The calculated base resistor of 83k is much lower than the previous 883k. We choose 82k from the list of standard values. The emitter currents with the 82k RB for β=100 and β=300 are:

What is going on with the denominator of the fraction in Beta=100 and Rb =82k,

Rb/B+Re.The quotient is not approximated to 1.01 Ma. Are we adding the Beta and Re or what.If done as is, then,.0009036 Amp.,instead of 1.01 Ma.http://www.allaboutcircuits.com/vol_3/chpt_4/10.html

 

No its all right the calculation is like this ,may be it could be corrected in the book
\(I_E = {V_{BB} - V_{BE} \over (R_B / \beta) + R_E}\)

Good Luck

 

You skipped over a lot of ground between 82K and 870KΩ.

Emitter-bias

Inserting a resistor RE in the emitter circuit as in Figure below causes degeneration, also known as negative feedback. This opposes a change in emitter current IE due to temperature changes, resistor tolerances, beta variation, or power supply tolerance. Typical tolerances are as follows: resistor— 5%, beta— 100-300, power supply— 5%. Why might the emitter resistor stabilize a change in current? The polarity of the voltage drop across RE is due to the collector battery VCC. The end of the resistor closest to the (-) battery terminal is (-), the end closest to the (+) terminal it (+). Note that the (-) end of RE is connected via VBB battery and RB to the base. Any increase in current flow through RE will increase the magnitude of negative voltage applied to the base circuit, decreasing the base current, decreasing the emitter current. This decreasing emitter current partially compensates the original increase.

Emitter-bias

Note that base-bias battery VBB is used instead of VCC to bias the base in Figure above. Later we will show that the emitter-bias is more effective with a lower base bias battery. Meanwhile, we write the KVL equation for the loop through the base-emitter circuit, paying attention to the polarities on the components. We substitute IB≅IE/β and solve for emitter current IE. This equation can be solved for RB , equation: RB emitter-bias, Figure above.

Before applying the equations: RB emitter-bias and IE emitter-bias, Figure above, we need to choose values for RC and RE . RC is related to the collector supply VCC and the desired collector current IC which we assume is approximately the emitter current IE. Normally the bias point for VC is set to half of VCC. Though, it could be set higher to compensate for the voltage drop across the emitter resistor RE. The collector current is whatever we require or choose. It could range from micro-Amps to Amps depending on the application and transistor rating. We choose IC = 1mA, typical of a small-signal transistor circuit. We calculate a value for RC and choose a close standard value. An emitter resistor which is 10-50% of the collector load resistor usually works well.

Our first example sets the base-bias supply to high at VBB = VCC = 10V to show why a lower voltage is desirable. Determine the required value of base-bias resistor RB. Choose a standard value resistor. Calculate the emitter current for β=100 and β=300. Compare the stabilization of the current to prior bias circuits.

An 883k resistor was calculated for RB, an 870k chosen. At β=100, IE is 1.01mA.

For β=300 the emitter currents are shown in Table below.

Emitter current comparison for β=100, β=300.

Bias circuitIC β=100IC β=300 base-bias1.02mA3.07mA collector feedback bias0.989mA1.48mA emitter-bias, VBB=10V1.01mA2.76mA

Table above shows that for VBB = 10V, emitter-bias does not do a very good job of stabilizing the emitter current. The emitter-bias example is better than the previous base-bias example, but, not by much. The key to effective emitter bias is lowering the base supply VBB nearer to the amount of emitter bias.

How much emitter bias do we Have? Rounding, that is emitter current times emitter resistor: IERE = (1mA)(470) = 0.47V. In addition, we need to overcome the VBE = 0.7V. Thus, we need a VBB >(0.47 + 0.7)V or >1.17V. If emitter current deviates, this number will change compared with the fixed base supply VBB,causing a correction to base current IB and emitter current IE. A good value for VB >1.17V is 2V.

The calculated base resistor of 83k is much lower than the previous 883k. We choose 82k from the list of standard values. The emitter currents with the 82k RB for β=100 and β=300 are:

Comparing the emitter currents for emitter-bias with VBB = 2V at β=100 and β=300 to the previous bias circuit examples in Table below, we see considerable improvement at 1.75mA, though, not as good as the 1.48mA of collector feedback.

Emitter current comparison for β=100, β=300.

Off the top of my head I don't see the problem. The point is the initial value was too high, creating major variation due to β differences between transistors (which is normal). The point it is making is you need to have much lower values to increase the stability.

So where is the problem again?

 

OK, I went through the exercise, the math works on my calculator. You did notice that VBB was 2V? The rest of the equation was the same, as it should be, and my answers matched the books.

 

So where is the problem again?

For the denominator he added \(\beta\) with \(R_E\) and then divided \(R_B\) with that result.

He did this
\( I_E = {V_{BB} - V_{BE} \over R_B / \beta + R_E}\)

\( I_E = {2 - 0.7 \over 82000 / 100 + 470}\)

\( I_E = {1.3 \over 82000 / 570}\)

\( I_E = {1.3 \over 143.86}\)

\( I_E = 0.0090 Amp \)

which is a wrong answer

As I already said this on my earlier post ,the book may use brackets for the equation like this

\( I_E = {V_{BB} - V_{BE} \over (R_B / \beta) + R_E}\)

\( I_E = {2 - 0.7 \over (82000 / 100) + 470}\)

\( I_E = {1.3 \over 820 + 470}\)

\( I_E = {1.3 \over 1290}\)

\( I_E = 0.00101 Amp \)

@Thevenin's Planet
Always use BODMAS method for order of operation
http://en.wikipedia.org/wiki/Order_of_operations

Good Luck

 

OK, but in this case the book is not incorrect.

Not Verified.

 

You skipped over a lot of ground between 82K and 870KΩ.

Off the top of my head I don't see the problem. The point is the initial value was too high, creating major variation due to β differences between transistors (which is normal). The point it is making is you need to have much lower values to increase the stability.

So where is the problem again?

The point is that you gave the wrong direction to the goal.Since people in this field depends on number representation, correct formulas,graphs and equations to get a clearer understanding of what is happening to the device or component not placing parenthesis,or brackets,ect can lead to a dispute. With that said,I assume your are expressing the idea that the higher the collector current,meaning lower values of resistors,the better the stability?

 

Not really, I worked the math. Order of operations is pretty basic, and has been discussed on this site. Debijt has shown your error is excruciating detail, if you can but open your mind long enough to look at it.

The denial is all yours. I understand transistors well, having derived the equations from scratch several decades ago to get them down. So if the equation is correct and the math is correct there is no problem.

Debjit is correct in that parenthesis might make it clearer for beginners, but they are not needed, since the math is correct either way.

Consider you have two people how have independently gone over your assertion, and have concluded you are wrong. Why not show your math as Debjit did if you are still not convinced?

Side note, I went through the exercise of calculating the real base resistors. I am not a fan of showing a smaller power supply as shown in the book.

 

It is interesting to note you changed your verbage after my reply. My comment about denial was in direct response to a comment you made about denial. It is now gone.

Changing history is easy, but there is always a time stamp left behind. Next time I will quote you to pin what you say down.

I think this thread has run its course.