# 6.6: Voltage Followers and Buffers

Unity gain noninverting buffers (voltage followers) are used in wide variety of applications. Any time a signal source needs to be isolated, a buffer is needed. As you have seen, connecting op amps in a follower configuration is a very straightforward exercise. This is not always the best choice. By optimizing the amplifier design for buffer operation, the manufacturer can increase performance, or in some cases, reduce the case size. This is possible because followers need very few connections: input, output, and power supply. Also, as the devices must, by nature, be unity gain stable, an external compensation connection is normally not desired.

One buffer amplifier optimized for high performance is the Maxim MAX4200-4205 series. This device is designed with video applications, high-speed drivers, and analog-to-digital converters in mind. Outside of the standard power supply bypass capacitors and perhaps a line termination resistor, no other parts are needed. Figure $$\PageIndex{1}$$ shows a minimum configuration for driving 50 $$\Omega$$ co-axial cable. This buffer is capable of supplying $$\pm$$90 mA to its load, so direct 50 $$\Omega$$ connection poses no problem. Slew rate is typically 4200 V/$$\mu$$s, and circuit bandwidth is 780 MHz.

Figure $$\PageIndex{1}$$: Video RF cable buffer. Reprinted courtesy of Maxim Integrated

Example $$\PageIndex{1}$$

Determine the approximate bandwidth required for a high-resolution video display amplifier. The display has a resolution of 1024 pixels wide by 1024 pixels high, with a refresh rate of 60 frames per second1. This specification is typical of a workstation or a high quality personal computer.

In its simplest form (monochrome, or black and white), each pixel is either on or off. As a controlling voltage, this means either a high or a low. An updated grid of 1024 by 1024 dots must be drawn on the display monitor 60 times each second. From this, we can determine the pixel, or dot, rate

$\text{Pixel Rate } = Height \times Width \times Refresh \notag$

$\text{Pixel Rate } = 1024 \times 1024 \times 60 \notag$

$\text{Pixel Rate } = 62,914,560 \text{ pixels per second} \notag$

This indicates that each pixel requires 1/62,914,560 seconds, or about 15.9 ns, to reproduce. For proper pulse shape, the risetime should be no more than 30% of the pulse width (preferably less). From the risetime/bandwidth relationship established in Chapter One,

$f_2 = \frac{0.35}{T_r} \notag$

$f_2 = \frac{0.35}{0.3\times 15.9 ns} \notag$

$f_2 = \frac{0.35}{4.77 ns} \notag$

$f_2 = 73.4MHz \notag$

Theoretically, a 73.4 MHz bandwidth is required. Once overhead such as vertical retrace is added, a practical circuit will require a bandwidth on the order of 100 MHz.

## References

1The term pixel is a contraction of picture element. The display is considered to be nothing more than a large grid of dots. The term refresh rate refers to how quickly the display is updated or redrawn.