

1. ### raghavanunm Thread Starter Member

Mar 21, 2007
14
0
Guys,
Please find attached, a copy of circuit for which, i have few questions:
- Can you tell me which one, turn ON first Q1 or Q2.

Thanks,

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2. ### Dodgydave AAC Fanatic!

Jun 22, 2012
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Q1 turns on first,that turns Q2 on.

3. ### c.marsh Active Member

May 16, 2009
100
0
isnt that a darlington pair?

no because the emitters are in different configurations?

4. ### raghavanunm Thread Starter Member

Mar 21, 2007
14
0
Does Q1 Turn ON first because , Base of Q1 is connected to ground ? ( or the PNP emitter is more positve than Base ?)

Also, how does the capacitor charge? ( the Path..)

5. ### crutschow Expert

Mar 14, 2008
21,366
6,121
That's a poorly designed circuit since it has DC going through the speaker voice coil.

6. ### studiot AAC Fanatic!

Nov 9, 2007
5,003
522
Is this a homework question?

You need to tell us your thoughts to receive help. I see that the notes believe you have already studied how a simpler similar circuit works. Did you understand that?

7. ### raghavanunm Thread Starter Member

Mar 21, 2007
14
0
I did understand a similar circuit..but i don't understand the capacitor's charging path.

8. ### raghavanunm Thread Starter Member

Mar 21, 2007
14
0
If q1 turns ON first, what can ever turn it off, isn't it connected to the ground visa the resistors ?

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,693
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At the beginning Q1 turns on first, because the base current finds a path to ground through resistors R1 and R2. Turns on Q1 opens the path for Q2 base current.
So Q2 is also ON. And C1 capacitor start charging. The C1 is quickly charge in this current path.
+4.5V ---> Q1 emitter-base junction ---> C1 ---> Q2 collector-emitter--->GND.

After C1 is charge to 4.5V - 0.7V = 3.8V capacitor end the charging phase.
This causes that the base current of Q1 is decreases. This also decreases IcQ1 and IbQ2, IcQ2 currents, this rise the Q2 collector-emitter voltages.
This small change in VceQ2 voltage is "feedback" by C1 capacitor to Q1 base. This will increase the voltage at Q1 base. So Q1 and Q2 will decrease his currents. This will increase VceQ2 and Q1 base voltage even faster.

So we have a very strong positive feedback in this circuit.
This cycle ends when Q1 and Q2 turns off. And capacitor start slowly discharge.
bottom C1 plate ---> R2 --> R1 --->Batter---> speaker--->R3 ---> upper C1 plate.
And after C1 is full discharge ( voltage at Q1 base drop form 8.3V to 3.8V).
Q1 and Q2 turns ON and C1 is quickly charge and speaker "play the sound".
And the whole cycle repeats itself.

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• ###### 3.PNG
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