Fundamentals of Amplifier Circuits

A sensor outputs 20 mV, the amplifier is labeled with a gain of 100, yet only 1.28 V appears across the load; when a larger signal is applied, the output clips. The first issue likely stems from loading, while the second arises from the operating point and swing limits. An amplifier must simultaneously satisfy gain, linear range, frequency response, and load-driving capability; it cannot be described by a single multiplication factor.

Amplification is Powered by the Supply; Gain Has Defined Ports

Amplifier circuits use input variations to modulate the energy flow from the power supply to the output. The output is approximately a linear transformation of the input signal within a certain range, but it is not linear for every input, frequency, and load. An emitter follower can have a voltage gain less than 1 while providing current or power gain; a higher voltage does not automatically imply increased power.

For small changes around the operating point, the voltage gain is $A_v=\Delta v_{out}/\Delta v_{in}$. The complete instantaneous voltage can be written as $v_{out}(t)=V_{OQ}+A_v\cdot v_{in,ac}(t)$; one cannot calculate signal gain by dividing the DC bias by the small AC input. In the frequency domain, $A_v(j\omega )$ contains both magnitude and phase.

Voltage gain in decibels is typically written as $20{\log }_{10}|A_v|$; the negative sign indicates phase inversion, not taking the logarithm of a negative number directly. Power gain is $10{\log }_{10}(P_{out}/P_{in})$. When the input and output reference resistances are not the same, the voltage ratio cannot directly replace the power ratio: for pure resistors using RMS voltage, $P_{out}/P_{in}=|V_{out}/V_{in}{|}^2\cdot R_{in}/R_{load}$.

Calculating Gain from Source to Load

Model the source as an open-circuit signal $v_s$ in series with $R_s$, the amplifier input resistance as $R_{in}$, the unloaded incremental voltage gain as $A_{v0}$, and the output resistance $R_{out}$ driving $R_L$. Under these simplified conditions:

$v_L/v_s=[R_{in}/(R_s+R_{in})]\cdot A_{v0}\cdot [R_L/(R_{out}+R_L)]$.

Take $R_s=10\text{ k}\Omega$, $R_{in}=40\text{ k}\Omega$, $A_{v0}=100$, $R_{out}=1\text{ k}\Omega$, and $R_L=4\text{ k}\Omega$. Input loading reduces 20 mV to 16 mV, the internal incremental output becomes 1.6 V, and output loading further reduces it to 1.28 V. The total gain is only 64, approximately 36.1 dB.

Increasing the input resistance or decreasing the source impedance can reduce front-end losses; decreasing the output resistance or increasing the load resistance can reduce back-end losses. However, real impedances vary with frequency, and feedback also alters them. A resistance model at a single frequency cannot guarantee performance across the entire band.

This calculation also illustrates that "high input impedance, low output impedance" is a common goal for voltage transmission. The goals for current input or transimpedance measurement may differ, so this combination should not be treated as a universal standard for all amplifiers.

Bias and Dynamic Range

Transistor amplifiers first require a suitable quiescent operating point, allowing the signal to vary nearby. BJTs typically use the forward active region, while MOSFETs typically use the appropriate saturation region; "swinging in the linear region" without specifying device terminology can easily reverse the understanding of the operating region.

Correct voltage-divider biasing must account for the loading of the base current on the divider, and the emitter resistor changes the operating point through feedback. One cannot treat $V_B=V_{CC}\cdot R_2/(R_1+R_2)$ as an unconditionally precise value under load. See BJT Bias for the complete circuit and calculations.

Consider a separate 5 V supply educational amplifier, assuming an output range of 0.2–4.8 V is allowed under the specified load, with the quiescent output set to 2.5 V. There is 2.3 V peak margin above and below; if the gain under these conditions is 20, the input peak can be at most about 115 mV to avoid clipping solely due to swing limits.

If the input peak increases to 200 mV, the gain formula requires the output to swing ±4 V around 2.5 V, which already exceeds the assumed capability. Adjusting the bias can only redistribute the upper and lower margins; it cannot expand the total available range out of thin air. Even if the waveform is not clipped, there may still be device nonlinearity, noise, or frequency distortion.

The output range typically varies with load current and temperature. A 5 V supply does not mean the output can precisely reach 0 and 5 V; the input common-mode range is also an independent constraint, and one cannot assume the input operating point is valid just because there is room at the output.

How does bias redistribute headroom?

At 200 mV input, moving bias changes which side clips first but cannot fit the full target waveform. Biasing at 2.5 V balances headroom and maximizes unclipped amplitude for this symmetric signal.

Purpose and Limits of Configurations

"Common" refers to the AC reference terminal shared by the input and output. A terminal being common for AC does not necessarily mean it is directly grounded in DC.

$-g_m{R}_C$ for common-emitter applies only when the emitter can be considered an AC ground, relevant output resistances are ignored, and the load meets simplified conditions. When an un-bypassed $R_E$ is retained in the emitter, feedback reduces gain and increases input resistance; when there is an output load, it should also be included in the collector equivalent load. Drawing $R_E$ in the circuit but directly writing the fully bypassed formula will significantly overestimate the gain.

The low input impedance of common-base/common-gate can sometimes reduce the time constant caused by input capacitance; the capacitance across collector-base or drain-gate is no longer connected across high-gain input/output terminals in the same way, but other capacitances, output poles, and device transit effects remain.

Bandwidth and Miller Effect

The low-frequency end may be affected by coupling capacitors, emitter bypass, and bias networks; the high-frequency end may be affected by device parasitics, output loads, and feedback networks. A DC-coupled amplifier does not necessarily have a non-zero low cutoff frequency; bandpass systems often mark the two boundaries where the gain drops by 3 dB relative to the midband gain as $f_L$ and $f_H$, with bandwidth $f_H-f_L$.

The magnitude of a single first-order pole is $1/\sqrt{1+(f/f_p{)}^2}$. At $f=f_p$, it drops by approximately 3.01 dB, corresponding to a magnitude of 0.707, which does not mean the signal disappears. Multiple poles, zeros, or response peaking will alter the shape; one cannot draw all amplifiers as completely flat and then suddenly dropping like a rectangle.

If a capacitor $C$ is connected between input and output, and $v_{out}=A_v{v}_{in}$, the current flowing into the capacitor from the input is $C\cdot d(v_{in}-v_{out})/dt$. In the frequency band where $A_v$ can be approximated as a real constant, the input side is equivalent to $C(1-A_v)$; for an inverting gain of $-49$, a 2 pF shunt capacitance can present an input burden of approximately 100 pF.

If this 100 pF is driven by a simplified 10 kΩ source, the single-pole estimate is approximately 159 kHz. Here, the device's physical capacitance has not actually expanded 50 times; rather, the output's reverse change causes a larger voltage change across the shunt capacitor. When $A_v$ varies with frequency, the equivalent relationship is no longer a fixed capacitor. MIT: High-Frequency Circuits and Miller Effect

Feedback Is Not Automatically Stable

In the linear feedback model, the error signal is $e=x-\beta y$, and the forward gain is $y=Ae$. Substituting this yields the closed-loop gain $A_f=A/(1+A\beta )$. If the low-frequency $A\beta$ is a large positive real number, $A_f$ approaches $1/\beta$, but the deviation is still determined by approximately $1/(1+A\beta )$; one cannot say the gain is "completely determined by external components."

For example, if $A=100,000$ and $\beta =0.1$, the closed-loop gain is approximately 9.999; when $A$ drops to 100 at the corresponding frequency, the closed-loop gain is only about 9.091. Feedback resistor errors, input offset, noise, and source/load conditions can also affect the results. For disturbances and nonlinearities within the appropriate loop, a large loop gain can reduce their output impact, but it cannot eliminate errors outside the loop or clipping that has already hit the power supply limits.

Actual $A(j\omega )$ and $\beta (j\omega )$ have phase. As frequency rises, feedback that was originally intended to cancel out may tend toward regeneration due to delay, causing peaking, ringing, or instability. One should analyze loop gain, crossover, and stability, rather than just confirming that feedback is connected to the "negative" terminal. Analog Devices: Loop Gain and Stability

These trends depend on negative feedback being effective and the feedback network loading being properly handled. In a single-pole dominant voltage feedback model, gain and bandwidth often have an approximate trade-off, but one cannot generalize a constant gain-bandwidth product to arbitrary feedback structures. Analog Devices: Gain and Bandwidth

Limits Beyond Small-Signal Bandwidth

Slew Rate (SR) limits the speed of output voltage change. For a sine wave with peak $V_p$, the maximum slope is $2\pi f{V}_p$, so the required SR must be at least this value. If the output peak is 2 V and the frequency is 100 kHz, the requirement is approximately 1.257 V/μs; an amplifier with an SR of only 0.5 V/μs cannot output this waveform without distortion, even if its small-signal bandwidth is sufficient. This necessary condition is still not a complete guarantee of low THD. TI: THD+N and Slew Rate

THD is usually calculated as the square root of the sum of squares of the RMS values of each harmonic, divided by the RMS of the fundamental; THD+N also includes components such as noise within the measurement bandwidth. Noise is not deterministic harmonic; a smooth appearance on a single oscilloscope trace does not prove low distortion.

from math import log10, pi, sqrt, isclose
gain = 40000/(10000+40000)*100*4000/(1000+4000)
assert isclose(gain, 64)
assert isclose(.02*gain, 1.28)
assert isclose((4.8-2.5)/20, .115)
closed = lambda a,b: a/(1+a*b)
assert closed(100000,.1) > 9.99
assert 9.09 < closed(100,.1) < 9.1
miller = 2e-12*(1-(-49))
fp = 1/(2*pi*10000*miller)
sr_required = 2*pi*100000*2/1e6
assert 159000 < fp < 160000
assert sr_required > .5
print(f"Total loaded gain {gain:.1f}, {20*log10(gain):.2f} dB; Miller single-pole estimate {fp/1000:.2f} kHz")
print(f"Large-signal required slew rate {sr_required:.3f} V/μs")

The analysis sequence can follow a practical signal chain: first confirm DC and load, then verify small-signal gain and frequency response, and finally increase the signal to check swing, slew rate, and stability. The next article Operational Amplifier applies these conditions specifically to op-amp feedback circuits.