A AMPLIFIERS circuit containing only capacitors, amplifiers (transistors) and resistors may resonate. A circuit containing only capacitors and resistors may not. Why does amplification permit resonance in a circuit with only one kind of storage element?
Amplification arises from static behavior, not dynamics and energy storage. Amplification is fundamentally dissipative. Poutput Psupply Poutput = Psupply + Pinput - Pdissipated Amplifiers are basically multiport resistors How does the addition of a (multiport) dissipator enable resonance?
Mod. Sim. Dyn. Sys.
Amplifiers
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WE WILL SEE THAT Amplification is fundamentally a non-equilibrium phenomenon. Resistors far from equilibrium may contain a "hidden" junction structure that includes a gyrator. This gyrator can cause resonance with only one kind of storage element.
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NODICITY Electrical amplifiers are nodic. Assuming conductance causality
the equations of the three-port resistor may be written as follows. f1 = 1(e1,e2,e3) f2 = 2(e1,e2,e3) f3 = 3(e1,e2,e3)
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Nodicity means that the efforts and flows at the ports are constrained so that they satisfy two conditions: (1) Continuity of flow: The sum of flows into the system is zero. This means that the element behaves as a node characterized by a Kirchhoff current law. (Note the implicit assumption of the "power positive in" sign convention) (2) Relativity of effort: Each flow depends only on the difference of applied efforts. The same effort may be added to all inputs without changing the output.
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The constitutive equations of a nodic three-port resistor may be written as follows. f1 = 1[(e1 e3),(e2 e3)] f2 = 2[(e1 e3),(e2 e3)] f3 = (f1 + f2) Nodicity implies that the behavior of the element is independent of any absolute reference frame. A nodic three-port (n-port) contains an "embedded" two-port (n-1-port)
characterized by two constitutive equations (not three).
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EXAMPLE: SEMI-CONDUCTOR DIODE. A semi-conductor diode has two ports (two wires) but, in common with all electronic devices it is well described as a nodic element and is characterized quite accurately by a single constitutive equation such as i = Is[e(e1e2)/Vt 1] where Vt: thermal voltage = kT/q = 25.3 mV at 20C k: Boltzmann's constant T: absolute temperature q: charge on the electron Is: reverse saturation current Note that the element satisfies the conditions for nodicity even though both the thermal voltage and the reverse saturation current depend on absolute temperature.
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Amplifiers
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EMBEDDED JUNCTION STRUCTURES In general a multi-port resistor may contain a "hidden" junction structure coupling the power flows on its ports. A two-port resistor may contain a "hidden" or "embedded" gyrator. Without loss of generality assume e3 = 0. The constitutive equations of the two-port f1 = 1(e1,e2) f2 = 2(e1,e2) f = (e) are restricted to the first and third quadrants.
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The general form may be replaced by f = G(e) e where G is a 2x2 conductance matrix. The matrix G may be decomposed into symmetric and anti-symmetric components. G = (G + Gt)/2 + (G Gt)/2 = Gs + Ga The anti-symmetric component describes an ideal power-continuous gyrator which is "embedded" in the resistive two-port.
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Power must be dissipated and not generated. The total power dissipated by the element is: Pdissipated = etf = etGe = etGse + etGae but etGae = 0 Pdissipated = etGse Power dissipation depends only on the symmetric component of G.
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EXAMPLE: NPN TRANSISTOR Consider a typical NPN transistor, Motorola's 2N4400. From the specification sheets, the AC small-signal characteristics at 1 KHz are as follows (operating conditions: ic = 1mA, vce = 10 Vdc) Forward current gain: hfe = 250 Voltage feedback ratio: hre = 8x10-4 Input impedance: hie = 7.5 K Output conductance: hoe = 30 mho The subscript e on the "h-parameters" refers to the common-emitter configuration used in a typical amplifier circuit. They are based on a "controlled-source" model of the transistor.
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The corresponding model equations are ic = hfe ib + hoe ec eb = hie ib + hre ec Rearranging into matrix form,
i c eb
hoe hfe ec = hre hie ib
Note that the gain terms, hfe and hre, appear in the off-diagonal positions. This matrix may be partially inverted and rearranged into a conductance form as follows. ic = (hoe - hfehre/hie) ec + (hfe/hie) eb ib = -(hre/hie) ec + (1/hie) eb
i c ib
Ycc Ycb ec = Ybc Ybb eb
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Equating terms: Ycc = hoe hfehre/hie Ycb = hfe/hie Ybc = hre/hie Ybb = 1/hie This is a two-port nodic resistor. NOTE: The controlled-source representation is widely used this conductance representation is not. The conductance representation reveals that the h-parameters are constrained the conductance matrix must be positive definite. The controlled-source model hides this constraint.
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Amplifiers
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EVALUATE PARAMETERS: Ycc = 3.33 mho Ycb =33,333 mho Ybc =0.107 mho Ybb =133.3 mho In mho units:
i c ib
3.33 33333 ec = 0.107 133.3 eb
Note the extreme asymmetry of the conductance matrix.
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Its symmetric and anti-symmetric components are:
i c ib
16667 ec 3.33 16666 ec 0 + = 16666 133.3 eb 16667 0 eb
The symmetric component is purely dissipative. The anti-symmetric component describes a gyrational coupling between input (base) and output (collector). Because of this gyrational coupling, a resonant circuit may be constructed using this transistor, two capacitors and no inductors.
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Consider connecting a capacitor Cc = 1000 pF across the output (collector grounded to emitter) and a capacitor Cb = 0.1 F across the input (base to grounded emitter) as shown.
Time-differentiating the constitutive equations for these capacitors results in the following. d 1 ec = C ic dt c d 1 dt eb = Cb ib
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Substituting for the collector and base currents using the transistor conductance matrix, state equations for this linear system may be written as follows. d ec Ycc/Cc Ycb/Cc ec = dt eb Ybc/Cb Ybb/Cb eb The characteristic equation for this system is s2 + (Ycc/Cc + Ybb/Cb) s + (YccYbb YbcYcb)/CcCb = 0 The parameters listed above yield n = 6.32 KHz and = 0.37
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BIASING The AC parameters of the transistor are only valid if the device is properly "biased" at DC. An appropriate network of resistors must be added. A typical common-emitter biasing circuit with "emitter degeneration" (an emitter resistor) is shown.
These resistors add to the damping in the circuit and may suppress resonance. but the central point still holds: two capacitors and a transistor can make a resonant circuit because the transistor contains a "hidden" gyrator. Why does "biasing" matter? The device only amplifies if operated far from equilibrium
Mod. Sim. Dyn. Sys.
Amplifiers
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LARGE-SIGNAL TRANSISTOR BEHAVIOR: THE EBERS-MOLL MODEL. (describes static characteristics only -- ignores important dynamics such as charge storage on the junction regions) The Ebers-Moll model describes a transistor as a pair of coupled back-toback diodes. The coupling is usually represented as a pair of current-controlled current sources.
ic
collector
ir
base
f if
ib if
r ir
ie
emitter
For the sign convention shown, current continuity yields ie = if r ir ic = f if - ir Mod. Sim. Dyn. Sys. Amplifiers page 18
The currents if and ir refer to the forward and reverse currents of the base-toemitter and base-to-collector diodes respectively. Each is related to the corresponding voltage drop through the usual exponential diode relation. if = Ies[e(ebee)/Vt 1] ir = Ics[e(ebec)/Vt 1] Physical constraints of the transistor require that r Ics = f Ies. For a typical transistor: f = 0.99 r = 0.01 Ies = 2.0 x 10-15 amps Vt = 25.3 x 10-3 volts
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An NPN transistor is usually operated in the "active" region with the emitter diode strongly forward biased (i.e. conducting with eb >> ee) and the collector diode strongly reverse biased (i.e. not conducting with ec >> eb). The collector diode current is small and insensitive to changes in the voltage drop from collector to base. As a result, ie if - r Ics if ie Ies[e(ebee)/Vt - 1] The collector current is primarily determined by its coupling to the emitter current ic f if f ie
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Applying current continuity to the entire network (i.e. treating it as a node) we may relate collector current to base current as follows. ie = ib + ic ic f (ib + ic) f i ib ic 1f b where is the transistor forward current gain. In practice may vary between 50 to 1000, but only changes from 0.98 to 0.999. Thus to a close approximation the collector current is ic Ies[e(ebee)/Vt 1]
NOTE: this model describes the transistor by a relation between an output (collector) current and an input voltage drop (from emitter to base) This again shows the fundamentally gyrational character of the transistor. Mod. Sim. Dyn. Sys. Amplifiers page 21
AMPLIFICATION IS A NON-EQUILIBRIUM PHENOMENON Linearizing the Ebers-Moll model reveals an interesting property of amplifiers in general: operated near equilibrium, amplifiers cease to amplify. Consider the grounded-emitter configuration described above. The elements of the locally linear conductance matrix are as follows. Ycc = ic/ec Ycb = ic/eb Ybc = ib/ec Ybb = ib/eb
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We may use the current continuity conditions to evaluate these. d ic = f d if d ir d ie = d if r d ir d ib = d ie d ic = (1 f) d if + (1 r) d ir From the diode equations, if/ec = 0 (ebee)/Vt ] 1 if/eb = Ies[e V
t
1 ir/ec = Ics[e(ebec)/Vt ] V
t
(ebec)/Vt ] 1 ir/eb = Ics[e V
t
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Substituting 1 ic/ec = Ics[e(ebec)/Vt ] V
t
(ebee)/Vt ] 1 Ics[e(ebec)/Vt ] 1 ic/eb = f Ies[e V V
t
t
1 ib/ec = (1 r) Ics[e(ebec)/Vt ] V (ebee)/Vt ] 1 ib/eb = (1 f) Ies[e V
t
t
1 + (1 r) Ics[e(ebec)/Vt ] V
t
Normal operating conditions (denoted by subscript n) are defined by ee = 0, eb > ee + Vt and ec >> eb. Under those conditions, e(ebec)/Vt is very close to zero, whereas e(ebee)/Vt is a positive number greater than unity (typically very large) whose exact value is strong function of eb. Mod. Sim. Dyn. Sys. Amplifiers page 24
Evaluating, Ycc = ic/ec 0 Ies Ycb = ic/eb f V eeb/Vt n
t
Ybc = ib/ec 0 Ies e /V Ybb = ib/eb (1 f) V e b t n t Thus, under normal operating conditions far from equilibrium, the locally linearized conductance matrix is strongly asymmetric as we saw before.
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Equilibrium conditions are defined by ee = eb = ec = 0. At equilibrium, both exponentials are equal to unity. Evaluating, Ics Ycc = ic/ec = V t Ies Ics Ycb = ic/eb = f V V t t Ics Ics Ics Ybc = ib/ec = (1 r) V = r V V t t t Ies Ics Ybb = ib/eb = (1 f) V + (1 r) V
t t
Using the relation between the diode saturation currents, r Ics = f Ies, we find that Ics Ics Ycb = r V V = Ybc t t
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CONCLUSION: If the transistor is operated about equilibrium, the locally linearized conductance matrix becomes symmetric. As a result, the transistor behavior is purely dissipative without any gyrational coupling between input and output. Relating this to the controlled-source model of the transistor, hfe = Ycb/Ybb hre = Ybc/Ybb -- the forward current gain is identical to the reverse voltage gain. OPERATED ABOUT EQUILIBRIUM, the embedded gyrator disappears the transistor fails to amplify
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