Author Topic: NPN transistor numerical  (Read 1459 times)

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Offline khatusTopic starter

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NPN transistor numerical
« on: November 08, 2018, 03:57:36 am »
Can anybody solve the following problem?? and posts each steps of solution if possible

 

Offline khatusTopic starter

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Re: NPN transistor numerical
« Reply #1 on: November 08, 2018, 04:42:10 am »


STEP-1: At first i apply KCL at node x. which give me



Then how i can proceed further??
« Last Edit: November 08, 2018, 04:58:48 am by khatus »
 

Offline IconicPCB

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Re: NPN transistor numerical
« Reply #2 on: November 08, 2018, 04:49:08 am »
Vc = 9- Rc*Ic
IC = 45 * Ib
Ib = Ira - Irb

Ira= (Vin-0.7)/Ra
Irb = 0.7/Rb
 

Offline khatusTopic starter

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Re: NPN transistor numerical
« Reply #3 on: November 08, 2018, 05:02:29 am »
So the node voltage

Vx = VBE (base emitter voltage) ??
 

Offline bson

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Re: NPN transistor numerical
« Reply #4 on: November 08, 2018, 07:31:39 am »
So the node voltage

Vx = VBE (base emitter voltage) ??
Well, yeah.  The only thing between it and ground is B-E, which is 0.7V.
 
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Offline khatusTopic starter

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Re: NPN transistor numerical
« Reply #5 on: November 08, 2018, 12:37:38 pm »
Same problem with an additional base resistor




I think the steps will be

STEP-1: 

Vx - VBE = IB×RE

STEP-2:

IB = Ira - Irb

STEP-3:

Vx = VIn×(RB/RB+RA)

Please tell me i am correct or not??
« Last Edit: November 08, 2018, 12:40:34 pm by khatus »
 

Offline LvW

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Re: NPN transistor numerical
« Reply #6 on: November 08, 2018, 03:45:59 pm »
No step 3 is not correct.
For the voltage Vx ALL THREE resistors and both voltages Vin and VB are to be considered.
And you are allowed to treat VB as a voltage source with VB=0.7V.
 


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