Emitter follower is "basically 1x amplifier" is there a "basicaly 2x amplifier"? - Page 1
Forum section: electronics / beginners

author: dusan
author_rank: Regular Contributor
timestamp: October 01, 2022, 01:07:19 pm
content: Emitter follower is 1x amplifier in wide range of settings, different supply voltage, different resistor values, etc. The only thing that changes is how close it is to 1x. The emitter follower "wants to be a 1x amplifier". It's it's nature to be 1x.Is there a topology such that it would make a natural 2x amplifier. I don't mean common emitter where you set resistors to have 2x amplification, I mean inherently 2x amplification. Something like a follower but 2x.

author: TimFox
author_rank: Super Contributor
timestamp: October 01, 2022, 01:20:39 pm
content: Vacuum triodes, such as the 6AS7 dual triode, designed for pass service in voltage regulators have low mu, approximately 2.https://frank.pocnet.net/sheets/093/6/6AS7GA.pdfIf you run a triode into a very high load impedance (or constant-current load), the voltage gain goes to -mu.Otherwise, 2 is not as fundamental number as 1.Of course, the emitter follower voltage gain with a finite load is < 1, since the output resistance from the emitter is finite and includes a term from the source impedance seen by the base.Similarly, the 6AS7 has a finite output impedance (plate resistance) of 280 ohms at 125 mA per triode after the (inverting) voltage gain of mu = 2.

author: MikeK
author_rank: Super Contributor
timestamp: October 01, 2022, 01:22:30 pm
content: An emitter follower takes its output from the EMITTER. How can the emitter be any value greater than the base?

author: golden_labels
author_rank: Super Contributor
timestamp: October 01, 2022, 05:48:52 pm
content: Emitter follower is a natural 1× “amplifier” forexactlythe same reason why a piece of wire is a natural 1× “amplifier”. It simply outputs approximately the same voltage as it sees on the input. In a BJT this is reduced by the voltage on BE junction, on a wire this is due to ohmic losses, but conceptually this is the same situation with respect to voltage on both ends. The difference lies only in impedances seen by both sides and the buffering function — but not in voltage amplification.With that picture in mind, realizing there is no actual amplification in here, rethink your question.The answer: no, there is no single transistor topology that has any particular amplification “baked in”. If you need amplification that is mostly independent of the parameters of the amplifier itself, the easiest choice is an operational amplifier.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: October 01, 2022, 06:08:02 pm
content: Quote from: edavid on October 01, 2022, 05:58:34 pmThat hints at the purpose of an emitter follower. It is a current amplifier, with current gain usually of tens or hundreds. Or thought of in a different way, an impedance converter.

author: golden_labels
author_rank: Super Contributor
timestamp: October 02, 2022, 05:34:06 am
content: Quote from: edavid on October 01, 2022, 05:58:34 pmHe did by mentioning “amplification” of 1×.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 02, 2022, 05:56:46 am
content: There's a family of amplifiers of arbitrary gain, of inverting sign, when common-emitter or common-base types are considered.For the CE case, there are two common approaches, or any combination thereof:If we have a resistor divider (in the AC equivalent circuit) from collector to base to input, we have shunt feedback, which reduces the input and output impedance, along with gain and distortion. In the high-gain limit (that is, gm or hFE --> infty), the overall gain equals the ratio of resistances.If we have a resistor divider the transistor is a part of, i.e. some resistance below the emitter (emitter degeneration resistance), and above the collector (load resistance, a parallel combination including any actual load), then the input and output impedances are increased, and gain and distortion are decreased. In the high-gain limit, overall gain equals the ratio of resistances.The impedance character of these two methods (shunt and series feedback) is important for any given application. Indeed, sometimes the shunt feedback case of equal resistances (thus giving nominal gain = -1) is sometimes called an "inverting follower", in analogy to the emitter follower, but any gain is possible, which seems an adequate example for the question.For common-base, feedback can be delivered to the base anyway, giving a hybrid approach. In this case, the input impedance is still fairly low (it's into the emitter, plus series degeneration resistance if applicable, or raised also by the effect of base feedback). Alternately, a fraction of output voltage can be fed back in parallel, series, or some combination thereof, by transformer action (sometimes applicable for low-noise RF amplifiers).The same is of course true of common-collector amplifiers, where some output can be fed back to the base or input, to partially cancel the "100%" degeneration of the emitter follower configuration.The main problem with the latter two kinds of feedback, is the requirement of transformer coupling, because signal must be subtracted, i.e. the transformer has to invert the feedback signal. So a mere resistor divider cannot provide the service, as can for the CE amplifier (which is inverting by itself).If we allow multiple transistors, we can consider circuits like the cascode, diff pair, Darlington, etc. which have different properties.Tim