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Here's another example. The base of the transistor is connected to a voltage divider

circuit between Vcc and ground. Also note that the emitter is biased with a supply

voltage Vee. In this example we will use a negative supply of Vee = -5V.

Figure 6.

The simplest way of solving this circuit is to just calculate the voltage Vb using the

voltage divider rule and proceed with a KVL loop from Vb, down to through the

transistor to Vee. However, this method assumes that the small base current Ib flowing

into the transistor is actually zero, and hence introduces a small error in the results.

Such an error can be negligible in circuit design. I'll step you through the process:

From inspection of the circuit you can see that the voltage Vb is actually just the

voltage across resistor R2 since R2 is connected to ground. I remember the voltage

divider rule by a phase that goes 'The one you want, over the sum of the other two'. Bythat I mean the resistors.

The equation for finding Vb using the voltage divider rule is:

Hence the voltage Vb, which is the voltage across R2, is Vcc multiplied by the one you

want (R2) divided by the sum of the other two (R1 + R2). For Vcc = 12V as before ,

and R1=33k, R2 = 10k.

Vkk

kVb 79.2

43

10.12

3310

,1012

Next you can write the loop through the transistor:

Vb = Vbe + Ie.Re +Vee, and solving for Ie,

R1

Re

Rc

Vcc

Ie

Ic

R2

Vee

Vb

21

2

RR

RVccVb

(9)

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from here it is simple to solve for the remaining currents and voltages as shown before.

However, note that Ve = Ie.Re +Vee = 7.09m 1k + (-5) = 2.09V. This answer could

have been negative if the emitter resistance was a little smaller.

Now, the second method for solving this circuit is to model the base divider as a

Thevenins equivalent circuit.

Figure 7.

The thevenins voltage source is just the voltage at the point of interest, which is Vb inseries with the equivalent resistance of the circuit, which is denoted Rb.

Vb is found by using the voltage divider rule as before, and Rb is just R1 in parallelwith R2. To keep these notes short I won't go into this but you can derive these results

from Thevenins theorem from previous subjects. So, from before:

and Rb = R1 || R2 = R1.R2/(R1+R2) or 1/( 1/R1 + 1/R2) = 7.67k for R2=10k and

R1=33k.

Writing the same loops as in the first example:

Vb = Ib.Rb + Vbe + Ie.Re + Vee

and substituting in equation 3:

Vb = Ib.Rb + Vbe +Ib(+1)Re + Vee

mAk

Ie

VeeVbeVbIe

09.71

)5(7.079.2

Re

R1

Re

Rc

Vcc

Ie

Ic

R2

Vee

Vb

Rb

Re

Rc

Vcc

Ie

Ic

Vee

Vb

21

2

RR

RVccVb

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AkkRb

VeeVbeVbIb

2.65

1).1100(67.7

)5(7.079.2

Re)1(

,

and hence Ie = (+1)Ib = 101 65.2 = 6.52mA.