Duave
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Jony130 said:Your answer is not correct. You forget about 150K resistor.
Maybe this circuit will be easy for you to solve, you can even use nodal analysis.
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Also notice that I1 = (Ib + I2)
Or you can replace voltage divider with his thevenin equivalent circuit.
You have error already in the the second expression. Are you sure that this is correct KVL/KCL for this circuit? Can you tell me where do you see RB resistor in this circuit ?Duave said:IE = IC + IB
............
VCC-(IB)(RB) -(IE)(RE)-(VBE) = 0
...................
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Jony130 said:You have error already in the the second expression. Are you sure that this is correct KVL/KCL for this circuit? Can you tell me where do you see RB resistor in this circuit ?
The KCL for VB node is I1 = I2 + Ib
Or why you don't try to use thevenin's theorem and replace R1 and R2 and Vcc with his equivalent circuit (Rth and Eth)?
Jony130 said:You have error already in the the second expression. Are you sure that this is correct KVL/KCL for this circuit? Can you tell me where do you see RB resistor in this circuit ?
The KCL for VB node is I1 = I2 + Ib
Or why you don't try to use thevenin's theorem and replace R1 and R2 and Vcc with his equivalent circuit (Rth and Eth)?
Yes, but what about Vcc ?Duave said:Could Rb= Rb1||Rb2?
I don't understand,Jony130 said:Two loops:
First loop
Vcc = I1*R1 + I2*R2
And the second one
I2*R2 = ??
Or we can use nodal equation
(15V - Vb)/130k = Vb/150k + Ib
Where Ib = ??Yes, but what about Vcc ?
No this is incorrect because Ib current flows only through R1 resistor.Duave said:I don't understand,
Wouldn't it be :
Vcc - Ib(R1||R2) - IeRe - Vbe = 0
and
Vcc = I1*R1 + I2*R2
Duave said:...
Jony130 said:No this is incorrect because Ib current flows only through R1 resistor.
But we can use Rb = R1||R2 if you replace R1 and R2 voltage divider with his thevenin's equivalent circuit.