A sample text widget

Etiam pulvinar consectetur dolor sed malesuada. Ut convallis euismod dolor nec pretium. Nunc ut tristique massa.

Nam sodales mi vitae dolor ullamcorper et vulputate enim accumsan. Morbi orci magna, tincidunt vitae molestie nec, molestie at mi. Nulla nulla lorem, suscipit in posuere in, interdum non magna.

Negative Feedback to compensate for Beta Variation in BJTs.

Negative FB Biasing.

Negative FB Biasing.

In the last post, “Using the Transistor Model – Our first circuit“, the wide tolerance of β (hFE) in the transistor specifications is making life complicated.  “Complicate our life will ya!” We will fix the problem by complicating the circuit to reduce the complications.   Seriously, we will add several more resistors to apply negative feedback to reduce problems caused by the variation in β.

The new resistor Re is the main source of the feedback.  When the transistor conducts and current flows out of the Emitter, Re will create a voltage drop across it.   This will cause the Emitter Voltage to be somewhat above ground voltage.   Now if we set the voltage divider so Vb is at the ideal voltage above Ve and Ve is raised because of the current flow, the current Ib will be reduced causing Ie to decrease.

I just described the negative feedback in words, but so far we do not have a way of calculating it.  Hopefully, it is like planning a road trip by looking at a map and knowing where you are headed.  Now all we have to do is drive there.  Unfortunately, we will have to make one detour along the way.

The transistor circuit of the first picture with the model replacing the transistor.

The transistor circuit of the first picture with the model replacing the transistor.

The first step we are going to do is replace the transistor in the circuit with our model for the transistor.   This gets us a step closer, but we do not have a good way to calculate the current flowing in the base leg.   What we will do is still leave Vcc connected across the output circuit, Rc to Re, but  we will replace the input circuit of Rb1, & Rb2 and Vcc on that side with a more simplified circuit.  We will use the Thevenin equivalent of that circuit and our final model diagram will look like the third picture.  This is the detour I was talking about earlier.  We now need to know how to determine Vth and Rth and how those relate to Rb1 and Rb2.
To do this we look at the circuit Vcc, Rb1, & Rb2 and disconnect that source from the transistor base terminal.

The input circuit replaced with the Thevenin Equivalent Circuit.

The input circuit replaced with the Thevenin Equivalent Circuit.

With the base disconnected, Rb1 and Rb2 is a simple voltage divider. Vth = Vcc X (Rb2 / (Rb1+Rb2)). This is following the exact rule I stated in the post “Thevenin and Norton Equivalent Sources“.  Determining the value for Rth is a little harder. There are two ways to do this.  I will show only one method for this specific circuit and we will revisit both Thevenin and Norton at a later date for more complete discussion.   The easiest way for now is to apply an imaginary short from the base terminal to ground to see how much current the original circuit can push through the short.   This would be Vcc / Rb1.  Now we want the same current to be pushed through our new circuit with the source having a value of Vth.

This becomes Vth / Rth = Vcc / Rb1 and if we solve for Rth it becomes:  Rth = Vth X Rb1 / Vcc.  This may have been hard for everyone to follow so I will make an example.

Imagine Rb1 = 8Ω and Rb2 = 2Ω and Vcc = 20V.

To find Vth:  Vth = Vcc* X Rb2 / (Rb1 + Rb2) = 20 X 2 / (8 + 2) = 40 /10 = 4 V

Now that we have Vth determined:  Rth = Vth X Rb1 /  Vcc = 4 X 8 / 20 = 32 / 20 = 1.6 Ω

We will double check our answer by placing a 2Ω load beween the base terminal and ground.

In the original circuit Rb2 and the new resistor in parallel would be equal to 1Ω and the voltage across it would be:
20 X 1 / (8 +1) or  20 / 9 or 2.22 Volts.    Our current through the new resistor would be V / R or 2.22 / 2 = 1.11 A.

Now we will check or equivalent circuit with a 2Ω load.   The voltage across the load would be Vth X 2 / (2 + Rth) = 4 X 2 / (2 + 1.6) = 8 / 3.6 = 2,22V and the current is 2.22 / 2 = 1.11 A.   IT AGREED!

Now that we have a way of analysing the input circuit we are ready to start designing the complete circuit.  Obviously, we have to worry about the maximum values stated on the datasheet for the transistor.  However, we have many more decisions on this circuit to concern ourselves with.  There are three general guidelines to help us make the decisions.

Guideline 1:  We want the current through Rb1 to be 10 to 100 times the current Ib.  The reason for this is to make the voltage set by the voltage divider Rb1 & Rb2 pretty much unaffected  by changes in the amount of Ib.   The voltage Ve will change due to the negative feedback and the goal of that is to change Vbe and therefore Ib.

Guideline 2:  Vcc should be 3 to 5 times the voltage of Vce.   Once we actually run a signal into the transistor the current Ic will change with the input signal.  That will cause changes in voltage drop on Rc and for now Re.   We need to have some “head room” so the Vce can increase and decrease.

Guideline 3:  The voltage drop across Re should be equal to or slightly less than the voltage drop Vce.   This guideline basically sets the amount of feedback.

Now that we have those guidelines,  we have a starting point for our design.  Design will actually end up being pretty much a cut and try situation until we get what we want.   That is the reason I put so much emphasis on understanding the purpose of each component.  After we get going on the design and we need to make modifications to the design we need to know which component to “tweek” and what the effect of that adjustment will be.

The next post will be using the information presented in this post to set the biasing on our transistor.

As I am working through all this there are things I choose to talk about and things I will have to put off until a later date.  The details of obtaining Thevenin and Norton equivalents is one of those.  Another one is working with FET’s. A third one is the configuration of the transistor amplifier circuit.   We are working with a configuration called Common Emitter, (CE).  There are two other configurations Common Collector, (CC), and Common Base, (CB). Again, we will get to those eventually but once we go through all the details of the CE it will be fairly easy to go to the other two.

Right now there is a lot of concepts being thrown at you and a lot for me to type.  I makes sense to just deal with the Common Emitter configuration for now and compare and contrast the others later.

Gary

Creative Commons License
Negative Feedback to compensate for Beta Variation in BJTs” by Create-and-Make.com is licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License.

If you found this post to be enjoyable and interesting please consider subscribing to this blog using one of the methods on the home page or the e-mail subscription form also found there and at the bottom of each page.


 

Print Friendly

Leave a Reply

You can use these HTML tags

<a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>