Common emitter amplifier

You have the basis for a great emitter follower here, but your biasing needs to be adjusted. Try changing the upper base resistor to around 1k, while the circuit is running. The upper base resistor does not see the lower base resistor. The upper base resistor sees mostly the emitter resistance to ground. (330ohms). Place the scope on the emitter(not the emitter current resistor) and watch what happens when you adjust the upper base resistor close to 1k ohms. Ignore the other current waveforms. Your most important waveform is the voltage at & after the emitter. Hope that helps. Cheers & Happy Holidays.

Hi unclerick. Happy holidays to you and Craig !! It's true there is a bias problem however your second statement is not quite true since the emiter resistor is amplified by beta other wise there could be no transistor action so emiter resistor X beta in parallel with base resistance is a good rule of thumb there's a bit more to it but that is it in the most basic nut shell ! So the emiter resistor is actually bigger than the base resistance

I must choose to differ with you, pip. :o) You may remember, from your electronic training, that the base-emitter gain is about .9 to nearly unity. A rule of thumb here is that the input impedance is effectively the emitter resistance and the output impedance is effectively the collector resistance to signal ground. (i.e. the power bus) Looking into the base of the transistor... all that is seen, is primarily, the emitter resistance. You can measure it, but it must be measured with a meter that has enough voltage, to forward bias the base-emitter junction. The amount of that meter current will influence the reading somewhat, but the beta of the transistor will not. Therefore with the base-emitter junction forward biased, the 330 ohms is seen, with no beta influence.

Yes thats right but thats voltage gain not current gain.  You seem to be forgetting the humble bjt is primarily a current amplifier not voltage and this current is transformed into a voltage using a resistor. Here's the formula for input impedance on an emiter follower I've kept it simple by canceling out the various dynamic characteristics and summing the junction capacitance's to infinity.  r in ~BoRe                                                  And  rout~Re||Rsorce × Bo If your statement is true it would not be possible to make a buffer stage since any circuit driving your circuit would simply see the output load × 0.9 so an even lower impedance. So more current is needed to drive the load than would otherwise be needed without the transistor.

And so transistor action is muted to simply less than zero aka attenuator. And no amplification is possible.

It is a bit more complicated for the common-emitter circuit. We need to know the internal emitter resistance REE. This is given by:   REE = KT/IEm   where:   K=1.38×10-23 watt-sec/oC, Boltzman's constant   T= temperature in Kelvins ≅300.   IE = emitter current   m = varies from 1 to 2 for Silicon   RE ≅ 0.026V/IE = 26mV/IE Thus, for the common-emitter circuit Rin is   Rin = βREE/IE As an example the input resistance of a, β = 100, CE configuration biased at 1 mA is:   REE = 26mV/1mA = 0.26   Rin = βREE = 100(26) = 2600Ω Moreover, a more accurate Rin for the common-collector should have included Re'   Rin = β(RE + REE) and this is as simple as I know how to show the maths on Ec.. And i have stripped out the frequency dependence associated with input and output loading extra as we'll end up with talking more about the calculus and the special operators involved. I'm not presuming to say that you cannot use calculus it's just unless you have studied mathematics to advanced level it's a bit of a leap and maybe impossible to show on Ec because of the lack of operands on the keyboard. I hope this helps to clarity things for you.

Wow! I can't argue with that, mainly because I have no idea what you are talking about, with all of those terms. You blew me out of the water with that jargon. You see, I am what you might call a seat-of-the-pants technician. I learned my trade when slide rules were all the rage. Then in the school of hard knocks for the next 40 years. I learned by doing, rather than by books. The practical method, one might say. Doing, measuring and seeing what works and what doesn't. Math had very little to do with it, beyond Ohm's Law. With that said, I still believe I am correct about the base current being sourced from the emitter. And with a half of a volt drop (forward biased) the resistance seen by the base is that which the emitter has to ground. (Common emitter or emitter follower) The transistor circuit I know the best. Massive current gain, no voltage gain. But in the collector circuit... just the opposite. Same configuration, just different resistor values for each type. But I could never see common base. Go figure. I'm sure that comes easy for you, but I might as well be trying to fly through a heavy fog. :o( When I get into a common base config. I am really lost. I could never get my head wrapped around that, without a lot of head scratching & still maybe never seeing the big picture. Count yourself lucky.

Oh, yeah... and don't get me started on calculus. I might as well try to fly to Alpha Centauri with my arms. It ain't gonna happen. ;o)

Lol. I have a lot of respect for the common sense hands on approach it's what built the modern world and gave the next generation a pair of shoulders to stand on and get our heads out of the mire. My apologies and respect. If it works it ain't broke and don't need fixing. .....

Thanks pip. I would hate to think I wasted all of those years since 1970. ;o) And there's nothing to apologize for.

Thanks unclerick !!!!