## Big Buffer

Ed Hitt of Louisville, Colorado writes [edited for clarity]:

What is the purpose of using a buffer-driver (EG 74LCX125) to drive a circuit with a large capacitive load?

Is it merely to make the rise time faster by decreasing the output impedance of the driving circuit, thereby increasing the timing margin, or is the low current drive capability of the non-buffer gate simply not adequate?

Is there any reason to think that the large current due to C dv/dt might damage a non-buffer type gate?

The output impedance and the current-driving capability of a driver are intimately related. Comparing two totem-pole CMOS drivers operated at the same power-supply voltage, a larger current-driving capability almost always implies a smaller output resistance.

Figure 1 illustrates the complete form of the output voltage-current relationship for a typical CMOS totem-pole driver. This type of plot is called a V-I diagram (or sometimes an I-V diagram). The vertical axis represents current emanating from your driver. The horizontal axis represents the output voltage. From this V-I diagram you can determine how the driver reacts to any static load.

The diagram displays two curves. The top curve (green) illustrates the relation of current to voltage for this driver when switched into the high state. This is a static curve, showing the behavior at DC. It is derived by connecting the driver to a load that draws progressively more and more current from the driver, over a very slow scale of time, and making a record of the driver output voltage at each particular current. Then the driver is switched to a low state, and the experiment repeated (purple curve).

This same V-I information may be encoded in a standard form
called an *IBIS model*. An IBIS model
holds, at a minimum, the high-state and low-state static V-I curves for a
driver, plus some information about how quickly the driver morphs from one
state to the other. Issues you may have heard about related to the accuracy of
IBIS modeling have to do with the form and quality of the information used to morph
the V-I curves from low to high (or vice versa) during a fast switching
transient, the effects of packaging parasitics, and the lack of information
within an IBIS model about cross-coupling between drivers within the same
package (SSO noise). Still, I like IBIS
models and find them useful.

A typical driver specification calls out only a couple of points along the V-I curves. If we are talking about a CMOS totem-pole driver, you are expected to know that the gate switches fully rail to rail in the absence of DC loading. In the V-I domain, this implies that the high-state curve includes the point [VCC,0], meaning the voltage floats all the way up to VCC when you draw zero current from the driver. In the diagram, the green curve crosses the horizontal axis at the point [VCC,0].

Starting from the point [VCC,0] on the green curve, as you pull progressively more current from the driver (moving up) the output droops (moving to the left), with the result that the high-state V‑I curves for CMOS totem-pole drivers always move up and to the left away from point [VCC,0].

Near VCC, the slope of the V-I curve (the localized delta-V divided by delta-I) is defined as the "output resistance" of the driver, R[ACTUAL]. For problems involving late reflections that return to the source long after the source has switched to a voltage above VOH, the value R[ACTUAL] determines the behavior of the driver at that point in time.

A typical datasheet specification calls out only one point in the high state. That point, [VOH,IOH], guarantees that the high-state V-I curve, on its way up and to the left away from [VCC,0], will always pass to the right side of the point [VOH,IOH] (marked "A" in the diagram).

The diagram illustrates the slope of the wimpiest possible driver (i.e., highest possible output resistance) that barely meets this specification. That resistance, marked R[MAX], can be calculated:

R[MAX] = (VCC–VOH) / IOH

When working through this formula use the minimum allowed value of VCC for your driver. The driver has to still meet VOH at the specified IOH current under this condition. The output resistance will likely be no greater than that amount under other conditions of VCC.

The maximum output resistance in the low state is simply –(VOL / IOL). Watch out for the polarity of current definitions on the specification sheet—a "sinking current" of 8 mA means an IOL of negative 8 mA.

Ed, your question had to do with the behavior of a driver when connected to a big clump of capacitance. I will assume the capacitance is all locally connected, at a distance of much less than one rise time from the driver, so we may treat it as a lumped-element load.

To determine the rise time of a large capacitive load we have to know how much current comes out of the driver at voltages other than VOH. For example, it would be nice to know the short-circuit current when switched to the high state. This current is the point where the high-state V-I curve crosses the vertical axis (VOUT=0). If you knew that, and assuming the V-I curve were at least concave down (no undulations), then a straight-line approximation drawn from [0,ISHORT] to [VCC,0] would yield a 90% rise time of:

TR = 2.3*(VCC/ISHORT)*CLOAD

**Where:**

- VCC is the minimum allowed value of VCC,
- ISHORT is the short-circuit current in Amps at that power-supply voltage, and
- CLOAD is the total load capacitance in farads.

This formula models the high-state V-I curve as a simple straight line passing from [VCC,0] to the short-circuit point [0,ISHORT]. In other words, for the purpose of first-order modeling, you may replace the driver with a single equivalent linear resistance having a value of VCC/ISHORT.

I happen to know that it takes 2.3 times the natural R-C time constant for the step response of an R-C circuit to pass from zero to the 90response point. The formula therefore reads, "2.3 times the product of the source resistance (VCC/ISHORT) times the load capacitance (CLOAD)."

In a real circuit, the actual V-I curve for your driver
exceeds the straight-line approximation everywhere (because a real high-state
V-I curve is concave-down). The real driver therefore pumps out *more* current than assumed by the
straight-line approximation, so the actual response time in a real circuit will
be less, *sometimes* *considerably* *less, *than* *the simple
estimate shown above. Any IBIS-model simulator can demonstrate that effect.

Now we come to an interesting facet of your question that concerns the maximum safe output current for a driver. First I will deal with the DC effects of massive current.

For many CMOS totem-pole drivers, you can short the outputs to ground for brief periods of time without damaging the driver. Damage to a driver under a static shorted condition is caused mostly by heating due to the excessive current. As long as the short circuit does not persist, no harm is done. Because CMOS drivers go into a current-limiting state as you pull more current from them (i.e., the output curve is concave down, flattening out as you approach short-circuit conditions), the CMOS output stage is somewhat self-protecting.

The output stage of a PECL driver, on the other hand, has no current-limiting feature. Without and special protection circuitry, PECL outputs are easily destroyed by unintentional grounding.

Finally, let's look at the effects of large AC currents. These effects have to do with simultaneously switching output noise (SSO noise). If you are not familiar with that topic check out these recent articles:

- Newsletter vol. 8 #3, "BGA Crosstalk (Xilinx Virtex-4 FPGA)"
- EDN, March 31, 2005, "SpreadYourReturns.htm"

In brief, SSO noise is property of every IC package. When the outputs of your IC switch, self-coupling within the IC package couples a certain amount of noise back into your IC inputs. If the noise is sufficiently large, it causes data errors at the inputs. The same SSO noise effect also produces undesirable glitches in your IC outputs.

The main factors that affect SSO noise are (1) the aggregate amount of current switched by your IC, (2) the rise time of that current, and (3) the number and quality of the power and ground connections provided in the IC package. Most relevant to this discussion is the total amount of current—more capacitive loading enlarges that current, increasing the amplitude of SSO noise.

An IC manufacturer is supposed to guarantee that, as long as you live within the loading guidelines published for their IC, you can switch any combination of outputs at will without causing SSO errors. This guarantee is enforced by incorporating an adequate number of power and ground pins, and possibly also solid plane layers, bypass capacitors and other features, into the IC package.

Do you suppose there is much SSO noise margin left in a typical IC package design? Can you safely exceed the loading guidelines without causing SSO errors? I doubt it. In a large package already crammed with oodles of power and ground pins, if there were any margin left for SSO noise the manufacturer would likely have shaved off a few of the power and ground pins to reduce the size and cost of the package. In fact, some packages carry specific limitations on how many outputs can be heavily loaded, or how many can be allowed to switch at any one given time.

If you exceed the loading guidelines, burdening each output with huge gobs of load capacitance, you will probably not damage the driver, but you may create excessive amounts of SSO noise, enough to induce data errors on your inputs.

My comments here refer to the undesirable practice of connecting gobs of capacitance locally to your driver, all located within a small fraction of one rise time of the driver. If you spread your many loads along a transmission structure with a uniform spacing of, say, 1/3 rise time between each load, then your driver need not charge all the loads at once. Spreading the loads in this manner reduces the peak current required of the driver (and thus limits the SSO noise). That is the secret to successfully driving many, many loads.

Best Regards,

Dr. Howard Johnson