Amplifier Frequency Response

Frequency response of RC-coupled amplifiers: decibels and cascading, the low-frequency poles set by the input coupling, output coupling and emitter bypass capacitors, a capacitor design to a specified cutoff, Miller capacitance and the high-frequency poles, gain-bandwidth product for n cascaded stages, and reading gain off the Bode asymptotes.

Each problem below is solved in full — the given data is listed first, the governing relation is stated, and the arithmetic is carried through to a boxed answer. Work through the statement yourself before opening the solution. The underlying theory is covered in the Electronic Devices lecture series.

A three-stage RC-coupled amplifier has mid-band voltage gains \(A_{v1} = 12\), \(A_{v2} = 30\) and \(A_{v3} = 45\).

1. Find the overall voltage gain, as a ratio and in decibels.

2. Show that the decibel gains of the individual stages add.

3. Find the output for a 1 mV rms input.

4. A 6 dB attenuator pad is inserted between stages 1 and 2. Find the new overall gain in dB and as a ratio.

• \(A_{v1} = 12\), \(A_{v2} = 30\), \(A_{v3} = 45\) (mid-band, loading already included in each figure)

• Voltage decibels: \(A_{v(\text{dB})} = 20\log_{10}|A_v|\).

Part (a) — overall gain

\(A_{vT} = 16200 = 84.19~\text{dB}\)

Part (b) — decibels add because gains multiply

\(21.58 + 29.54 + 33.06 = 84.19~\text{dB}\), matching part (a).

Part (c) — output level

\(V_o = 16.2~\text{V rms}\)

Part (d) — inserting a 6 dB pad

A pad is a negative gain in decibels, so it simply subtracts:

\(A'_T = 78.19~\text{dB}\), i.e. a ratio of 8119 — the pad halves the voltage gain.

6 dB is a factor of 2 in voltage and 4 in power; 20 dB is exactly a factor of 10. If source and load resistances are equal, the power gain in dB is the same number as the voltage gain in dB — here 84.190 dB — which is why the two are so often quoted interchangeably.

A voltage-divider-biased common-emitter stage has \(V_{CC} = 20~\text{V}\), \(R_1 = 40~\text{k}\Omega\), \(R_2 = 10~\text{k}\Omega\), \(R_C = 4~\text{k}\Omega\), \(R_E = 2~\text{k}\Omega\) and \(\beta = 100\). It is driven from a source of internal resistance 1 k\(\Omega\) through \(C_1 = 10~\mu\text{F}\) and drives a 2.2 k\(\Omega\) load through \(C_2 = 1~\mu\text{F}\); the emitter is bypassed by \(C_E = 20~\mu\text{F}\).

1. Find the mid-band \(r_e\), \(Z_i\) and \(A_v\).

2. Find the lower cutoff frequency contributed by each of the three capacitors.

3. State which is dominant and estimate the overall \(f_L\).

• \(V_{CC} = 20~\text{V}\), \(R_1 = 40~\text{k}\Omega\), \(R_2 = 10~\text{k}\Omega\), \(R_C = 4~\text{k}\Omega\), \(R_E = 2~\text{k}\Omega\)

• \(\beta = 100\), \(V_{BE} = 0.7~\text{V}\), \(r_e = 26~\text{mV}/I_E\) (simplified model, 26 mV used for \(V_T\))

• \(R_{\text{sig}} = 1~\text{k}\Omega\), \(R_L = 2.2~\text{k}\Omega\)

• Each capacitor is treated one at a time, the other two being short circuits.

Part (a) — mid-band values

\(r_e = 15.76~\Omega\), \(Z_i = 1316~\Omega\), \(A_{v(\text{mid})} = -90.1\)

Part (b) — one pole per capacitor

Each coupling or bypass capacitor sees its own Thevenin resistance; the cutoff is \(f = 1/(2\pi RC)\) with that resistance.

The bypass capacitor sees the resistance looking into the emitter node, which is \(R_E\) in parallel with the base circuit reflected down by \(\beta\):

\(f_{L(C_1)} = 6.87~\text{Hz}\), \(f_{L(C_2)} = 25.7~\text{Hz}\), \(f_{L(C_E)} = 327~\text{Hz}\)

Part (c) — the dominant pole

The highest of the three breaks first as the frequency falls, so it dominates:

A closer estimate adds the poles in quadrature, \(f_L \approx \sqrt{f_1^2+f_2^2+f_3^2} = 327.9~\text{Hz}\) — barely different, because the other two are more than a decade away.

\(f_L \approx 327~\text{Hz}\), set by the emitter bypass capacitor.

\(C_E\) always dominates in this topology because it works into \(r_e + R'_s/\beta\), only about 24.3 \(\Omega\) — two orders of magnitude smaller than the resistances the coupling capacitors see. Bypass capacitors are therefore always the largest in the circuit.

Design. The amplifier of the previous problem is to be used down to 50 Hz: its overall lower cutoff must not exceed 50 Hz. The Thevenin resistances are unchanged — \(R_{\text{sig}} + Z_i = 2316~\Omega\) at the input, \(R_C + R_L = 6200~\Omega\) at the output, and \(R_e = 24.35~\Omega\) at the emitter.

1. Size \(C_E\) for a 50 Hz break, and choose the nearest standard value that meets the specification.

2. Size \(C_1\) and \(C_2\) so that their breaks sit a decade below, at 5 Hz, and choose standard values.

3. Confirm the resulting overall \(f_L\).

• Specification: \(f_L \le 50~\text{Hz}\)

• \(R\) for \(C_1\): 2316 \(\Omega\); for \(C_2\): 6200 \(\Omega\); for \(C_E\): 24.35 \(\Omega\)

• Standard electrolytic values (E6/E12); a larger capacitor gives a lower break, so round up.

Step 1 — the emitter bypass capacitor

The nearest standard values are 120 \(\mu\)F and 150 \(\mu\)F. 120 \(\mu\)F would give \(f = 54.5~\text{Hz}\), above specification, so take 150 \(\mu\)F:

\(C_E = 150~\mu\text{F}\), breaking at 43.6 Hz.

Step 2 — the coupling capacitors, a decade lower

Putting the other two poles at 5 Hz keeps them from adding to the 3 dB point:

\(C_1 = 22~\mu\text{F}\), \(C_2 = 6.8~\mu\text{F}\)

Step 3 — the resulting cutoff

Overall \(f_L = 43.9~\text{Hz}\) — the 50 Hz specification is met with 6.14 Hz to spare.

Rounding a capacitor up is the safe direction for a low-frequency cutoff, and it is the only direction available here: rounding \(C_E\) down to 120 \(\mu\)F would have broken the specification by 4.5 Hz on its own.

The same common-emitter stage (\(R_{\text{sig}} = 1~\text{k}\Omega\), \(R_1 = 40~\text{k}\Omega\), \(R_2 = 10~\text{k}\Omega\), \(\beta r_e = 1576~\Omega\), \(R_C = 4~\text{k}\Omega\), \(R_L = 2.2~\text{k}\Omega\), mid-band \(A_v = -90\)) has junction capacitances \(C_{be} = 36~\text{pF}\), \(C_{bc} = 4~\text{pF}\), \(C_{ce} = 1~\text{pF}\), plus wiring capacitance \(C_{Wi} = 6~\text{pF}\) at the input and \(C_{Wo} = 8~\text{pF}\) at the output.

1. Find the Miller input capacitance \(C_{Mi} = C_{bc}(1 - A_v)\) and the total input capacitance.

2. Find the input-circuit cutoff \(f_{H_i}\).

3. Find \(C_{Mo} = C_{bc}(1 - 1/A_v)\), the total output capacitance and \(f_{H_o}\).

4. State the overall \(f_H\) and the bandwidth, given \(f_L = 327~\text{Hz}\) from the low-frequency analysis.

• \(A_{v(\text{mid})} = -90\) (the value that multiplies \(C_{bc}\))

• \(C_{be} = 36~\text{pF}\), \(C_{bc} = 4~\text{pF}\), \(C_{ce} = 1~\text{pF}\), \(C_{Wi} = 6~\text{pF}\), \(C_{Wo} = 8~\text{pF}\)

• Coupling and bypass capacitors are short circuits at high frequency.

Part (a) — Miller multiplication at the input

\(C_{bc}\) bridges input and output. Because the two ends move in antiphase by a factor \(|A_v|\), the current drawn from the input is \((1-A_v)\) times larger than a grounded \(C_{bc}\) would draw:

\(C_{Mi} = 364~\text{pF}\), \(C_i = 406~\text{pF}\) — the 4 pF collector-base capacitance dominates the input.

Part (b) — input cutoff

\(f_{H_i} = 690~\text{kHz}\)

Part (c) — output side

\(C_o = 13.0~\text{pF}\), \(f_{H_o} = 8.60~\text{MHz}\)

Part (d) — dominant pole and bandwidth

The lower of the two high-frequency poles rolls the response off first:

\(f_H = 690~\text{kHz}\), \(BW = 689~\text{kHz}\)

The input pole wins by a factor of 12.5 purely because of Miller: 4 pF becomes 364 pF. Halving the mid-band gain would roughly halve \(C_{Mi}\) and nearly double \(f_H\) — the gain-bandwidth trade in its most literal form, and the reason cascode and common-base stages exist.

An RC-coupled stage has a mid-band gain of 20, \(f_L = 100~\text{Hz}\) and \(f_H = 1~\text{MHz}\).

1. Find the gain-bandwidth product, and the bandwidth that would result if negative feedback reduced the gain to 5.

2. Three such stages are cascaded. Using \(f_H^* = f_H\sqrt{2^{1/n}-1}\) and \(f_L^* = f_L/\sqrt{2^{1/n}-1}\), find the overall cutoffs.

3. Tabulate \(f_H^*\) for \(n = 1\) to 4.

4. Find the overall gain of the three-stage cascade in dB and compare its gain-bandwidth product with that of one stage.

• Per stage: \(A_v = 20\), \(f_L = 100~\text{Hz}\), \(f_H = 1~\text{MHz}\)

• Identical, non-interacting stages, each with a single dominant pole at each end of the band.

Part (a) — gain-bandwidth product

For a single-pole roll-off the product of mid-band gain and \(f_H\) is a constant of the device:

\(GBW = 20.0~\text{MHz}\); dropping the gain to 5 buys \(f_H = 4.00~\text{MHz}\).

Part (b) — three identical stages

Each stage is 3 dB down at its own \(f_H\), so three stages are 9 dB down there. The overall \(-3\) dB point must therefore move inwards:

\(f_H^{*} = 509.8~\text{kHz}\), \(f_L^{*} = 196.1~\text{Hz}\)

Part (c) — how the band shrinks with n

Cascading four identical stages costs 56.5% of the upper cutoff.

Part (d) — gain bought, bandwidth paid

Three stages give 78.06 dB with a 509.6 kHz bandwidth: the gain rises by \(20^2\) while the bandwidth falls only to 51.0%.

That is the payoff of cascading — the product \(A_{vT}\times BW\) is far better than one stage can achieve, even though each individual stage still obeys its own fixed \(GBW\).

An amplifier has a mid-band gain of 80, a single dominant low-frequency pole at \(f_L = 200~\text{Hz}\) and a single dominant high-frequency pole at \(f_H = 500~\text{kHz}\). The normalised responses are

1. Express the mid-band gain in dB, and find the gain in dB at \(f = 200~\text{Hz}\) and at \(f = 100~\text{Hz}\).

2. Find the gain in dB at \(f = 20~\text{Hz}\) and compare with the \(-20~\text{dB/decade}\) asymptote.

3. Find the gain in dB at \(f = 2~\text{MHz}\) and compare with its asymptote.

4. Two such amplifiers are cascaded. What is the gain in dB at 2 MHz, and what is the roll-off rate there?

• \(A_{v(\text{mid})} = 80\), \(f_L = 200~\text{Hz}\), \(f_H = 500~\text{kHz}\)

• One dominant pole at each end; the other poles are far enough away to ignore.

Part (a) — mid-band level and the corner

Mid-band 38.06 dB; 35.05 dB at 200 Hz and 31.07 dB at 100 Hz.

Part (b) — a decade below \(f_L\)

The asymptote predicts exactly \(-20.0~\text{dB}\) one decade below the break; the exact value is 0.043 dB below that.

\(A_v(20~\text{Hz}) = 18.02~\text{dB}\), i.e. -20.0 dB relative to mid-band — the \(-20\) dB/decade asymptote is accurate to 0.043 dB here.

Part (c) — two octaves above \(f_H\)

Asymptotically, four times \(f_H\) is two octaves, i.e. \(2 \times (-6.02) = -12.04~\text{dB}\); the exact figure is 0.26 dB lower.

\(A_v(2~\text{MHz}) = 25.76~\text{dB}\)

Part (d) — two cascaded amplifiers

Two stages give 51.51 dB at 2 MHz and roll off at \(-40\) dB/decade (\(-12\) dB/octave).

Each pole contributes \(-20\) dB/decade and \(-45^\circ\) of phase at its own corner. Counting poles is therefore enough to sketch the whole Bode plot: the asymptotes are exact more than a decade from any break, and never more than 3 dB wrong at the break itself.