Trabalho 2. Theoretical Values

P. Stallinga

30.11.2004

1: Common Emitter

Assume: β = 200 (α = 1), V_{A} = 120 V, C_{μ} = 8 pF, C_{π} = 25 pF

Actual values: R_{S} = 1 kΩ C_{S} = 10 μF R_{B1} = 10 kΩ R_{B2} = 20 kΩ R_{C} = 4.7 kΩ C_{L} = 10 μF R_{E} = 3.3 kΩ C_{E} = 100 μF R_{L} = 10 kΩ V_{CC} = +10 V

Bias conditions: V_{B} = +10 V · 10 kΩ / (20 kΩ + 10 kΩ) = 3.33 V V_{E} = V_{B} − 0.7 V = 2.63 V I_{E} = 2.63 V / 3.3 kΩ = 0.797 mA V_{C} = 10 − 0.797 10^{−3} * 4700 = 6.25 V r_{e} = 26 mV / 0.798 mA = 32.58 Ω r_{π} = (β + 1)r_{e} = 6.5449 kΩ r_{o} = 120 V / 0.798 mA = 150 kΩ g_{m} = 1/r_{e} = 30.7 mS

Gain:

Typeset with L^{A}TEX 2ε

The gain is defined by the standard formula for a CE amplifier:

A_{v} = −\frac{R_{C}}{R_{E} + r_{e}}

In the intermediate frequency regime all the condensators (of the LPFs) can be considered shortcircuited. Thus, the resistor R_{E} is bypassed and the gain increases to

A_{v} = −\frac{R_{C}}{r_{e}} = −\frac{4700}{32.58} = −144.26

To calculate the input resistance of the transistor and the stage we also consider R_{E} bypassed:

r_{in}(transistor) = (β + 1)r_{e} = 201 · 32.58 Ω = 6.549 kΩ

r_{in}(stage) = R_{B1}∥R_{B2}∥r_{in}
= 10 kΩ∥20 kΩ∥6.549 kΩ = 3.304 kΩ

\frac{v_{o}}{v_{i}} = A_{v} [\frac{r_{in}(stage)}{r_{S} + r_{in}(stage)}] [\frac{R_{L}}{R_{L} + R_{C}} ]
= −144.26 (\frac{3.304}{1 + 3.304}) (\frac{10}{10 + 4.7})
= −144.26 · 0.7677 · 0.6803 = −75.33

lower cut-off frequency: Using the short-circuit-time-constants method (all other C’s are short circuited) 1: The condensator (C_{S}) at the input is part of a high-pass filter (HPF). To calculate the cutoff frequency of this filter we have to calculate the RC time and hence we have to determine the resistance this condensator sees. Figure 2 summarizes this. The part after the collector can be neglected, because it is much larger. The resistor R_{E} is bypassed by the capacitor C_{E}.

R = R_{S} + R_{B1}∥R_{B2}∥(β + 1)r_{e}
= 1kΩ + 10 kΩ∥20 kΩ∥6.549 kΩ
= 4.304 kΩ
C_{S} = 10 μF
τ_{S} = RC = 43.04 ms
f_{LS} = \frac{1}{2πτ_{S}} = 3.7 Hz

2: For the condensator at the load we have:

R = R_{L} + R_{C} = 10 kΩ + 4.7 kΩ = 14.7 kΩ

(note: the resistance after the collector is again neglected). With the capacitance C = C_{L} = 10 μF we have

τ_{L} = RC = 147 ms
f_{LL} = \frac{1}{2πτ_{L}} = 1.08 Hz

3: Finally, for the emittor condensator we have (everything at the collector ignored):

R = R_{E}∥(r_{e} + \frac{R_{B1}}{β + 1}∥\frac{R_{B2}}{β + 1}∥\frac{R_{S}}{β + 1})
= 3300 Ω∥(32.58 Ω + 49.75 Ω∥99.50 Ω∥4.975 Ω) = 36.5 Ω
C = C_{E} = 100 μF
τ_{E} = RC = 3.65 ms
f_{LE} = \frac{1}{2πτ_{E}} = 43 Hz

The cut-off frequency (f_{L}) is determined by the higher of the three f_{LS}, f_{LL} and f_{LE}, namely by condensator at the emittor, f_{L} = f_{LE}. More exact, f_{L} = f_{LS} + f_{LL} + f_{LE} = 3.7 Hz + 1.08 Hz + 43 Hz = 48 Hz. Below this frequency the attenuation is 20 dB / decade. The phase slowly shifts 90^{o} and is 45^{o} at f_{L}. When the frequency is further lowered it hits the second cut-off frequency (approximately f_{LS}) from where the attenuation will be 40 dB / decade, etc. Each filter introduces an attenuation of 20 dB / decade and a phase shift of 90^{o}.

higher cut-off frequency: Using the open-circuit-time-constants method (all other high frequency C’s are open circuit). The higher cut-off frequencies are determined by the shunt capacitances that connect the input or the output to ground. We have C_{π} (between base and emittor) and C_{μ} (between base and collector). Note that in this configuration the latter is connected between the input and the output (with negative gain) and is therefore susceptible to feedback effects (resulting in so-called Miller capacitance, C_{M}). For the higher cut-off frequencies it is most convenient to decompose the system into an input filter and an output filter. This is allowed because the resistance between the input and the output is so high that there is virtually no contact between these points. At the input we have:

R = R_{S}∥R_{B1}∥R_{B2}∥(β + 1)r_{e}
= 1 kΩ∥10 kΩ∥20 kΩ∥6.549 kΩ

= 767.6 Ω
C = C_{π} + C_{M} = C_{π} + C_{μ}(1 − A_{v})  (1)
= 25 pF + 8 pF(1 + 98.14)
= 818.12 pF
τ_{i} = RC = 627.99 ns
f_{Hi} = \frac{1}{2πτ_{i}} = 253 kHz

Note that for the amplification A_{v} we have to use the voltage amplification factor between the collector and the base because the condesator C_{μ} is connected between those points. It does not include the voltage devider at the source:

A_{v} = −\frac{R_{C}∥R_{L}}{r_{e}} = −98.14

At the output we have:

R = r_{c}∥R_{C}∥R_{L}
= 150 kΩ∥4.7 kΩ∥10 kΩ
= 3.131 kΩ
C = C_{μ}(1 − 1/A_{v})
= 8 pF · 1.0102
= 8.08 pF
τ_{o} = RC = 25.3 ns
f_{Ho} = \frac{1}{2πτ_{o}} = 6.3 MHz

The cut-off frequency is therefore determined by the filter at the input. Using the open-circuit-time-constants method τ = τ_{i} + τ_{o} = 627.99 ns + 25.3 ns = 655 ns and f_{H} = 1/2πτ = 243 kHz.

Finally, the internal cut-off frequency of the transistor is determined by (see p. 437 of Bogart)

f_{T} = \frac{1}{2πr_{e}(C_{π} + C_{μ})} = \frac{1}{2π(32.58 Ω)(12 pF)} = 407 MHz

(According to the specifications (datasheets), for the 2N2222 the unitary current amplification frequency is f_{T} = 250 MHz.) This cut-off frequency is not important for the circuit. The gain and cut-off frequencies are summarized in the Bode plot on the next page.

phase diagram: The overall gain of the amplifier at mid-band frequencies is negative. This is equal to a phase shift of 180^{o}. The filter at the source with f_{LS} introduces a phase shift of +90^{o} (see p. 33 of Sedra) at low frequencies. In fact, every HPF introduces a phase shift of 90^{o}. On the other side of the spectrum, every LPF introduces a phase shift of −90^{o} at high frequencies. The figures show the voltage gain (in dB) at the load and the phase relative to the source voltage:

Without C_{E}:

When the shunt capacitance C_{E} is pulled, the gain of the amplifier reduces to

A_{v} = −\frac{R_{C}}{r_{e} + R_{E}} = −\frac{4700}{32.58 + 3300} = −1.410

The voltage gain between the base and the collector (to which C_{μ} is connected; we will use this value to calculate the Miller capacitance) when a load resistance is connected is

A_{v} = −1.410 · \frac{R_{L}}{R_{L} + R_{C}} = −1.410 · \frac{10}{10 + 4.7} = −0.959

The voltage divider at the input:

r_{in} = (β + 1)(r_{e} + R_{E}) − 669.8 kΩ
(r_{in})_{stage} = r_{in}∥R_{B1}∥R_{B2} = 669.8 kΩ∥10kΩ∥20kΩ = 6.6 kΩ
\frac{v_{i}}{v_{S}} = \frac{(r_{in})_{stage}}{(r_{in})_{stage} + R_{S}} = \frac{6.6}{6.6 + 1} = 0.868

The ratio of the voltage at the output and the voltage at the input then becomes (taking into account the voltage dividers at the input and the output):

\frac{v_{o}}{v_{i}} = −0.832

The filter at the emittor is removed and the highest cut-off frequency is then f_{1}(S) whose value is also slightly changed by the elimination of the shunt capacitance:

R = R_{S} + R_{B1}∥R_{B2}∥(β + 1)(r_{e} + R_{E})
= 7.601 kΩ
C = 10 μF
τ_{S} = RC = 76.01 ms
f_{LS} = \frac{1}{2πτ_{S}} = 2.09 Hz
f_{LL} = f_{LL} = 1.08 Hz

For the higher cut-off frequency. 1: At the entrance of the transistor:

R = R_{S}∥R_{B1}∥R_{B2}∥(β + 1)(R_{E} + r_{e})
= 1kΩ∥10kΩ∥20kΩ∥669.8kΩ = 868 Ω
C = C_{π} + (1 − A_{v})C_{μ} = 25 pF + (1 + 0.959) · 8 pF = 40.7 pF
τ_{i} = RC = 35.3 ns
f_{Hi} = \frac{1}{2πτ_{i}} = 4.5 MHz

2: At the exit of the transistor:

R = R_{L}∥R_{C} = 3.131 kΩ
C = (1 − 1/A_{v})C_{μ} = (1 + 1/0.959) · 8 pF = 16.3 pF
τ_{o} = RC = 51.0 ns
f_{Ho} = \frac{1}{2πτ_{o}} = 3.12 MHz

3: internal filter:

f′_{T} = f_{T} = 407 MHz

The cut-off frequency is determined by the exit of the transistor.

2: Cascode

Actual values: R_{S} = 1 kΩ C_{S} = 10 μF R_{B1} = 8.2 kΩ R_{B2} = 8.2 kΩ R_{B3} = 20 kΩ R_{C} = 4.7 kΩ C_{L} = 10 μF R_{E} = 3.3 kΩ C_{E} = 100 μF R_{L} = 10 kΩ V_{CC} = 15 V C_{B} = 10 μF

Bias conditions: V_{B1} = +15 V · 8.2 kΩ / (20 kΩ + 8.2 kΩ + 8.2 kΩ) = 3.38 V V_{E1} = V_{B}− 0.7 V = 2.68 V I_{E1} = 2.68 V / 3.3 kΩ = 0.8119 mA (3.3 kΩ choosen so that I_{E1} will be approximately 0.8 mA) I_{C2} = I_{E2} = I_{C1} = I_{E1} (α=1) V_{C2} = +15 − R_{C}I_{C2} = 11.18 V r_{e} = 26 mV / 0.8119 mA = 32.024 Ω g_{m} = 1/r_{e} = 31.2 mS r_{π} = (β + 1)r_{e} = 6.437 kΩ r_{o} = 120 V / 0.8119 mA = 150 kΩ The output resistance of Q1 is r_{o}. The output resistance of Q2 is much higher (reasoning is similar to current source of TP1) Gain:

(r_{in})_{stage} = (β + 1)r_{e1}∥R_{B1}∥R_{B2}
= 6.437 kΩ∥8.2 kΩ∥8.2 kΩ
= 2.505 kΩ

(r_{o})_{stage} = R_{C}∥r_{out,2} = R_{C}
= 4.7 kΩ
A_{v}^{1} = −\frac{r_{out,1}∥r_{e2}}{r_{e1}} = −\frac{r_{e2}}{r_{e1}} = −1
A_{v}^{2} = \frac{R_{C}∥r_{out,2}}{r_{e2}∥r_{c1}} = \frac{R_{C}}{r_{e2}} = \frac{4700}{32.024} = +147.76
A_{v} = A_{v}^{1} · A_{v}^{2} = −1 · 147.76 = −147.76
\frac{v_{L}}{v_{S}} = \frac{r_{in}}{r_{in} + r_{S}} · A_{v} · \frac{r_{L}}{r_{L} + r_{o}}
= (\frac{2505}{1000 + 2505}) (−146.76) (\frac{10000}{10000 + 4700})
= 0.7147 · (−146.764) · 0.6803
= −71.36

At low frequencies nothing much changes, though there is an extra filter caused by the condensator C_{B}:

C = C_{S} = 10 μF
R = R_{S} + R_{B1}∥R_{B2}∥(β + 1)r_{e} = 1 + 8.2∥8.2∥6.437 = 3.505 kΩ
τ_{S} = RC = 35.05 ms
f_{LS} = 4.54 Hz
C = C_{B} = 10 μF
R = R_{B3}∥(R_{B2} + R_{S}∥R_{B1}∥(β + 1)r_{e}) = 20∥(8.2 + 1∥8.2∥6.437) = 6.199 kΩ
τ_{B} = RC = 61.99 ms
f_{LB} = 2.57 Hz
C = C_{E} = 100 μF
R = R_{E}∥ (r_{e} + \frac{R_{B1}∥R_{B2}∥R_{S}}{β + 1} ) = 3.3∥ (0.032024 + \frac{8.2∥8.2∥1}{201}) = 35.63 Ω
τ_{E} = RC = 3.563 ms
f_{LE} = 44.67 Hz
C = C_{L} = 10 μF
R = R_{C} + R_{L} = 4.7 + 10 = 14.7 kΩ
τ_{L} = RC = 147 ms
f_{LL} = 1.08 Hz

The cut-off frequency is the largest of the four: 45 Hz. Or using the short-circuit-time-constants theory: f_{L} = ∑_{i} f_{i} = 53 Hz. The advantage is in the high frequencies, namely the virtual elimination of the Miller capacitance because the capacitance C_{μ} at the entrance of the stage is connected to two points with a voltage gain close to −1 as described above. The Miller capacitance therefore reduces to

C_{M} = C_{μ} (1 − A_{v}) = 8 pF (1 − (−1)) = 16 pF

and the total capacitance becomes

C = C_{π} + C_{M} = 25 pF + 16 pF = 41 pF

The resistance is approximately the same as in Equation 1.

R = R_{S}∥R_{B1}∥R_{B2}∥(β + 1)r_{e} = 1∥8.2∥8.2∥6.537 = 714.7 Ω
τ_{i} = RC = 29.3 ns
f_{Hi} = 5.43 MHz

At the output the capacitor C_{μ} ow has one leg connected to ground (V_{B2}) and the Miller effect disappears (1-0 = 1)

R = R_{C}∥R_{L} = 10∥4.7 = 3.197 kΩ
C = C_{μ} = 8 pF
τ = RC = 25.6 ns
f_{Ho} = 6.2 MHz

In-between the two transistors we have another filter:

C = C_{π2} + (1 − (−1/1))C_{μ1} = 41 pF
R = r_{e} = 32.58 Ω
τ = RC = 1.34 ns
f_{Hm} = 119 MHz

Using the theory of open-circuit time constants we find a total time constant of

τ = 29.3 ns + 25.6 ns + 1.34 ns = 56.2 ns
f_{H} = 1/2πτ = 2.8 MHz

Conclusion: there is a significant increase in high-frequency response by approximately one decade, basically caused by the elimination of the Miller effect at the entrance.

3: Differential Pair with Current Source

Actual values: R_{C} = 3.3 kΩ R_{S} = 1 kΩ I_{0} = 3 mA

bias: I_{0} = 3 mA, I_{E} = 1.5 mA. r_{e} = 26 mV / 1.5 mA = 17.33 Ω. r_{π} = 201 x 17.3 Ω = 3.48 kΩ. V_{o2} = V_{o1} = 10 V - 1.5 mA x 3.3 kΩ = 5.05 V. r_{in} = 2 x r_{π} = 7.0 kΩ.

gain:

A_{v}^{l} = −\frac{αR_{C}}{2r_{e}} = −95.4
A_{v}^{r} = +95.4
\frac{v_{o}}{v_{s}} = \frac{r_{in}}{r_{in} + R_{S}} A_{v} = \frac{(β + 1)2r_{e}}{(β + 1)2r_{e} + R_{S}} A_{v} = 0.8742 · 95.2 = +83.4

high-frequencies: The high-frequency response is not improved compared to a single-stage amplifier. At the input exists C_{π} = 25 pF and C_{μ} = 8 pF. Both suffer from the Miller effect and are therefore multiplied by the respective voltage gains between the two points that are bridged by the capacitance. From Trabalho 1 we know that in differential mode, the gain between base and collector of the left transistor (Q1) is given by

A_{v} = −\frac{αR_{C}}{2r_{e}}
= −\frac{1 · 3300}{2 · 17.33}
= −95.2

The gain felt by C_{π} is exactly 0.5. The total capacitance at the input is therefore

C = (1 − 0.5)C_{π} + C_{M} = 0.5C_{π} + C_{μ} (1 − A_{v})
= 12.5 pF + 8 pF (1 − (−95.4)) = 783.7 pF

The resistance-to-ground at the input is

R = R_{S}∥(β + 1)2r_{e}

= 1 kΩ∥201 · 17.33 Ω = 777 Ω
τ_{i} = RC = 608.9 ns
f_{Hi} = \frac{1}{2πτ_{i}} = 261 kHz

At the output (collector of right transistor Q2) we have C_{μ} = 8 pF to the base. Without any Miller effects, because the gain between emittor and base is zero. The total capacitance is therefore

C = C_{μ} = 8 pF

At the output the resistance is R_{C}∥r_{out,2} ≈ R_{C} = 3.3 kΩ. Therefore

τ_{o} = RC = 26.4 ns
f_{Ho} = \frac{1}{2πτ_{o}} = 6.0 MHz

Note: with a load resistance of 10 kΩ, the equivalent resistance at the output would be 2.48 kΩ and the frequency would further increase to 8.0 MHz. The internal cut-off frequency remains at

f_{T} = f_{T} = 407 MHz

low frequencies: Since there are no coupling capacitors at the entrance or the exit, there is no theoretical lower cutoff frequency. If we do introduce them (at the source and the load only) we can say the following: Since there is no need for any bypass condensator at the emittor we can expect an improvement of the lower frequency response, since in the single-stage amplifiers the lower-frequency response is limited by this condensator (see first part). There still remain the filters at the source and at the load. For the source we have C_{S} = 10 μF and the resistance is here R_{S} + (β + 1)2r_{e} = 4.48 kΩ. The cut-off frequency is then f_{LS} = 3.6 Hz. At the load the resistance is R_{L} + R_{C}∥r_{o} ≈ 4.3 kΩ and with a load-condensator of 10 μF this gives a cut-off frequency of f_{LL} = 3.7 Hz.

In conclusion: For a differential pair we see a significant improvement of the low-frequency performance compared to a single-stage amplifier (about one decade), mainly due to the elimination of the emittor condensator.

Further information: Chapter 10 of “Electronic Devices and Circuits”, T. F. Bogart, Jr., 4th edition, Prentice Hall, 1997. Chapter 7 of “Microelectronic Circuits”, A. S. Sedra, K. C. Smith, 2nd edition, Oxford University Press, 1982.