Tough BJT? Is for me.

Thread Starter

Sleepcakez

Joined Jun 26, 2010
13
I have the following circuit.
No values are given for R1 or R2, and we do not know Vth or Rth.
I cannot figure out how to find Rth.

I've seen this equation used but I cannot derive it myself so I don't really know what is going on.
They say Rth=.1(Beta+1)Re

Edit: I'm using Thevenin to find values for R1 and R2
 

Attachments

Last edited:

t_n_k

Joined Mar 6, 2009
5,448
You find the Thevenin equivalent for the voltage divider combination of R1, R2 and Vcc.

Vth=R2/(R1+R2)*Vcc

Rth=R1||R2=R1*R2/(R1+R2)

Looking into the base you see a series DC voltage VBE≈0.6 to 0.7 V, and an equivalent resistance of (1+β)(re+RE). Resistance re is the dynamic emitter resistance which may be neglected when RE is sufficiently large.

Given these conditions you can then find the base current IB and so on.

\(I_B=\frac{V_{th}-V_{BE}}{R_{th}+(1+\beta)(r_e+R_E)}\)
 

Thread Starter

Sleepcakez

Joined Jun 26, 2010
13
The thing is I'm not given any values for R1 and R2.
I was told to find values for R1 and R2 to obtain a bias stable circuit with the Q-point in the center of the load line.
We calculated Icq and Ibq no problem.
I'll include an excerpt of the solution manual.
 

Attachments

Wendy

Joined Mar 24, 2008
21,848
What are you using as the DC bias for beta? There won't be any one correct answer, there will be a range of correct answers.
 

t_n_k

Joined Mar 6, 2009
5,448
The thing is I'm not given any values for R1 and R2.
I was told to find values for R1 and R2 to obtain a bias stable circuit with the Q-point in the center of the load line.
We calculated Icq and Ibq no problem.
I'll include an excerpt of the solution manual.
Then you need to provide whatever other information the question includes. Simply giving the general schematic as shown would not enable the calculation of any values - there is no explicit design criteria.
 

Thread Starter

Sleepcakez

Joined Jun 26, 2010
13
My bad Beta is 150.

I'm not really looking to solve anything. I want to derive the formula they used to find R thevenin. I absolutely cannot figure it out.
 

t_n_k

Joined Mar 6, 2009
5,448
The thing is I'm not given any values for R1 and R2.
I was told to find values for R1 and R2 to obtain a bias stable circuit with the Q-point in the center of the load line.
We calculated Icq and Ibq no problem.
I'll include an excerpt of the solution manual.
The solution method you provided is what you should follow - once you are given the specific details for the design criteria. Is the problem that you don't understand the method given?
 

Thread Starter

Sleepcakez

Joined Jun 26, 2010
13
I don't understand how you calculate R Thevenin without resistor values for R1 and R2. The easy way out would be Rth = R1||R2, but thats not the case.

Basically I want to know how they derived
Rth = (.1)(B+1)Re
 

t_n_k

Joined Mar 6, 2009
5,448
The "trick" in the solution is (as you say in your first post) to make a guess at a reasonable value of Rth in relation to the resistance looking into the base. In the case of your solution the suggestion is to make Rth equal to 1/10th of (1+β)RE which is probably a reasonable approach. As Bill Marsden points out there is a range of values for R1 & R2 you could work on to set the required bias condition.

Another (similar) starting estimate for the total bias resistor series value is to set the current flowing in the divider chain (R1 & R2 in series) to about 10 times the base current.

Sometimes there may be other constraints such as having a minimum input resistance value for the amplifier stage - which could have some bearing on the slection of the values of R1 & R2.
 

Thread Starter

Sleepcakez

Joined Jun 26, 2010
13
So it goes something like this

10RbIb = IeRe >
RbIb = .1(IeRe) >
Rb = .1(IeRe/Ib)>
Rb = .1((IbRb(Beta+1))/Ib) >
Rb = .1(Beta+1)Re

So the normal assumption is that the voltage in the base is 10 times the voltage in the emitter?
 

Wendy

Joined Mar 24, 2008
21,848
Have you figured out the midpoint you are after?

Figure the voltage when the transistor is fully on, figure the voltage when the transistor is fully off, figure out the mid point from that.
 

Thread Starter

Sleepcakez

Joined Jun 26, 2010
13
For finding Icq
10V = 5V + Ic(Rc+Re)
Icq = 3.57143 mA
Ibq = 3.57143/150 = .0238mA

From there it was finding Rth and Vth and then solving down into R1 and R2. I just really didnt understand the equation
Rth = .1(Beta+1)Re.
Escpecially when I saw elsewhere
Rb<= .1(Beta)Re

I think I'm cleared up on what I need to do so far.

I appreciate the help.
 

Jony130

Joined Feb 17, 2009
4,983
Given these conditions you can then find the base current IB and so on.
\(I_B=\frac{V_{th}-V_{BE}}{R_{th}+(1+\beta)(r_e+R_E)}\)
Well, I think that if we talk about dc current we cannot mix it with ac parapets. And certainly re is ac parapets.
So equation for Ib should look like this:

\(I_B=\frac{V_{th}-V_{BE}}{R_{th}+(1+\beta)(R_E)}\)

Or if we do the math
Vcc - I2• R2 - I1• R1=0
I2=IB+I1
I1•R1 - Ube - Ie•RE=0


\(I_B=\frac{R_1*V_{cc}-V_{BE}*(R_1+R_2)}{R_E*(1+\beta)*(R_1+R_2)*(R_1*R_2)}\)

So the normal assumption is that the voltage in the base is 10 times the voltage in the emitter?
No, current that is flow through voltage divider mus be at least 10 times large then base current.
Or with slightly different point of view but withe the same result, we make impedance of a voltage divider (impedance looking into the voltage divider) Rth = (R1||R2) small (at least 10 times) compared with the dc impedance looking into the base (β+1)*RE
 

t_n_k

Joined Mar 6, 2009
5,448
Well, I think that if we talk about dc current we cannot mix it with ac parapets. And certainly re is ac parapets.
So equation for Ib should look like this:

\(I_B=\frac{V_{th}-V_{BE}}{R_{th}+(1+\beta)(R_E)}\)
Yeah - I really was asleep on the job there!
 
Top