Help me understand a paragraph from a book (001)

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samy555

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Hi
From this book:
**broken link removed**
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


==================================================

After discussion with the gentlemen: Audioguru, SunnySkyguy, LvW and FvM
The answer to the first question is:

The answer to the second question is to connect 43 ohm resistance, as shown above.
 
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The effective dynamic internal emitter resistance Re is 26mV/4.4mA= 5.9 ohms. The minimum hfe of the 2N3904 is 100 and the source resistance is 200 ohms. The 200/100= 2 ohms which is in series with the Re of 5.9 ohms to create a 7.9 ohms output impedance.

A 43 ohm resistor is added in series with the 7.9 ohm output impedance of the transistor so the 50 ohm filter is fed from an impedance of 43 ohms + 7.9 ohms= 50.9 ohms (almost 50.0 ohms).
 
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    samy555

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Thank you Audioguru
Can I understand from your words that the output impedance of an emitter followe = re + (Rs/beta); Rs is the source output resistance?
If that were true, I understand that the source impedance is reflected to the output divided by beta, so why did not reflect the base bias resistors?
Also, I'm surprised this is the first time in my life I hear these words. I have many books in the electronics, no one of them mention it, for example:
Book: Semiconductor Devices and Basic Applications
Chapter 6 Page 428
**broken link removed**
**broken link removed**
Excuse me Audioguru I hope to be patient with me as you've always.
thank you again
 

samy555 said:
Thank you Audioguru
Can I understand from your words that the output impedance of an emitter follower = re + (Rs/beta); Rs is the source output resistance?
Click to expand...
Yes, RS is the source resistance.

If that were true, I understand that the source impedance is reflected to the output divided by beta, so why did not reflect the base bias resistors?
Click to expand...
The value of each base bias resistor is 500 times the value of the source resistance so they make the actual input 100k//100k//200= 199.2 ohms.

Also, I'm surprised this is the first time in my life I hear these words.
Click to expand...
I learned it about 51 years ago but I have never used it.
 
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    samy555

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Audioguru said:
The value of each base bias resistor is 500 times the value of the source resistance so they make the actual input 100k//100k//200= 199.2 ohms.
Click to expand...
the actual input 3.3k//3.3k//200= 178.4 ohms,,, about the same........ no problem its OK

I learned it about 51 years ago but I have never used it.
Click to expand...
I wish you continued good health and wellness and longevity
Can I ask you why you have never used it?

A 43 ohm resistor is added in series with the 7.9 ohm output impedance of the transistor so the 50 ohm filter is fed from an impedance of 43 ohms + 7.9 ohms= 50.9 ohms (almost 50.0 ohms).
Click to expand...
Are they actually doing it in practice? why not using say, pi impedance matcheng network?

thank you Audioguru
 

samy555 said:
the actual input 3.3k//3.3k//200= 178.4 ohms,,, about the same........ no problem its OK
Click to expand...

I was wrong and you are wrong.
The bias resistors in the first schematic are 50k (I assumed they were 100k) and you think they are only 3.3k.

I wish you continued good health and wellness and longevity
Click to expand...
Thank you. I had a heart attack 5 years ago and survived luckily with no damage. Since then I eat better and exercise every day. Recently my cardiologist (heart specialist doctor) told me I am fine and he does not need to see me anymore. He said to discontinue the medications. I am 69 years old but I feel like 25.

Can I ask you why you have never used it?
Click to expand...
I never used an emitter follower. I have done mostly audio circuits in my career. Most used opamps.

Are they actually doing it in practice? why not using say, pi impedance matcheng network?
Click to expand...
Many RF filters need to be driven with a 50 ohms source resistance and have a 50 ohms load.
 
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I am very happy to hear you are in good health, you do not imagine the amount of your respect in my heart. You always help me and I will never forget it. (Excuse me on the thinness of the language I'm sure you understand me). In my country, electicity is cut off most of the day, so I printed many of the subjects that interest me. One of the main topics of this was the discussion with you about transistor circuit design, nine years ago. What surprised me is that you're patient with me for a degree unimaginable. I felt extremely embarrassed of myself when I repeat the same question and you're trying to answer without getting bored, and at last I understood, and this was the most important step for me in trying to understand electronics:thinker:.

 

I've used the same formula as audioguru, Z out = Rsource / ( β +1)

the "+1" term is always ignore since β is >>1
also β has at least +/-50% tolerance even for binned parts. 2:1 ratio

In Op Amps, Zout is the open loop resistance (in hundreds of ohms ) divided by gain reduction by feedback (Similar to Rs/β) . except .... since your example was 5MHz , where most Op Amps had no gain ( until more recent decades), so an Emitter Follower is a good choice >1MHz treated like a "step-down impedance transformer." with unity gain, although Op Amps with buffers may have complementary Emitter Followers built in or similar but at lower currents.

The lowest source impedance dominates , in this case 200Ω previous stage rather than the two 3.3k or Re.
If they say the Zout=7.9Ω then they must have previously stated the β=200/7.9 = 32 which may be the worst case minimum.

Adding a small series resistor also reduces the Q of the emitter, Zo when nH parasitic inductances drive a capacitive load and leads to ringing or oscillation at the bandwidth of the transistor (like 500MHz)
 

First I thank you for trying to help
Second: I could not understand any paragraph of explanation above. I do not know why. The reason may be that you explain a high level. If graciously tried to re-annotation simplest manner
 

SunnySkyguy said:
Source impedance / current gain = Output impedance

The source includes the generator, and bias resistors in parallel
Click to expand...
Thank you for this excellent and direct answer
Now I understand that:
Output impedance = (3.3k//3.3k//200)/beta + (re//RE)
Is that true?
 

Not quite.. I simplified it to the most significant parts

Re is in parallel not added in series but negligible since Re >> Zout, Re can be neglected.
rπ. should include all sources and the base emitter resistance can be neglected in series the source. since the base-emitter diode bulk resistance is much smaller the generator and bias resistors in most applications.
 

Unfortunately, I do not understand
Ok, why do not you try to write me a full expression (without rounding) for the Output Impedance.
rπ = (beta+1) re ??
Re >> Zout,
Click to expand...
Are you mean that Re is re which = 0.026/IE(mA), or Re the bias emitter resistor (RE)?
thank you
 

samy555 said:
Output impedance = (3.3k//3.3k//200)/beta + (re//RE)
Is that true?
Click to expand...

Gentlemen, may I jump into the discussion?

* At first, the above equation is correct (except it should be (beta+1) instead of beta).

* Secondly, I propose to use the term 1/gm instead of re only. Otherwise, misunderstandings and misinterpretations can occur (example: in samy`s post appears a quantity Re. What does it mean ? re or RE ? There are THREE symbols used for only two quantities).
Remember, the quantity you call re is NOT a resistive element. It is another expression for 1/gm - and gm is the transconductance which connects a small-signal input parameter (vbe) with a small-signal output parameter (ic).
Hence, it is not a resistive element between two nodes. And it is also a small-signal parameter which is identical to the slope of the Ic=f(Vbe) curve.

- - - Updated - - -

SunnySkyguy said:
Source impedance / current gain = Output impedance

The source includes the generator, and bias resistors in parallel
Click to expand...

SunnySkyguy - please, can you explain? I cannot follow this statement.
Where is the transconductance?
What if the source impedance is zero (voltage injection)?

As another help for understanding:
The output resistance of the emitter follower (node E, without emitter resistor RE and for Rsource=0) is identical to the input resistance of the common base stage which is simply 1/gm.
 
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    samy555

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LvW said:
Gentlemen, may I jump into the discussion?
Click to expand...
You are always welcome

Thank you for this clear answer, I understand completely
Greetings to you
Thank you all

- - - Updated - - -

Hello
For the second part of the question:

**broken link removed**

Can you show me how to connect the 43 ohm resistor to the output?
**broken link removed**
Thanks
 

It says, "adding a series 43 ohm resistor to the output". It will be in series with the output of the transistor that has an impedance of 7.9 ohms so the total resistance feeding the load is 43 ohms + 7.9 ohms= 50.9 ohms.
 
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    samy555

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FvM said:
In series with the 0.1 uF output capacitor.
Click to expand...

Audioguru said:
It says, "adding a series 43 ohm resistor to the output". It will be in series with the output of the transistor that has an impedance of 7.9 ohms so the total resistance feeding the load is 43 ohms + 7.9 ohms= 50.9 ohms.
Click to expand...
Like this:
**broken link removed**

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 ?
 
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Output impedance = (3.3k//3.3k//200)/beta + (re//RE)
Is that true?
No.... test it for zero values of RE.
If you consider the transistor Rπ as part of the source impedance where (β+1)*Rbe= Rπ

Then you get my generalized solution Zout = all source impedances/current gain

Changing labels for RE = Re and re=Rbe and beta=β
I see it as ...

Zo=(3.3k//3.3k//200)/(β+1) // Re + Rbe
= 178Ω/(β+1) // Re + Rbe
= 1.8Ω // Re + Rbe
= 1.8Ω + 25Ω /Ie
= 1.8 + 5 = 6.8Ω ( they got 7.8)

Note for Rbe above from Ebers-Moll equation
I = I s [e^(qV/ kT) -1] you can derive Rbe for T=25'C where Rbe=25 [Ω]/Ie [mA] of emitter current.
thus if Vc=6V across 1.2k Ic=5mA then Rbe=5Ω

Where is the transconductance?
Click to expand...
Don't need to use it, but can if you insist
 
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    samy555

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I think that this explanation is true 100%
I am very convinced
Thank you
**broken link removed**
 
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