Common-Emitter Amp doubts and questions

Thread Starter

adam555

Joined Aug 17, 2013
858
PS. We always need RB resistor
I see; I will always use one then.

My next problem is with the voltage divider to bias the base. I find it a bit hard choosing the right values taking into account the Ib current. I mean, getting the right bias voltage is easy; where I struggle is in calculating the current that will go into the base.

At this point I have to confess that I'm a bit rusty in maths; and this is probably slowing me down a lot.
 

Thread Starter

adam555

Joined Aug 17, 2013
858
What I mean is... it would not be the same using 8466 and 1533, than using 16932 and 3066, or 4233 and 766 for R2 and R3; even though all three pairs would give the same bias voltage.

V1*R3/(R3 + R2) = 15*1533/(1533+8466) = 15*3066/(3066+16932) = 15*766/(766+4233) = 2.3V

So, the voltage is always the same, but the current to the base is not...

 

Jony130

Joined Feb 17, 2009
5,487
.
I find it a bit hard choosing the right values taking into account the Ib current.
Why ? I don't know Kirchhoff's laws?

I mean, getting the right bias voltage is easy; where I struggle is in calculating the current that will go into the base.
I don't see any problem here. What you should do is select your collector current to meet your desired goals. Then look up the minimum β in data sheet and find Ib_max = Ic/βmin.
Next select your bias divider resistors so that the current in the divider is larger than Ib_max.
In practice 5 to 30 times larger then Ib _max. Why ?
Because if the current drawn from voltage divider is a small in comparison with the current that flows through voltage divider resistors, then the voltage divider output voltage is not upset very much.

So if we choose Idv >> Ib_max We can ignore the upsetting effect of the base current on the divider output voltage.
 

Thread Starter

adam555

Joined Aug 17, 2013
858
Why ? I don't know Kirchhoff's laws?

I don't see any problem here. What you should do is select your collector current to meet your desired goals. Then look up the minimum β in data sheet and find Ib_max = Ic/βmin.
Next select your bias divider resistors so that the current in the divider is larger than Ib_max.
In practice 5 to 30 times larger then Ib _max. Why ?
Because if the current drawn from voltage divider is a small in comparison with the current that flows through voltage divider resistors, then the voltage divider output voltage is not upset very much.

So if we choose Idv >> Ib_max We can ignore the upsetting effect of the base current on the divider output voltage.
It's not that I find a problem with it -I understand all that- it's that I struggle a bit with the maths to find the current. :(

I can do it with Kirchhoff, but it would take me quite a while. So... well, I feel a bit ashamed to ask... I was kind of wondering if you have a quick and easy formula to calculate the current coming out of the middle node and into the base of a voltage divider.

It's been many years since I studied maths, and barely touched them since; so I really need to get a good maths book to review everything that I forgot.
 
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Thread Starter

adam555

Joined Aug 17, 2013
858
For which circuit?
Thanks.

This one, for example. Imagine that the middle node of the voltage divider connects to the base of a BJT; what's the simplest way to calculate the base current?



I hope I'm not abusing your kindness...
 
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Thread Starter

adam555

Joined Aug 17, 2013
858
Thanks again; that's exactly what I was looking for.

You see, I wouldn't be able to do that...

getting from here:

Vcc = I1*R1 + I2*R2 (1)

I1 = Ib + I2 (2)

I2*R2 = Vbe + Ie*Re (3)

And Ib = Ie/(β+1) (4)
To here:

And we can solve this for Ib.

\(Ib = \frac{R2Vcc - Vbe(R1+R2)}{(\beta + 1)Re(R1+R2) +(R1R2) } = 419.047\mu A\)
Most often than not I get stuck in electronics because of the maths; specially with anything involving algebra or calculus.
 

Thread Starter

adam555

Joined Aug 17, 2013
858
And with that I finished the list. I really appreciate all your help and patience with me. If I can help you in anything; just ask.

I'm going to review all the material I have on BJTs again, just to make sure I left no loose ends; so I might come back with more questions... I hope you'll be around. :)
 
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screen1988

Joined Mar 7, 2013
310
The second doubt is about a particular exercise that I saw in a book; which takes into account the input resistance to an amplifier, to calculate how much current goes into the base of the transistor along with the base resistor current.

My question is: how do you know in practice what's the input resistance, when you don't have a book telling you?... and, do you really need to know this?
We need to know input resistance, because we don't want put too much loading on the input source (most of a signal source will have low current capability). So we need high input resistance.
Can you give more detail this? Did you mean that we do that to avoid large voltage dropping in the internal resistance of the source?

most of a signal source will have low current capability
Not really sure, my thought is that if the load impedance is too low => the current flowing through is too high => destroy the source.

We use small signal analysis and find Rin. For example for CE amplifier without Re resistor.
Rin = (Hfe + 1) * 26mV/Ic
Is there a reason for using small signal analysis in finding Rin?
Is there a way to do that using large signal analysis?

EDIT: After review LvW's post, I see that input impedance here is small signal dynamic resistance.
But is this the only resistance that the voltage source see? I wonder what is the resistance that the source sees in large signal model?
 
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LvW

Joined Jun 13, 2013
1,752
Adam, in spite you have "finished the list", I have one comment/question:

With regard to your post#46 (see the voltages in red), did you combine ac and dc voltages? I am a bit confused about the figures.
 

Jony130

Joined Feb 17, 2009
5,487
You are an expert, Jony. :)
Just curious, where did you learn these detail knowledge, from books or in practice?
Am far away from being expert in the subject. An my knowledge comes from the book, but building an practical circuit is also very important in the learning process.
My first very poor one amp that i design and build was very long time ago (1998).
And I build it on a piece of carton board.



Can you give more detail this? Did you mean that we do that to avoid large voltage dropping in the internal resistance of the source?
Not really sure, my thought is that if the load impedance is too low => the current flowing through is too high => destroy the source.
Yes, we try to avoid large or any voltage dropping in the internal resistance of the source. Because our signal is a voltage. So the internal resistance of the signal together with input resistance of a amp form a voltage divider. And of course we want as much voltage from the source at the input of a amp. And this can be done only if Rin >>Rsource.

Is there a reason for using small signal analysis in finding Rin?
Is there a way to do that using large signal analysis?

EDIT: After review LvW's post, I see that input impedance here is small signal dynamic resistance.
But is this the only resistance that the voltage source see? I wonder what is the resistance that the source sees in large signal model?
Small signal analysis gives us the simplest way to find input impedance, without the need to solve any differentials equations. For the large signal we have to solve differentials equations. But in the end both of this method will give the same result.
 
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LvW

Joined Jun 13, 2013
1,752
Is there a reason for using small signal analysis in finding Rin?
Is there a way to do that using large signal analysis?

EDIT: After review LvW's post, I see that input impedance here is small signal dynamic resistance.
But is this the only resistance that the voltage source see? I wonder what is the resistance that the source sees in large signal model?
Adam, perhaps it helps to define "large signal". For example, it is not possible to give any number for "large" (mikro-, millivolts or so).

* Let`s start with "small signal". In this case all the changes in current and voltage are so small that the non-linearity of the transistor parameters (input,transfer and output charcteristics) can be neglected. Thus, we have to deal with LINEAR relationships only.
* In contrary, this is NOT the case for large signals (larger than those signals that allow to be treated as "small"). Here we must consider the mentioned non-lineraities - and this is a problem:
Because of the non-linearity we have distortions.
* Now, imagine you want to calculate the input resistance for "large" signals. You have a sinusoidal voltage and a current that is NOT sinusoidal (distorted). Can you compute the ratio of two signals that have NOT the same shape?
* Of course, you are able to compute the momentary (instantenious) ratios. But for which purpose? In such a case (if the signals are not too large) I still would use the samll-signal values knowing that they are not correct by 100%.
* To illustrate the scene: That is the reason you cannot give a reasonable voltage gain for push-pull class-C amplifiers.
 

Thread Starter

adam555

Joined Aug 17, 2013
858
Thank you for asking those questions screen1988...

Am far away from being expert in the subject. An my knowledge comes from the book, but building an practical circuit is also very important in the learning process.
My first very poor one amp that i design and build was very long time ago (1998).
And I build it on a piece of cardboard.
I'm going to try doing the same exercise as soon as I finish reviewing the material.

So, and please correct me if I'm wrong, what you did there was first to pre amplify the voltage with T1, then the current with the Darlington pair, and then amplify both positive and negative semi-cycles with the pull-push NPN and PNP transistors respectively.

Yes, we try to avoid large or any voltage dropping in the internal resistance of the source. Because our signal is a voltage. So the internal resistance of the signal together with input resistance of a amp form a voltage divider. And of course we want as much voltage from the source at the input of a amp. And this can be done only if Rin >>Rsource.

Small signal analysis gives us the simplest way to find input impedance, without the need to solve any differentials equations. For the large signal we have to solve differentials equations. But in the end both of this method will give the same result.
You explained how to calculate Rin, but I'm not sure how you get Rsource; this value is normally given in the analysis exercises of the books, but I don't remember ever calculating it before. For example: how would I know what's the Rsource of my guitar?... also curious what Rsource would be for the function generator I built.
 
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Thread Starter

adam555

Joined Aug 17, 2013
858
Adam, in spite you have "finished the list", I have one comment/question:

With regard to your post#46 (see the voltages in red), did you combine ac and dc voltages? I am a bit confused about the figures.
Hi LvW,

That exercise is not mine, I took it from one of Jony's posts in one of the linked threads as an example. In any case, I think the 0.8V and the 3.8V are the bias voltage plus the signal peak voltages; which should swing between those 2 values at the node of the voltage divider.

Adam, perhaps it helps to define "large signal". For example, it is not possible to give any number for "large" (mikro-, millivolts or so).

* Let`s start with "small signal". In this case all the changes in current and voltage are so small that the non-linearity of the transistor parameters (input,transfer and output charcteristics) can be neglected. Thus, we have to deal with LINEAR relationships only.
* In contrary, this is NOT the case for large signals (larger than those signals that allow to be treated as "small"). Here we must consider the mentioned non-lineraities - and this is a problem:
Because of the non-linearity we have distortions.
* Now, imagine you want to calculate the input resistance for "large" signals. You have a sinusoidal voltage and a current that is NOT sinusoidal (distorted). Can you compute the ratio of two signals that have NOT the same shape?
* Of course, you are able to compute the momentary (instantenious) ratios. But for which purpose? In such a case (if the signals are not too large) I still would use the samll-signal values knowing that they are not correct by 100%.
* To illustrate the scene: That is the reason you cannot give a reasonable voltage gain for push-pull class-C amplifiers.
Thanks.
 
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Thread Starter

adam555

Joined Aug 17, 2013
858
You explained how to calculate Rin, but I'm not sure how you get Rsource; this value is normally given in the analysis exercises of the books, but I don't remember ever calculating it before. For example: how would I know what's the Rsource of my guitar?... also curious what Rsource would be for the function generator I built.
Found it... forgot for a moment that most things are easy to find on the internet.
 
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