The follower that lets go on the way down

Four essays have taken the emitter follower apart as a small-signal circuit. The buffer that is not a buffer found its output impedance inductive and peaking in the middle of its range; the input that pushes back found a negative resistance at its base when its emitter is loaded by a capacitor; the resistance that is below zero found the same at its output; and what the cure at the base costs priced the repairs. Every one of them linearised the transistor about a fixed operating point, as the small-signal model does, and so every one of them assumed the transistor was conducting.

A follower driving a capacitor does not always conduct. Its transistor can source current into the load as hard as its junction’s exponential allows — a few tens of millivolts more on the base doubles the current — but it cannot sink any: when the input falls, the only thing that can take charge off the load is whatever pulls the emitter down, a current source or a resistor. If the input falls faster than that can discharge the capacitor, the base falls below the emitter, the junction turns off, and the output stops following and simply falls at the rate the pull-down allows. The follower has two slew rates, one for each direction, and only one of them is its own.

The circuit, marched #

The model is the smallest one that has the asymmetry. The base is driven by a stiff source at the signal voltage; the emitter current is the junction’s $I_s\exp((v_{in} - v_{out})/V_T)$, with the base current ignored; the emitter carries a 1 nF load to ground and a 2 mA current sink. The capacitor’s charge balance is then one equation,

$$C\,\frac{dv_{out}}{dt} = I_s\,e^{(v_{in} - v_{out})/V_T} - I,$$

and it is marched by backward Euler: at each step the new output is bisected so that the capacitor’s current, the junction’s current and the sink’s balance, which is robust however stiff the exponential becomes. The input is a sine on a 2 V bias; four cycles are marched and the last is drawn, so the start-up has gone.

On the rising half the junction does what the small-signal model says. The output sits a base–emitter voltage below the input and rises with it; to charge the capacitor at the input’s slope the transistor carries the sink’s 2 mA plus $C\,dv/dt$, and a few tens of millivolts of extra drive provide it. On the falling half, the capacitor must lose charge at $C\,dv/dt$, and the only path is the sink, which takes exactly 2 mA. When $C\,|dv_{in}/dt|$ exceeds that, there is nothing for the transistor to do: its current would have to be negative, and a junction cannot conduct backwards, so it turns off. From then on the output falls at $I/C$ — exactly, the march gives −2.000 V/µs — while the input falls faster and opens a gap below it. The gap closes when the input turns and comes back up through the output, and the transistor turns on again.

A slope, not a bandwidth #

The condition for the transistor to stay on is that the input’s steepest fall be no faster than the sink can discharge the load: for a sine of amplitude $A$ and frequency $f$,

$$2\pi f A \le \frac{I}{C}.$$

The marched boundary — the largest amplitude at which the emitter current never falls below a tenth of its quiescent value — follows one over frequency: 2.87 V at 100 kHz, 0.592 at 500 kHz, 0.309 at 1 MHz. At 100 kHz it sits at 0.903 of $I/(2\pi fC)$, which is the 0.9 a tenth of reserve current gives: the current reaches a tenth when $C\,dv/dt$ is nine tenths of the sink. At higher frequencies it creeps above the line — 1.06 at 2 MHz, 1.34 at 5 MHz — because as the current falls the junction’s own logarithm lowers its base–emitter voltage, which lets the output fall a little behind the input before the current has reached its tenth. The time constant of that lag is $CV_T/I = 12.9$ ns, and once the period is only a few dozen of it the follower can no longer be described by a slope alone.

At low frequency, though, the boundary is a slope and only a slope. It is not a bandwidth — the follower’s small-signal bandwidth into 1 nF is tens of megahertz, far above anything on this figure — and it is not visible to any model that linearises the transistor. How small is small signal measured where a junction’s exponential stops being its tangent, at 7.3 mV; here the departure is not a gradual curvature but a switch, the junction going from conducting to off, and it happens at whatever amplitude makes the slope too steep.

Off, then all at once #

The emitter current over a cycle makes the mechanism plain. For 57 per cent of the cycle at 1 V and 1 MHz the transistor carries less than a hundredth of its quiescent current: it is off, and the sink alone is discharging the load. When the input comes back up to meet the output, the transistor turns on hard — up to 8.3 mA, four times the sink — because now it has to recharge the capacitor at the input’s rising slope and supply the sink as well. Averaged over the cycle, the emitter supplies 2.003 mA, the sink’s current, as it must: the capacitor ends the cycle with the charge it began with, so whatever the sink took out, the transistor put back.

The current pulse is worth reading for its size as well as its shape. A capacitor that the sink discharged at 2 V/µs for over half a cycle must be recharged in the rest of it, and the input’s rising slope near the end of the gap is close to its steepest, 6.28 V/µs at 1 V and 1 MHz, so the transistor must supply roughly $C$ times that plus the sink — 8.3 mA by the march, and more at larger drives. A follower whose quiescent current was chosen for a comfortable dissipation thus delivers peaks several times larger, and its supply decoupling, its emitter’s wiring and its thermal design have to be sized for the peak, not the quiescent value.

That current waveform is a large-signal distortion of the worst kind. A follower is usually inserted to isolate a source from a load, and a source driving its base now sees a base current that switches between almost nothing and a pulse of four times the quiescent value, a load that changes with the signal. The negative input resistance the input that pushes back found is a small-signal property of the conducting follower; the switching shown here is what the same circuit does when the signal is no longer small.

The follower rectifies #

Past the boundary the asymmetry has a direct-current consequence, which is the one most likely to be mistaken for something else.

The output keeps up with every rise and lags every fall, so it spends more of each cycle near the top of the swing than near the bottom, and its average climbs. Below the boundary the shift is a few millivolts — the junction’s own even-order term, which what a pair cancels found a single transistor keeping and a pair removing. Above it the shift grows rapidly: 68 mV at 0.5 V of drive, 415 mV at 1 V, 1.30 V at 2 V, where the output barely falls at all before the input comes back. The follower has turned a sine into a direct-current offset proportional to how far past the boundary it is driven — a peak detector, which is in fact how a peak detector is built: a follower, a capacitor and a weak discharge.

A direct-coupled stage downstream reads the shift as an offset that depends on the signal’s amplitude and frequency, which is exactly the kind of error that survives a calibration made with a small or slow test signal. The step that is too big found an amplifier’s linear model failing abruptly once a step exceeded a slew-limited size; the follower fails in the same way and in one direction only, which is what turns the failure into a level shift rather than a distortion symmetric about the mean.

A resistor is a slew rate that varies #

Many followers are pulled down by a resistor to a negative rail rather than by a current source. With the transistor off, the resistor discharges the load at $(v + V_{EE})/(RC)$ — a slope proportional to how far the output is above the negative rail, so it slows as the output falls. Sized to give the same 2 mA at the operating point, a 3.134 kΩ resistor to −5 V discharges the load at 2.41 V/µs near the top of a 1.5 V swing, where the transistor switches off, and at 1.99 V/µs near the bottom; the march’s off-state slope matches $(v + V_{EE})/(RC)$ to three per cent at every point.

So a resistor-loaded follower’s slew boundary is set at the bottom of its swing, where its pull-down is weakest, not at its operating point. A designer who sizes the resistor for the quiescent current and computes $I/C$ from it has computed the boundary at the wrong voltage, optimistically, by the ratio of the rail-to-operating-point distance to the rail-to-minimum distance. A larger negative rail reduces the variation, and a current source removes it, which is one reason current sinks are used where the swing is large.

Edges, as well as sines #

The sine makes the boundary easy to state, but the asymmetry is starkest on edges. A falling step of a volt at the input leaves the output to fall the whole volt at $I/C$, which for 2 mA into 1 nF takes 500 ns, the transistor off throughout. A rising step of a volt meets a transistor that will supply whatever current the capacitor asks for, limited only by the junction’s exponential and whatever resistance is in series with the base, and the output follows within a few multiples of $CV_T/I$, the 12.9 ns time constant that appeared above — an order of magnitude faster. The same follower, the same load, the same quiescent current, and a pair of edges whose durations differ by more than an order of magnitude.

How long a large step takes to settle measured an amplifier’s large-step settling as a slew-limited ramp followed by a linear tail. The follower’s falling edge is the same shape, and its rising edge has no ramp at all; a measurement of settling time on a follower is therefore two measurements, and quoting one of them — usually the rising edge, because it is the one a pulse generator’s positive edge exercises first — understates the other by the ratio of the transistor’s available current to the sink’s.

The asymmetry also explains a familiar bench symptom. A follower driven by a symmetric triangle wave whose slopes exceed $I/C$ produces a sawtooth: the rising ramps follow the input and the falling ramps are replaced by the sink’s slower slope, so the output’s peaks arrive late and its troughs never reach the input’s. Viewed on an oscilloscope at low sweep speed it looks like a frequency-dependent gain loss with a direct-current shift, and it is often blamed on the bandwidth of the stage, which is tens of megahertz and has nothing to do with it. The test that distinguishes the two is to invert the drive: a bandwidth limit treats rising and falling edges alike, and this mechanism does not. A buffer data sheet that quotes a single slew rate for a single-ended output stage is quoting one of the two, usually the faster.

Sizing the sink #

Read as a design rule, the boundary is a requirement on the pull-down: to carry a sine of amplitude $A$ at frequency $f$ into a capacitance $C$ without the transistor turning off, the sink must carry at least $2\pi fAC$, and a margin for the junction to stay well on — a tenth of the sink in reserve needs ten ninths of that. A 1 V, 1 MHz signal into 1 nF needs 6.3 mA of sink, three times the 2 mA of this essay’s examples; into 10 nF, 63 mA, which is a power stage rather than a buffer. The cost is quiescent power, dissipated whether the signal is there or not, and it is the reason a single-ended follower is a poor way to drive a large capacitance at speed — the capacitance of a long cable, a converter’s input, the gate of a power transistor. The load that gets inside the loop found a capacitive load costing an amplifier its phase margin; the follower’s version of the same load costs it quiescent current, in proportion to the slew it is asked for.

Where the small-signal essays sit #

None of this overturns the earlier essays, and it says where they stop. The output impedance, the negative input resistance and the oscillation conditions were all computed for a conducting transistor, and they hold as long as the signal is small enough that the transistor never turns off — below the slope boundary, which for 2 mA into 1 nF is a third of a volt at a megahertz. A follower that oscillates, as the earlier essays found possible with a capacitive load, may well drive itself past that boundary on its own, and then the oscillation’s amplitude is set not by any small-signal gain but by the point at which the transistor starts switching off each cycle.

The large-signal follower also has a capacitive-load story of its own that the feedback field met in an amplifier. The step too large to have an impedance found an amplifier driving a capacitor through a current limit, where no impedance is defined for the duration of the slew. The follower’s falling edge is that regime exactly, with the limit set by the sink rather than by an output stage, and only in one direction.

The numbers #

A follower with a 2 mA sink and 1 nF on its emitter can fall no faster than I/C = 2.00 V/µs; a 1 V, 1 MHz sine asks for 6.28, so the transistor is off for 57 per cent of each cycle and the output falls as a straight line of −2.000 V/µs, while on the rise the transistor peaks at 8.3 mA. The amplitude at which the current stays above a tenth of the sink’s is 2.87 V at 100 kHz and 0.309 V at 1 MHz — 0.903 of I/(2πfC) at low frequency, more as the period approaches the junction’s 12.9 ns. Past the boundary the output’s average rises by 415 mV at 1 V of drive and 1.30 V at 2 V. Pulled down by a resistor to −5 V, the off-state fall runs from 2.41 V/µs at the top of the swing to 1.99 at the bottom.

Still open: a push–pull stage, a source that is not stiff, and the oscillation that switches off #

A complementary pair. A push–pull follower adds a second transistor to pull as the first pushes, which removes the asymmetry and replaces it with a crossover region where neither conducts. Whether the crossover distortion of a complementary pair driving a capacitor is smaller than a single follower’s slew distortion at the same quiescent current, as a function of the drive, is the comparison that decides when the second transistor is worth its bias network.

A source with resistance. The base here is driven stiffly. With a source resistance, the base current the switching transistor draws — near nothing, then a pulse — flows through it, and the base voltage dips exactly when the follower needs it most. Whether the boundary moves in, and by how much per kilohm of source, is a march with one more node.

An oscillation that limits itself. The resistance that is below zero found a band of load capacitance in which a follower oscillates. A small-signal model cannot say how large the oscillation grows. Marched as here, with the inductance and capacitance that made it unstable, the amplitude at which the transistor begins to switch off each cycle should be the amplitude the oscillation settles at, and it would be a number in the same units as this essay’s boundary.