The Transistor, A Semi-Conductor Triode

How to read the original

1. Use the publisher’s article-level record and PDF link, not the entire journal issue.
2. Identify the device’s three terminals, biasing, and reported amplification; do not substitute a modern MOSFET diagram for the historical device.
3. Use the independent idealized circuit exercise below to connect amplification with logic, while keeping its model separate from the original experiment.

A device can control energy without creating it

This short Bardeen and Brattain communication concerns the early semiconductor transistor, not a modern CMOS gate. Its paper-specific publisher record replaces an issue-wide link that made the assigned reading difficult to locate. The following explanation uses an explicitly invented small-signal model to teach why a three-terminal active device matters. Its numerical parameters are not measurements from the 1948 paper, and no modern transistor equation is being attributed to that experiment.

An amplifier uses a small input variation to control a larger output variation powered by an external supply. The output signal’s energy does not come from nothing. A passive resistor can change voltage and current relationships, but an active circuit can direct supply energy according to an input. Terminal count alone is insufficient to prove useful amplification; the operating point, load, transfer characteristic, and power balance matter.

Work through a controlled-current model

Imagine a device with an input voltage v and an output current i. Around a chosen bias point, approximate the change in current as delta_i=g_m*delta_v. Let g_m=0.002 amperes per volt and connect a 1,000-ohm load resistor to an ideal supply. A positive input change of 0.01 volt increases current by 0.00002 ampere, or 20 microamperes. The resistor’s voltage drop increases by 0.02 volt. If the output is measured at the device side of that resistor, the output voltage decreases by 0.02 volt.

The local voltage gain is therefore -g_m*R=-2. The minus sign records inversion. This calculation assumes the device remains in the region where the linear approximation applies and the supply leaves enough voltage headroom. Doubling the input indefinitely cannot yield an unbounded output from a finite supply. Near a rail, the model must change and the waveform may clip.

From analog response to digital restoration

Digital logic needs more than a nonzero gain. Choose two input voltage regions interpreted as low and high, then require the circuit to produce corresponding output regions with enough margin for the next gate. For an inverter, low input should produce a valid high output and high input a valid low output. Intermediate voltages may be deliberately unspecified. A transfer curve and its loading behavior determine whether cascaded stages restore distinctions or gradually lose them.

Suppose a teaching gate accepts low inputs below 0.3 volt and high inputs above 0.7 volt, while guaranteeing outputs below 0.1 or above 0.9 volt on those valid inputs. The nominal margins are 0.2 volt at each side. These invented thresholds illustrate a contract; they are not universal semiconductor constants. Connecting an excessive load may invalidate the output guarantee even if the unloaded curve looked excellent.

Time and fabrication re-enter the model

A gate must charge and discharge capacitance. A crude resistor-capacitor approximation gives a characteristic time R*C, connecting smaller loads or stronger drive to faster transitions. Real devices also have leakage, temperature dependence, noise, and manufacturing variation. Static truth tables omit these effects, while a small-signal gain calculation describes only a local operating region.

The conceptual chain is physical control, an electrical input/output contract, stable logic levels, and then composed circuits. The 1948 device is historically important within that chain, but it is not itself a complete explanation of integrated circuits or modern transistor fabrication. Read the experimental claim at its actual scale, then use separate models to understand the additional engineering required for reliable digital computation.

Reconstruction exercise

Use the invented controlled-current model with g_m=0.003 amperes per volt and a 2,000-ohm load. Calculate the local voltage gain and the output change for a positive 5-millivolt input perturbation. Explain where the output energy comes from and why the approximation cannot hold for arbitrarily large input swings. Then assess a gate that accepts high inputs above 0.8 volt but guarantees high outputs only above 0.7 volt. State why one such gate cannot be assumed to drive another reliably without additional information.

Worked solution

The local gain is -g_m*R=-6. A positive input perturbation of 0.005 volt produces delta_i=0.000015 ampere and a load-drop increase of 0.03 volt, so the measured output change is -30 millivolts under the stated circuit convention. The supply provides the energy directed by the device; gain does not imply energy creation. Finite supply rails and changes in device behavior limit the range where the linear approximation applies. In the gate chain, an output permitted to be 0.7 volt does not meet the next gate’s 0.8-volt high-input requirement. The interface lacks a guaranteed positive high-state margin. A successful unloaded example is insufficient; a valid cascade needs compatible output and input guarantees under the specified load, supply, temperature, and timing conditions. These conclusions concern the toy models, not numerical measurements from the historical article.

Test the limits and transfer the idea

1. Measurement and calculation · allow 45 minutes

Separate current gain, voltage gain, and power gain

Read the actual short letter through the publisher's PDF. It defines current amplification using a change in collector current with collector voltage held constant, and says the factor depends on bias. Use these invented measurement points at one fixed collector voltage: emitter currents 1,2,3 mA produce collector currents 1.5,3.0,4.6 mA.

Estimate incremental current gain in each adjacent interval. For a separate simplified amplifier, assume current gain magnitude 1.5, input resistance 200 ohms, and effective load resistance 5000 ohms. With a 2 mV RMS input, calculate signal currents, output voltage, voltage gain, power gain, and power gain in decibels. Explain why a large voltage gain does not equal the power gain, and why changing load invalidates a gain measurement unless the operating conditions are controlled.

Get a starting hint

Use differences for incremental gain. Signal powers are V_rms squared divided by resistance; the two resistances are unequal.

Compare your reasoning with the worked solution

The adjacent current slopes are (3.0-1.5)/(2-1)=1.5 and (4.6-3.0)/(3-2)=1.6. These finite differences approximate local derivatives; their difference shows that one global constant would conceal bias dependence. They are invented data, not values extracted from the historical graph.

In the separate toy amplifier, input current is 0.002/200=10 microamperes RMS. Output current magnitude is 15 microamperes RMS and load voltage is 15×10^-6×5000=0.075 V RMS. Voltage gain magnitude is 0.075/0.002=37.5. Input signal power is (0.002)²/200=20 nanowatts. Output signal power is (0.075)²/5000=1.125 microwatts. Power gain is 56.25, or 10log10(56.25), about 17.50 dB. Squaring voltage gain alone would give 1406.25, but the correct conversion includes Rin/Rload=200/5000.

The supply provides output energy; amplification does not violate conservation. The current slope measured at fixed collector voltage is not automatically the loaded circuit's gain, because a resistor load lets collector voltage vary with current. Bias, output impedance, frequency, and signal amplitude affect that transfer. The letter's report of power amplification and its current-gain definition should therefore be read as distinct measurements. The PDF also contains adjacent letters on the journal pages; evidence from those neighbors must not be silently attributed to this article.

Check your reasoning

• Computes both incremental slopes and all RMS quantities with units.
• Uses the resistance ratio and the correct 10log10 power convention.
• Distinguishes fixed-bias device measurements from loaded-circuit behavior.

2. Model derivation · allow 45 minutes

Find the stability condition hidden by static gain

The letter discusses a positive-feedback contribution and notes possible instability. Build an original scalar model rather than assigning modern device equations to the experiment. Let an emitter-current perturbation i obey taudi/dt=-(1-L)i+gv, where tau>0, input perturbation v is constant, g=0.002 A/V, and L=gaRf. Assume a=0.8 and Rf=300 ohms. The output-current perturbation is a*i.

Compute L, steady-state i for v=1 mV, and output current. Derive the condition for disturbances about equilibrium to decay. Find the critical Rf and explain what happens to the mathematical model as that value is approached or exceeded. Why does a formally large static gain fail to establish a usable amplifier?

Get a starting hint

Subtract the equilibrium equation. The remaining disturbance obeys a first-order homogeneous equation.

Compare your reasoning with the worked solution

The loop factor is L=0.002×0.8×300=0.48. At equilibrium, (1-L)i=g*v, so i=(0.002×0.001)/0.52, approximately 3.846 microamperes. The output-current perturbation is 0.8i, about 3.077 microamperes. Without the positive-feedback term, emitter perturbation would be 2 microamperes, so feedback increases the local response.

Write i=i_star+delta. Subtracting the equilibrium relation gives taud(delta)/dt=-(1-L)delta. Thus delta(t)=delta(0)exp[-(1-L)t/tau]. It decays only when L<1, remains constant when L=1 and the homogeneous model is used, and grows when L>1. The critical resistance is Rf=1/(ga)=625 ohms. With nonzero input at L=1 there is no finite static equilibrium in this linear model; the differential equation instead has a constant forcing term. Just below the threshold, both static gain g/(1-L) and settling time tau/(1-L) become large.

A real device cannot produce unbounded current or voltage. Nonlinear behavior, supply limits, heating, and circuit parasitics eventually invalidate the local model, and growing disturbances may lead to saturation or oscillatory behavior in a richer model. This single-state equation itself does not model oscillation. The paper motivates concern about feedback, while these invented dynamics demonstrate why local amplification, stability, and operating range require separate checks.

Check your reasoning

• Calculates loop factor, equilibrium, and output current correctly.
• Derives disturbance decay and the 625-ohm threshold.
• Explains the singular limit and avoids attributing oscillation to a one-state model.

Check your understanding

An amplifier delivers a larger signal power than its input signal supplies. What explains the energy?

• The semiconductor creates energy.
• An external supply contributes energy controlled by the input.
• The output resistor removes energy conservation.

Explanation

An active device controls energy drawn from the supply.

A gate guarantees VOH >= 0.7 V but the next requires VIH >= 0.8 V. Is a high-level connection guaranteed?

• No, a permitted output fails the receiving input specification.
• Yes, because both voltages are positive.
• Yes, because the gates are digital.

Explanation

Digital compatibility requires numerical interface guarantees, not merely labels.

Original paper and sources

John Bardeen and Walter H. Brattain. The Transistor, A Semi-Conductor Triode. Physical Review 74, 230–231, published 15 July 1948. DOI: 10.1103/PhysRev.74.230. The publisher PDF contains the full letter and portions of adjacent articles on the same journal pages.

Read the original paper →

Read the primary paper: American Physical Society

• The Transistor, A Semi-Conductor Triode: American Physical Society