When modeling a transistor in a circuit by hand, the specific model and its level of complexity that you use depends on the questions one wants answered.
A simple small signal gm
model will suffice in answering the original question of feedback as a follower. More elements need to be added if you are also interested in input impedance. More for frequency dependence. Yet more for output impedance.
Other models are used for large signal behavior where beta based models are needed.
Raw voltage gain of the transistor is not as useful for analysis as it is almost always limited by the gm and load resistance. Hence the approximation that A=gm*Rload for the feedback equation I gave earlier.
Please attempt assembling the 2 node small signal circuit model using only gm and Rload and deriving the equation for circuit gain.
A simple small signal gm
model will suffice in answering the original question of feedback as a follower. More elements need to be added if you are also interested in input impedance. More for frequency dependence. Yet more for output impedance.
Other models are used for large signal behavior where beta based models are needed.
Raw voltage gain of the transistor is not as useful for analysis as it is almost always limited by the gm and load resistance. Hence the approximation that A=gm*Rload for the feedback equation I gave earlier.
Please attempt assembling the 2 node small signal circuit model using only gm and Rload and deriving the equation for circuit gain.
Hi Alessandro,
I wanted to solve the equations for a push-pull follower output stage.
I had already solved the equations for the Common Emitter version.
The Emitter Follower version must be given by the gain equation with 100% voltage feedback - right?
So I checked this in LTspice using a Common Emitter stage that has the output voltage subtracted from the input voltage (below).
This provides 100% voltage feedback to an Emitter Follower stage. The output has to be inverted since a CE stage inverts.
The wingspread plots for Out1 and Out2 are plotted below:
The two wingspread plots are pretty much perfectly the same. The two current plots are the same as well.
So I concluded the Emitter Follower is EXACTLY the same as 100% voltage feedback to the Common Emitter stage when voltage driven.
BTW I used this to generate plots for a DoubleCross follower stage here https://www.diyaudio.com/forums/sol...ordells-power-amplifier-book-post6535473.html using a spreadsheet by applying 100% feedback to the CE equations.
If this is a question of whether the transistor is controlled by base current or by base voltage then I vouch for the BJT as voltage controlled. The strongest evidence for this can be seen in the Gummel plot at low currents (Region1 below):
In Region 1 the collector current is a straight line with Vbe (exponential in base-emitter voltage) whereas the base current is nonlinear. So it appears it is the base-emitter voltage voltage that controls the collector current in an exponential fashion.
In Region 2 you cannot tell whether base current or base-emitter voltage is controlling the collector current. But then why would it change from voltage drive in Region 1 to current drive in Region 2?
In Region 3 the base current is increasing exponentially with base-emitter voltage while the collector current is now no longer a pure exponential with base-emitter voltage. Again, no way to tell which is in control. But then why would it change from Region 2?
Using Occam's Razor the most likely one is the BJT collector current is, at its core, controlled by the base-emitter voltage in all regions.
BTW the VBE in Gummel plots is the internal base-emitter voltage after the parasitic emitter resistance and base resistance voltage drops have been removed. If the parasitic emitter resistance is not removed then the Ic plot in Region 3 does not continue as a straight line at half the slope of Region 2 but becomes more curved (due to degeneration feedback from the emitter resistance).
So I don't think your question "if the transistor does not amplify in voltage ..." is relevant.
Cheers,
Ian Hegglun
I have wanted an answer to that question for many many years. So thanks for asking this.
I wanted to solve the equations for a push-pull follower output stage.
I had already solved the equations for the Common Emitter version.
The Emitter Follower version must be given by the gain equation with 100% voltage feedback - right?
So I checked this in LTspice using a Common Emitter stage that has the output voltage subtracted from the input voltage (below).
This provides 100% voltage feedback to an Emitter Follower stage. The output has to be inverted since a CE stage inverts.
The wingspread plots for Out1 and Out2 are plotted below:
The two wingspread plots are pretty much perfectly the same. The two current plots are the same as well.
So I concluded the Emitter Follower is EXACTLY the same as 100% voltage feedback to the Common Emitter stage when voltage driven.
BTW I used this to generate plots for a DoubleCross follower stage here https://www.diyaudio.com/forums/sol...ordells-power-amplifier-book-post6535473.html using a spreadsheet by applying 100% feedback to the CE equations.
If this is a question of whether the transistor is controlled by base current or by base voltage then I vouch for the BJT as voltage controlled. The strongest evidence for this can be seen in the Gummel plot at low currents (Region1 below):
In Region 1 the collector current is a straight line with Vbe (exponential in base-emitter voltage) whereas the base current is nonlinear. So it appears it is the base-emitter voltage voltage that controls the collector current in an exponential fashion.
In Region 2 you cannot tell whether base current or base-emitter voltage is controlling the collector current. But then why would it change from voltage drive in Region 1 to current drive in Region 2?
In Region 3 the base current is increasing exponentially with base-emitter voltage while the collector current is now no longer a pure exponential with base-emitter voltage. Again, no way to tell which is in control. But then why would it change from Region 2?
Using Occam's Razor the most likely one is the BJT collector current is, at its core, controlled by the base-emitter voltage in all regions.
BTW the VBE in Gummel plots is the internal base-emitter voltage after the parasitic emitter resistance and base resistance voltage drops have been removed. If the parasitic emitter resistance is not removed then the Ic plot in Region 3 does not continue as a straight line at half the slope of Region 2 but becomes more curved (due to degeneration feedback from the emitter resistance).
So I don't think your question "if the transistor does not amplify in voltage ..." is relevant.
Cheers,
Ian Hegglun
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The physics isn't really with this one - the BJT is voltage driven, the output current being an exponential function of Vbe. It also happens that base current is typically directly proportional to collector current (when in the active region) i.e. beta is approximately constant, but it is Vbe that controls carrier injection from the emitter to the base. Minority carrier recombination in the base is what makes the base current proportional to the collector current - this is really a secondary effect, the main point of transistor action is the Vbe control of collector current, independent of collector voltage (ignoring the Early effect).
Even if a transistor was such that the base and collector currents were 1:1 (no net current gain) it would still be a voltage amplifier as the collector voltage swing can be much larger than the Vbe swing. However good current gain makes a device much more useful, and is often sought-after (its hard to achieve as it requires very precise control of the doping profile of the thin base region during manufacture, and its harder to achieve for high voltage devices).
So in summary a BJT is a very non-linear voltage controlled dependent current source combined with current gain. About 20mV of Vbe change doubles/halves the current, so raw voltage gains of several 100 are possible even in low voltage circuits, independent of beta. The common-base configuration uses only the voltage gain, the converse of emitter-follower which only uses the current gain.
In the common emitter config. both the full current and voltage gain can be used, giving power gains upto 10^5 or so (consider a VAS transistor for instance, often boosting a couple of microwatts to a sizable fraction of a watt - at low frequencies at least).
A BJT can be thought of as a non-linear voltage amplifier, and a roughly-linear current amplifier. Its voltage-controlled as the EB junction potential barrier has to be overcome to allow carrier injection, but in many circuits you can treat it as current-controlled since a current source will allow the Vbe to reach the correct point via local feedback (or put another way the low Vbe impedance dominates when the input is a current source).
So based on your tests you both agree that BJT is actually a voltage driven device and not a current driven device, and agree that Emitter Follower is actually 100% negative voltage feedback ?
Ok I'll wait for their confirmation so I can close this thread. I trust their knowledge.
Sorry for my english but I'm not native speaker..
An emitter follower runs with nearly 100% series feedback, which is negative feedback.
New member here.
Emitter resistance is negative feedback. It is considered "series feedback" and it tends to increase the input resistance of the transistor stage.
"Shunt feedback" is also negative feedback, but is different in that some of the collector output voltage is fed back to the base through a feedback resistor.
Shunt feedback tends to REDUCE the amplifier's input resistance. Shunt feedback is a subject for a different discussion.
A bipolar transistor is a voltage controlled device. I know that people usually say that a bipolar transistor is a current controlled device, but it is ALSO a voltage controlled device in that there must be some minimum voltage between the base and the emitter in order to drive the required base current.
Here's a simple example that might help:
Let's say we're talking about a plain old 2N3904 transistor.
Lets hook it up as shown in this diagram, which I drew using a free simulation program called LT-Spice. You can get it from the Analog.com website:
So, we have a 1V input signal, and an Emitter resistor of ~0 ohms or directly shorted to ground. OK?
With a DC input of 1V, the base-emitter voltage will also be 1V, which will exceed the 2N3904's base current limit.
Running the simulator with this set of conditions, and measuring the collector current will show that the device will try to draw 5A from the power supply.
As you might guess, the transistor will immediately be destroyed.
Now lets increase the emitter resistance to 1 ohm, and we now have what we call an emitter follower circuit.
With 1V on the base, the 1 ohm emitter resistor develops a voltage drop that limits the differential voltage between the base & the emitter.
The emitter voltage rises to about 0.17 volts, which limits the emitter current to 0.17/1.0 == 0.17Amps, which is well within the 2N2904's current capabilities, but it will still fail from power dissipation exceeding 1.5 Watts.
Before it dies though, we can calculate the emitter follower's input resistance:
The 2N3904 datasheet shows that at 150 milliamps collector-emitter current, the base-emitter voltage will be about 0.8V-0.9V, and the current gain is roughly about 50 (typical).
The input resistance is roughly the value of the emitter resistor multiplied by the current gain, or about 50 ohms.
So, let's continue increasing the emitter resistance to 8 ohms, OK? It's a convenient value for audio work.
With the same 1V from the base bias supply, the 8 ohm emitter resistor limits the output current to about 31 milliamps, which means the base-emitter voltage is now lower, at less than 0.75V, which yields an emitter voltage of about 250 mV.
Again, 31 mA is well within the current handling capabilities of the device, and the power dissipation is now reduced to about 1/4 Watt. It'll still be toasty hot, but it will survive.
What is the input resistance now? At 30 mA collector current, the current gain is now typically 100, so the input resistance is roughly 8 * 100 == 800 ohms.
Let's do one more iteration:
With a 100 ohm emitter resistor, the output current is limited to ~3mA, which gives a base- emitter differential of about 0.65V.
At 3mA, the current gain is still about 100, so the input resistance of the emitter follower is now 100 * 100 == 10K ohms. We're now getting into the range of buffering guitar and line level signals.
We can continue with higher emitter resistance values, but it should be apparent that an emitter follower gets negative feedback from the emitter resistance, because the presence of the emitter resistor reduces the effect of the input drive by limiting the emitter - base voltage differential. Since the base-emitter voltage is logarithmically related to the input current, less variation here results in a more linear amplifier.
If this isn't quite clear, questions are welcome.
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you have to consider both voltage and current. The two cannot be separated.
😎 👍
Simple, short, 100% correct answer. This is the way to analyze a feedback circuit.
Jan
Hi Alessandro,
In textbooks the emitter follower (source follower, cathode follower) are classified into one of the 4 canonical feedback classes: as series voltage feedback. And yes, there is 100% negative voltage feedback. It comes from 100% the output voltage at the emitter, and the input voltage is to the base.
Re: "BJT is actually a voltage driven device and not a current driven device"
The term "voltage driven" and "current driven" have special meanings in circuit analysis. (These terms depend on the circuit configuration and not usually used to understand the physics of the BJT itself). What I was saying was the BJT appears to be a voltage controlled phenomena that's happening inside the transistor.
The term a "voltage driven device" is used to describe how a signal is applied to a circuit (or a sub-circuit of the main circuit). The term a "voltage driven device" is where there is a negligible effect of series resistances (often lumped together and called a Thevenin equivalent series resistance). The term "current driven device" is the Norton dual - a current source with a negligible effect of shunt resistances.
As Mark pointed out the base current is a secondary effect arising out of the application of a base-emitter voltage, very similar to a diode where current flows as a result of applied voltage. This may seem pedantic but it was a major step forward in modelling the BJT for computer analysis, giving rise to the Ebers-Moll model (1954) and later the Gummel-Poon model (1970) and later still the VBIC model (1996).
Unfortunately, the maths of these large signal models cannot be used for hand calculations because the maths for even the simplest of circuits, like the emitter follower, is hugely complicated because of the presence of at least two exponentials per transistor and a few associated resistors. That's even without secondary nonlinear effects like Early effect and high and low level injection effects. That's why we have to use simulators to solve these nonlinear transcendental equations in matrix form with numerical methods.
Of course we can do small signal analysis by hand by linearizing the BJT at it's operating point, and this type of analysis is covered in the textbooks since year dot for all the basic BJT arrangements - common emitter, common base, and emitter follower.
BTW it wasn't until year 2000 that the basic common emitter configuration was solved analytically using a basic large signal model of the BJT by Robert Banwell here. I gave one example in Post 22 for a DoubleCross follower output stage using Banwell's method. But not much more since then.
Question
if I connect the emitter leg directly to the primary of a transformer, therefore I do not use an emitter resistance, is 100% negative feedback still there?
I quote what the manufacturer states "Current signal transformer feeding. Feeding cannot control the current flowing into the primary transformer coil as consequence no local feedback is applied."
What do you think?
In this amp the voltage gain is done only with transformers and 2 emitter follower stages are used as impedance adaptors and for the current gain. The output transformer is used for matching the speakers impedance.
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The transformer is an audio impedance. If appropriate to the transistor's operating point, then there is large NFB for audio.
The ideal transformer is zero DC impedance. There will be no NFB at DC.
Separating AC and DC is only a mind-game. You must control the DC condition to make a good audio amplifier.
Voltage-controlled or Current-controlled? It has been convincingly argued that vacuum tubes, BJTs, and FETs are all Charge controlled. This truth is muddled in BJTs because to get maximum gain they leak horribly in the (uninsulated) Base.
Amplifying Devices and Low-pass Amplifier Design, Daryl E. Hooper and E. M. Cherry
https://worldradiohistory.com/BOOKSHELF-ARH/Technology/Amplifying-Devices-andLow-Pass-Amplifier Design-Cherry-Hooper-1968-RR.pdf
This is a simplified schematic diagram, and so several details are missing: for instance, DC biasing of both transistor stages.
Assuming such necessary measures, such a connection will provide negative feedback, but only within the operating frequency range of the transformer.
I would personally never hook up a circuit like this without separate DC biasing measures.
I worked for a few years for a large semiconductor manufacturer. I was mildly surprised (at first) that the designers tended to think in terms of transconductance of BJT transistors, but it made sense in that the BJT has a much higher transconductance than do FETs. Very small transistors such as used to populate ICs require little in terms of base current.
Current depends of voltage. Ib depends from Vbe. That's why bipolar transistors are voltage controlled devices, they are transconductors like other active amplifiers devices, tubes, FETs.