How Does a Transistor Amplify? It's a Valve, Not a Source • The Science 360

Q: What is the gain of a common-emitter amplifier?

With the emitter bypassed, the voltage gain is about minus gm times RC, where gm equals the collector current divided by the thermal voltage of 25.7 mV at room temperature. That simplifies to the DC voltage across RC divided by 25.7 mV. With 6 V across RC the gain is about 230. Current gain beta does not appear in the formula.

Put a 1 mV signal into the base of the small amplifier in Figure 3 and a 220 mV copy comes out of the collector, upside down. Measured as power, the output is about 24,000 times stronger than the input. The transistor doing this is a 3-cent part with three legs and no battery inside, which raises the obvious question: where did the extra power come from?

Most explanations on the first page of search results say "a small base current controls a larger collector current," and stop. That is true as far as it goes, but it hides the two things that actually make an amplifier work: the power supply does all the lifting, and the control is really by voltage, through an exponential law that sets the gain without any reference to the transistor's current gain. This post builds an amplifier from those two ideas, with real numbers at every step.

Quick answerA transistor amplifies by acting as a valve on the power supply. A few millivolts on the base-emitter junction change the collector current exponentially, by a factor of 10 per 59 mV. A collector resistor turns that current change into a large voltage change. The voltage gain is about

g_m R_C = (I_C R_C)/25.7\ \mathrm{mV}

, typically 100 to 250.

Infographic: How does a transistor amplify? A three-legged transistor turns a small input wave on its base into a large, inverted wave at its collector. Its voltage gain at 1.1 mA is about 239: the DC voltage across the collector resistor divided by 25.7 mV, with no β in the formula. A line chart of collector current against base-emitter voltage shows a steep exponential, with the operating point at 1.095 mA and 0.644 V and a tangent slope of 42.6 mS. Every 59.2 mV more multiplies the current by 10. All 15.7 mW of power comes from the supply. Tiles: β of a BC547 spans 110 to 800, and distortion reaches 9.7% at a 10 mV input. — Tap the image to view it full size. View the accessible, editable SVG.

Does a transistor create energy?

No. A transistor is closer to a tap than a pump. The water pressure comes from the mains (the DC supply); the tap only decides how much flows. Turning a tap takes a few newtons of finger force, and the water it releases can carry far more power than your finger supplied. Nobody thinks the tap made the energy.

The electrical version is easy to check with the circuit in Figure 3, a textbook common-emitter stage on 12 V that we will design properly below. Here is the power budget with a 1 mV peak signal at the input:

DC power from the supply, with 1.095 mA collector current plus 0.217 mA through the bias divider:

\begin{aligned}P_{\mathrm{DC}} &= 12\ \mathrm{V} \times (1.095 + 0.217)\ \mathrm{mA} \\ &= \mathbf{15.7\ mW}\end{aligned}

of which: R_C 6.7 mW, transistor 5.2 mW, bias divider 2.6 mW, R_E 1.2 mW

Signal power in (2.8 kΩ input resistance) and out, power gain, and signal out as a share of supply power:

\begin{aligned}P_{\mathrm{in}} &= \frac{(1\ \mathrm{mV})^2}{2 \times 2.8\ \mathrm{k\Omega}} = \mathbf{0.18\ nW} \\ P_{\mathrm{out}} &= \frac{(220\ \mathrm{mV})^2}{2 \times 5.6\ \mathrm{k\Omega}} = \mathbf{4.3\ \mu W} \\ G_P &= \frac{4.3\ \mu\mathrm{W}}{0.18\ \mathrm{nW}} \approx 24{,}000\ (44\ \mathrm{dB}) \\ \frac{P_{\mathrm{out}}}{P_{\mathrm{DC}}} &= \frac{4.3\ \mu\mathrm{W}}{15.7\ \mathrm{mW}} = \mathbf{0.03\%}\end{aligned}

The signal is a ripple on a large, steady flow of supply current. The transistor's job is to make that ripple follow the input, and most of the supply power ends up as heat in the resistors and the transistor whether or not any signal is present. That is why a class A amplifier like this one is so inefficient, and why audio power amplifiers use other arrangements.

Log-scale horizontal bar chart. Signal in, 1 millivolt peak: 0.18 nanowatts. Signal out, 220 millivolts peak across RC: 4.3 microwatts. DC from the 12 volt supply: 15.7 milliwatts. Text below splits supply power into RC 6.7, transistor 5.2, bias divider 2.6 and RE 1.2 milliwatts and notes a power gain of about 24,000. — Figure 1. Power in, power out and power from the supply for the amplifier of Figure 3, on a log scale. The output signal is 24,000 times the input but a tiny fraction of the supply power.

Is a transistor current-controlled or voltage-controlled?

The page-one answer is "current-controlled": base current in, β times as much collector current out. It is a useful bookkeeping rule, and it is the right way to size a base resistor for a switch. But it is not the physics. In a bipolar transistor the collector current is set by the voltage across the base-emitter junction, through the same exponential law as a diode:

\begin{aligned}I_C &= I_S \exp(V_{BE}/V_T) \\ V_T &= kT/q \\ &= \frac{1.380649{\times}10^{-23}\ \mathrm{J/K} \times 298.15\ \mathrm{K}}{1.602176634{\times}10^{-19}\ \mathrm{C}} \\ &= \mathbf{25.69\ mV}\ \text{at } 25\ ^\circ\mathrm{C}\end{aligned}

Raising V_BE lowers the energy barrier at the emitter junction, and electrons pour from the emitter into the thin base. Almost all of them diffuse straight across it and are swept into the collector by the reverse-biased collector junction. A small fraction recombine in the base or are matched by holes flowing back into the emitter, and that leakage is the base current. Base current is the by-product, not the cause. The ratio $\beta = I_C/I_B$ just measures how leaky the base is, which is why it varies so much from part to part (onsemi lists the BC547 at 110 to 800).

The exponential is steep. One consequence is a number you can check on any bench:

Change in V_BE for a 10× change in I_C, and for a 2× change:

\begin{aligned}\Delta V_{BE}(10\times) &= V_T \ln 10 \\ &= 25.69\ \mathrm{mV} \times 2.3026 \\ &= \mathbf{59.2\ mV} \\ \Delta V_{BE}(2\times) &= V_T \ln 2 = 17.8\ \mathrm{mV}\end{aligned}

Collector current from 0 to 3 milliamps against base-emitter voltage from 0.58 to 0.70 volts, rising exponentially. An operating point at 0.644 volts and 1.10 milliamps is marked with a tangent line of slope 42.6 millisiemens. Purple dashed lines show that plus and minus 10 millivolts of base voltage move the current up 0.52 milliamps and down 0.35 milliamps. — Figure 2. Collector current against base-emitter voltage for a small NPN (Shockley law, I_S = 14.34 fA). The orange tangent is the transconductance at the operating point. A ±10 mV input moves the current by +0.52 / −0.35 mA: already visibly lopsided.

What is transconductance, and why is it IC/VT?

Amplifier designers care about how much the collector current changes per millivolt of input, the slope of Figure 2. That slope is called transconductance, g_m. Differentiate the exponential and something neat drops out:

\begin{aligned}g_m &= \frac{dI_C}{dV_{BE}} = \frac{I_S \exp(V_{BE}/V_T)}{V_T} \\ &= \mathbf{I_C / V_T}\end{aligned}

At I_C = 1.095 mA, with the equivalent emitter resistance r_e:

\begin{aligned}g_m &= 1.095\ \mathrm{mA} / 25.69\ \mathrm{mV} \\ &= \mathbf{42.6\ mA/V}\ (42.6\ \mathrm{mS}) \\ r_e &= 1/g_m = 23.5\ \Omega\end{aligned}

The slope depends only on the operating current and on temperature. Not on the transistor type, not on its β, not on its size. Every silicon bipolar transistor ever made, biased at 1 mA, has a transconductance of about 39 mS at room temperature. MIT's 6.012 course notes give the same relation, $g_m = I_C/(kT/q)$ , as the starting point of the small-signal model. (For comparison, a MOSFET in saturation has $g_m = 2I_D/V_{ov}$ , where V_ov is how far the gate is above threshold. At a typical 0.2 V overdrive that is 10 mS per mA, about a quarter of the bipolar figure. That is one reason bipolar transistors still turn up in low-noise and high-gain front ends.)

How does a current change become a voltage gain?

Put a resistor R_C between the supply and the collector, and take the output at the collector. Collector current flows through R_C, so the collector voltage is $V_{CC} - I_C R_C$ . A small input v_in changes I_C by $g_m v_{in}$ , which changes the collector voltage by $-g_m R_C v_{in}$ :

\begin{aligned}A_v &= v_{out}/v_{in} = -g_m R_C \\ &= -(I_C / V_T) R_C \\ &= \mathbf{-V_{RC} / V_T}\end{aligned}

That last form is the most useful sentence in small-signal design. The voltage gain of a common-emitter stage with its emitter grounded for signals is just the DC voltage across the collector resistor divided by 25.7 mV. A Walla Walla University lab handout asks students to compare exactly this, $A_v = -I_C R_C/V_T$ , with what they measure. The minus sign is the inversion: input up, current up, more drop across R_C, collector down.

Look at what is missing from the formula. β is not there. The current gain story says amplification comes from β, yet the voltage gain is the same for a transistor with β = 100 as for one with β = 400, provided both run at the same collector current. β matters only for the input resistance ( $r_\pi = \beta/g_m$ , 4.7 kΩ here) and for how hard the bias network has to work.

The formula also sets a ceiling. On a 12 V supply you cannot put much more than half the supply across R_C and still leave room for the collector to swing, so a single resistor-loaded stage tops out around $6\ \mathrm{V}/25.7\ \mathrm{mV} \approx 230$ . The transistor's own limit is set by its output resistance $r_o = V_A/I_C$ (V_A is the Early voltage): the "intrinsic gain" $g_m r_o = V_A/V_T$ , roughly $74\ \mathrm{V}/25.7\ \mathrm{mV} \approx 2{,}900$ for a 2N2222-class model. Op-amps reach gains of 100,000 or more by stacking stages and replacing resistors with current sources, then trade most of it back for accuracy, as described in our post on op-amp gain-bandwidth product.

Worked example: designing a common-emitter amplifier

Common-emitter amplifier schematic on a 12 volt rail. R1 47 kilohms and R2 8.2 kilohms form the base divider at 1.75 volts. Input through a 10 microfarad capacitor to the base of NPN Q1. RC 5.6 kilohms from 12 volts to the collector at 5.87 volts, output through a 10 microfarad capacitor. Emitter at 1.10 volts through RE 1 kilohm to ground, bypassed by a 100 microfarad capacitor. Box: IC 1.10 milliamps, gm 42.6 millisiemens. — Figure 3. The worked example: divider-biased common-emitter stage with a bypassed emitter resistor. Green tags are the DC node voltages from our calculation with β = 200.

Target: about 1 mA, the collector sitting near the middle of the supply, and a bias point that does not care much about β. Supply 12 V.

Step 1, emitter resistor. Put about 1 V across R_E to stabilise the current:

R_E = 1\ \mathrm{V} / 1\ \mathrm{mA} = \mathbf{1\ k\Omega}

Step 2, collector resistor. For V_C near 6 V, V_RC ≈ 6 V:

\begin{aligned}R_C &= 6\ \mathrm{V} / 1\ \mathrm{mA} = 6\ \mathrm{k\Omega} \\ &\to \mathbf{5.6\ k\Omega}\ (\text{E12})\end{aligned}

Step 3, base divider. The divider current should be at least 10× the base current (5 µA at β = 200).

V_B = V_E + V_{BE} \approx 1.0 + 0.65 = 1.65\ \mathrm{V}

With R1 = 47 kΩ and R2 = 8.2 kΩ, the open-circuit V_TH and R_TH are:

\begin{aligned}V_{TH} &= 12 \times 8.2/55.2 = 1.783\ \mathrm{V} \\ R_{TH} &= 47\mathrm{k} \parallel 8.2\mathrm{k} = 6.98\ \mathrm{k\Omega}\end{aligned}

Divider current: 0.217 mA.

Step 4, solve the loop (V_BE from the exponential, iterated):

\begin{aligned}I_B &= \frac{V_{TH} - V_{BE}}{R_{TH} + (\beta+1)R_E} \\ &= \frac{1.783 - 0.644}{6.98\mathrm{k} + 201\mathrm{k}} = 5.48\ \mu\mathrm{A} \\ I_C &= 200 \times 5.48\ \mu\mathrm{A} = \mathbf{1.095\ mA} \\ V_E &= 1.10\ \mathrm{V} \\ V_C &= 12 - 1.095 \times 5.6 = \mathbf{5.87\ V} \\ V_{CE} &= 4.77\ \mathrm{V}\end{aligned}

Step 5, small-signal gain.

\begin{aligned}g_m &= 42.6\ \mathrm{mS} \\ A_v &= -42.6\ \mathrm{mS} \times 5.6\ \mathrm{k\Omega} \\ &= \mathbf{-239} \\ \text{check: } V_{RC}/V_T &= 6.13\ \mathrm{V} / 25.69\ \mathrm{mV} \\ &= 239\end{aligned}

Including r_o in parallel with R_C, and the input resistance:

\begin{aligned}r_o &= 74\ \mathrm{V} / 1.095\ \mathrm{mA} = 67.6\ \mathrm{k\Omega} \\ A_v &= \mathbf{-220} \\ R_{TH} \parallel r_\pi &= 6.98\mathrm{k} \parallel 4.69\mathrm{k} \\ &= \mathbf{2.8\ k\Omega}\end{aligned}

The emitter resistor is there for stability. Run the same loop with β = 100 and with β = 400, a 4 to 1 spread typical of a reel of small-signal transistors:

\begin{aligned}\beta = 100:\ I_C &= 1.055\ \mathrm{mA} \\ V_C &= 6.09\ \mathrm{V},\ \text{gain } {-230} \\ \beta = 400:\ I_C &= 1.116\ \mathrm{mA} \\ V_C &= 5.75\ \mathrm{V},\ \text{gain } {-243}\end{aligned}

Same transistors with a single 2.07 MΩ base resistor and no R_E (fixed-bias, tuned for 1.1 mA at β = 200):

\begin{aligned}\beta = 100:\ I_C &= 0.55\ \mathrm{mA} \\ \beta = 400:\ I_C &= 2.19\ \mathrm{mA}\end{aligned}

At β = 400, V_C would be:

V_C = 12 - 12.3\ \mathrm{V}

Saturated, no amplifier.

That is the real reason for the slightly fussy four-resistor circuit: the emitter resistor makes the collector current depend on a voltage you set with resistors, not on β. The divider is the same kind of loaded divider discussed in why a voltage divider is a poor power supply, and the rule of "divider current at least 10× the load current" is doing the same job here.

The 100 µF bypass capacitor shorts R_E for signals, so the gain is set by $g_m R_C$ instead of the much smaller $R_C/R_E$ . Capacitors only short above some frequency, and the corner frequencies follow from the familiar $1/(2\pi RC)$ , as in our RC filter cutoff post:

Input coupling:

f = \frac{1}{2\pi \times 10\ \mu\mathrm{F} \times 2.8\ \mathrm{k\Omega}} = 5.7\ \mathrm{Hz}

Emitter bypass: resistance seen by C_E, then the corner:

\begin{aligned}R &\approx (r_e + R_{TH}/(\beta+1)) \parallel R_E \\ &= (23.5 + 34.7) \parallel 1000 = 55\ \Omega \\ f &= \frac{1}{2\pi \times 100\ \mu\mathrm{F} \times 55\ \Omega} = \mathbf{29\ Hz}\end{aligned}

This sets the low-frequency end of this amplifier.

What does the load line show?

Figure 4 plots the transistor's output characteristics, collector current against collector-emitter voltage at fixed base-emitter voltages, with the circuit's constraints drawn on top. The blue DC load line is every combination the resistors allow: 12 V shared between the transistor and $R_C + R_E = 6.6\ \mathrm{k\Omega}$ . The bias network picks the point Q on it.

Output characteristics of an NPN: seven nearly flat curves of collector current from 0.25 to 1.75 milliamps against collector-emitter voltage from 0 to 12 volts, each labelled with its base-emitter voltage from 604 to 655 millivolts, rising slightly with voltage. A blue DC load line runs from 1.82 milliamps at 0 volts to 12 volts at zero current. A steeper dashed orange AC load line passes through the quiescent point at 4.8 volts and 1.10 milliamps. A narrow purple band near 0 volts marks saturation. — Figure 4. Output characteristics from our 2N2222-class model, with the DC load line ( $R_C + R_E$ ) and the AC load line (R_C only, since C_E shorts R_E for signals). Note how a few millivolts of V_BE separate each curve.

Three things are visible at a glance. The curves are nearly flat, so the transistor behaves like a current source controlled by V_BE; their slight upward tilt is the Early effect that limits intrinsic gain. Adjacent curves, 0.25 mA apart, are separated by only 5 to 18 mV of V_BE, which is the transconductance seen from the other side. And the signal actually moves along the orange AC load line, whose slope is set by R_C alone. Along it, the collector can swing from 4.8 V down to roughly 0.3 V (about 4.5 V of room) before saturation, and up by $I_C R_C = 6.1\ \mathrm{V}$ before the transistor cuts off. The smaller of those, about 4.5 V peak, is the most this stage can deliver, and the next section shows why you rarely get near it.

Connect a load and the AC load line steepens further. A 10 kΩ load in parallel with R_C and r_o leaves 3.4 kΩ, and the gain falls from −220 to −145. This is why a high-gain stage is usually followed by a buffer.

Why does a transistor amplifier distort?

The gain formula assumed the curve in Figure 2 is a straight line over the input swing. It is an exponential, and the error grows fast. For an input of peak amplitude v_p, the collector current is $I_Q \exp(v_{in}/V_T)$ , and expanding that exponential gives a second harmonic whose size relative to the fundamental is about $v_p/(4V_T)$ :

\begin{aligned}\exp(x) &\approx 1 + x + x^2/2 \\ x &= (v_p/V_T) \sin \omega t \\ x^2/2 &\ \text{contains}\ \tfrac14 (v_p/V_T)^2 \cos 2\omega t \\ \mathrm{HD2} &\approx v_p / (4 V_T) \\ 1\ \mathrm{mV\ peak}&:\ 1/102.8 \approx \mathbf{1.0\%} \\ 10\ \mathrm{mV\ peak}&:\ 10/102.8 \approx \mathbf{9.7\%}\end{aligned}

Our FFT of the exact waveform at 10 mV peak: 9.7%.

Collector voltage change against time over two milliseconds for a 10 millivolt peak, 1 kilohertz input. The exact response of the bypassed stage, in blue, swings down 2.9 volts and up only 2.0 volts, while the dashed small-signal prediction is a symmetric 2.39 volt sine. A green curve for the unbypassed stage, multiplied by 40, is a clean 55 millivolt sine with 0.005 percent second harmonic. — Figure 5. A 10 mV input through the Figure 3 amplifier. With R_E bypassed the output is lopsided (−2.9 V, +2.0 V) and carries 9.7% second harmonic. Remove the bypass capacitor and the output shrinks to ±55 mV but is almost perfectly clean.

So the high-gain version of this amplifier is only a good amplifier for inputs of a millivolt or so. That suits a dynamic microphone or a magnetic pickup, and it is useless for a 1 V line signal, which would slam the collector from rail to rail.

The usual fix is emitter degeneration: leave some or all of R_E unbypassed. The input voltage now has to drive the junction and raise the emitter, so most of v_in appears across R_E rather than across the junction, and the junction sees only a fraction of a millivolt:

\begin{aligned}A_v &= -R_C / (r_e + R_E) \\ &= -5.6\ \mathrm{k\Omega} / (23.5 + 1000)\ \Omega \\ &= \mathbf{-5.5}\end{aligned}

Junction share of v_in: $23.5/1023.5 = 2.3\%$ , so 10 mV in puts only 0.23 mV across V_BE.

Result (Figure 5, green): ±55 mV out, second harmonic 0.005%.

You trade a factor of 44 in gain for a factor of almost 2,000 in distortion. Negative feedback in general works this way, and the gain now depends on a resistor ratio instead of on temperature and V_T. A compromise is to split R_E into a small unbypassed part and a bypassed part, setting the gain anywhere between the two extremes.

Source note The Early voltage (74 V) and saturation current (14.34 fA) used for r_o, Figure 2 and Figure 4 come from the commonly shared Q2N2222 SPICE model rather than a manufacturer file I opened; onsemi's 2N2222A and BC547 datasheets do not list an Early voltage. Real parts vary, so treat r_o, the −220 figure and the 2,900 intrinsic gain as illustrative. The formulas

g_m = I_C/V_T

and

A_v = -g_m R_C

do not depend on these parameters.

So is the “small current controls a big current” explanation wrong?

It is incomplete in a way that misleads. Three corrections, in order of how often they trip people up.

First, energy. The bigger output is paid for by the supply, every joule of it, and an amplifier with no supply amplifies nothing. Feed a sine wave into the base of an unpowered transistor and the "output" is at best a diode-clipped copy of the input.

Second, what controls what. Base current and collector current are both consequences of V_BE. Treating β as the gain mechanism suggests that a high-β transistor gives a high-gain amplifier, and it does not: at a fixed collector current, gain is $V_{RC}/V_T$ for any bipolar transistor. β only buys a higher input resistance.

Third, linearity. A transistor is a strongly nonlinear device that we coax into acting linearly by biasing it, keeping signals small, and wrapping it in feedback. The same exponential that gives 59 mV per decade gives 10% distortion at 10 mV. The same transistor driven hard is a switch, with all the gain-independence that implies, which is the subject of our base resistor calculation post. Noise-cancelling headphones, which have to amplify microphone signals cleanly before they can subtract anything, rely on exactly these trade-offs; see how noise-cancelling headphones work.

The one-line version: the supply provides the power, V_BE steers it with a slope of

g_m = I_C/25.7\ \mathrm{mV}

, and R_C converts current to voltage, so the gain is the DC voltage across R_C divided by 25.7 mV. Bias with an emitter resistor so β stops mattering, and keep the signal at the base below a few millivolts or degenerate the emitter.

One practical footnote: a stage with a gain of 200 will happily amplify ripple and noise on its own supply rail too. Local decoupling at the collector resistor's supply end matters, for the reasons in why 100 nF is the standard decoupling capacitor.

FAQ

Does a transistor create energy when it amplifies?

No. All the output power comes from the DC power supply. The transistor acts as a valve: a small input signal changes how much supply current flows through the load. In a typical small common-emitter stage the supply delivers about 16 mW and less than 0.03% of it leaves as signal.

Does a transistor amplify current or voltage?

Either, depending on the circuit. Physically, a bipolar transistor's collector current is set exponentially by the base-emitter voltage, and the base current is a small by-product. A collector resistor converts the current change into a voltage change, so a common-emitter stage gives both current gain and voltage gain.

What is the gain of a common-emitter amplifier?

With the emitter bypassed, the voltage gain is about $-g_m R_C$ , where g_m equals the collector current divided by the thermal voltage of 25.7 mV at room temperature. That simplifies to the DC voltage across R_C divided by 25.7 mV. With 6 V across R_C the gain is about 230. Current gain β does not appear in the formula.

Why is a transistor amplifier output inverted?

In a common-emitter stage the output is taken at the collector, below the collector resistor. When the input rises, collector current rises, the voltage drop across the resistor grows, and the collector voltage falls. So the output moves opposite to the input, a 180° phase shift.

Why does a transistor amplifier need biasing?

A silicon transistor conducts almost nothing until the base-emitter voltage reaches about 0.6 V, and the exponential relation is only close to straight over a few millivolts. Biasing sets a steady operating current and collector voltage so that the signal rides on a nearly linear part of the curve with room to swing both ways.

Sources

NIST CODATA, Boltzmann constant $k = 1.380649 \times 10^{-23}\ \mathrm{J/K}$ (exact): physics.nist.gov/cgi-bin/cuu/Value?k; elementary charge $e = 1.602176634 \times 10^{-19}\ \mathrm{C}$ (exact): physics.nist.gov/cgi-bin/cuu/Value?e
MIT OpenCourseWare 6.012 Microelectronic Devices and Circuits, Spring 2009, Recitation 19 ( $g_m = I_C/(kT/q)$ , small-signal model, loaded transconductance gain): ocw.mit.edu 6.012 rec19 PDF
Walla Walla University ENGR 356 Lab 8 ( $g_m = I_C/V_t$ , compare measured gain with $A_v = -I_C R_C/V_t$ ): gab.wallawalla.edu lab8 PDF
onsemi, BC546/BC547/BC548 datasheet, Rev. 8 (hFE 110 to 800 at 2 mA): onsemi.com/pdf/datasheet/bc546-d.pdf
onsemi, P2N2222A datasheet, Rev. 7 (hFE ranges, no Early voltage listed): onsemi.com/pdf/datasheet/p2n2222a-d.pdf
Typical page-one explanation ("small current controls a larger current"): All About Circuits forum, "How does a transistor amplify voltage?": forum.allaboutcircuits.com

Method note: all numbers for the Figure 3 circuit, Figures 1, 2, 4 and 5 and the β sweep are our own calculations in Python from the Shockley law, a simplified Gummel-Poon model and the formulas shown, at 25 °C with ideal capacitors.

Header image: NASA (image jsc2025e015685), regraded for this article.