# Transistor Base

#### Jony130

Joined Feb 17, 2009
5,241
What type of a BJT have you use in your simulation?
And your simulation program is ??

#### gerases

Joined Oct 29, 2012
177
The simulation program is iCircuit. In the simulation program the tansistor has a beta of 1000. The BE forward voltage is about 0.7. I don't have much control over the type of the transistor I think.

I just wanted to know how to arrive at the base voltage they indicated (4.32V) without looking at the drop over the emitter resistor first, which has the drop of 3.6V. Once we know the drop over the emitter resister, it's easy: 3.6V + 0.72V = 4.32V. But how do I know what the drop over that resistor will be in advance?

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#### Jony130

Joined Feb 17, 2009
5,241
As printout by WBahn hfe = 1000 is very unrealistically.
Voltage at base is equal to
Vb = Vbe + Ie *Re

so first we solve for Ie

Vth = 4.5V and Rth = 500Ω

Ie = (Vth - Vbe)/( Re + Rth/(hfe+1) ) = (4.5V - 0.73V)/( 10Ω + 500Ω/1001) = 3.77V/(10Ω + 0.5Ω) = 359mA .

so Vb = 0.73V + 359mA*10Ω = 0.73V+ 3.59V = 4.32V

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#### WBahn

Joined Mar 31, 2012
26,398
The simulation program is iCircuit. In the simulation program the tansistor has a beta of 1000. The BE forward voltage is about 0.7. I don't have much control over the type of the transistor I think.

I just wanted to know how to arrive at the base voltage they indicated (4.32V) without looking at the drop over the emitter resistor first, which has the drop of 3.6V. Once we know the drop over the emitter resister, it's easy: 3.6V + 0.72V = 4.32V. But how do I know what the drop over that resistor will be in advance?
The drop across the emitter resistor is what determines the base voltage because the base voltage is one diode drop above the emitter voltage and the emitter voltage is established by the voltage drop across the emitter resistor.

Use the technique I used but with β=1000 and you get:

Vcc/1kΩ +0.7V/(1000*10Ω) = Vb( 1/1kΩ + 1/1kΩ + 1/(1000*10Ω))
(10Vcc+0.7V)/10kΩ = Vb(21)/10kΩ
Vb = (10Vcc+0.7V)/21 = 90.7V/21 = 4.32V