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I am currently studying BJTs but am having trouble analysing the four resistor bias circuit. My textbook gives the following circuit and says we can use voltage divider across R1 and R2 to find VB. enter image description here

But I feel like we can't ignore RE when we try to find VB. Shouldn't RE (and the 0.7V drop across VBE) combine with R2 in parallel and then use the voltage divider with that parallel combination and R1?

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  • \$\begingroup\$ There are two methods of solving it. Considering base current and neglecting base current. Both are given in most of the text books.. \$\endgroup\$ – nidhin May 20 '14 at 6:43
  • \$\begingroup\$ Beta is not given, so you cannot come up with a numerical answer. \$\endgroup\$ – Spehro Pefhany May 20 '14 at 11:05
  • \$\begingroup\$ It's often assumed that beta is 100 in textbook problems like this one. \$\endgroup\$ – krs013 May 20 '14 at 11:31
  • \$\begingroup\$ What's with those annoying white lines thru three of the four resistors? I have no idea what you are trying to communicate with them. Stick to standard symbols here. \$\endgroup\$ – Olin Lathrop May 20 '14 at 12:18
  • \$\begingroup\$ That schematic is taken directly from the textbook so those lines through the resistors must mean something but I don't know what \$\endgroup\$ – Sam May 20 '14 at 23:16
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I suppose, your design starts with a given value for Ic, correct? Hence, you know the corresponding value for the base current Ib (based on a given current gain value). More than that, I assume you have fixed already the values for Rc and Re.

Because of Ie=Ic+Ie you know the voltage at the emitter node as well as on the base node (0.65...07 V larger). With these information you can design the voltage divider at the base. Don´t forget that it is a LOADED divider (load=base node).

During calculation you have to make some assumptions:

1.) Vbe=(0.65...0.7): Choice of the value is of less importance because of Re feedback, which stabilizes the DC operating point,

2.) Resistance niveau of the voltage divider: It is common practice to select resistances which allow a current through the divider chain which is approximately (6...10) times the base current.

(Explanation: You have to find a trade-off. Small resistors give an unwanted small signal input resistance, but allow a good dc stabilizatin - and vice versa).

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  • \$\begingroup\$ We aren't given Ic so we can't find Ib like that \$\endgroup\$ – Sam May 20 '14 at 7:30
  • \$\begingroup\$ I just want to find Vb \$\endgroup\$ – Sam May 20 '14 at 7:54
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But I feel like we can't ignore RE when we try to find VB

In fact, the DC base voltage is given by

$$V_B = V_{BB} - I_B R_{BB} = V_{CC}\frac{R_2}{R_1 + R_2} - I_B R_1||R_2$$

The first term on the right-most side is \$V_{BB}\$, the open-circuit base voltage; the voltage on the base if the base current were zero.

Of course, the base current will depend on the value of \$R_E\$ and the base-emitter voltage \$V_{BE}\$.

Indeed, it's easy to show that the collector current is given by (ignoring the Early effect)

$$I_C = \frac{V_{BB} - V_{BE}}{\frac{R_{BB}}{\beta} + \frac{R_E}{\alpha}}$$

and

$$I_B = \frac{I_C}{\beta}$$

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Consider this: R1 and R2 can be expressed as a Thevenin equivalent of 5 volts and about 3k (I leave it to you to find the real number). With an infinite beta, Ve will be maintained with no drop across the Thevenin resistance. With a beta, for instance, of 100, Ie will be ~ 4.3 mA, and this will require 43 uA of base current. The drop across the Thevenin resistance will 43 uA x 3k, or about 120 mV. This is a change from the unloaded voltage of less than 3%. So, for reasonable values of beta, you can ignore loading effects, and the Vbc drop of .7 volts doesn't matter much either. Note that this changes if you try to use high resistances for the bias network, and it changes if the load resistors are too low. However, low load resistances imply high power dissipation with no signal, and this is obviously not a normally desired design.

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Sam - I don´t know why, however up to now I was of the opinion that the values of all 4 resistors are NOT known. Therefore, I have removed some of my comments because they are not relevant anymore.

In the following, I give you the way to an exact solution, which consists of 3 equations for 3 unknown paameters:

  • VB=k*R2*IB (k: unknown factor)
  • R1=(Vcc-VB)/[(k+1)*IB]
  • VE=VB-0.7=(B+1)*IB*RE (B: DC current gain).

Now you have the 3 equations to solve for the three unknown parameters k, IB and VB.

Please note that you have to make a choice for the current gain B and the voltage VBE=0.7 volts.

Comment: There is another solution which is based on the superposition theorem and the substitution theorem: In this case, you can perform two separate calculations for the two "fixed" voltages in your circuit: Vcc and VBE=0.7Volts (superposition). That means: You can treat the voltage VBE as a voltage source (substitution theorem). However, for my opinion, the solution with 3 equations is more "logical".

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If you ignore the base current you can first solve the voltage divider to get the base voltage. This calculation will be ok as long as base current will be much smaller than current through the voltage divider (R2).

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  • \$\begingroup\$ Since the question is for homework purpose, you have to say that this is an approx method. There is an exact method by replacing \$R_1\$ and \$R_2\$ with \$R_{th}\$. :-) \$\endgroup\$ – nidhin May 20 '14 at 6:25
  • \$\begingroup\$ I have some examples that use the thevenin equivalent. Are you saying that if I use this method, the value I get for V<sub>B</sub> will take R<sub>E</sub> into consideration? \$\endgroup\$ – Sam May 20 '14 at 7:47
  • \$\begingroup\$ You can solve this circuit exactly including base current. To do that you need to write few equations more. Using Thevenin (or Norton) equivalent will make this circuit easier to solve by reducing the number of equations. As nidhin pointed: ignoring base current will give you an approximate result. \$\endgroup\$ – Szymon Bęczkowski May 20 '14 at 13:56
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You can safely ignore the current flowing through the base. Yes, there is some, but it combines with the current coming through the collector. Since those currents together have to pass through the 1K resistor on the bottom right, the transistor is not saturated; it only "opens" enough to pull the emitter voltage to about 0.6V below the base voltage. Thus, the right side of the circuit is just an emitter follower. If that concept confuses you, it's ok, because it's a confusing concept, but essentially think that any extra current through the base would let a lot more current through the collector, increasing the voltage across the resistor and shutting the transistor again.

So the transistor is in equilibrium where the current flowing through the base is $\frac{1}{\beta}$ of the current through the collector, and thus $\frac{1}{\beta+1}$ of the total current through the emitter. Since values of $\beta$ are usually about 100 if not greater, you're safe to assume that this current is small enough compared to the voltage divider that you can ignore it. Then again, you can calculate it, if you know \beta.

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