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http://i67.tinypic.com/a9mkv7.jpg

I think the steps will be enter image description here

STEP-1:

\$V_x - V_{BE} = I_B\times R_E\$

STEP-2:

\$I_B = I_{ra} - I_{rb}\$

STEP-3:

\$V_x = V_{in}\times R_B/(R_B+R_A)\$

Please tell me i am correct or not??

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  • \$\begingroup\$ You talk about a Vx, but you don't define what Vx is anywhere. \$\endgroup\$
    – Hearth
    Commented Nov 8, 2018 at 13:18
  • \$\begingroup\$ Now its included \$\endgroup\$
    – khatus
    Commented Nov 8, 2018 at 13:48
  • \$\begingroup\$ Nope. Step 3 is not correct. \$\endgroup\$
    – jonk
    Commented Nov 8, 2018 at 13:56
  • \$\begingroup\$ What is Re? And VB is not identified anywhere? Is VX the same as VB? \$\endgroup\$ Commented Nov 8, 2018 at 14:37

3 Answers 3

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Your "Vx = VIn×(RB/RB+RA)" is incorrect as that is only valid if Ib = 0, which it is not.

You're applying the voltage divider formula to this but it cannot be used here as it ignores that there is current flowing from the output of that voltage divider, that current is Ib.

How I would solve this:

I analyze the circuit of Ra, Rb, Re and the NPN as if it is:

schematic

simulate this circuit – Schematic created using CircuitLab

I prefer to use Thevenin for that, once for \$V_{in}\$ and \$R_a\$ and once for \$V_{BE}\$ and \$R_e\$ which would give:

schematic

simulate this circuit

Now I can combine the currents \$I_{tot} = V_{in}/R_a + V_{be}/R_e\$

and all resistors in parallel: \$R_{tot} = R_a ||R_b||R_e\$

Then \$V_1 = I_{tot} * R_{tot}\$

Then the current through Re: \$I(R_e) = V_1 / R_e\$

That current is the \$I_b\$ I need to know.

Multiply that by \$\beta\$ and you have \$I_c\$

Then it is easy to determine \$V_c\$ which is probably the collector potential (the question should have made that more clear).

Assuming that \$V_b\$ is the base potential then that's easy as the emitter is grounded and \$V_{BE}\$ is given.

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  • \$\begingroup\$ Although, with RA at 200 ohms and RB at 100k, the error will be very small for likely values of IB. \$\endgroup\$ Commented Nov 8, 2018 at 15:22
  • \$\begingroup\$ @WhatRoughBeast I agree but "for educational purposes" I decided to ignore the component values. I believe that only when you're able to find the exact solution are you allowed to "cut corners" like that. And even then, you must have a good reason, which 200 ohms against 100k ohm of course is. \$\endgroup\$ Commented Nov 8, 2018 at 15:24
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khatus - it is a rather simple calculation when you are aware that you are allowed to treat VBE=0.7 volts as a ideal voltage source! This is due to the substitution theorem.

Then the easiest way is to use the voltage division procedure twice (superposition rule for Vin and VBE)

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There is a simple general solution for the common node voltage when you have a bunch of resistors connected to various voltage sources:

schematic

simulate this circuit – Schematic created using CircuitLab

Vx = (V1/R1 + V2/R2 ... Vn/Rn)* (R1 || R2 ||..||Rn)

So in this case,

V1 = Vin, R1 = Ra V2 = 0, R2 = Rb V3 = 0.7, R3 = Re

R1||R2||R3 = 1/(1/Ra+1/Rb+1/Re) = k

Vx = k*(Vin/Ra + 0.7/Re)

Ib = (Vx-0.7)/Re

Ic = \$\beta \cdot\$Ib (assuming transistor in linear region)

Vc = Vcc- Rc*Ic (check that Vce is sufficiently positive for the transistor to be in linear region).

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