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NPN transistor numerical
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Topic: NPN transistor numerical (Read 1459 times)
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khatus
Regular Contributor
Posts: 154
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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
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khatus
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Posts: 154
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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
»
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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
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khatus
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Re: NPN transistor numerical
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Reply #3 on:
November 08, 2018, 05:02:29 am »
So the node voltage
Vx = VBE (base emitter voltage) ??
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bson
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Re: NPN transistor numerical
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Reply #4 on:
November 08, 2018, 07:31:39 am »
Quote from: khatus on November 08, 2018, 05:02:29 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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khatus
khatus
Regular Contributor
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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:
V
x
- V
BE
= I
B
×R
E
STEP-2:
IB = Ira - Irb
STEP-3:
V
x
= V
In
×(R
B
/R
B
+R
A
)
Please tell me i am correct or not??
«
Last Edit: November 08, 2018, 12:40:34 pm by khatus
»
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LvW
Frequent Contributor
Posts: 282
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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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