Self biasing a transistor

[boylesg]

Assuming R1 is the collector resistor and R2 is the base resistor then the voltage at the collector would be R2 / (R1 + R2).

Wouldn't the voltage at the base then be the rail voltage? How is that supposed to work? I thought you usually bias the base to be roughly 0.7V above the emitter?

 

[LvW]

Assuming R1 is the collector resistor and R2 is the base resistor then the voltage at the collector would be R2 / (R1 + R2).

Please, can you show the corresponding calculation?

 

[boylesg]

Please, can you show the corresponding calculation?

An example I have seen have seen was 2.2M and 220k which would mean the base would be sitting at (2200 / (2200+220)) * 6 = 0.9 * 6 = 5.45V
Isn't the base supposed to sit midway between rail voltage and gnd to provide for equal swing for the negative and positive phase of the signal?

 

[betwixt]

It's more likely you want the output connection to be mid way between supply and ground to get most equal output swing. This could be the collector or emitter, depending on the circuit configuration. Holding the base at mid point implies you want to feed the circuit with maximum swing, in which case the amplifier isn't necessary anyway.

We need to see your circuit to confirm resistor values but I'm guessing you are ignoring base current and Hfe in your assumptions.

Brian.

 

[LvW]

I'm guessing you are ignoring base current and Hfe in your assumptions.

I am afraid, boylesg is not familiar with transistor basics (operation principles).

 

[boylesg]

It's more likely you want the output connection to be mid way between supply and ground to get most equal output swing. This could be the collector or emitter, depending on the circuit configuration. Holding the base at mid point implies you want to feed the circuit with maximum swing, in which case the amplifier isn't necessary anyway.

We need to see your circuit to confirm resistor values but I'm guessing you are ignoring base current and Hfe in your assumptions.

Brian.

I was looking at these at the time: http://www.talkingelectronics.com/p...ier/TheTransistorAmplifier-P2.html#BaseBias-1

See "Biasing the base"

 

[boylesg]

I am afraid, boylesg is not familiar with transistor basics (operation principles).

Yes you are correct.

I sort of get bits and pieces of it but I would in no way be confident in designing an amplier without the help of that software I downloaded.

With the help of that software I at least know where to begin with biasing but I am still rather hazy on the calculations at present.

If some one could point me in the direction of a website that has a worked example of transistor biasing, including all the impedance calculations that transistoramp has, I would be very greatful. I have found a some of the transistor biasing calculations in worked exampled but not all of them.

I need one worked example with everything rather than trying to piece together different aspects of the calculations from seperate examples.

I think I partially get the base biasing now. Tell me if this is correct please.

1) Using that graph in the datasheets of collector current vs colector-emitter voltage, you select a collector current you want to use.

2) That willl be associated with a particular level base current above the cut-off collector emitter voltage and the base bias resistors set the base at a voltage level to acheive this required base current.

3) You then want to fix the collector voltage at a point half way between GND and Vcc so that the signal current can swing evenly between those two voltage levels.

OK so for a common emitter amplifier the ratio of the collector resistor and the emittter resistor sets tha voltage gain of the amplifier doesn't it? So what sets the collector voltage at the half way point between Vcc and GND?

- - - Updated - - -

I intend to keep sticking at this until I gain at least a little confidence in designing these things without the software. But without formal classes this is going to take time and patience.

 

[betwixt]

There are many reasons why different currents/biasing are used so it's difficult to give a single example of how it's done.
Your 123 steps miss an important first stage, you have to design to a specification so there is more to it than just reading graphs. I'm grossly over-simplifying the steps below but they are more in line with designing for purpose than for for a specific operating condition:

1. Know what you are driving. The amplifier stage has to be able to deliver the required output voltage into the load of the following stage so before anything else you work out how much current it has to deliver into the load. Using Ohms law, I = V/R where V is the maximum voltage to be delivered and R is the load impedance. The collector current should be at least as much as the load needs.

2. Now you know Ic, and you know how much supply voltage you have available, work out the collector load resistor needed to set Vc to about half the supply. Rc = (V/2) * Ic.

3. The base current needed is Ic/Hfe so using Hfe from the data sheet, work out what Ib will be.

4. The simplest way to provide base current is with a single resistor to supply, so assuming the emitter is grounded and hence has 0V on it, Rb = (Vsupply - 0.7)/Ib.

The problem with this is that the value of Rb is highly dependent on the transistors Hfe and it varies greatly from one transistor to another, even in the same batch. There are various ways to reduce the variability, one is to add resistance in the emitter so as Ic increases, Ve increases and because it's a fairly constant Vbe, so does Vb. As Vb increases, less base current flows and the transistor conducts less. This gives a simple self stabilizing effect. The down side to this method is the output voltage swing can only possibly go from supply to Ve so it is reduced. The other way is to make Ib have less effect on the value of the bias resistor by adding a resistor from the base to ground and reducing the resistor to supply accordingly. Now, instead of all the base current coming through one resistor, it is taken from a potential divider. The more current flows directly through the potential divider, the less effect Ib will have. Typically, a designer will set the current through the potential divider to around 5x to 10x Ib. This is probably the most used method. If you work out resistor values, don't forget the top resistor carries the base current as well as the current through the bottom resistor.

I think you can see why some of your circuits trying to feed loudspeakers from low current sources don't work very well. They simply can't provide enough current into a few Ohms load to produce any appreciable output and to scale down the resistor values to allow enough output would cook the transistor. Professional class A audio amplifiers use huge heat sinks and several Amps collector current to deliver only a handfull of Watts.

Brian.

 

[LvW]

Hi boylesg,

apparently, you have learned that there are two completely different currents flowing through both resistors as shown in the diagram you have linked.
Therefore, the voltage divider rule does not apply and the "formula" you gave in your first posting was completely wrong.

Now to your steps to bias the BJT:

1) Using that graph in the datasheets of collector current vs colector-emitter voltage, you select a collector current you want to use.

For selection of the collector current you do not need this diagram. You select the current according to your requirements regarding power consumption, transistor type, desired gain (the gain depends on the transconductance gm which is proportional to Ic).

2) That willl be associated with a particular level base current above the cut-off collector emitter voltage and the base bias resistors set the base at a voltage level to acheive this required base current.

"base current above the ... voltage"??? . Instead, the base current is directly dependent on the selected Ic (Ic=B*Ib).
And the base resistor (if you choose the configuration as shown in your attachement) is calculated using Ohms law (RB=Vcb/IB) with Vcb=V(collector)-0.7 Volts.

3) You then want to fix the collector voltage at a point half way between GND and Vcc so that the signal current can swing evenly between those two voltage levels.

Yes, but how do you "fix"? You select a proper Rc which causes a voltage drop Ic*Rc of approx. Vcc/2.

OK so for a common emitter amplifier the ratio of the collector resistor and the emittter resistor sets tha voltage gain of the amplifier doesn't it?

Approximately - yes. But it`s better to understand what`s going on. If you select the base resistor as shown in your diagrams you do NOT need an emitter resistor. (Therefore, 0.7 volts as mentioned above).
But there are other schemes for biasing with a voltage divider (insread of RB) and an emitter resistor.

The gain formula depends on this scheme. In your case (with RB) the gain is NOT simply determined by Rc but also by the internal resistance of your source (feedback action) .
You see, it`s not so simple as you might think.

So what sets the collector voltage at the half way point between Vcc and GND?

As mentioned - the drop across Rc and - if existent - across Re.

I recommend to study a book on BJT basics. Otherwise you are using formulas without knowing if applicable or not.

 

[boylesg]

I have been watching this series of lectures and it has been really helpful. The lecturers indian accent can't be a little distracting at times but I have never the less gotten more out of it. He goes through 3 different ways of biasing a common emitter class A amplfier and in particular I now 'get' the single base resistor which I was not previously connecting with the voltage divider and self biasing methods.

This lecture has most of what I need with fully worked examples of each biasing method. The impedance calculations I will have to piece together from other websites.

https://www.youtube.com/watch?v=PSdHf6yozyc

 

[LvW]

I have been watching this series of lectures and it has been really helpful. The lecturers indian accent can't be a little distracting at times but I have never the less gotten more out of it. He goes through 3 different ways of biasing a common emitter class A amplfier and in particular I now 'get' the single base resistor which I was not previously connecting with the voltage divider and self biasing methods.
This lecture has most of what I need with fully worked examples of each biasing method. The impedance calculations I will have to piece together from other websites.
https://www.youtube.com/watch?v=PSdHf6yozyc

OK - so now you know three different ways how to bias a transistor.
By the way - did the lecturer also compare these three alternatives? Advantages vs. disadvantages?

 

[LvW]

* The arrow directions for Ie and Ic are false.
* After choosing Ic you select Re so that Ve=Ie*Re=0.1*Vcc (approximately).
* Then: Vb=Ve+0.7V
* Use Vb to calculate the corresponding voltage divider R1-R2.
* Select Current I2 through resistor R2 with I2=k*Ib and with k=6...10 (select k not too small because of the resulting input resistance).
* Consider the current through R1 is I1=I2+Ib=(k+1)*Ib.

 

[jeffrey samuel]

* Then: Vb=Ve+0.7V
.

Does this condition hold good for what ever Ic current flowing in the CE amplifier

Cause I thought that Ic was proportional to Vbe or Ib and so to increase Ic I must increase Ib

If I am wrong pls clarify me here Sir

And I am using potential divider biasing But If you can clear it any config all the other follow the same idea so no problem with it

 

[LvW]

Hi boylesg, why have you completely removed your last posting?
This was not a good idea because it destroys the logical sequence question-answer, don`t you think so?



Hi Jeffrey Samuel

Does this condition hold good for what ever Ic current flowing in the CE amplifier
Cause I thought that Ic was proportional to Vbe or Ib and so to increase Ic I must increase Ib

This "equation" is an assessment - because it is assumed that Vbe will be approximately in the range 0.6...0.7 volts.
Try to calculate the bias divider based on 0.7V and then based on 0.6 V - and you will notice that the difference is rather small.
In most cases the "error" (better: uncertainty) will be not larger than other uncertainties and tolerances.
For example, think of the value of B resp. beta. This value is given in the data sheet with tolerances of 100% or even larger.
More than that, it is the task of the emitter resistor to reduce the influence of uncertain input parameters (Vbe, Ib) on the dc operating point.

Therefore, it is common practice to use Vbe=0.6...0.7 for calculating the base divider circuit (perhaps 0.6 volts for some milliamperes and 0.7 volts for some tenth of milliampere).
Of course, this approach does not ignore the fact that Ic is controlled by Vbe - but it uses the knowledge that Vbe will be in the range as mentioned above.