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The Rheostat as a Transistor
The transistor can be thought of as a device that is like a rheostat
(potentiometer). If you think of a pot tied to a fixed resistor as a transistor
amplifier: the pot is working against the fixed resistor--the collector
load resistor. This means the transistor cannot generate a positive and
a negative signal, it can only draw more or less current, e.g., the pot
decreases its resistance, causing more current through the "load" resistor,
thus causing the voltage dropped across that resistor to increase; the
pot increases its resistance, causing less current through the load resistor,
and this causes less voltage to be dropped across the load resistor. If
we think of the extremes of current as being the equivalent of the positive
and negative alternations of a sine wave, then it follows that the equivalent
of zero is some current equidistant between the two. |
There's an Echo in Here
A NPN transistor connected as a common emitter amplifier: the base
needs current to do its thing.
The collector cannot output voltage, it can only cause more or less
current to be drawn through its load resistor. If a voltage is applied
to the base resistor a current now flows into the base (base emitter junction).
If a resistor is connected between the collector and a positive supply
voltage: the collector current flowing through the collector or load resistor
causes a voltage to be dropped across said load resistor. |
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Simulated Transistor
We can simulate a NPN transistor using two diodes and connecting both
anodes together. One cathode is tied to common (the emitter); the other
cathode (the collector) goes to a load resistor tied to the positive supply.
Now connect a 1k resistor to the junction of the two anodes (the base),
and using a signal generator, apply a 0 to 2 volt P-P sine wave to the
other end. Using a dual beam oscilloscope, observe the signal at both ends
of the resistor, i.e., the generator and the "base."
The results should resemble the figure: the diode
signal starts up unimpeded until it reaches ~ 0. 6 volts peak (1.2 volts
P - P), at which point the voltage at the "base" appears to stop increasing,
even though the signal generator is still increasing in amplitude. No matter
how much the voltage applied from the generator increases (within reason),
the "base" voltage appears to not increase. However, the current into that
junction (two anodes) increases linearly: I = [E - 0.6]/R.
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Now at this point, the analogy falls apart: these two diodes have no
gain, as the transistor we are trying to simulate would have. However,
let us pretend that it does: the "collector" is a high impedance current
source and if a resistor (the load resistor) is connected between the "collector"
and the positive supply, a voltage is seen at the collector. This changing
voltage drop across the resistor--caused by the changing collector current--will
change correspondingly to the "base" current.
Now follow me, just a few more words, and You've got it!
As the voltage at the generator goes more positive; the base current increases;
the collector current increases; the voltage drop across the collector
resistor increases; and the voltage at the collector goes less positive
or lower. |
Hang on! Stay with me!
Conversley, when the voltage at the generator goes less positive; the
base current decreases; the collector current decreases; the voltage drop
across the collector resistor decresses; and the voltage at the collector
goes more positive or higher. Feel better now OK, So I Lied: There is just
a little more to the story. Remember when the base reached ~0.6 volts?
well the collector output is only that part of the signal that caused the
base to conduct current. In other words: until the base rises to ~ 0.6
volts and there is base current, there is no change at the collector--no
collector output. |
Voltage to Current Convertor
First, you must convert the input voltage to a
current by using a voltage to current convertor
--a resistor. |
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Current to Voltage Convertor
Next, you convert the output current into a voltage by using a Current
to Voltage convertor in the collector circuit--you guessed it--a
resistor. |
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Make a List
The following list of attributes may, at first
glance, seem confusing and contradictory, however they are all true and
are offered as clues to the puzzle of: how does a transistor really work?
Abstractly, here are some Characteristics:
1.. An equivalent circuit of a NPN transistor is two diodes tied
anode to anode; one cathode being the emitter, the other the collector,
and the junction of the anodes is the base.
2.. When a NPN transistor is doing-its-thing, there is always
a constant 0.6 volt drop between the base and emitter, i.e., the base is
always ~ 0.6 volts more positive than the emitter--always!
3.. There is no output at the collector, until the base has reached
~ 0.6 volts and the base in drawing current, i.e., any signal that appears
at the base that is not up to ~ 0.6 volts (and not drawing base current),
is never seen at the collector.
4.. The base requires a current, not a voltage to control the
collector current.
5.. The collector is a current source: it does not source a voltage.
6.. The collector appears to outputs a voltage when a resistor
is connected between it and power.
7.. The collector is a high impedance when compared to the emitter.
8.. The transistor can output an amplified signal either form
the collector or the emitter (or both).
9.. When operating with a collector resistor (RL): the output
voltage from the collector is an amplified voltage.
10. When operating with only an emitter resistor (Re):
the output voltage from the emitter is not an amplified voltage, because
it is always ~ 0.6 volts, below the input (base) voltage--hence the name
voltage follower. But because the emitter can source
large amounts of current to the "LOAD," it can be said, there was CURRENT
amplification.
11. The collector--being high impedance--cannot drive a low impedance
load.
12. The emitter--being a low impedance--can drive a low impedance
load.
13. The voltage gain from the collector is greater than one (Gv
> 1).
14. The voltage gain from the emitter is less than one (Gv <
1).
15. Both the collector and the emitter: output ~ the same power:
E x I = P.
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One More Explanation
of How a Transistor Works.
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So, as in the figure: First thing is to realize that a transistor is
a current device: if you cause some current to flow in the base, a larger
amount of current is caused to flow in the collector. There's that pesky
echo again. Looking at the common emitter circuit in the figure: first,
while measuring the voltage and the current, one starts to apply a voltage
to the base of the transistor through the base resistor. As the voltage
increases from, zero there is no current flowing. At 0.1 volt, no current;
0.2 volt, no current; 0.5 volts, still no current; as the voltage at the
base approaches 0.6 volts--where there was no current--all of a sudden
a small current starts to be drawn by the base, and the voltage at the
base stops increasing--and remains at ~ 0.6 volts. As the voltage from
the generator continues to increase, the voltage at the base remains ~
0.6 volts, and the current continues to increase--as well as the collector
current. At some point, as the currents continue, the increase in the collector
current starts to slow, until it stops increasing altogether: it is said
now to be in saturation (if this transistor was being used as a switch
or as part of a logic element, then it would be considered to be switched
on). |
What have we learned?
First, as the input voltage is increased from 0 volts towards 0.6 volts,
there is an abrupt change in current, i.e., from zero current to some small
current flow. Just below this point where there is no current flow, the
device is said to be in cutoff. This low end region is considered a nonlinear
part of the operating curve (see the curves). Next, consider the other
extreme: as the currents in the base and collector are increasing (base
and collector are tracking), and the collector current is starting to no
longer track the input base current: this too is considered a nonlinear
part of the operating curve, and is in saturation (again refer to the curves). |
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Now, to the heart of the matter!
We have an operating curve consisting of a fairly linear segment bounded
by two nonlinear ends: cutoff and saturation.
Operating in the Middle
The transistor will operate very nicely if one could insure that no
input voltage, i.e., signal voltage--would cause the collector current
to ever operate beyond either end of the linear portion of the operating
curve. |
Sorry, But I Have a Bias
Along comes bias. You have heard about it, you've read about it, you
may have even dreamed about it: now is your chance to see-for-yourself--up
close and personal. Before one applies a signal voltage to the base circuit,
an arrangement for a steady voltage to be applied to the base, such that--with
no input signal--the collector current is the same as when it is about
half way up, or center of--the linear part of the curve. Now if we apply,
say, an AC sinusoid to the base circuit (through a capacitor), the collector
current--when seen as a large AC signal voltage at the collector--will
be linear and undistorted. |
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Hit 'em Again, While He's Down
To further beat a point into the ground: if one increased the input
signal beyond this level, the output signal would now start to "clip" and
cause distortion (sine wave gets flat on top and bottom). If the bias point
were set either too low or too high, then the sine wave would start to
clip on the top before the bottom, or visa versa (asymmetric clipping). |
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Hint #31, Active in Only One Direction
The transistor can be thought of as a device that is active in only
one direction: it can draw more or less current through its load resistor.
In the case of a NPN transistor tied as a common emitter amplifier: the
device can only actively sink current through the load resistor (otherwise
known as a pull-up resistor) it cannot source current.
Effects of different bias settings
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Let Me Count The Ways
By now you have probably guessed that there are several other ways
to "hook-up" the transistor. In the previous 3 volumes we have discussed
using the, so called, common emitter amplifier: where the only output is
at the collector. Now we will introduce you to an interesting arrangement:
the common collector, otherwise known as an emitter
follower, or voltage follower.
Now gang, this is where it gets sticky:
The definition of an "ideal" voltage source is a source having zero
output impedance, i.e., infinite current can be drawn, and the voltage
stays the same.
Where the common emitter amplifier required a voltage to current convertor
for its current input requirement, this configuration requires voltage
input only.
And because there is always a ~ 0.6 volt offset between the base/emitter
junction (as did the common emitter), the emitter sources a voltage that
reflects the input voltage, minus this offset, times the voltage gain:
Vout = [Vin - 0.6volts] x [Gv = .95]. |
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Lets see if I have this right: "Voltage in, voltage out; and it's
a current Amplifier?"
Bingo! Think about it:
1) The voltage-in is not amplified (Gv ~ .95);
2) There is impedance transformation--high to low; there is power amplification:
Therefore there must be current amplification.
Some other attributes are:
It has a voltage gain of less than one (Gv ~ .95); it is not easy to
cutoff, or saturated the transistor. Unlike the common emitter, it does
not invert the polarity of the input signal; it is among the most stable
of amplifiers--yes it is an amplifier, even if it has a voltage gain below
one. Because it has high input impedance, and low output impedance, it
is often used for transforming a high impedance output, to a low impedance
output: it is often used to drive transmission lines, e.g., video cable
from camera to monitor. Also, it is often used as the output stage (pass
transistor) of linear voltage regulators. If a 5.6 volt voltage
source (low impedance) is connected to the base, the emitter output will
try to maintain that voltage minus 0.6 volts: 5.60 - 0.6 = 5.0 volts (how
well it maintains this voltage is dependant on the transistor's gain: Hfe
= large number).
Another attribute is its excellent high frequency response. Because
there is no voltage gain, or because it has a gain of ~ 1, the bandwidth
is equal to the cutoff frequency of the transistor, Ft(where
Ft = [Hfe = 1]: BW = Ft). |
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Note: because
there is no voltage gain, there is no multiplication of the base/collector
capacitance (Co) which reduces the high frequency response of common emitter
amplifiers; see, also: Miller effect. |
Why a Voltage Gain of Less-Than-One? Good question.
Here goes! In an emitter follower configuration, as voltage equal to--or
greater than--0.6 volts is applied directly to the base, a current is caused
to flow through the the emitter resistor resulting in a commensurate voltage
drop. This voltage drop is always equal to the input minus ~0.6 volts multiplied
by some value slightly less than one. e.g., .95.
In the previous common emitter amplifier the current into the base was
determined by the relative difference between the base and emitter--above
0.6 volts.
In the case of the emitter follower, as the base voltage is increased,
there is a corresponding tracking of the base/emitter differential: the
emitter rises to--or follows--the base's change. If the output follows
the input, there can never be enough current drawn by the base to cause
a voltage drop across the emitter which exceeds the input voltage--hence
no voltage gain. This is an elegant case of (internal) negative feedback.
The amount of base current required to cause some larger current to
flow through the emitter resistor (and corresponding voltage drop) is dependant
on the gain--Hfe--of the transistor and the emitter load (emitter
resistor and load).
Another way of thinking about this relationship, is as input impedance:
if the transistor had infinite gain, there would be no base current, resulting
in infinite input impedance.
If the transistor had zero gain, the input impedance would be directly
dependant on the emitter resistor, i.e., base current = emitter current.
If the transistor had some finite gain, the input impedance would be
finite, i.e., base current would be dependant on the emitter resistor modified
by the transistor's finite gain (Hfe), i.e., base current ~=
emitter current/Hfe.
In all of this, one can think of it as a sort of internal feedback,
or bootstrapping of the input impedance.
Why is an emitter follower so stable? Another good question. Easy to
answer: As long as the gain is 1 or less than 1, it can never oscillate.
Oscillation requires a positive feedback and a gain of greater than
1 to sustain oscillation (of which instability is a precursor). |
That Pair's DARLINGton
The maximum input impedance one can expect from an emitter follower,
is limited by the finite gains of individual transistors (~ 50 to ~ 350).
However, there is a way to increase the effective gain or transistors by
using two transistors. The total gain of this transistor pair is Gv1 x
Gv2 = Gv total (Gv ~ 2k - 100k). This is achieved by arranging the transistors
such that the emitter of one is driving the base of the next and connecting
the collectors together. This is known as a Darlington pair, and can be
used as any single transistor would be: common emitter, emitter follower,
etc.
The down side of this arrangement, is reduced speed: because of the
very high gain's effect on the collector to base capacitance, Co (Ctotal
= Co x Hfe). |
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An Ideal Amplifier
An ideal amplifier is one that is made up of some gain device (transistors)
that has very much more gain than the finished amplifier. If this gain
device had infinite gain, then the amplifier's gain would be completely
dependent on the gain setting resistors: which set the gain by determining
the amount of feedback used to overcome the amplifier's open loop
gain (e.g., Op Amps). In the case of simple single transistor gain stages,
the control exerted by the gain setting resistors is limited and has less
effect on the stage's overall performance, i.e., the transistor's inherent
gain is dominant. However, realize that the greater the ratio of final
amplifier gain to the maximum possible gain (no feedback) of the transistor,
the less vulnerable the gain of the amplifier is to variations of the individual
transistor's gain (within limits).
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A Common Emitter Amplifier Without Feedback
A simple common emitter transistor amplifier--having no negative feedback--is
not an ideal amplifier. This is because of the variability of gain from
one transistor to another: making uniform gain, from amplifier to amplifier,
impossible. Also, without feedback some amplifiers--having transistors
with excessive gain--might be unstable and prone to be oscillate, as well
as, poor signal to noise and distortion ratios (S/N+D); low input impedance
(poor impedance matching between stages, etc), and susceptibility to temperature
extremes. Without negative feedback, high ambient temperatures can raise
the operating point, thus heating the device further; ending with this
positive (thermal) feedback, bringing
on the transistor's permanent failure. |
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So That's Feedback, ah...
When (negative) feedback is introduced, most of these problems diminish
or disappear, resulting in improved performance and reliability. There
are several ways to introduce feedback to this simple amplifier, the easiest
and most reliable of which is accomplished by introducing a small value
resistor in the emitter circuit. The amount of feedback is dependent on
the relative signal level dropped across this resistor, e.g., if the resistor
value approached that of the collector load resistor, the gain would approach
unity (Gv ~ 1). |
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And to beat a simple point into Terrafirma: with no emitter feedback
(no Re), the gain would be essentially that of the transistor.
Another feedback technique is the introduction of some fraction
of the collector signal back to the base circuit. This is most easily done
via the positive biasing resistor (Rb1) --as in the figure. A third --but
by no means last --approach is to use a combination of feedback techniques. |
The Miller Effect
In a gain stage (common emitter) there is a limit to the achievable
bandwidth at some set gain: i.e., the higher the gain, the lower the bandwidth;
conversely, the lower the gain, the wider the bandwidth. This is the now
famous, Gain Bandwidth Product. The dominant mechanism for this is found
in the intrinsic feedback capacitance, Ccb, between the collector and the
base. The effect--as frequency increases--is to increase feedback via Ccb's
capacitive reactance, XCcb, thus reducing the overall gain. To compound
this problem: XCcb is dependent on the intrinsic capacitance, Ccb, multiplied
by the gain, i.e., as the gain is reduced, the bandwidth is increased.
There are ways of reducing this effect, such as peaking coils in the collector
(Xl cancels Xc); pre-emphasis of the signal's higher frequencies at the
input; frequency selective feedback, etc...
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Gain Bandwidth Product
Using several lower gain stages in cascade is a strategy that also
works. And, a very direct and effective solution is a common base
configuration, in which the input signal drives the emitter, and the base
is grounded, which has the effect of breaking the collector/base feedback
path. Frequency dependent feedback In the figure, the capacitor, Ce, across
the emitter resistor, Re, causes the gain of this device to be greater
at higher frequencies. As capacitive reactance, Xc, approaches the value
of Re, a rapid increase in gain occurs. The effect, of course, is to reduce
the negative feedback at higher frequencies. This is often done to compensate
for the limited bandwidth of the transistor stage. |
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Common Base Stage
Because the base is "grounded",
this configuration does not suffer from the Miller Effect, thus yielding
the widest bandwidth of all configurations. Note that the drive is to the
Emitter, and there is no signal inversion. |
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Video Amplifiers
A video amplifier is used to amplify video from TVs, cameras, computer
graphic devices, etc. Aside from having sufficient bandwidth and the ability
to drive long cables: they cannot invert the signal's polarity; if they
did: unless you were using an even number of amplifiers in cascade, the
image would end up a negative. If you wanted a gain stage, but didn't want
the signal to be inverted, you would drive the emitter instead of the base.
This works, but as you might imagine, the input impedance is quite low.
So by using what we learned about emitter followers back in chapter 219,
we can "transform impedances," and now the noninverting video amplifier
looks better. |
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The Differential Amplifier
Differential amplifiers are everywhere: input stages of Op Amps; comparator
inputs; some video amps; balanced line receivers for digital data transmission
; etc... It is not one of the more easily understood combinations of transistors,
however, I shall attempt to explain this "not-so-little-bugger."
A differential amplifier is an amplifier that has two inputs, each of
which is sensitive to the opposite polarity of the other, i.e., if the
inverting input has a positive going signal, and the non-inverting input
has the negative version, then there is an output equal their difference
(multiplied by some gain, Gv). Conversely, if both inputs happen to be
at the same value, then there is no output signal: they cancel one another,
i.e., both signals (being the same polarity and amplitude) make no change
is the shared emitter resistor's current, therefore, neither signal affects
the other: there is "cancellation," otherwise known as Common Mode Rejection,
CMR. Another way of saying the same thing is: if both inputs have the opposite
polarity (or phase) signal, the shared emitter resistor draws current equal
to the algebraic summation of both transistors.
Dejavu All Over Again
You may have noticed that the configuration of the transistors in a
differential amplifier are a combination of common emitter and emitter
follower. OK? OK.
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OK, Point #1:
A signal into either input's base, causes an inverted signal at its
collector, and simultaneously, a smaller, non-inverted output at the (shared)
emitter resistor.
OK, Point #2:
Any signal at the emitter will appear at the collector as a non-inverted
version of this signal--but amplified (remember the video amp?).
OK, Point #3:
Therefore, any signal at one transistor's input is not only seen at
its collector, but is also seen at the other transistor's collector, enabled
by the action of the shared emitter resistor (Points #1 & #2). |
Click Me!
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What, Not a Restatement of the Same Old Thing!
This amplifier consists of two or three transistors (two in the simple
version, three or more in the more precision version). These two input
transistors are coupled to each other, via each's emitter, and share the
same emitter resistor . At this common connection each input transistor
affects the output of itself, as well as, the other transistor's output.
"I Lied." Or Did He? Now that you think you understand how a
"Differential Pair" works, there is just a little more to the story. Previously
I said that the two input transistors share the same emitter resistor,
leaving the impression that a signal voltage was at the junction of the
emitters and Re. If you think about it, when one transistor is increasing
in current, e.g., positive alternation of a sine wave; the other transistor
is decreasing in current, by an equal amount, for the negative alternation.
Since the pair is sharing the one resistor, one can deduce that, ideally,
there is always a constant current in that resistor. Ideally, it is desired
that the emitters transfer all of their signal to the other transistor's
emitter. |
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Enter the Oft-Maligned Constant Current Source
Because of a non-ideal world and the non-ideal transistors that co-habitat
it, a constant current source (generator) is substituted for Re. A constant
current generator is a circuit in which a fixed voltage source (zener diode)
is applied to the base, along with some current determining resistor in
the emitter circuit. The result is a collector that will furnish a constant
current over a wide range. |
On the Level ----------------------------------
Differential stages are also useful for level translation. Either input
can be driven (biased) to affect the operating point of both transistors
in a complementary fashion, and therefore the output (collector) offset
voltage. This is what allows the Op Amp's offset voltages to be trimmed
to zero. (See figures)
Don't lose your Temperature
As mentioned, one of the features of a differential amplifier is its
ability to reject common mode signals (CMRR), i.e., if the same signal
is on both inputs in equal amounts the output does not change. This works
because of "common" signal cancellation that occurs within that first diff
stage, between the inverting & non-inverting inputs. The degree of
precision of this effect is dependent directly on how closely the two transistors
are matched (gain, etc.). Typically both transistors share the same substrate
and/or package; these appear as one transistor but are, in fact, a pair--sometimes
refereed to as a "differential pair."
As you might guess, when packaged like this, they also share the same
temperature gradients. However, if the two transistors are separated, the
slightest change in temp that is not shared can cause a large shift in
offset voltages as seen at the output (e.g., between both collectors).
This might appear as a change in gain, but it is really more a "shift"
in its quiescent voltages. However, if there is any cancellation going
on, this shift might reduce the cancellation which would appear as a change
in gain...
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