Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

current and voltage distribuition in the circuit

Status
Not open for further replies.

erece

Junior Member level 3
Joined
Mar 17, 2011
Messages
28
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,470
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 ??
 

Attachments

  • untitled.JPG
    untitled.JPG
    107.7 KB · Views: 34

chuckey

Advanced Member level 5
Joined
Dec 26, 2010
Messages
4,863
Helped
1,308
Reputation
2,622
Reaction score
1,283
Trophy points
1,393
Location
Southampton and holiday cottage in Wensleydale (UK
Activity points
31,695
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
 
  • Like
Reactions: erece

    erece

    Points: 2
    Helpful Answer Positive Rating

erece

Junior Member level 3
Joined
Mar 17, 2011
Messages
28
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,470
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.
 

erece

Junior Member level 3
Joined
Mar 17, 2011
Messages
28
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,470
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
Joined
Aug 11, 2009
Messages
13
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,283
Location
US
Activity points
1,359
Q2 is fully on which means it's saturated, in which case, Vce ≈ 0, almost like a straight wire down from collector to emitter.
 
  • Like
Reactions: erece

    erece

    Points: 2
    Helpful Answer Positive Rating

erece

Junior Member level 3
Joined
Mar 17, 2011
Messages
28
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,470
thank you to both of you... My concepts are weak. i.e why it took me a lot of time..
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top