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Figure 1046 Diode OR gate
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iC VCC RC 6 4 2
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Saturation Collector characteristic iB = 50 A B 1 RC iB = 40 A iB = 20 A A
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In discussing large-signal models for the BJT, we observed that the i-v characteristic of this family of devices includes a cutoff region, where virtually no current ows through the transistor On the other hand, when a suf cient amount of current is injected into the base of the transistor, a bipolar transistor will reach saturation, and a substantial amount of collector current will ow This behavior is quite well suited to the design of electronic gates and switches and can be visualized by superimposing a load line on the collector characteristic, as shown in Figure 1047 The operation of the simple BJT switch is illustrated in Figure 1047, by means of load-line analysis Writing the load-line equation at the collector circuit, we have vCE = VCC iC RC and vout = vCE (1067) (1066)
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VCE sat 02 V
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Thus, when the input voltage, vin , is low (say, 0 V, for example) the transistor is in the cutoff region and little or no current ows, and vout = vCE = VCC (1068)
iC RB vin + vBE
RC + vCE vout
Elementary BJT inverter
Figure 1047 BJT switching characteristic
so that the output is logic high When vin is large enough to drive the transistor into the saturation region, a substantial amount of collector current will ow and the collector-emitter voltage will be reduced to the small saturation value, VCE sat , which is typically a fraction of a volt This corresponds to the point labeled B on the load line For the input voltage vin to drive the BJT of Figure 1047 into saturation, a base current of approximately 50 A will be required Suppose, then, that the voltage vin
Part II
Electronics
could take the values 0 V or 5 V Then, if vin = 0 V, vout will be nearly equal to VCC , or, again, 5 V If, on the other hand, vin = 5 V and RB is, say, equal to 89 k (so that the base current required for saturation ows into the base: iB = (vin V )/RB = (5 06)/89,000 50 A), we have the BJT in saturation, and vout = VCE sat 02 V Thus, you see that whenever vin corresponds to a logic high (or logic 1), vout takes a value close to 0 V, or logic low (or 0); conversely, vin = 0 (logic low ) leads to vout = 1 The values of 5 V and 0 V for the two logic levels 1 and 0 are quite common in practice and are the standard values used in a family of logic circuits denoted by the acronym TTL, which stands for transistor-transistor logic2 One of the more common TTL blocks is the inverter shown in Figure 1047, so called because it inverts the input by providing a low output for a high input, and vice versa This type of inverting, or negative, logic behavior is quite typical of BJT gates (and of transistor gates in general) In the following paragraphs, we introduce some elementary BJT logic gates, similar to the diode gates described previously; the theory and application of digital logic circuits is discussed in 13 Example 1011 illustrates the operation of a NAND gate, that is, a logic gate that acts as an inverted AND gate (thus the pre x N in NAND, which stands for NOT)
EXAMPLE 1011 TTL NAND Gate
Problem
Complete the table below to determine the logic gate operation of the TTL NAND gate of Figure 1048
v1 0V 0V 5V 5V
v2 0V 5V
State of Q1
State of Q2
vout
R1 v1 R2 R3 vout Q1 Q2 Q3 R4
0V 5V
Solution
Known Quantities: Resistor values; VBE
and VCE sat for each transistor
Figure 1048 TTL NAND gate
Find: vout for each of the four combinations of v1 and v2 Schematics, Diagrams, Circuits, and Given Data: R1 = 57 k ; R2 = 22 k ; R3 = 22 k ; R4 = 18 k ; VCC = 5 V; VBE on = V = 07 V; VCE sat = 02 V Assumptions: Treat the BE and BC junctions of Q1 as offset diodes Assume that the transistors are in saturation when conducting
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