Measuring gain of common emiter stage

[PRR]

> not a follower to have a gain of 3 ?

"Follower" generally means it follows the leader. There is a large class of circuits with "almost unity voltage gain" but substantial current gain.

That's not going to give you Voltage Gain. I never realized, but the only place you get Voltage gain is at the Collector.

> getting 3.9v at the emitter not 9v, which I am finding confusing.

Where is there 9V? Sure, way over at a battery. But the emitter resistor can NOT "see" the battery. It sees what the emitter lets it see. Generally a bit less than where ever we hung the base. In a 9V world, around 4V. Then a 10K resistor demands 0.4mA.

> you saying the input impedence was 1M

I think the words were "Counting on my thumbs I am seeing >1Meg". Yes, the ">" sign may have been overlooked, especially because _I_ habitually use it to set-off quote-backs (from another world where we did little math).

> total input impedence 1.6M , did I do that right ?

Stunningly close to what my idiot assistant gives for slightly different transistors:



The fall at high frequency is generally true of everything (there is capacitance everywhere and caps go to zero impedance at infinite frequency.... transistors amplify this.)

> Gus ,...as I understand bootstrapping that would make the input impedance 108K.

I got a different number. I do not follow your bootstrap thinking.

The bootstrap comes from a follower. The gain at the emitter is almost unity.

If it were EXACTly unity, the voltage across the 10K would be identical at both ends. No signal current would flow. The 10K would appear to be infinite.

If the gain is really 9/10th, then only about 1/10th input voltage appears across the 10K, and to signal it acts 10 times bigger than it looks.

I do think that bootstrapping is Advanced and Treacherous. If you wanted general transistor theory, I would tell you to stay off that path. How-Ever.... since you want simple hi-Z circuits, it "IS" something you will need in the tool-bag.

> I am still confused

Hell no. I think you came further in days than I did in years. Further than many EE students really get in two semesters. (Solving quiz-puzzles is a different skill than actually understanding random real-world stuff.)

[Brymus]

Thank you Paul,I do feel better I was able to solve for the top two resistors of the Darlington example you gave me.
And that I am starting to get the base impedence/input impedence relationship.

Could the Darlington above be wired without the 4k5 on Q2 collector and then take the output from the emitter of Q2 for a "super buffer" ?

Oh and you were right,I overlooked the > before the 1M in your post,I thought it was pointing and not meaning "greater than"
But it made me think harder at least.
I let my brain settle today...I will do some more reading and then some breadboard work,and probably have some more questions in the days to come.
I really want to get all this down,and then move on to FETs
I "REALLY" appreciate you taking the time to help me learn and understand this,so I can stop hacking/splicing and start applying math and proven method to my work.
Bryan

[Brymus]

Next work out what is going on in the beginner project.  Set to min gain what is it's input resistance?

look at this
http://....org/viewtopic.php?f=7&t=9255&p=99794&hilit=bootstrap#p99794

Using Z in = Rg/1-Av where Rg is the 100k bootstrap resistor and Av is the voltage gain off the follower.
Is it then Z in = 100k/ 1 - .95 = 2M input resistance ?

And it doesnt matter what the gain is because the bootstrap is open to DC (due to the 22uf) and taken from the emitter,which has a less than unity gain.

[Brymus]

Check if I am understanding the bias right ?
The base in Gus' NPN boost sits at 2.87V the divider is the 100k-47k junction which in turn makes the available current to the base .000287 amps ?

[PRR]

> Rg is the 100k bootstrap resistor

The input does not "see" the 100K, it "sees" the 10K. One end at input, other end at follower.

0.95 may not be a good number for the follower's gain. Try other numbers: bootstrapped impedance  is very sensitive to actual gain.

In addition to whatever the 10K "looks like", we still have to drive the base and the 5K emitter resistor reflected through Beta. This will be OTOO 500K-2,000K, so the total imput impedance won't be as high as 2Meg.

[Brymus]

Duh Rg is the 10K resistor 
man that was a bad/stupid, mistake/oversight...
Would .98 be a better gain for the follower ? ( Z in = 500k ) Or do I have to calculate the gain exactly ?
How would I calculate the Av at the emitter ?

[PRR]

> do I have to calculate the gain exactly ?

Bootstrapped impedance.... gain of 0.5 doubles it, gain 0.9 makes it 10X, gain of 0.99 makes it 100X.

Any of these are possible.

> How would I calculate the Av at the emitter ?

I like to see it as a Voltage Divider. If the active device were "ideal" it would have zero emitter impedance. We measure how non-ideal a device is with a figure of merit called Transconductance, Gm(*). We can also invert a Conductance to get a Resistance. This resistance and the total emitter loading is a voltage divider.

Vacuum tube with Gm=1,000uMho (1,000 ohms) and 1K cathode resistor, the gain to the cathode is 0.5. Hot FET with Gm=3,000 and 3K source resistor, the follower gain is near 0.9. BJT at 1mA has Gm near 30,000uMho, 30 ohms. Say we take 3V across the cathode resistor, it will be 3K. The follower gain can be seen as a voltage divider, 3,000/(30+3,000)= 0.99. However if we bypass the emitter resistor (turn the Boost pot in the Beginner Project) with a "zero"-impedance cap, gain is 0/(30+0) or just zero. Or apply a 8 ohm load (weak loudspeaker amp), gain is 8/(30+8) or nearly 0.2.

So you shouldn't just pull a number out of your hat.

Often you don't need an "exact" answer. If the resistor being bootstrapped is 47K, and the application wants over 100K total, then if the follower gain is clearly much-more than "2" (say over 5) then it is probably fine, or at least not getting in the way.

(*) When values got nickname letters, Conductance couldn't have "C" because that was capacitance. It got "G". A simple resistor is just a G. (Anyway it looks cool in German typography.) In a tube (any 3-leg amplifying device) we are interested in the "mutual" conductance, a phantom "resistor" which is the ratio of "input voltage" to "output current". Hence Gm.

[Brymus]

Thanks Paul I will look it over again in a day or two.
I actually have freakin dreams about all this math I dont understand 
Lately my neck injury has been acting up
TOO much pain to even sit at the PC and work the mouse,the last few days.But I do appreciate the input.
Its amazing what you can miss in these forums in a few days (lots of good info)

[PRR]

OK, I was worried that you lost interest.

Stay off your neck. Walking helps my lower back, but that won't do much for a neck.