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Section D3: The Emitter-Resistor

The common-emitter amplifier of the previous section, in which the emitter resistor is bypassed for ac operation, can provide a very high gain. However, has a low input impedance, is noisy and does not possess the stability of the common-emitter with emitter resistor (emitter-resistor) amplifier configuration.

The emitter-resister (ER) configuration of Figure 5.4a in your text and reproduced to the right, is simply the common- emitter with the emitter bypass removed. Since the are open circuits to dc, the dc operation and strategies between the CE and ER are identical.

The difference between the two arises during ac operation. In the CE amplifier, the emitter resistor is shorted by the bypass capacitor and does not contribute to ac behaviors. Conversely, in the ER amplifier, the emitter resistor takes part in ac operation through a feedback mechanism. Very simply, the across the emitter resistor (vRE=ieRE) modifies the base-emitter voltage. We will be talking about feedback extensively later but if you‘ll take it on faith for now, this form of negative feedback decreases the sensitivity to β variations, thereby increasing the amplifier stability. The cost (and there‘s always a cost) is a reduced voltage gain due to RE being included in the ac equivalent resistance.

Note that Figure 5.4b in your text is incorrect for the emitter-resistor configuration in that it does not show the emitter resistor RE. Also, the output resistance rO is shown to make the circuit as generic as possible.

An ac small-signal model for the ER amplifier is shown below on the left, labeled as Figure 5.4b Version 1. Using the technique of impedance reflection, an equivalent small-signal model is shown below on the right, labeled as Figure 5.4b Version 2.

Be sure that you are completely comfortable that these two circuits do indeed represent the small-signal equivalent circuit for the emitter-resister amplifier configuration! Remember that we are assuming that iC≈iE and iE≈(β+1)ib and that these relationships determine how an resistance is reflected up (or down).

Using Version 2 of the small- signal model above, and making the assumption that β>>1, so that β+1≈β, the small-signal equivalent with the input and output circuits separated is shown to the right (Figure 5.4c). Once we have gotten to the point where the input and output circuits can be separated (i.e., they have nothing in common but a ground), we can use the strategies developed in the previous section, based on the —black box“ two-port network, to solve for what we‘re interested in…

I know I promised a not-so-mind-numbing derivation after last time, but there are a some intermediate steps that I feel you should see for some of these characteristics, so please bear with me! While we‘re going through this, keep in mind the comments of the previous section with respect to the effect of a practical input source that possesses a nonzero source resistance. Also, don‘t forget the comments about the purely resistive representation we‘re using and the possibility of having to deal with complex impedances œ same strategy, different animal.

Input Resistance, Rin

We always follow SOP (standard operating procedure) to find the input resistance; that is, —look into“ the input side after the source and find the equivalent resistance. The difference in this circuit and that for the common- emitter is the impedance reflected up from the emitter (remember that the emitter resistor is shorted for ac in the CE amplifier configuration). Using Figure 5.4c, the input resistance for the ER configuration is

RB (rπ + β RE ) Rin = RB || (rπ + β RE ) = . (Equation 5.15) RB + rπ + β RE

Once again, note that your text derives this equation in a slightly different manner, but ends up the same place.

Another representation for the input resistance is found by using the relationship between re and rπ, or

R (β r + β R ) R (r + R ) R = B e E = B e E . (Equation 5.16) in R RB + β re + β RE B + r + R β e E

Output Resistance, Rout

Unfortunately, finding the output resistance for the ER configuration is not quite as simple as for the CE amplifier. If you look at Figure 5.4c, you can see that the emitter resister RE appears in both the input and output circuits. This indicates that, although we have drawn them as separate, something of the input side can be —seen“ by looking in at the collector terminal. This means that the output resistance will be RC in with some Req1, where Req1 is the resistance looking back into the collector of the BJT. Modifying the output side to reflect this effect, the circuit (based on Figure 5.5a in your text) presented to the right is obtained.

In the next sequence of figures, I‘m going to take a slightly different approach than your author to (hopefully) clarify what‘s going on. Note: as each step is introduced, the point(s) of discussion will show in a different color. We‘re going to concentrate on getting an expression for Req1 in this derivation, then we‘ll back up and put it in parallel with RC for the total output resistance, Rout.

1. The first step will be to remove (short) the input source, like we did for the CE configuration. At the same time, we‘re going to apply a to the output to the collector terminal. To maintain as much consistency as possible with your text, we‘ll call this test source i1 and the voltage across it vtest. Using this representation, the equivalent resistance we‘re looking for may be found by: Req1 = vtest/i1.

2. A couple of things are going to happen now. Setting vin to zero

a. shorts out the resistance RB, effectively setting the base terminal to ground, and b. turns off the dependent current source with respect to the base current ib. However, there may still be a voltage across the base- emitter terminals due to the test source, so we need to leave in the dependent current source with the gmvbe notation.

3. In the next step, the parallel combination of rπ and RE may be combined into a single resistor. We can further simplify this circuit by using source transformations. The parallel current source resistor combination may be turned into a series voltage source resistor combination, yielding the circuit to the right. Note that the polarity of the dependent voltage source follows that of vbe and that the parameters gm and rO simply serve as scaling factors.

4. We can finally get to where we want to be by noticing that the current source i1 is also the loop current. The magnitude of vbe may be expressed as the loop current times the parallel resistor combination, or

vbe = −i1 (RE || rπ ) .

Note that the negative sign arises from the discrepancy between the current direction and the polarity of vbe. Writing the KVL equation for the last circuit presented above allows us to express vtest as a function of circuit components, transistor parameters and the base-emitter voltage:

vtest = i1rO − g mvbe rO − vbe

Substituting in for vbe:

v = i r + g i (R || r )r + i (R || r ) = i [r + g r (R || r ) + (R || r )] = test 1 O m 1 E π O 1 E π 1 O m O E π E π . i1[ro + (RE || rπ )(g m rO +1)]

Solving for Req1:

vtest i1[rO + (RE || rπ )(1+ g mrO ) Req1 = = = rO + (RE || rπ )(1+ gm rO ) . (Equation 5.25) i1 i1

And, finally, solving for Rout:

Rout = Req1 || RC = [rO + (RE || rπ )(1+ gmrO )]|| RC . (Equation 5.26: Modified)

Note that, just as in the common-emitter configuration, if rO is much larger than RC the parallel combination will be approximately equal to RC, or

Rout = RC if rO >> RC . (Equation 5.26: Modified)

If we have a practical source (vS with RS), the resistance RB would not be shorted as in step 2(a) above. Instead, we would have a parallel combination of RB and RS that is in series with rπ. Everywhere there is a rπ in the equations above, we would have to substitute rπ+(RB||RS), yielding an alternate representation of Equation 5.26:

Rout = Req1 || RC = [rO + (RE || (rπ + (RB || RS )))(1+ gmrO )]|| RC

The nice thing is that, if rO is very large with respect to RC, this whole mess is still dominated by RC and we still have Rout=RC!

Current Gain, Ai

We have already defined the current gain by the relationship:

iL Ai = . (Equation 5.7) iin

Using current division (refer to Figure 5.4c and assuming rO is very large), the load current is

− RC β ib iL = . (Equation 5.18) RL + RC

Note that this is the same relationship as for the common-emitter configuration.

The difference between the CE and ER configurations shows up in the expression for ib. The base current is still found by using current division, but now we have the reflected resistance βRE that must be included:

RBiin ib = . (Equation 5.17) RB + rπ + β RE

Solving Equation 5.17 for iin and substituting the expressions for iL and iin into Equation 5.7:

iL − RB RC β Ai = = , iin (RB + rπ + β RE )(RL + RC ) or, by factoring β out of the numerator and rearranging,

− R R A = B C . (Equation 5.19) i ≈ R ’ ∆ B + r + R ÷(R + R ) « β e E ◊ L C

Voltage Gain, Av

Making use of the Gain Impedance Formula, the voltage gain is

vout RL Av = = Ai . (Equations 5.4 & 5.21) vin Rin

Using Equations 5.16 and 5.19, as well as a fair bit of algebraic manipulation (and recognizing the parallel resistor relationship):

− RL || RC Av = . (Equation 5.22: Modified) re + RE

An alternative representation for the voltage gain in terms of the transconductance parameter gm is found by replacing re with 1/gm and simplifying:

− g m (RL || RC ) Av = . (Equation 5.22: Modified) 1+ g m RE

Before we leave this section, keep in mind the —words of caution“: be careful to include all relevant components in all intermediary expressions and check that the β>>1 assumption is valid (if β is not much larger than one, all β terms become β+1).