BJT Based Logic Gates

Bipolar Junction Transistors

A bipolar junction transistor is a three-terminal device which acts as a current-controlled switch. If we put a small current into one of the terminals, called the base, then the switch is “on”—current may flow between the other two terminals, called the emitter and the collector. If no current is put into the base, then the switch is “off”—no current flows between the emitter and the collector.

Consider the operation of a pair of diodes connected as shown in Fig.6 (a), the current can flow from node B to node C or node E, when the appropriate diode is forward biased. However, no current can flow from C to E, or vice versa, since for any choice of voltages on nodes B, C, and E, one or both diodes will be reverse biased. The pn junctions of the two diodes in this circuit are shown in fig.6 (b).

If the back-to-back diodes are fabricated so that they share a common p-type region, they appear as shown in Fig.6 (c). The resulting structure is called an npn transistor and has an amazing property (introduced in 1950s). If we put current across the base-to-emitter pn junction, then current is also enabled to flow across the collector-to-base np junction and from there to the emitter. The circuit symbol for the npn transistor is shown in Fig.6 (d).

The symbol contains a subtle arrow in the direction of positive current flow. As the base-to-emitter junction is a pn junction, the same as a diode whose symbol has an arrow pointing in the same direction. It is also possible to fabricate a pnp transistor, as shown in Fig.7. However, pnp transistors are seldom used in digital circuits.

The current Ie flowing out of the emitter of an npn transistor is the sum of the currents Ib and Ic flowing into the base and the collector. A transistor is often used as a signal amplifier, because over a certain operating range (the active region) the collector current is equal to a fixed constant times the base current (I_c = βI_b).

Fig.8 shows the common-emitter configuration of an npn transistor most often used in digital switching applications, which uses two discrete resistors, R1 and R2, in addition to a single npn transistor. In this circuit, if VIN is 0 or negative, then the base-to-emitter diode junction is reverse biased, and no base current (Ib) can flow. If no base current flows, then no collector current (Ic) can flow, and the transistor is said to be cut off (OFF). Since the base-to-emitter junction is a real diode, as opposed to an ideal one, V_IN must reach at least +0.6 V (one diode-drop) before any base current can flow. Once this happens, Ohm’s law tells us that

By ignoring the forward resistance R_f of the forward-biased base-to-emitter junction, which is usually small compared to the base resistor R1. When base current flows, then collector current can flow in an amount proportional to I_b, i.e., 

The constant of proportionality, β, is called the gain of the transistor, and is in the range of 10 to 100 for typical transistors.

Although the base current I_b controls the collector current flow Ic, it also indirectly controls the voltage V_CE across the collector-to-emitter junction, since V_CE is just the supply voltage V_CC minus the voltage drop across resistor R₂: 

In an ideal transistor V_CE can never be less than zero (the transistor cannot just create a negative potential), and in a real transistor V_CE can never be less than V_CE(sat), a transistor parameter that is typically about 0.2 V.

If the values of V_IN, b, R₁, and R₂ are such that the above equation predicts a value of V_CE that is less than V_CE(sat), then the transistor cannot be operating in the active region and the equation does not apply. Instead, the transistor is operating in the saturation region, and is said to be saturated (ON). No matter how much current I_b we put into the base, V_CE cannot drop below V_CE(sat), and the collector current I_c is determined mainly by the load resistor R2:

where, R_CE(sat) is the saturation resistance of the transistor. Typically, R_CE(sat) is 50 W or less and is insignificant compared with R2.

Transistor Logic Inverter

Fig.9 shows that we can make a logic inverter from an npn transistor in the common-emitter configuration. When the input voltage is LOW, the output voltage is HIGH, and vice versa.

  

In digital switching applications, bipolar transistors are often operated so they are always either cut off or saturated. That is, digital circuits such as the inverter in Fig.9 are designed so that their transistors are always in one of the states depicted in Fig.10.

 When the input voltageVIN is LOW, it is low enough that Ib is zero and the transistor is cut off; the collector-emitter junction looks like an open circuit. When VIN is HIGH, it is high enough (and R1 is low enough and b is high enough) that the transistor will be saturated for any reasonable value of R2; the collector - emitter junction looks almost like a short circuit. Input voltages in the undefined region between LOW and HIGH are not allowed, except during transitions.

Alternatively, the operation of a transistor inverter is shown in Fig.11. When VIN is HIGH, the transistor switch is closed, and the output terminal is connected to ground, definitely a LOW voltage. When VIN is LOW, the transistor switch is open and the output terminal is pulled to +5 V through a resistor; the output voltage is HIGH unless the output terminal is too heavily loaded (i.e., improperly connected through a low impedance to ground).

NAND GATE using BJT

A	B	OUT
0	0	1
0	1	1
1	0	1
1	1	0

Fig.12: NAND Gate using BJT

The Fig.12 shows the circuit diagram and Truth Table of NAND Gate using BJT, Let Q1 be the transistor connected to VA and Q2 be the transistor connected to VB. If VA=VB=0 or VA=0 and VB=1 or VA=1 and VB=0 then VOUT=1, If VA=VB=1 then VOUT =0. The corresponding potentials are Vout=6V – V_4.7K for Logic 1 at the output and V_OUT=V_CE(sat)Q1 + V_CE(sat)Q2

NOR Gate using BJT

 

A	B	OUT
0	0	1
0	1	0
1	0	0
1	1	0

  

Fig.13: NOR Gate using BJT

The NOR gate using BJT is shown in fig.13, let Q1 be the transistor connected to VA and Q2 be the transistor connected to VB. If VA=VB=0 then VOUT=1, If VA=VB=1 or VA=0 and VB=1 or VA=1 and VB=0 then VOUT =0. The corresponding potentials are Vout=6V – V_4.7K for Logic 1 at the output and V_OUT=V_{CE(sat)(Q1 or Q2)}