#### Theory of Static Characteristics IV and Drive Capability

Drive Capability:

We have seen that in order of assurance that the transistor in the simple inverter remains in saturation whenever needed, the base is overdriven by injecting more base current than is required to bring the transistor to the edge of saturation. A usual base overdrive factor of σu = 5 has been utilized. There is another significant reason for such a high degree of overdrive and that is to provide the inverter output drive capability. Let consider the inverter with a load joined to the output as shown in figure below.

Figure: Transistor Inverter Driving a Multiple Resistive Load

In this case, the collector current flowing via the transistor whenever it is turned on is determined not just by the collector resistor, RC, however as well by the effective load in parallel with it. This effective load, in practice, might represent N similar loads, each representing an equal resistance, RL. Of interest is the maximum number of such loads which can be driven by the inverter.

In this case, the overall transistor collector current is as:

iC = iRC + iLTOT

In saturation, iC << βFiB, that is guaranteed by overdrive. The limiting condition occurs whenever so many phases are added that the transistor comes out of saturation into forward active region since the collector current is no longer limited by the RC. Critical point is if the transistor is at the edge of saturation. At this point:

iC = βFiB

iC = (VCC – VCESAT)/RC + (VCC – VCESAT)/RLOT

And, iB = (VCC – VCESAT)/RB

And hence,

(VCC – VCESAT) [(1/RC) + (1/RLOT)] = βF (VCC – VBESAT) (1/RB)

For N phases of load RL, in parallel: RLOT = RL/N

In the limit, with maximum number of load phases, NMAX and taking:

VCC – VCESAT ≈ VCC – VBESAT then,

(1/RC) + (NMAX/RL) = βF/RB

This gives:

(RL/RC) + NMAX = (βF RL)/RB

NMAX = [(βF RL)/RB] - (RL/RC)

By slightly modifying it,

NMAX = [(βF RL)/RB] . (RC/RC) - (RL/RC)

NMAX = [{(βF RC)/RB} - 1] (RL/RC)

But as βF (RC/RB) = σu then,

NMAX = (σu - 1) (RL/RC)

Notice that the number of load phases which can be driven by inverter depends on the value of load resistor RL, in relation to RC however as well on the degree of base overdrive applied to unloaded inverter phase. When no overdrive was present, then σu – 1 and NMAX = 0 that would mean no extra loading could be tolerated. The conclusion is that overdrive of transistor is as well essential to enable it to drive loads.

Fan-Out:

Fan-Out is a direct, practical measurement of drive capability of a logic gate.

Fan-Out is generally defined as the number of gate inputs similar to its own, joined in parallel, that a gate output is capable to drive devoid of malfunction.

Let consider the single transistor inverter loaded by numerous identical phases as shown in figure below. The question is how many of such phases can be joined in parallel as load before an inaccurate value of VO outcomes. Whenever the output VO is LO, the load transistors are OFF and therefore there is no demand for current therefore this condition doesn’t limit the fan-out. Whenever the output VO is HI, the load transistors are ON and each one demands a base current adequate to saturate it. The question is how many load transistors can be tolerated prior to VO falls to the minimum level which will be seen by the load transistor as HI input, ViH MIN. In another words, how many load transistors can be joined in parallel before:

Vo = VIH MIN = VBE SAT + [RB/(βFRC)] (VCC - VCESAT)

Figure: A Bipolar Transistor Inverter Loaded by Multiple similar Inverters

Figure: Equivalent Circuit of Loaded Transistor Inverter

An equivalent circuit can be drawn symbolizing the N load transistors in parallel as shown in figure above. Each and every load transistor consists of a base-emitter potential drop VBE SAT and draws a base current:

IB = (VO – VBESAT)/RB

Whenever the driving inverter output is high, then from equivalent circuit of figure shown above:

Vo = VBESAT + (RB/N)/{(RB/N) + RC} (VCC - VBESAT)

In limit Vo = ViH MIN and N = NMAX therefore equating the two expressions provides:

VBESAT + [(RB/NMAX)]/[(RC + RB)/NMAX] (VCC - VBESAT) = VBESAT + (RBFRC) (VCC - VCESAT)

When we consider, VCC – VBESAT ≈ VCC - VCESAT then,

RB/(RB + NMAX RC) = RBF RC

Therefore, RB + NMAX RC = βF RC

And hence,

NMAX = (βF RC - RB)/RC

NMAX = βF (1 - RB/ βF RC)

However as βF RC/RB = σu, we can state the Fan-Out, F = NMAX is:

F = βF (1 – 1/σu)

Again if σU = 1 then F = 0.

This can be seen that a high transistor βF gives a high gate Fan-out.

When βF = 50, σU = 5 then F = 50(1 – 0.2) = 50 x 0.8 = 40 that is a little higher than manufacturers assurance in practice. A usual value is of the order of 10.

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