imported>John R. Brews |
imported>John R. Brews |
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| | In [electronics]], the '''mode''' of an electrical device refers to its steady-state bias condition or '''operating point''' in the absence of signals. In [[analog circuits]] the so-called ''active mode'' mode of the device is chosen by the circuit designer to allow adequate signal amplitude and adequate voltage or current gain, along with acceptable signal distortion. In [[digital circuits]], devices toggle between modes that are either in '''cutoff mode''' (''off'') or in '''ohmic mode''' (''on''), and visit the active mode only briefly during the transition between the ''on'' and ''off'' modes. |
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| In the theory of electrical circuits, '''Norton's theorem''' allows the replacement of a two-terminal portion of a linear circuit by a simplified circuit consisting of a current source, called the '''Norton voltage source''', in parallel with an impedance, called the '''Norton impedance'''. Norton's theorem is the ''dual'' of [[Thévenin's theorem]], which replaces a two-terminal portion of a linear circuit by a simplified circuit consisting of a voltage source in series with an impedance.<ref name=Sedra/>
| | For historical reasons, some confusing nomenclature has arisen that is only slowly being replaced. For both bipolar and MOSFET devices, it is common to refer to the ''off'' mode as ''cutoff''. However, the ''active mode'' of the bipolar transistor often is called the ''saturation mode'' of the MOSFET, while the ''on'' mode of the bipolar transistor often is called its ''saturation mode''. This confusion of terminology does nothing to clarify the discussion of circuitry. |
| | | <div class="wikitable" style="float:left"> |
| {{Image|Norton equivalent circuit.PNG|right|300px|The result of applying Norton's theorem.}}
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| The upshot of using Norton's theorem is shown in the figure: the circuit behavior for small, time-varying signals is boiled down to a simple current divider.
| | !MOS transistor!!G-S <br /> Bias !!G-D <br /> Bias !!S-B <br /> Bias !!D-B <br /> Bias !! Mode |
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| It will be noted that when the Norton impedance ''Z<sub>N</sub>'' is much larger than the original parallel source resistance ''Z<sub>S</sub>'', the output current across the load ''Z<sub>L</sub>'' benefits from a favorable current division when contrasted with direct attachment of the source to the load. In any event, the Norton circuit makes this current-division role of the more complex circuit clear. In general, this theorem is a tremendous aid to an intuitive understanding, essential for the creative design of circuits .
| | |Channel at source end only || ≥ V<sub>T</sub> at source || ≤ V<sub>T</sub> at drain || Zero or Reverse||Reverse, more than S-B|| Saturation |
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| ==Finding the Norton components==
| | |Channel at both ends || ≥ V<sub>T</sub> at source || > V<sub>T</sub> at drain || Zero or Reverse||Reverse, more than S-B || Ohmic |
| To determine the Norton current, short circuit the two terminals and measure or calculate the short-circuit current between the terminals.
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| | | |No channel ||< V<sub>T</sub> at source || ≤ V<sub>T</sub> at drain || Zero or Reverse||Zero or reverse|| Cutoff (Subthreshold) |
| To determine the Norton impedance, set all independent sources between the two terminals to zero (leave the ''dependent sources'' active), and determine the impedance seen looking into the two terminals. For example, apply a known voltage across the port terminals and measure the current drawn, or apply a known current and determine the voltage developed. The Norton impedance is then the ratio of the voltage across the terminals to the current passing through the terminals.
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| | | |Channel at drain end only ||< V<sub>T</sub> at source || ≤ V<sub>T</sub> at drain || Reverse ≥ V<sub>DB</sub>||Reverse || Reverse Saturation |
| In the case of a frequency dependent applied voltage or current, the impedance may be dependent upon frequency and exhibit a phase dependent upon frequency, as normally expressed using an impedance represented by a complex number.
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| | | </div> |
| ==Example==
| | <div class="wikitable" style="float:left"> |
| {{Image|Current follower.PNG|left|200px|Bipolar current buffer.}}
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| {{Image|Current follower; small-signal.PNG|right|300px|Small-signal circuit to find output current.}}
| | !Bipolar transistor!!B-E Junction <br /> Bias !!B-C Junction <br /> Bias !! Mode |
| The bipolar circuit in the figure at left represents a current buffer. The operation of this circuit is as follows. The DC current source ''I<sub>E</sub>'' determines the current leaving the emitter in the quiescent state when no signal is applied. By appropriate selection of the DC current source it can be arranged that the transistor is configured to operate usefully, that is, it is in ''active mode''.
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| | | | || Forward || Reverse || Forward active |
| A signal voltage ''i<sub>S</sub>'' is applied to the emitter. The DC current source ''I<sub>E</sub>'' does not permit any time varying current to pass through it, so at the signal frequency it behaves as an open circuit. No signal is lost through it. As a dual to this behavior, DC base voltage ''V<sub>BB</sub>'' will not support a time-varying voltage across it, so it acts as an AC short circuit. The signal causes no change in base voltage. The time-varying current drawn through the base of the transistor is very small compared to that in the emitter (lower by a factor of the transistor β). As the signal alters the emitter current, the collector current follows suit, and the output current ''i<sub>out</sub>'' in the load resistor ''R<sub>L</sub>'' is a copy of the input signal ''i<sub>in</sub>''.
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| | | | || Forward || Forward || Saturation |
| As this discussion indicates, the current gain of this circuit is very nearly unity, so amplification is not achieved. The purpose of this circuit is to isolate the signal source from the load ''R<sub>L</sub>'', with the object of preventing the load from affecting the performance of the source. In particular, if the source is forced to draw a large current, that may adversely impact its performance, and this circuit insures that only a small current is demanded of the source. Because the object of this circuit is not current amplification, but isolation of the driver, it is called a ''current buffer''.
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| | | | || Reverse || Reverse || Cut-off |
| The transistor is a very nonlinear device, so the output signal current in the load resistor is not a faithful amplification of the input signal at the emitter unless the signal amplitude is small. For purposes of estimating the amplifier behavior, the assumption is made that the signal amplitudes are very small so the amplifier appears to behave as a linear circuit. Under these circumstances, the transistor can be approximated by a linear circuit composed of resistors called the [[hybrid pi model]]. The linear small-signal approximation to the circuit using the transistor hybrid pi model is shown in the figure with a few algebraic expressions derived by application of [[Kirchhoff's current law]]. This circuit is used to determine the Norton current (the short-circuit current) of the buffer.
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| | | | || Reverse || Forward || Reverse-active |
| The value of the Norton current is the output current through a short-circuit load, that is, with ''R<sub>L</sub>'' = 0. From the small-signal circuit on the right, this current is ''i<sub>O</sub>'', which is found by equating the values of the emitter voltage found from the output and the input sides of the circuit:
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| :<math>v_E=(i_S-i_O)(r_\pi//R_S) \ , </math>
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| and:
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| :<math>v_E = i_OR_L + (i_O-g_mv_\pi)r_O \ . </math>
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| Equating these expressions, the Norton current is found as:
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| :<math>i_N = i_O = i_s \left(\frac{r_\pi//R_S}{r_\pi//R_S + \frac{r_O+R_C//R_L}{1+g_mr_O}}\right) \approx i_s \left(\frac {g_m (r_\pi//R_S)}{1+g_m (r_\pi // R_S)} \right) \ , </math>
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| where the approximation holds because ''R<sub>L</sub>'' is taken as zero for a short-circuit output condition, and because ''g<sub>m</sub>r<sub>O</sub>'' >> 1. The input resistance ''R<sub>S</sub>'' is likely to be quite large (if it were not the signal driver would be better represented as a [[Thévenin source]] with a small series ''R<sub>S</sub>'') so the parallel resistance ''{{nowrap|R<sub>S</sub>//r<sub>π</sub>}}'' is approximately ''r<sub>π</sub>'', and the Norton current is approximately:
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| :<math>i_N = i_S \ \frac{\beta}{\beta+1} \ , </math>
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| where ''g<sub>m</sub>r<sub>π</sub> = ß''. Inasmuch as ß is probably over 100, the Norton current is very nearly the same as the signal current (current gain is nearly unity).
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| {{Image|Current follower output resistance.PNG|right|300px|Small-signal circuit with test current ''i<sub>X</sub>'' to find Norton resistance.}}
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| To apply Norton's theorem, the resistance of the circuit looking back into the circuit from the output node is needed. That is accomplished using the second small-signal circuit. To determine the Norton impedance, the signal source is turned off (open-circuited), a test current ''i<sub>x</sub>'' is applied to the output, and the resulting voltage at the output ''v<sub>x</sub>'' is determined. Applying Ohm's law to obtain the voltage ''v<sub>E</sub>'' at the emitter node, one finds:
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| :<math>v_E=i_O (r_\pi//R_S )\ , </math>
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| and
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| :<math>v_E=v_X-(i_O+g_mv_E)r_O \ . </math>
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| Equating these two expressions the corresponding resistance is:
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| :<math>Z_O=\frac{v_X}{i_O} = (1+g_mr_O)(r_\pi//R_S) + r_O \ . </math>
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| The impedance ''Z<sub>O</sub>'' is seen at the collector looking into the circuit at the point where current ''i<sub>O</sub>'' is injected, after the parallel resistors ''(R<sub>C</sub>//R<sub>L</sub>'').
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| The output resistance is found looking backward from the load, so ''R<sub>L</sub>'' is removed from the circuit, leaving only ''R<sub>C</sub>'' in parallel with ''Z<sub>O</sub>''. Therefore, the Norton impedance is:
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| :<math> Z_N = Z_O//R_C \ . </math>
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| If this circuit is realized using a simple resistor ''R<sub>C</sub>'', the likelihood is that ''Z<sub>O</sub>'' >> ''R<sub>C</sub>'', in which case the current division predicted by the Norton equivalent circuit places ''R<sub>C</sub>'' in parallel with the load. The success of this circuit thus depends upon how large ''R<sub>C</sub>'' is compared to the driver resistance ''R<sub>S</sub>'', and limitations imposed upon signal swings limit how large ''R<sub>C</sub>'' can be. To avoid losing signal current to ''R<sub>C</sub>'' and thus losing signal current to the load, a large ''R<sub>C</sub>'' is desirable. For this reason, this circuit is better implemented using a load circuit that involves more transistors, an ''[[active load]]''. Such techniques can greatly increase the effective ''R<sub>C</sub>''; as the effective ''R<sub>C</sub>'' increases, the upper limit upon the Norton resistance is ''Z<sub>O</sub>'', which is likely to be considerably larger than the driver impedance ''R<sub>S</sub>'' because:
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| :<math>(g_mr_O)(r_\pi//R_S) + r_O = r_O \left( 1+ g_m \frac{r_\pi R_S}{r_\pi+R_S} \right) = r_O \left(1+\beta\frac{R_S}{r_\pi+R_S} \right)\ , </math>
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| which is considerably larger than ''r<sub>O</sub>'', itself a large resistance. Thus, favorable current division can be achieved with this buffer circuit modified to use an active load.
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| ==References==
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| {{reflist|refs=
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| <ref name=Sedra>
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| {{cite book |title=Microelectronic circuits |author=Adel S Sedra and Kenneth C Smith |publisher=Oxford University Press |chapter=Appendix E: Some useful network theorems |pages=pp. E-1 ''ff'' |edition=4rth ed |year=1998 |isbn=0-19-511690-9 |url=http://www.amazon.com/Microelectronic-Circuits-Smith-Kenneth-Sedra/dp/0195116909/ref=sr_1_1?s=books&ie=UTF8&qid=1305993614&sr=1-1#reader_0195116909}}
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| </ref> | |
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