Design Article
Low-distortion discrete buffer amplifier handles bipolar signals
Peter Demchenko, Vilnius, Lithuania
11/30/2012 2:06 PM EST
Sometimes, the need arises for a low-distortion buffer amplifier capable of handling bipolar signals. You can use an op amp or integrated buffer for these applications, but for more flexibility, a discrete design may prove useful. Applications include buffering the input of an ADC or the output of a DAC, or an audio line driver.
The buffer in Figure 1 provides unity gain, low output impedance, and low distortion. It uses two emitter followers configured as symmetrical class-A amplifiers; current sources replace the usual emitter resistors (Figure 2). To obtain the best results, you should use complementary transistors (Q1 and Q2) with closely matched dc gain (beta).

| Figure 1 The buffer provides unity gain, low output impedance, and low distortion; see Figure 2 for details of the current sources in the transistors’ emitters. | Figure 2 Shown are details of the current sources used in the emitters of the transistors in Figure 1. |
This topology has advantages over a conventional emitter follower. It produces a lower level of even-order harmonics and lower noise, it can provide low IBIAS and VBIAS at the input and low offset voltage at the output, and it exhibits a high power-supply rejection ratio. The circuit doesn’t require temperature compensation and is dc stable. Like conventional voltage followers, it has local feedback only. This setup is advantageous in some applications where a long feedback loop can introduce additional distortions or instability.
Resistors R1 and R2 sum the two outputs. For even harmonics cancellation, their values should be matched. Preferred devices—metal film/foil, for example—should be stable and linear, and should produce low noise.
The voltage drop across R1 is equal to the base-emitter voltage, VBE, of Q1; thus, R1=K×VBE/I1, where K is in the range of 3 to 20.
R2 is set equal to R1. The same resistors also provide stability when driving a capacitive load, so the value of K depends on this capacitance. For the ac-equivalent circuit, these resistors appear to be connected in parallel, thus providing low output impedance. Diode D1 protects the emitter junctions of both transistors from excess input voltages. When the buffer is used as an output stage, you can eliminate D1.
The dc gains of the two transistors usually are not perfectly matched, resulting in a slight output offset voltage. To compensate, note the addition of base-emitter resistors R5A and R5B in Figure 1. To reduce the output offset voltage to almost zero, you can add R5A or R5B, not both. As an example, assume that β2>β1; R5B is then used at Q2. If Q1 has the higher beta, R5A would be used at Q1. You can estimate R5’s value from the following equation: R5=β1×β2×VBE/ (I1×(β2−β1)), where β1 and β2 refer to the beta of Q1 and Q2.
When the output is balanced with the help of R5, input bias current is also minimized because the currents I3 and I4 cancel out each other.
The circuit shown in Figure 3 is a version of the circuit shown in Figure 1 that will automatically servo the output to a voltage close to zero. The integrator, IC1, averages the output voltage but does not pass the ac signal, because it is acting like a high-pass filter; its corner frequency, fC, can be calculated from this equation: fC=1/(2×π×R3×C3). In this circuit, fC is approximately 1.6 Hz.

Figure 3 This circuit is a version of the circuit shown in Figure 1 that will automatically servo the output to a voltage close to zero.
The output of the integrator drives an optocoupler that uses a photoresistive element on the output side. This resistor replaces the upper and lower R5 resistors. The circuit in Figure 3 provides an output offset voltage of almost zero even with an input offset voltage applied, as long as it isn’t too high. The op amp, IC1, should have low noise, low bias current, and low offset voltage; also, resistor R3 and capacitor C3 should be high-quality, stable devices.
One of the optocouplers in Figure 3 will always be inactive, but unless you know in advance which of the two beta values is higher, you won’t know which optocoupler is not active. High-quality photoresistor optocouplers can be rather expensive, so if you know the transistors’ beta values, you can replace one device with a diode, D2, as shown in Figure 4. In this version, β2>β1, so the photoresistor shunts Q2. R4 also can be omitted if the optocouplers’ LEDs can tolerate the maximum output current from the integrator.

Figure 4 If you know which beta value is higher, you can replace one device with a diode, D2.
Incidentally, an optocoupler with an
incandescent (filament type) lamp can
be used; in this case, the integrator is not
needed, because the filament acts as an
integrator. Change the integration capacitor
to 1M and the input resistor value to
1k (Figure 5). The last circuit has low dc
gain (compared with the integrator), so
the output dc offset can be rather high—tens of millivolts. Diode D2 prevents possible
“latching” of the circuit.

Figure 5 An optocoupler with an incandescent lamp can be used; in this case, the integrator is not needed, because the filament acts as an integrator.


Walt Morrey
12/4/2012 2:27 AM EST
Cute idea. Have you actually built it?
Initial problem is that the output impedance cannot be low without wasting a tremendous amount of power.
R1 has around 0.6V - base-emitter voltage of Q1. The current source supplies 3mA. Q1 needs to have some of that current to keep the distortion low - say 1.5mA. That requires R1 to be 400 Ohms. With R2 also 400 Ohms, the lowest output impedance of the buffer is 200 Ohms. For many things, that is just fine. Additional effects - there will be some distortion due to modulation of the base-emitter resistances of Q1 & Q2. You can correct the distortion by taking 2 more high-value, equal resistors from the emitters of Q1 & Q2 and using the junction for negative feedback to the opamp driving the buffer. You then do have a nice buffer with 200 Ohm output impedance and a lot of complexity.
A low-distortion, rail-to-rail, low-output-impedance buffer is the holy grail of designs, and unless we get some new devices to play with, the circuitry is likely to be complex - which is why rail-to-rail opamps were developed. Unfortunately these parts are not available at a reasonable price everywhere in the world -- yet.
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Nanoman2012
12/5/2012 1:06 PM EST
You can make an open loop near rail to rail buffer but it's difficult to prevent offset changes as the input approaches the rail (current sources will turn off). A simple thing to do would be to swap R1 and R2 for diode strapped transistors (low offset), for Q1 add another pnp Q3 and connect their emitters. Connect Q3's base to say 1V below the positive rail and its collector to the output. As the input voltage approaches the rail and Q1 turns off the current will divert through Q3 and keep alive Q2 and its diode. Apply the same idea to Q2.
The drawbacks are the keep alive current will be twice the value needed causing distortion and the input current will change with input voltage.
You can improve performance with scaled current mirrors but complexity increases quickly and it's still class A.
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Peter Demchenko
12/8/2012 11:51 AM EST
The circuit is no way rail2rail; my approach was also rather minimalistic - but without sacrificing THD level.
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Nanoman2012
12/10/2012 6:54 AM EST
If the current source bias is replaced by a current mirror (no degeneration) the output will swing within two Vce(sat) from the rail. Looks ok in Simetrix.
Image here:
http://img233.imageshack.us/img233/5530/r2rbuffer.jpg
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Peter Demchenko
12/8/2012 11:44 AM EST
Yes, I've built it:) The load was not so small - more than 10kOhm, but variable. One of the goals was to eliminate this.
So the circuit as it is - isn't for heavy loads and if it's the case it should be redesigned.
Peter
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Walt Morrey
12/4/2012 2:38 AM EST
One more limit to the circuit, if the output current exceeds 3mA, Q1 or Q2 will be completely cut-off, and you have a big increase in output impedance -- to 400 Ohms.
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Peter Demchenko
12/8/2012 11:58 AM EST
This was not a problem, the load was light. But you surely can redesign the circuit increasing currents within some limitations.
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jan m didden
1/7/2013 1:40 PM EST
Peter, interesting design. I wonder why you decided to use the opto couplers?
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Peter Demchenko
1/9/2013 2:54 AM EST
You see, optoresitors can provide lowest THD level, so because my app required low output offset I used them. Surely, it would be better without them..
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Guru of Grounding
1/7/2013 2:40 PM EST
If low THD is this circuit's "claim to fame", why don't you include some typical measured data?
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