The differential amplifier
The differential pair does not saturate abruptly: it splits a fixed current between two sides. Being symmetric, it generates no second harmonic, and having a current tail, it ignores whatever arrives equally at both inputs.
01The differential pair
Two identical transistors with their emitters joined and fed by a current source \( I_{EE} \) form the most important circuit in analog electronics. The source fixes the sum of the two currents; all the input can do is split it between the two sides according to the voltage difference between the bases, \( v_{id} = v_{B1} - v_{B2} \). With the exponential relationship of each transistor, the split is exact:
That property makes it ideal as the input of an operational amplifier, a comparator or an instrument: it amplifies the signal of interest and rejects the noise or the DC that arrives equally at both inputs. It uses the current sources and active loads from the previous unit.
02Small signal: differential and common-mode gain
For small \( v_{id} \), the hyperbolic tangent is a straight line and each transistor has \( g_m = I_{EE}/(2 V_T) \). With collector resistors \( R_C \):
| Quantity | Differential output | Output from a single collector |
|---|---|---|
| Differential gain \( A_d \) | \( g_m R_C \) | \( g_m R_C / 2 \) |
| Common-mode gain \( A_{cm} \) | Zero if everything is symmetric | \( \approx -R_C / (2 R_{EE}) \) |
| Differential input resistance | \( 2\, r_\pi \) | |
\( R_{EE} \) is the output resistance of the tail current source. The ratio between differential gain and common-mode gain is the common-mode rejection ratio:
With a mirror as the active load, the output from a single collector recovers the full differential gain: \( A_d = g_m (r_{o2} \parallel r_{o4}) \), the input structure of almost all operational amplifiers.
03Large signal: the pair saturates gracefully
The hyperbolic tangent flattens out: with \( v_{id} = 2 V_T \approx 52 \) mV the pair already delivers 76 % of its maximum current, and with about 100 mV practically all the current flows through one side only. Unlike the common emitter, the pair neither clips abruptly nor saturates: it limits smoothly, always between \( \pm I_{EE} \).
Being an odd function, the hyperbolic tangent generates no even harmonics: the second harmonic of the common emitter disappears and the third remains:
The difference of collector currents as a function of the differential input voltage, calculated with the exponential of each transistor and an optional emitter resistor. The band marks the swing of the chosen signal; its gain and its distortion are calculated.
04Mismatch and offset voltage
If the two transistors are not identical, with zero input the currents do not split in half and the output is not zero. The differential voltage that must be applied to balance them is the offset voltage. A 1 % difference in the saturation current of the transistors produces \( V_T \ln 1.01 \approx 0.26 \) mV; a 1 % difference in the load resistors, the same again. That is why pairs are built with the transistors close together, in symmetric layouts, as you will see in the unit on integrated circuit design.
05The MOS differential pair
With MOS transistors the idea is the same, but the transfer characteristic follows the square law instead of the exponential. The linear range is no longer set by \( V_T \) but by the overdrive voltage: the pair stops splitting when \( |v_{id}| \) reaches \( \sqrt{2}\, V_{ov} \), a few hundred millivolts. It is more linear and has zero input current, at the cost of less gain and more offset.
06In the lab
With a pair of matched transistors, a 1 mA tail source and 4.7 kΩ resistors, sweep the differential voltage from −150 to +150 mV and measure the difference of collector voltages. Fit it with a hyperbolic tangent.
Apply the same signal to both bases and measure the output with a 10 kΩ tail resistor and then with a current source. Calculate the rejection in each case.
With the inputs grounded, measure the differential output voltage and refer it to the input. Compare loose transistors with a matched pair.
07Common mistakes
- Using a resistor as the tail. The common-mode rejection ends up poor.
- Exceeding the linear range. Without degeneration, beyond about 20 mV the distortion grows quickly.
- Building the pair from transistors of different batches. The offset grows.
- Confusing differential output with single-ended output. The gain is half with a single collector and resistors.
08Self-assessment
What is the \( g_m \) of each transistor with \( I_{EE} = 2 \) mA?
\( 1 \text{ mA} / 25.85 \text{ mV} \approx 38.7 \) mS.
Why does the differential pair not produce a second harmonic?
Because its transfer characteristic, a hyperbolic tangent, is an odd function.
What CMRR is obtained with \( g_m = 20 \) mS and a 500 kΩ tail?
\( 20 \times 10^{-3} \times 5 \times 10^5 = 10\,000 \), 80 dB.
What fraction of \( I_{EE} \) is steered with \( v_{id} = 2 V_T \)?
\( \tanh 1 \approx 0.76 \): the current difference is 76 % of \( I_{EE} \).
09Further reading
- Paul R. Gray, Paul J. Hurst, Stephen H. Lewis and Robert G. Meyer. Analysis and Design of Analog Integrated Circuits. 5th ed., Wiley, 2009. The bipolar and MOS differential pair, with the large-signal transfer characteristic and the offset.
- Adel S. Sedra and Kenneth C. Smith. Microelectronic Circuits. 7th ed., Oxford University Press, 2014. The differential amplifier with an active load, step by step.
- Barrie Gilbert. A Precise Four-Quadrant Multiplier with Subnanosecond Response. IEEE Journal of Solid-State Circuits, 1968. The differential pair taken to the extreme: the multiplier that uses its hyperbolic tangent on purpose.