Hi
From this book:

I read the following in page (2-11):

​

I did not understand the sentences that are underlined.
For the first red sentence, I do not understand where the value 7.9 ohm is came from?
The same for the second blue sentence 43 ohm?

​

thank you

 

Hi
For the first red sentence, I do not understand where the value 7.9 ohm is came from?

It's the output resistance of the emitter follower, calculated from the intrinsic emitter resistance (dependent on the emitter current), the emitter resistor, and the 3 resistors on the input of the transistor (including the 200 ohm source resistance) divided by the transistor beta.

The same for the second blue sentence 43 ohm?

Since it's supposed to drive a 50 ohm circuit, it must be for the purpose of impedance matching (7.9 + 43). I suppose this would be to reduce signal reflection or something, since there's not a lot of power involved.

 

It's the output resistance of the emitter follower, calculated from the intrinsic emitter resistance (dependent on the emitter current), the emitter resistor, and the 3 resistors on the input of the transistor (including the 200 ohm source resistance) divided by the transistor beta.

Please show me how the 3 resistors on the input of the transistor (including the 200 ohm source resistance) divided by the transistor beta = 2 ?? (Let beta = 100)
Thank you very much

 

(200||3.3k||3.3k)/100 = 1.8. Close enough.

 

(200||3.3k||3.3k)/100 = 1.8. Close enough.

Thank you
3.3k||3.3k is ok, but I think that the 200 is in series with them!!! from where I look at the base??

Why many of basic electronic books dont mention that?

 

Actually, the 200 ohm resistor is in parallel with the 3.3k resistors. A good estimate of resistance is to consider the 200 ohm resistor by itself, since it's resistance is so much lower than the others. The reason books don't mention it is because source resistance is usually so high that it doesn't need to be considered.

 

Actually, the 200 ohm resistor is in parallel with the 3.3k resistors.

I can not imagine it in parallel. Can you help me so that I can be convinced it is in parallel?

A good estimate of resistance is to consider the 200 ohm resistor by itself, since it's resistance is so much lower than the others.

It's resistance is not so much lower than the others: 1650/200 = 8.25 . In electronics, to be so much it must be ten times lower!

The reason books don't mention it is because source resistance is usually so high that it doesn't need to be considered.

200/100 = 2 ohm added to 5.9
if it is usually so high In this case, it becomes unreasonable ignored. I mean, it's become big to the point that ignoring it makes the calculations inaccurate
thank you very much

 

I can not imagine it in parallel. Can you help me so that I can be convinced it is in parallel?

In the analysis, set all independent sources to zero. Redraw the circuit, and you'll see why it's in paralle.

It's resistance is not so much lower than the others: 1650/200 = 8.25 . In electronics, to be so much it must be ten times lower!

The other two resistors are 3.3k. 3.3k/200 = 16.5

200/100 = 2 ohm added to 5.9
if it is usually so high In this case, it becomes unreasonable ignored. I mean, it's become big to the point that ignoring it makes the calculations inaccurate
thank you very much

Source resistance is typically several tens to hundreds of K ohms. But is this case, 200 ohms must be considered.

 

In the analysis, set all independent sources to zero. Redraw the circuit, and you'll see why it's in paralle.

Yes, now I understand, thank you

The other two resistors are 3.3k. 3.3k/200 = 16.5

The other two resistors are (3.3k//3.3K) = 1.65K
1.65k/ beta = 16.5
16.5//(200/beta) = 16.5//2 =1.8 ,,,,,,,,,, it is ok

Source resistance is typically several tens to hundreds of K ohms. But is this case, 200 ohms must be considered.

Let Source resistance (Rs) = 6k
(3.3k//3.3K//6k) =1.294k
1.294k/beta = 12.9 ohm, so the output resistance = 12.9 + (re//RE)........ is it true?
thank you very much

 

Close, it's (12.9 + re) || Re.

 

Close, it's (12.9 + re) || Re.

thank you very much

What about the addition of the 43 ohm resistor to the output, do they actually do that in real circuits?

 

They do when output impedance needs to match load impedance.

BTW, in post #9, you wrote:

The other two resistors are (3.3k//3.3K) = 1.65K
1.65k/ beta = 16.5
16.5//(200/beta) = 16.5//2 =1.8 ,,,,,,,,,, it is ok

But (1.65k/beta)/(200/beta) = 1.65k/200 = 8.25. Further 3.3k || 3.3k || 200 = 178.4. The error in estimating it at 200 is a little over 10%.

 

They do when output impedance needs to match load impedance.

BTW, in post #9, you wrote:




But (1.65k/beta)/(200/beta) = 1.65k/200 = 8.25. Further 3.3k || 3.3k || 200 = 178.4. The error in estimating it at 200 is a little over 10%.

Thank you very much

Can you please show me how to connect the 43 ohm resistor to the output?

 

Simply put a 43 ohm resistor in series with 100nF output capacitor.

 

 

View attachment 72903

Simply put a 43 ohm resistor in series with 100nF output capacitor.

Thank you very much
Still I have a last question:
Is adding a series 43 ohm resistance is the only solution to accomplish the impedance matching?
If not, the only solution, does it the best solution ?

 

No reason to complicate things so just use the resistor.

 

No reason to complicate things so just use the resistor.

Who said I want to complicate things!!!!
I just put a question, you have the option to either answer or not answer.
Thank you

 

That was proper answer.In practice you would go with something that is simple but at the same time does the job properly.That's why IC exist.

 

That was proper answer.In practice you would go with something that is simple but at the same time does the job properly.That's why IC exist.

Ok thanks