Current sources and active loads
Replacing a collector resistor with a transistor takes the gain of a stage from a few hundred to almost two thousand. It is the trick that makes an operational amplifier possible.
01Transistors instead of resistors
In an integrated circuit, resistors are expensive —they take up a lot of area—, imprecise —they vary by 20 % from one wafer to another— and cannot be very large. Transistors, on the other hand, are cheap and, most importantly, identical to each other: two neighboring transistors on the same wafer have almost identical parameters. This gives rise to the technique that distinguishes integrated analog design: using transistors as current sources for biasing, and as loads for amplifiers.
An ideal current source delivers the same current at any voltage, that is, it has infinite output resistance. Real ones are described by two numbers: their output resistance \( R_{out} \) and their useful voltage range (the minimum voltage at which they still work, called the compliance voltage).
02The current mirror
Two identical transistors with the same base-emitter voltage conduct the same current. If one of them is connected as a diode (base tied to collector) and a reference current is forced through it, the other one copies it. In the bipolar:
In MOS there is no gate current and the current ratio is set by the geometry: \( I_{out} = I_{ref}\,(W/L)_2 / (W/L)_1 \), with the same error from channel-length modulation. With several output transistors connected to the same diode, a single reference biases an entire chip.
03The Widlar current source
To obtain small currents without enormous resistors, Bob Widlar added a resistor in the emitter of the output transistor. The two base-emitter voltages are no longer equal, and their difference drops across that resistor:
The emitter resistor also provides feedback and raises the output resistance to approximately \( r_o (1 + g_m R_E) \). Widlar used this idea to design the first successful integrated operational amplifiers, such as the μA702 and the LM101.
04The cascode
Stacking a second transistor on the output, in common base, leaves the bottom transistor with its collector voltage fixed and the Early effect almost disappears. The output resistance is multiplied:
| Source | Output resistance | Minimum operating voltage |
|---|---|---|
| Simple mirror | \( r_o \) | \( V_{CE(sat)} \approx 0.2 \) V |
| Widlar | \( \approx r_o (1 + g_m R_E) \) | \( I_{out} R_E + V_{CE(sat)} \) |
| Bipolar cascode | \( \approx \beta\, r_o / 2 \) | \( V_{BE} + V_{CE(sat)} \approx 0.9 \) V |
| MOS cascode | \( \approx g_m r_o^2 \) | \( V_{th} + 2 V_{ov} \) |
The price of the cascode is range: it consumes more voltage, a serious problem when the supply is 1 V. That is why modern design looks for low-voltage cascode configurations.
The output current of three bipolar sources as a function of the voltage across their output, with the same reference. The chosen curve is highlighted; the others stay in the background. An ideal source would be a horizontal line.
05The active load
In a common emitter with a collector resistor, the gain \( g_m R_C = I_C R_C / V_T \) is limited by the supply: \( I_C R_C \) cannot exceed a few volts, and the gain ends up at a few hundred. If the resistor is replaced by a current source —a mirror PNP transistor—, the load seen by the collector is its output resistance, which is enormous, while the DC drop is small. The gain becomes the intrinsic gain of the transistors:
The active load is the reason an integrated operational amplifier reaches gains of one hundred thousand with two or three stages, and it appears again in the differential amplifier.
06In the lab
Build a mirror with two BC548s and a 1 mA reference. Measure the output current for voltages from 1 to 10 V and estimate the output resistance. Repeat with the two transistors bonded together with thermal grease to equalize their temperature.
Add to the mirror an emitter resistor calculated for 50 µA and verify the current and the increase in output resistance.
Repeat the mirror with transistors from an integrated array (such as the CA3096 or a matched pair) and compare the error with that of the loose transistors.
07Common mistakes
- Building mirrors with loose transistors and expecting accuracy. The copy depends on matching and temperature.
- Forgetting the minimum voltage. Below it, the output transistor saturates and the source stops being a source.
- Ignoring the base currents. The \( 2/\beta \) error accumulates if many outputs are copied from the same diode.
- Measuring the gain with an active load using a 1 MΩ probe. The probe loading is lower than the output resistance and the measured gain drops.
08Self-assessment
What error does a simple mirror with β = 50 have?
\( 1/(1 + 2/50) = 0.96 \): the copy comes out 4 % low.
What emitter resistance does a Widlar source need to go from 1 mA to 100 µA?
\( 25.85 \text{ mV} \times \ln 10 / 100 \text{ µA} \approx 595 \) Ω.
What is the output resistance of a simple mirror at 100 µA with \( V_A = 80 \) V?
\( 80 / 100\text{ µA} = 800 \) kΩ.
Why does an active load give more gain than a resistor?
Because it presents a very high small-signal resistance with a small DC drop.
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. Chapter 4 is the reference on mirrors, Widlar sources and cascodes, in bipolar and MOS.
- Behzad Razavi. Design of Analog CMOS Integrated Circuits. 2nd ed., McGraw-Hill, 2017. Mirrors and active loads in CMOS, with low-voltage cascodes.
- Robert J. Widlar. Some Circuit Design Techniques for Linear Integrated Circuits. IEEE Transactions on Circuit Theory, 1965. The paper in which the source that bears his name appears.