Impedance Reflection Rules

Impedance reflection rules Andrea Pacelli Department of Electrical and Computer Engineering SUNY at Stony Brook [email protected] First edition, February 2002 Copyright c 2002 Andrea Pacelli All Rights Reserved

In these notes we present a set of so-called impedance reflection rules. The name is really misleading. Reflection means a change of course, where something goes back to whence it came. Here, no such thing occurs. In the language of optics, we should rather speak of refraction, where an image is magnified or reduced when passing through a lens. This is what happens to the impedance when going through a transistor: It is magnified or reduced, according to the current and voltage gain of the stage under consideration. For example, the base terminal of a BJT carries much less current than the emitter, while sustaining sustaining a similar similar voltage. Corresponding Correspondingly ly,, an impedance connected connected to the emitter appears appears to be magnified magnified when seen from the base. We will examine impedance impedance reflection reflection rules for BJTs first, then for MOSFETs. MOSFETs. Although performing similar functions, the two devices differ in that the base current of a BJT is usually usually nonnegligible, nonnegligible, while the gate current of a MOSFET is always always zero. Also, MOSFETs MOSFETs display display the body effect, effect, which complicates complicates calculations calculations slightly slightly.. Finally Finally,, the lower transconductance of MOSFETs makes some effects apparent, which are usually negligible negligible for BJTs. As a result, the impedance impedance reflection reflection rules for MOSFETs are quite different different from those for BJTs.

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Bipol Bipolar ar transi transisto stors rs

The bipolar junction junction transistor has three current-carrying current-carrying terminals. terminals. We will compute compute the impedance seen from each of them, when a generic load is connected to one of the remaining remaining two terminals, and the other is grounded. grounded. The astute reader will figure out that one could construct six possible possible combinations combinations of that kind. However However,, it turns out that three of them are not very interesting, interesting, so we are left with only three, three, which we proceed proceed to discuss in the following. 1.1

Magnifi Magnificat cation ion of emit emitter ter load load

Let us consider a BJT with an impedance connected between the emitter and ground, and the collector grounded grounded (Fig. 1a). We want to compute compute the impedance seen ‘into the

1

i x

i x v x

r π

g m v be

v x Z E Z E

a)

b)

Figure 1: Large-signal (a) and small-signal (b) ac circuits for impedance reflection rule #1.

base.’ To this purpose we connect a probe source to the base and take the ratio of probe voltage to the probe current . Note that it is irrelevant whether the probe source is a voltage or current source: The result is the same. Replacing the BJT by its hybrid-pi model, we obtain the circuit of Fig. 1b. Note that the output resistance is connected from the emitter to ground, i.e., parallel to . Therefore we can lump it together with into a single impedance

Note that in most cases, , so that . In order to compute the impedance at the base side, we only need one equation relating and . However, the emitter voltage, , is unknown. It is convenient and instructive to use the base-emitter voltage as an unknown, rather than . To relate the two, let us write the KVL for the base-emitter-ground loop:

so that, We now write the KCL for the emitter:

from which we can obtain

The input impedance is now found as

2

Z B Z B

r π

g m v be

i x i x v x v x

a)

b)


or, substituting

,

Eq. (1) is the first impedance-reflection rule: The impedance seen when looking into the base is the base-emitter resis tance , plus times the impedance connected to the emitter. The bipolar transistor is like a ‘magnifying glass’ for impedances. Whatever is connected to the emitter ‘appears’ magnified by when seen through the base. 1.2

Reduction of base load

Let us now consider the symmetric case of an impedance connected between the base and ground, where the impedance is measured from the emitter side (Fig. 2a). Replacing again the BJT by a small-signal model, we obtain the circuit of Fig. 2b. Let us start again from computing as a function of . From the partition between and , we obtain immediately

Let us now write the KCL for the emitter:

from which, substituting

, we obtain

and the impedance is

3

i x

i x

v x

r π

gm v be

i x

v x

r o

g m v be

r o

v x

Z E Z E

a)

Z E ’

b)

c)

Figure 3: Large-signal (a), small-signal (b), and simplified small-signal (c) ac circuits for impedance reflection rule #3.

or

The second impedance-reflection rule is then: The impedance seen when looking into the emitter is the inverse transcon ductance , plus the impedance connected to the base divided by . The ‘magnifying glass’ analogy works also the other way around: When looking into the emitter, impedances connected to the other side ‘appear’ to be smaller. 1.3

Magnification of output resistance

Let us consider the third and final case of Fig. 3a, which becomes Fig. 3b for small-signal analysis. Since both and are connected between emitter and ground, we can lump them together in a single obtaining Fig. 3c. The reader will note that the probe current flows through the entire circuit into ground, therefore the current through is also equal to , and the emitter voltage is Since the base is grounded, We then write the KCL at the collector:

This is the wanted equation that allows us to compute

4

:

i x

i x

v x

v gs

g m v gs

g mb v s

r o

v x

Z S Z S

a)

b)


Since in most cases

, we can simplify the equation:

We can summarize the third impedance-reflection rule as follows: The emitter degeneration impedance the factor .

2

boosts the output resistance

by

MOSFETs

The MOSFET has only two current-carrying terminals, therefore we derive only two impedance-reflection rules. One is very similar to the third BJT rule. The other (reduction of the drain-side impedance) is peculiar to MOSFETs with their low product. 2.1

Magnification of output resistance

Let us consider the same circuit of Fig. 3a, but with a MOSFET replacing the BJT (Fig. 4a). The gate is grounded. However, remember that the gate of a MOSFET never carries any dc current. Since we are working at low enough frequencies that the MOSFET can be assumed to be operating at dc, the results would not change even if we connected a finite resistance to the gate. This consideration also applies to the rule in the next section. The two differences with respect to the BJT case are: (a) Resistor is missing, (b) The body-effect transconductance has appeared. The analysis, however, is completely analogous to the case of the BJT. Since the gate current is zero, the source voltage is so that the KCL at the drain becomes

5

Z D

v gs

g m v gs

g mb v s

Z D

i x i x v x v x

a)

b)


from which we obtain the drain-side impedance

or, neglecting

in comparison with

,

We repeat the summary for preceding result, rewording it in MOSFET terms: The source degeneration impedance the factor . 2.2

boosts the output resistance

by

Reduction of drain load

In this configuration, an element is connected on the drain side, while the impedance is measured at the source side, with the gate grounded (Fig. 5a). Note that for a BJT, according to Eq. (2), the measured impedance is expected to be . However, the lower transconductance of MOSFETs, along with the high load impedances sometimes present, can let the impedance on the drain side ‘contaminate’ the impedance seen on the source side. Consider for example the case of a drain connected to an ac open circuit (an ideal current source). The drain current is simply constant, because it cannot change. Since the source current is equal and opposite to the drain current, it cannot change either, and the impedance measured at the source side is infinite. To analyze the effect, let us consider the small-signal circuit of Fig. 5b. The probe current traverses the transistor and flows into ground through , therefore the drain voltage is The source voltage is

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The KCL at the source is

so that the source-side impedance is

and multiplying numerator and denominator by

,

We then conclude that The source-side impedance in the presence of a drain load the sum of and , attenuated by the factor

is equal to .

Note that, when is small, the apparent impedance is , consistently with the BJT result. However, when is larger than , the MOSFET works as a reverse ‘magnifying glass’, reducing the apparent impedance by a factor of . When , the source-side impedance is also infinite. Let us now extend the same rule to bipolars. The reader can verify that, for a BJT, the corresponding impedance reflection rule (replacing by ) is

When

, a value of is recovered, as expected. However, for , , not . Moreover, the product for BJTs is always rather high (at least 500) so that even when , the resulting impedance is usually moderate. On the other hand, for MOSFETs the product can be as low as 10, with resulting high values of emitter-side impedance. That is why the rule for the attenuation of the drain load has a special importance for MOSFETs. In the case of BJTs, this effect must be taken into account only if a very low impedance is desired, and a large load is connected to the collector. The table on the next page summarizes the impedance-reflection rules for BJTs and MOSFETs.

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Summary of impedance reflection rules Load at

Probe at

E/S

B/G

B/G

E/S

E/S

C/D

C/D

E/S

BJT

E = emitter, B = base, C = collector S = source, G = gate, D = drain

MOS

Impedance Reflection Rules

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