Technical Article

How to Buffer an Op-Amp Output for Higher Current, Part 2

March 17, 2016 by Robert Keim

In this article, we will look at two variations on the basic BJT current-buffering circuit.

In this article, we will look at two variations on the basic BJT current-buffering circuit.

Supporting Information

Previous Article

If Your Op-Amp Output Stage Can’t Handle the Pressure

The basic BJT buffering circuit discussed in the previous article is great for many applications, but it suffers from two limitations that need to be addressed: first, high load currents may require too much output current from the op-amp; second, it is not compatible with negative load voltages. We’ll start with the first concern.

As mentioned in the previous article, the output current required from the op-amp will be approximately equal to the load current divided by the transistor’s active-region current gain (aka beta or hFE). In some situations, it might be problematic to incorporate this much output-current capacity into your design. One possibility is the following: You are using an op-amp part that includes multiple amplifiers in one package. If you already have a low-output-current part that is just right for your system and you are using three out of the four amplifiers in the package, you might be determined to put that fourth amplifier to work. Or let’s say you have an op-amp integrated into the board’s microcontroller. That op-amp is not likely to offer much in the way of output current, but you don’t want to bring in an external part just because you need another 20 or 30 mA from the integrated op-amp. The solution to situations such as these is additional current gain from the BJT. The first thing to do is simply look for a transistor with higher hFE, but if you need much more current gain—because your load current is so high or your op-amp is so puny, or both—it’s time for Darlington.

The Darlington Pair


The circuit symbol tells most of the story. A Darlington pair is two BJTs sharing a common collector, integrated into a single package. The result is a device that functions very much like a normal BJT, but with extremely high hFE—the overall current gain is approximately equal to hFE of the first transistor multiplied by hFE of the second transistor. At this point you might be thinking, “I’ve got plenty of 2N2222 transistors lying around in the basement, I’ll just connect them Darlington-style and call it good.” Well, it’s not quite that simple. Take a look at the equivalent circuit for the TIP142T Darlington transistor from Fairchild (PDF):



In addition to the BJTs, we have a protection diode and two resistors. The resistors improve turn-off time by providing a discharge path for the right-hand transistor’s base-emitter junction capacitance, and they ensure a defined state for the right-hand transistor’s base node, which otherwise would be floating when the Darlington pair is in cutoff. They also result in lower hFE, because some of the base current is shunted around the base-emitter junctions. This reduction in gain is actually beneficial in many situations because it reduces the effects of leakage current—and the fact is you really shouldn’t need the full current gain, which would be something like 10,000 if we assume that each BJT has hFE of 100. So the point is, it’s probably better to buy a Darlington device rather than make your own out of two discrete BJTs.

Here is an LTspice schematic with a Darlington pair instead of a single BJT.


LTspice doesn’t come with a Darlington part, but you can go here to download subcircuit and symbol files for the TIP142.

Here is a plot with VIN, VOUT, and the voltage applied to the base of the Darlington.



As with the single-BJT circuit, the output voltage tracks the input voltage (the VIN trace is hidden underneath the VOUT trace). Note that VBASE is about 1.3–1.4 V above the load voltage; this is because now we have two base-to-emitter voltage drops instead of one. Thus, you have to be especially careful to ensure that your Darlington and op-amp supply voltages are high enough to allow for your full load-voltage range (for more details on this, see the “Simple, but Not Foolproof” section at the end of the previous article).

This next plot shows the load current and the current flowing into the base of the Darlington.



So with a load current of 360 mA, the base current is 169 µA, corresponding to hFE ≈ 2130. The datasheet indicates that the current gain should be closer to 1000; perhaps this particular SPICE model is not as accurate as it could be. In any event, we have succeeded in greatly reducing the output current required from the op-amp.

Another way to deal with an op-amp that can’t supply enough output current is to use a MOSFET instead of a BJT. We’ll cover MOSFET implementations in the next article.

Going Below Ground

Op-amps are often used with negative output voltages. An obvious example is sinusoidal signals, such as we find in audio, video, and RF applications. When the op-amp generates a positive output voltage, the output current flows “out of” the op-amp and “into” the ground node through the load. Hence, when the output is positive, the op-amp sources current. With a negative output voltage, current flows “out of” the ground node, through the load, and “into” the op-amp, such that the op-amp is now sinking current. Thus, to support signals that extend above and below ground, we need an output-current buffer that can both sink and source current. Voilà:



The general idea is the same: the BJTs provide higher current capacity, and the feedback configuration makes the op-amp modify its output in whatever way is needed to ensure that the load voltage VOUT is equal to VIN. The difference is the addition of a PNP transistor, which accomplishes for negative load voltages what the NPN transistor accomplishes for positive load voltages. In other words, when the input voltage is positive, the op-amp output swings positive to turn on the NPN transistor, and current is sourced from the NPN to the load; when the input voltage is negative, the op-amp output swings negative to turn on the PNP, and the PNP sinks the load current. Here is an LTspice schematic:



Note that I have chosen the PNP part number recommended as a complementary device in the datasheet for the 2SCR293P:



Here is a plot with VIN and VOUT. As usual, the input voltage is hidden underneath the output voltage.


This next zoomed-in plot includes the op-amp’s output voltage. Note how the negative-feedback action causes the op-amp to automatically bypass the “dead band,” i.e., the voltage range (from about –0.7 V to 0.7 V) in which both transistors are in cutoff.

Is This an Audio Amplifier?

At this point you might be wondering if you can use this circuit as a power amplifier for audio signals. You certainly can, but the sound quality will not be the best. Actually, the NPN-plus-PNP arrangement in this circuit is referred to as a Class B output stage, and combining the Class B stage with an op-amp and some negative feedback yields a power amplifier with minimal crossover distortion (the Class B stage by itself has major distortion problems created by the large dead band). However, even with negative feedback, sound quality is still somewhat degraded by the alternate turning on and off of the NPN and PNP transistors. This is why the preferred circuit for audio is the Class AB amplifier, in which the transistors are biased such that both conduct for small input voltages above or below the crossover point.


We have covered three simple, low-cost circuits that can greatly amplify an op-amp’s output current. These three configurations cover most situations in which a high-output-current amplifier is needed—just don’t forget to double-check your supply voltages, current limitations, and power dissipation.

Next Article in Series: How to Buffer an Op-Amp Output for Higher Current, Part 3

  • Z
    zombywoof March 21, 2016

    Quad made an audio version many years ago with the 405…

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  • Navendu Pottekkat July 10, 2018

    I am trying to do this circuit with 741 op amp and cn100/cp100 transistors but it does not seem to work. I used a simulation in multisim using 3904/3906 transistors and this is working . but i don’t have 3904 transistor with me and multisim doesnt have cn100/cp100 transistor. my input voltage is 6vrms 6khz sine wave and the load is a 30ohm resistance
    i hope you help

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    • RK37 August 06, 2018
      I'm not familiar with the CN100/CP100 transistors. I suggest that you ask this question in our forum.
      Like. Reply