# Help me understand a paragraph from a book (001)

Discussion in 'General Electronics Chat' started by samy555, Sep 10, 2014.

1. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
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

2. ### Veracohr Well-Known Member

Jan 3, 2011
600
86
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.

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.

3. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
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

4. ### Veracohr Well-Known Member

Jan 3, 2011
600
86
(200||3.3k||3.3k)/100 = 1.8. Close enough.

5. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
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?

6. ### Brownout Well-Known Member

Jan 10, 2012
2,374
999
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.

7. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
I can not imagine it in parallel. Can you help me so that I can be convinced it is in parallel?
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!
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

8. ### Brownout Well-Known Member

Jan 10, 2012
2,374
999
In the analysis, set all independent sources to zero. Redraw the circuit, and you'll see why it's in paralle.
The other two resistors are 3.3k. 3.3k/200 = 16.5

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

9. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
Yes, now I understand, thank you

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
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

10. ### Brownout Well-Known Member

Jan 10, 2012
2,374
999
Close, it's (12.9 + re) || Re.

11. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
thank you very much

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

12. ### Brownout Well-Known Member

Jan 10, 2012
2,374
999
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%.

Last edited: Sep 11, 2014
samy555 likes this.
13. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
Thank you very much

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

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,168
1,188
Simply put a 43 ohm resistor in series with 100nF output capacitor.

15. ### Brownout Well-Known Member

Jan 10, 2012
2,374
999

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16. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
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 ?

17. ### ISB123 Well-Known Member

May 21, 2014
1,240
537
No reason to complicate things so just use the resistor.

18. ### samy555 Thread Starter Active Member

May 24, 2010
116
3
Who said I want to complicate things!!!!
I just put a question, you have the option to either answer or not answer.
Thank you

19. ### ISB123 Well-Known Member

May 21, 2014
1,240
537
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.

May 24, 2010
116
3
Ok thanks