# current and voltage distribuition in the circuit

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#### erece

##### Junior Member level 3 In the given figure , when base voltage of Q2 is made 3.2 V then why Q1 reaches cut off and Q1 in active region and when base voltage of Q2 is made 3.5 V the why Q2 reaches active region and Q1 in cut off ??

#### chuckey 1. Its a Shmidtt trigger
2. You have you Q1s and Q2s mixed up.
3. How it works, with Q1 off (Q1 Ie = 0). Q2 Vb = (1/1+1.5 )X12V = 4.8V. Q2 Ie = (4.8 - .6)/ 160 +330 = 4.2/490 = 8.5 mA. Therefore voltage at resistor tap = 330 X 8.5 mV = 2.8 V
So Q1 is sitting there with 2.8V on its emitter. When its base reaches 2.8 + .6 V it conducts ( 3.4V), this pulls the Q2 Vb down which switches off Q2, the current though it decreases, the voltage on its emitter falls, the tapping point falls so the emitter voltage on Q1 falls, this increases the current through Q1 which turns on fully.
The voltage across the 330 is 330/ (1.5k +330 ) X 12 = 2.16 so when Q1s base falls to this voltage you are back to square 1.
Frank

• erece

### erece

Points: 2

#### erece

##### Junior Member level 3 1. Its a Shmidtt trigger
2. You have you Q1s and Q2s mixed up.
3. How it works, with Q1 off (Q1 Ie = 0). Q2 Vb = (1/1+1.5 )X12V = 4.8V. Q2 Ie = (4.8 - .6)/ 160 +330 = 4.2/490 = 8.5 mA. Therefore voltage at resistor tap = 330 X 8.5 mV = 2.8 V
So Q1 is sitting there with 2.8V on its emitter. When its base reaches 2.8 + .6 V it conducts ( 3.4V), this pulls the Q2 Vb down which switches off Q2, the current though it decreases, the voltage on its emitter falls, the tapping point falls so the emitter voltage on Q1 falls, this increases the current through Q1 which turns on fully.
The voltage across the 330 is 330/ (1.5k +330 ) X 12 = 2.16 so when Q1s base falls to this voltage you are back to square 1.
Frank

thanks for the reply............ I want to know the impact of base voltage of Q2 on Q2 and Q1. It is like base voltage of Q2 is the input voltage to the circuit... And you wrote that when Q1 is off Ie of Q2 will be 8.5 mA , but why did you take resistor 160 ohms. With Q1 off , no current will flow through 160 ohms resistor. Correct me if i am wrong.

#### amangupta009

##### Newbie level 3 #### chuckey Sorry , it was me that got my Q1s and Q2s mixed up, change every one around!!!! :-( . Does that make more sense now?
Frank

• erece

### erece

Points: 2

#### erece

##### Junior Member level 3 1.
The voltage across the 330 is 330/ (1.5k +330 ) X 12 = 2.16 so when Q1s base falls to this voltage you are back to square 1.
Frank
I think you didn't take voltage drop from collector to base of Q2 in the last line to calculate voltage across 330 ohms resistor. Why so?

#### qdnboy

##### Newbie level 6 Q2 is fully on which means it's saturated, in which case, Vce ≈ 0, almost like a straight wire down from collector to emitter.

• erece

### erece

Points: 2

#### erece

##### Junior Member level 3 thank you to both of you... My concepts are weak. i.e why it took me a lot of time..

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